Fundamentals of Power Electronics Instructor's slides
Fundamentals of Power Electronics Instructor's slides
Fundamentals of Power Electronics R. W. Erickson Accompanying material for instructors The materials below are intended to be used by instructors of power electronics classes who have adopted Fundamentals of Power Electronics as a text. These instructors may download and use the files for educational purposes free of charge. Students and others who have purchased the text may also use the slides as an educational supplement to the text. Other uses of these materials is prohibited. All slides copyright R. W. Erickson 1997. The slides for each chapter are contained in a .pdf file. These files can be read using the Adobe Acrobat viewer, available free from the Adobe Acrobat web site. Slides and overhead transpariencies covering the material of the entire book can be produced using these files.
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Part 4: Modern Rectifiers and Power System Harmonics ● ● ● ●
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Chapter 19. Resonant conversion 325kB Chapter 20. Quasi-resonant converters 177kB
Appendices
Introduction ● ●
Chapter 15. Power and harmonics in nonsinusoidal systems 91kB Chapter 16. Line-commutated rectifiers 130kB Chapter 17. The ideal rectifier 235kB Chapter 18. Low harmonic rectifier modeling and control
Part 5: Resonant Converters
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Back
Chapter 12. Basic magnetics theory 196kB Chapter 13. Filter inductor design 67kB Chapter 14. Transformer design 175kB
Chapter 1. Introduction 98kB
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Appendix 1. RMS values of commonly-observed converter waveforms 26 kB Appendix 2. Magnetics design tables 26kB Appendix 3. Averaged switch modeling of a CCM SEPIC 41kB
Part 1: Converters in Equilibrium ● ● ● ● ●
Chapter 2. Principles of steady-state converter analysis 126kB Chapter 3. Steady-state equivalent circuit modeling, losses, and efficiency 98kB Chapter 4. Switch realization 201kB Chapter 5. The discontinuous conduction mode 96kB Chapter 6. Converter circuits 283kB
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Part 2: Converter Dynamics and Control ● ● ● ● ●
Chapter 7. Ac equivalent circuit modeling 422kB Chapter 8. Converter transfer functions Chapter 9. Controller design 365kB Chapter 10. Ac and dc equivalent circuit modeling of the discontinuous conduction mode 218kB Chapter 11. The current programmed mode 236kB
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Revision to Fundamentals of Power Electronics
Revision to Fundamentals of Power Electronics
Fundamentals of Power Electronics First Edition R. W. Erickson Power Electronics Group, University of Colorado at Boulder
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Index 101kB -searchable with Adobe Acrobat Errata, first printing
Supplementary material for instructors ● ●
Slides Solutions to selected problems
About the second edition Other supplementary material A new textbook on power electronics converters. This book is intended for use in introductory power electronics courses at the senior and first-year graduate level. It is also intended as a source for professionals working in power electronics, power conversion, and analog electronics. It emphasizes the fundamental concepts of power electronics, including averaged modeling of PWM converters and fundamentals of converter circuits and electronics, control systems, magnetics, lowharmonic rectifiers, and resonant converters.
Proximity effect: computer functions 70kB Ferrite toroid data: Excel 5 spreadsheet Derivation of Gg0, Eqs. (11.84) and (11.85)
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Publisher and vitals New York: Chapman and Hall, May 1997. Hardback ISBN 0-412-08541-0 TK7881.15.E75 1997 7"x10", 791 pages, 929 line illustrations. Note: Chapman and Hall has recently been acquired by Kluwer Academic Publishers Note to instructors: how to obtain a copy
More information regarding contents of book ● ● ●
Complete Table of Contents Abridged Table of Contents: Chapter titles only Preface
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CoPEC
Revision to Fundamentals of Power Electronics
CoPEC Colorado Power Electronics Center
Fundamentals of Power Electronics Second Edition
University of Colorado, Boulder
About CoPEC Research Publications Students Faculty Courses Textbook: Fundamentals of Power Electronics
Authors: R. W. Erickson and D. Maksimovic University of Colorado, Boulder Publisher: Kluwer Academic Publishers 912 pages ISBN 0-7923-7270-0 ●
Power Electronics in the CU Boulder Electrical and Computer Engineering Department Links to Other Power Electronics Sites Updated May 21, 2001.
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First edition web site To order directly from the publisher Note to instructors: how to obtain desk copies Errata, second edition, first printing Errata, second edition, second printing New Certificate Program in Power Electronics PSPICE circuit files and library Courses at the University of Colorado that use the second edition ECEN 5797 Power ❍
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Revision to Fundamentals of Power Electronics
Revision to Fundamentals of Power Electronics
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Electronics 1 ECEN 5807 Power Electronics 2 ECEN 5817 Power Electronics 3
Major Features of the Second Edition ●
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New material on converter simulation using averaged switch models Major revision of material on current mode control, including tables of transfer functions of basic converters Major revision of material on averaged switch modeling New material covering input filter design and Middlebrook's extra element theorem Improved explanations of the proximity effect and MMF diagrams New section on design of multiplewinding magnetics using the Kg method, including new examples New material on soft switching, including active clamp snubbers, the ZVT full bridge converter, and ARCP Major revision of material on lowharmonic rectifiers, to improve flow and readability. New material on critical conduction mode control Major revision and simplification of the chapter on ac modeling of the discontinuous conduction mode Revised problems, and a solutions manual
Detailed description of revisions ● ● ●
Contents Preface to the Second Edition Chapter 1 Introduction
Part 1. Converters in Equilibrium There are no substantial changes to the chapters of Part 1. ● ● ● ● ●
Chapter 2 Principles of Steady-State Converter Analysis Chapter 3 Steady-State Equivalent Circuit Modeling, Losses, and Efficiency Chapter 4 Switch Realization Chapter 5 The Discontinuous Conduction Mode Chapter 6 Converter Circuits
Part 2. Converter Dynamics and Control ●
Chapter 7 AC Equivalent Circuit Modeling
Chapter 7 has been revised to improve the logical flow, including incorporation of the First Edition Appendix 3 into the chapter. The treatment of circuit averaging and averaged switch modeling (Section 7.4) has undergone major revision. Other changes include Fig. 7.4 and the related text, and Sections 7.2.2, 7.2.7. ●
Chapter 8 Converter Transfer Functions
Major revisions to Chapter 8 include a new introduction, a new input filter example in Section 8.1.8, and substantial changes to the buck-boost converter example of Section 8.2.1 and the material of Sections 8.3 and 8.4. ●
Chapter 9 Controller Design
Only minor changes to Chapter 9 were made. ●
Chapter 10 Input Filter Design
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converter, and how to design an input filter that is adequately damped. The approach is based on Middlebrook's Extra Element Theorem (EET) of Appendix C, although it is possible to teach this chapter without use of the EET. ●
Chapter 11 AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode
This chapter has been entirely revised and simplified. ●
Chapter 12 Current Programmed Control
Revision to Fundamentals of Power Electronics ●
Chapter 16 Power and Harmonics in Nonsinusoidal Systems
Information on harmonic standards has been updated. ●
Chapter 17 Line-Commutated Rectifiers
There is little change to this chapter. ●
Chapter 18 Pulse-Width Modulated Rectifiers
Treatment of the "more accurate model" in Section 12.3 has undergone a major revision. The explanation is more straightforward, and results are summarized for the basic buck, boost, and buck-boost converters. The results of simulation are used to illustrate how current programming changes the converter transfer function. The treatment of discontinuous conduction mode in Section 12.4 has been shortened.
Chapter 18 is a consolidation of Chapters 17 and 18 of the First Edition. The material has been completely reorganized, to improve its flow. A new section 18.2.2 has been added. Section 18.3.3 has been expanded, to better cover critical conduction mode control. The material on three-phase rectifier topologies has been streamlined.
Part 3. Magnetics
Part 5. Resonant Converters
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Chapter 13 Basic Magnetics Theory
The material on the skin and proximity effects has undergone a major revision, to better introduce the concepts of the proximity effect and MMF diagrams. The summary of operation of different magnetic devices has been moved from the filter inductor design chapter into this chapter. ●
Chapter 14 Inductor Design
A new section on design of multiple-winding inductors using the Kg method has been added, including two new examples. The summary of different magnetic devices has been moved to the previous chapter, and the material on winding area optimization (previously in the transformer design chapter) has been moved into this chapter. ●
Chapter 15 Transformer Design
Notation regarding maximum, peak, and saturation flux density has been made more clear. The section on winding area optimization has been moved to the previous chapter.
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Chapter 19 Resonant Conversion
The order of the sections has been changed, to improve readability. Section 19.4 has been modified, to include better explanation of resonant inverter/electronic ballast design, and two examples have been added. The material on the ZVT converter has been moved to Chapter 20. ●
Chapter 20 Soft Switching
A new Section 20.1 compares the turn-on and turn-off transitions of diode, MOSFET, and IGBT devices under the conditions of hard switching, zero-current switching, and zero-voltage switching. The material on quasi-resonant converters is unchanged. Coverage of multi-resonant and quasi-squarewave switches has been exapanded, and includes plots of switch characteristics. A new Section 20.4 has been added, which covers soft-switching techniques. Included in Section 20.4 is an expanded explanation of the ZVT full-bridge converter, new material on active-clamp snubbers, and a short treatment of the auxiliary resonant commutated pole. The material on ac modeling of ZCS quasi-resonant converters has been dropped.
Appendices Part 4. Modern Rectifiers, Inverters, and Power System Harmonics ●
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Appendix A RMS Values of Commonly Observed Converter Waveforms
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This appendix is unchanged. ●
Appendix B Simulation of Converters
Appendix B is completely new. It covers SPICE simulation of converters using averaged switch models, including CCM, DCM, and current-programmed converters. This material complements the discussions of Chapters 7, 9, 11, 12, and 18. It has been placed in an appendix so that the chapter narratives are not interrupted by the details required to run a simulation program; nonetheless, the examples of this appendix are closely linked to the material covered in the chapters.
Fundamentals of Power Electronics
Fundamentals of Power Electronics Second Edition Up To instructors of Power Electronics courses: how to obtain an examination copy Evaluation copies are available on a 60-day approval basis.
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PSPICE circuit files and library used in Appendix B Please submit requests in writing on department letterhead and include the following course information:
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Appendix C Middlebrook's Extra Element Theorem ●
This is a completely new appendix that explains the Extra Element Theorem and includes four tutorial examples. This material can be taught in conjunction with Chapter 10 and Section 19.4, if desired.
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Appendix D Magnetics Design Tables
This appendix is unchanged.
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course name and number estimated enrollment semester date the text currently used your decision date
Please direct all requests to the Textbook Marketing Department at Kluwer Academic Publishers at the Norwell (Americas) or Dordrecht (all other countries) offices:
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[email protected] All other countries: PO Box 989 3300 AZ Dordrecht The Netherlands Telephone: (0) 31 78 6392 392 Fax: (0) 31 78 6546 474 E-mail:
[email protected]
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Fundamentals of Power Electronics
Fundamentals of Power Electronics Table of Contents
Fundamentals of Power Electronics R. W. Erickson
Table of Contents Back 1. Introduction 1.1. Introduction to power processing 1.2. Several applications of power electronics 1.3. Elements of power electronics
Part I. Converters in Equilibrium 2. Principles of steady state converter analysis 2.1. Introduction 2.2. Inductor volt-second balance, capacitor charge balance, and the small-ripple approximation 2.3. Boost converter example 2.4. Cuk converter example 2.5. Estimating the output voltage ripple in converters containing two-pole low-pass filters 2.6. Summary of key points 3. Steady-state equivalent circuit modeling, losses, and efficiency 3.1. The dc transformer model 3.2. Inclusion of inductor copper loss 3.3. Construction of equivalent circuit model 3.4. How to obtain the input port of the model 3.5. Example: Inclusion of semiconductor conduction losses in the boost converter model 3.6. Summary of key points 4. Switch realization 4.1. Switch applications 4.1.1. Single quadrant switches 4.1.2. Current-bidirectional two-quadrant switches 4.1.3. Voltage-bidirectional two-quadrant switch http://ece-www.colorado.edu/~pwrelect/book/copies.html (2 of 2) [25/04/2002 16:42:07]
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Fundamentals of Power Electronics Table of Contents
4.1.4. Four-quadrant switches 4.1.5. Synchronous rectifiers
Fundamentals of Power Electronics Table of Contents
6.4.1. Switch stress and utilization 6.4.2. Design using computer spreadsheet
4.2. A brief survey of power semiconductor devices 4.2.1. Power diodes 4.2.2. Metal-Oxide-Semiconductor Field-Effect Transistor (MOSFET) 4.2.3. Bipolar Junction Transistor (BJT) 4.2.4. Insulated Gate Bipolar Transistor (IGBT) 4.2.5. Thyristors (SCR, GTO, MCT)
6.5. Summary of key points
Part II. Converter Dynamics and Control 7. AC modeling
4.3. Switching loss 4.3.1. Transistor switching with clamped inductive load 4.3.2. Diode recovered charge 4.3.3. Device capacitances, and leakage, package, and stray inductances 4.3.4. Efficiency vs. switching frequency
4.4. Summary of key points 5. The discontinuous conduction mode 5.1. Origin of the discontinuous conduction mode, and mode boundary 5.2. Analysis of the conversion ratio M(D,K) 5.3. Boost converter example 5.4. Summary of results and key points 6. Converter circuits 6.1. Circuit manipulations 6.1.1. Inversion of source and load 6.1.2. Cascade connection of converters 6.1.3. Rotation of three-terminal cell 6.1.4. Differential connection of the load
7.1. Introduction 7.2. The basic ac modeling approach 7.2.1. Averaging the inductor waveforms 7.2.2. Discussion of the averaging approximation 7.2.3. Averaging the capacitor waveforms 7.2.4. The average input current 7.2.5. Perturbation and linearization 7.2.6. Construction of the small-signal equivalent circuit model 7.2.7. Results for several basic converters
7.3. Example: A nonideal flyback converter 7.4. State-space averaging 7.4.1. The state equations of a network 7.4.2. The basic state-space averaged model 7.4.3. Discussion of the state-space averaging result 7.4.4. Example: State-space averaging of a nonideal buck-boost converter
7.5. Circuit averaging and averaged switch modeling 6.2. A short list of converters 6.3. Transformer isolation 6.3.1. Full-bridge and half-bridge isolated buck converters 6.3.2. Forward converter 6.3.3. Push-pull isolated buck converter 6.3.4. Flyback converter 6.3.5. Boost-derived isolated converters 6.3.6. Isolated versions of the SEPIC and the Cuk converter
7.5.1. Obtaining a time-invariant circuit 7.5.2. Circuit averaging 7.5.3. Perturbation and linearization 7.5.4. Averaged switch modeling
7.6. The canonical circuit model 7.6.1. Development of the canonical circuit model 7.6.2. Example: Manipulation of the buck-boost converter model into canonical form 7.6.3. Canonical circuit parameter values for some common converters
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Fundamentals of Power Electronics Table of Contents
Fundamentals of Power Electronics Table of Contents
7.7. Modeling the pulse-width modulator
functions
7.8. Summary of key points
9.4. Stability
8. Converter transfer functions 8.1. Review of Bode plots 8.1.1. Single pole response 8.1.2. Single zero response 8.1.3. Right half-plane zero 8.1.4. Frequency inversion 8.1.5. Combinations 8.1.6. Double pole response: resonance 8.1.7. The low-Q approximation 8.1.8. Approximate roots of an arbitrary-degree polynomial
8.2. Analysis of converter transfer functions 8.2.1. Example: Transfer functions of the boost converter 8.2.2. Transfer functions of some basic dc-dc converters 8.2.3. Physical origins of the RHP zero
8.3. Graphical construction of converter transfer functions 8.3.1. Series impedances: addition of asymptotes 8.3.2. Parallel impedances: inverse addition 8.3.3. Another example 8.3.4. Voltage divider transfer functions: division of asymptotes
9.4.1. The phase margin test 9.4.2. The relation between phase margin and closed-loop damping factor 9.4.3. Transient response vs. damping factor
9.5. Regulator design 9.5.1. Lead (PD) compensator 9.5.2. Lag (PI) compensator 9.5.3. Combined (PID) compensator 9.5.4. Design example
9.6. Measurement of loop gains 9.6.1. Voltage injection 9.6.2. Current injection 9.6.3. Measurement of unstable systems
9.7. Summary of key points 10. Ac and dc equivalent circuit modeling of the discontinuous conduction mode 10.1. DCM averaged switch model 10.2. Small-signal ac modeling of the DCM switch network
8.4. Measurement of ac transfer functions and impedances 10.3. Generalized switch averaging 8.5. Summary of key points 9. Controller design 9.1. Introduction 9.2. Effect of negative feedback on the network transfer functions 9.2.1. Feedback reduces the transfer functions from disturbances to the output 9.2.2. Feedback causes the transfer function from the reference input to the output to be insensitive to variations in the gains in the forward path of the loop
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10.3.1. DCM buck converter example 10.3.2. Proof of generalized averaged switch modeling
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Fundamentals of Power Electronics Table of Contents
11.2.2. Averaged switch modeling
11.3. A more accurate model 11.3.1. Current programmed controller model 11.3.2. Example: analysis of CPM buck converter
11.4. Discontinuous conduction mode 11.5. Summary of key points
Fundamentals of Power Electronics Table of Contents
13.1. Several types of magnetic devices, their B-H loops, and core vs. copper loss 13.2. Filter inductor design constraints 13.2.1. Maximum flux density 13.2.2. Inductance 13.2.3. Winding area 13.2.4. Winding resistance
13.3. The core geometrical constant Kg
Part III. Magnetics
13.4. A step-by-step procedure
12. Basic magnetics theory
13.5. Summary of key points
12.1. Review of basic magnetics 12.1.1. Basic relations 12.1.2. Magnetic circuits
12.2. Transformer modeling 12.2.1. The ideal transformer 12.2.2. The magnetizing inductance 12.2.3. Leakage inductances
12.3. Loss mechanisms in magnetic devices 12.3.1. Core loss 12.3.2. Low-frequency copper loss
14. Transformer design 14.1. Winding area optimization 14.2. Transformer design: basic constraints 14.2.1. Core loss 14.2.2. Flux density 14.2.3. Copper loss 14.2.4. Total power loss vs. Bmax 14.2.5. Optimum flux density
14.3. A step-by-step transformer design procedure 14.4. Examples
12.4. Eddy currents in winding conductors 12.4.1. The skin effect 12.4.2. The proximity effect 12.4.3. Magnetic fields in the vicinity of winding conductors: MMF diagrams 12.4.4. Power loss in a layer 12.4.5. Example: power loss in a transformer winding 12.4.6. PWM waveform harmonics
14.4.1. Example 1: single-output isolated Cuk converter 14.4.2. Example 2: multiple-output full-bridge buck converter
14.5. Ac inductor design 14.5.1. Outline of derivation 14.5.2. Step-by-step ac inductor design procedure
14.6. Summary 12.5. Summary of key points 13. Filter inductor design
Part IV. Modern Rectifiers, and Power System Harmonics 15. Power and harmonics in nonsinusoidal systems
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Fundamentals of Power Electronics Table of Contents
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16.4. Harmonic trap filters 15.1. Average power 16.5. Transformer connections 15.2. Root-mean-square (rms) value of a waveform 16.6. Summary 15.3. Power factor 17. The ideal rectifier 15.3.1. Linear resistive load, nonsinusoidal voltage 15.3.2. Nonlinear dynamic load, sinusoidal voltage
17.1. Properties of the ideal rectifier
15.4. Power phasors in sinusoidal systems
17.2. Realization of a near-ideal rectifier
15.5. Harmonic currents in three-phase systems
17.3. Single-phase converter systems incorporating ideal rectifiers
15.5.1. Harmonic currents in three-phase four-wire networks 15.5.2. Harmonic currents in three-phase three-wire networks 15.5.3. Harmonic current flow in power factor correction capacitors
15.6. AC line current harmonic standards 15.6.1. US MIL STD 461B 15.6.2. International Electrotechnical Commission standard 555 15.6.3. IEEE/ANSI standard 519
16. Line-commutated rectifiers 16.1. The single-phase full wave rectifier 16.1.1. Continuous conduction mode 16.1.2. Discontinuous conduction mode 16.1.3. Behavior when C is large 16.1.4. Minimizing THD when C is small
16.2. The three-phase bridge rectifier 16.2.1. Continuous conduction mode 16.2.2. Discontinuous conduction mode
16.3. Phase control 16.3.1. Inverter mode 16.3.2. Harmonics and power factor 16.3.3. Commutation
17.4. RMS values of rectifier waveforms 17.4.1. Boost rectifier example 17.4.2. Comparison of single-phase rectifier topologies
17.5. Ideal three-phase rectifiers 17.5.1. Three-phase rectifiers operating in CCM 17.5.2. Some other approaches to three-phase rectification
17.6. Summary of key points 18. Low harmonic rectifier modeling and control 18.1. Modeling losses and efficiency in CCM high-quality rectifiers 18.1.1. Expression for controller duty cycle d(t) 18.1.2. Expression for the dc load current 18.1.3. Solution for converter efficiency 18.1.4. Design example
18.2. Controller schemes 18.2.1. Average current control 18.2.2. Feedforward 18.2.3. Current programmed control 18.2.4. Hysteretic control 18.2.5. Nonlinear carrier control
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Fundamentals of Power Electronics Table of Contents
Fundamentals of Power Electronics Table of Contents
20.1. The zero-current-switching quasi-resonant switch cell 18.3.1. Modeling the outer low-bandwidth control system 18.3.2. Modeling the inner wide-bandwidth average current controller
20.1.1. Waveforms of the half-wave ZCS quasi-resonant switch cell 20.1.2. The average terminal waveforms 20.1.3. The full-wave ZCS quasi-resonant switch cell
18.4. Summary of key points 20.2. Resonant switch topologies
Part V. Resonant converters
20.2.1. The zero-voltage-switching quasi-resonant switch 20.2.2. The zero-voltage-switching multi-resonant switch 20.2.3. Quasi-square-wave resonant switches
19. Resonant Conversion 19.1. Sinusoidal analysis of resonant converters 19.1.1. Controlled switch network model 19.1.2. Modeling the rectifier and capacitive filter networks 19.1.3. Resonant tank network 19.1.4. Solution of converter voltage conversion ratio M = V/Vg
20.3. Ac modeling of quasi-resonant converters 20.4. Summary of key points
Appendices 19.2. Examples Appendix 1. RMS values of commonly-observed converter waveforms 19.2.1. Series resonant dc-dc converter example 19.2.2. Subharmonic modes of the series resonant converter 19.2.3. Parallel resonant dc-dc converter example
19.3. Exact characteristics of the series and parallel resonant converters 19.3.1. Series resonant converter 19.3.2. Parallel resonant converter
19.4. Soft switching 19.4.1. Operation of the full bridge below resonance: zero-current switching 19.4.2. Operation of the full bridge above resonance: zero-voltage switching 19.4.3. The zero voltage transition converter
A1.1. Some common waveforms A1.2. General piecewise waveform Appendix 2. Magnetics design tables A2.1. Pot core data A2.2. EE core data A2.3. EC core data A2.4. ETD core data A2.5. PQ core data A2.6. American wire gauge data Appendix 3. Averaged switch modeling of a CCM SEPIC
19.5. Load-dependent properties of resonant converters
Index 19.5.1. Inverter output characteristics 19.5.2. Dependence of transistor current on load 19.5.3. Dependence of the ZVS/ZCS boundary on load resistance
19.6. Summary of key points 20. Quasi-resonant converters
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Fundamentals of Power Electronics Table of Contents Chapter titles only
Fundamentals of Power Electronics
Fundamentals of Power Electronics Table of Contents Chapter titles only
17. The ideal rectifier 18. Low harmonic rectifier modeling and control
R. W. Erickson
Part V. Resonant Converters
Table of Contents
19. Resonant conversion 20. Quasi-resonant converters
Chapter titles only
Appendices
Back 1. Introduction
Part I. Converters in Equilibrium
Appendix 1. RMS values of commonly-observed converter waveforms Appendix 2. Magnetics design tables Appendix 3. Averaged switch modeling of a CCM SEPIC
Index
2. Principles of steady state converter analysis 3. Steady-state equivalent circuit modeling, losses, and efficiency 4. Switch realization 5. The discontinuous conduction mode 6. Converter circuits
Part II. Converter Dynamics and Control 7. AC modeling 8. Converter transfer functions 9. Controller design 10. Ac and dc equivalent circuit modeling of the discontinuous conduction mode 11. Current programmed control
Part III. Magnetics 12. Basic magnetics theory 13. Filter inductor design 14. Transformer design
Part IV. Modern Rectifiers, and Power System Harmonics 15. Power and harmonics in nonsinusoidal systems 16. Line-commutated rectifiers
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Fundamentals of Power Electronics Preface
Fundamentals of Power Electronics Preface
needed. This material can then be directly applied to power electronics examples. This approach is suitable for a course in the fourth or fifth year, in which fundamentals are stressed.
Preface Fundamentals of Power Electronics R. W. Erickson
Approach (2) is employed here. Thus, the book is not intended for survey courses, but rather, it treats fundamental concepts and design problems in sufficient depth that students can actually build converters. An attempt is made to deliver specific results. Completion of core circuits and electronics courses is the only prerequisite assumed; prior knowledge in the areas of magnetics, power, and control systems is helpful but not required.
Back In many university curricula, the power electronics field has evolved beyond the status of comprising one or two special-topics courses. Often there are several courses dealing with the power electronics field, covering the topics of converters, motor drives, and power devices, with possibly additional advanced courses in these areas as well. There may also be more traditional power-area courses in energy conversion, machines, and power systems. In the breadth vs. depth tradeoff, it no longer makes sense for one textbook to attempt to cover all of these courses; indeed, each course should ideally employ a dedicated textbook. This text is intended for use in introductory power electronics courses on converters, taught at the senior or first-year graduate level. There is sufficient material for a one year course or, at a faster pace with some material omitted, for two quarters or one semester. The first class on converters has been called a way of enticing control and electronics students into the power area via the "back door". The power electronics field is quite broad, and includes fundamentals in the areas of ● ● ● ● ●
Converter circuits and electronics Control systems Magnetics Power applications Design-oriented analysis
This wide variety of areas is one of the things which makes the field so interesting and appealing to newcomers. This breadth also makes teaching the field a challenging undertaking, because one cannot assume that all students enrolled in the class have solid prerequisite knowledge in so many areas. Indeed, incoming students may have individual backgrounds in the power, control, or electronics areas, but rarely in all three. Yet it is usually desired to offer the class to upper-division undergraduate and entering graduate students. Hence, in teaching a class on converters (and in writing a textbook), there are two choices: 1. Avoid the problem of prerequisites, by either (a) assuming that the students have all of the prerequisites and discussing the material at a high level (suitable for an advanced graduate class), or (b) leaving out detailed discussions of the various contributing fields. 2. Attack the problem directly, by teaching or reviewing material from prerequisite areas as it is http://ece-www.colorado.edu/~pwrelect/book/contents/Preface.html (1 of 4) [25/04/2002 16:42:32]
In the power electronics literature, much has been made of the incorporation of other disciplines such as circuits, electronic devices, control systems, magnetics, and power applications, into the power electronics field. Yet the field has evolved, and now is more than a mere collection of circuits and applications linked to the fundamentals of other disciplines. There is a set of fundamentals that are unique to the field of power electronics. It is important to identify these fundamentals, and to explicitly organize our curricula, academic conferences, and other affairs around these fundamentals. This book is organized around the fundamental principles, while the applications and circuits are introduced along the way as examples. A concerted effort is made to teach converter modeling. Fundamental topics covered include: ●
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Fundamentals of PWM converter analysis, including the principles of inductor volt-second balance and capacitor charge balance, and the small-ripple approximation (Chapter 2). Converter modeling, including the use of the dc transformer model, to predict efficiency and losses (Chapter 3). Realization of switching elements using semiconductor devices. One-, two-, and four-quadrant switches. A brief survey of power semiconductor devices (Chapter 4). An up-to-date treatment of switching losses and their origins. Diode stored charge, device capacitances, and ringing waveforms (Chapter 4). Origin and steady-state analysis of the discontinuous conduction mode (Chapter 5). Converter topologies (Chapter 6). The use of averaging to model converter small-signal ac behavior. Averaged switch modeling (Chapter 7). Converter small-signal ac transfer functions, including the origins of resonances and right half-plane zeroes. Control-to-output and line-to-output transfer functions, and output impedance (Chapter 8). A basic discussion of converter control systems, including objectives, the system block diagram, and the effect of feedback on converter behavior (Chapter 9). Ac modeling of the discontinuous conduction mode. Quantitative behavior of DCM small-signal transfer functions (Chapter 10). Current-programmed control. Oscillation for D > 0.5. Equivalent circuit modeling (Chapter 11). Basic magnetics, including inductor and transformer modeling, and loss mechanisms in high-frequency power magnetics (Chapter 12). An understanding of what determines the size of power inductors and transformers. Power inductor and transformer design issues (Chapters 13 and 14). Harmonics in power systems (Chapter 15). A modern viewpoint of rectifiers, including harmonics, power factor, and mitigation techniques in conventional rectifiers, and operation of sophisticated low-harmonic rectifiers (Chapters 16-18). Analysis and modeling of low-harmonic rectifiers (Chapters 17-18). Resonant inverters and dc-dc converters: approximate analysis techniques, characteristics of basic converters, and
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Fundamentals of Power Electronics Preface
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load-dependent properties (Chapter 19). Zero voltage switching, zero current switching, and the zero-voltage-transition converter (Chapter 19). Resonant switch converters, including basic operation, efficiency and losses, and ac modeling (Chapter 20).
On teaching averaged converter modeling: I think that this is one of the important fundamentals of the field, and hence we should put serious effort into teaching it. Although we in the academic community may debate how to rigorously justify averaging, nonetheless it is easy to teach the students to average: Just average all of the waveforms over one switching period. In particular, for the continuous conduction mode, average the inductor voltages and capacitor currents over one switching period, ignoring the ripple. That's all that is required, and I have found that students quickly and easily learn to average waveforms. The results are completely general, they aren't limited to SPDT switches, and they can easily be used to refine the model by inclusion of losses, dynamics, and control variations. To model dynamics, it is also necessary to linearize the resulting equations. But derivation of small-signal models is nothing new to the students --they have already seen this in their core electronics classes, as well as in numerous math courses and perhaps also in energy conversion. It isn't necessary to teach full-blown state-space averaging, but I have included an optional (with asterisk) section on this for the graduate students. I personally prefer to initially skip Sections 7.4 and 7.5. After covering Chapters 8 and 9, I return to cover Sections 7.4 and 7.5 before teaching Chapters 10 and 11. Averaging aside, it is also important to teach modeling in a pedagogically sound way. The object is to describe the important properties of the converter, in a simple and clear way. The dc transformer represents the basic function of a dc-dc converter, and so the modeling process should begin with a dc transformer having a turns ratio equal to the conversion ratio of the converter. For example, the model of the buck-boost converter ought to contain a buck transformer cascaded by a boost transformer, or perhaps the two transformers combined into a single D : D' transformer. This first-order model can later be refined if desired, by addition of loss elements, dynamic elements, etc. The design-oriented analysis methods of R. D. Middlebrook have been well accepted by a significant portion of the power electronics community. While the objective of this text is the introduction of power electronics rather than design-oriented analysis, the converter analyses and examples are nonetheless done in a design-oriented way. Approximations are often encouraged, and several of the techniques of design-oriented analysis are explicitly taught in parts of Chapters 8 and 9. We need to teach our students how to apply our academic theory to real-world, and hence complicated, problems. Design-oriented analysis is the missing link. Chapter 8 contains a "review" of Bode diagrams, including resonant responses and right half-plane zeroes. Also included is material on design-oriented analysis, in the context of converter transfer functions. The Bode diagram material is covered in more depth than in prerequisite classes. I have found that the material of Chapter 8 is especially popular with continuing education students who are practicing engineers. I recommend at least quickly covering this chapter in lecture. Those instructors who choose to skip some or all of Chapter 8 can assign it as reading, and hold students responsible for the material. In a similar manner, Chapter 9 contains a "review" of classical control systems, in the context of switching regulators. This chapter explicitly makes the connection between the small-signal converter models http://ece-www.colorado.edu/~pwrelect/book/contents/Preface.html (3 of 4) [25/04/2002 16:42:32]
Fundamentals of Power Electronics Preface
derived in other chapters, and their intended application. Many power area students are unfamiliar with this material, and even control-area students comment that they learned something from the designoriented approach. Parts III, IV, and V can be covered in any order. Part III includes a review of basic magnetics, a discussion of proximity loss, and an introduction to the issues governing design of magnetic devices. The inclusion of step-by-step design procedures may be somewhat controversial; however, these procedures explicitly illustrate the issues inherent in magnetics design. Student tendencies towards cookbook mentality are mitigated by the inclusion of homework problems that cannot be solved using the given step-by-step procedures. Part IV, entitled "Modern rectifiers," covers the issues of power system harmonics, generation of harmonics by conventional rectifiers, and low-harmonic rectifiers. Chapters 17 and 18 cover low-harmonic rectifiers in depth, including converter analysis and modeling, and rectifier control systems. Resonant converters are treated in Part V. There have been a tremendous number of papers written on resonant converters, most of which are very detailed and complicated. Indeed, the complexity of resonant converter behavior makes it challenging to teach this subject in depth. Two somewhat introductory chapters are included here. State-plane analysis is omitted, and is left for an advanced graduate class. In Chapter 19, resonant inverters and dc-dc converters are introduced and are analyzed via the sinusoidal approximation. Soft switching is described, in the context of both resonant converters and the zero-voltage transition converter. Some resonant network theorems are also presented, which yield insight into the design of resonant inverters with reduced circulating currents, with zerovoltage switching over a wide range of load currents, and with desired output characteristics. Resonant switch converters are introduced and modeled in Chapter 20. Most chapters include both short analysis problems, and longer analysis and/or design problems. References are given at the end of each chapter; these are not intended to be exhaustive bibliographies, but rather are a starting place for additional reading. This text has evolved from course notes developed over thirteen years of teaching power electronics at the University of Colorado, Boulder. These notes, in turn, were heavily influenced by my previous experience as a graduate student at the California Institute of Technology, under the direction of Profs. Slobodan Cuk and R. D. Middlebrook, to whom I am grateful. In addition, I appreciate the numerous helpful technical discussions and suggestions of my colleague at the University of Colorado, Prof. Dragan Maksimovic. I would also like to thank the following individuals for their suggestions: Prof. Arthur Witulski (University of Arizona, Tucson), Prof. Sigmund Singer (Tel-Aviv University, Israel), Dr. Michael Madigan, and Carlos Oliveira. Robert W. Erickson Boulder, Colorado
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http://ece-www.colorado.edu/~pwrelect/book/errata.html
http://ece-www.colorado.edu/~pwrelect/book/errata.html
Fundamentals of Power Electronics ●
Errata, first edition Back
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Errors can be reported for inclusion on this list by sending email to Bob Erickson. Please include page number and a short description.
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Page ii, small-print disclaimer near bottom of page contains two misspelled words Page 10, paragraph beginning "Design of the converter..." "line-to-transfer functions" should read "line-to-output transfer functions" Page 49, Eq. (3.19) RHS of equation should be divided by R Page 86, Fig. 4.35(a) Units of horizontal axis should be A Page 172, Fig. 6.36 The bottom of the figure is cut off. Transistor Q4 conducts during the first subinterval of every switching period. Page 180, third line under heading 6.4.2 typo "votlage" should read "voltage" Page 181, Table 6.2 Values of L and C are missing. The values are: L = 25.96 uH, C = 25.0 uF. Page 188, problem 6.8(d) Forward voltage drops of diodes D1 and D2 are equal to 0.7 V. Page 237, second line following Eq. (7.126) "small" should read "smaller" Page 241, third paragraph, first line reads "valiables" -- should read "variables" Page 250, Equation (7.144) should read: e(s) = (Vg - V) - sLI/DD' (plus sign should be minus sign) Page 275, Figure 8.15 Phase asymptote for frequencies between 10*f1 and f2/10 should be +90 deg instead of -90 deg. Pages 289-293, Equation (8.112) and Table 8.2 Missing factor of D' in Eq. (8.112). Mistake propagates to Eqs. (8.113), (8.121), (8.123), (8.124), (8.127), and Fig. 8.26. Buck-boost Gd0 entry in Table 8.2 is also incorrect. Corrected Section 8.2.1 and Table 8.2 are in attached pdf file. Page 300, line following Eq. (8.144) Should read: "Subsitution of Eq. (8.142) into Eq. (8.144) leads to" Page 311, last sentence
http://ece-www.colorado.edu/~pwrelect/book/errata.html (1 of 2) [25/04/2002 16:42:57]
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is missing a period Page 333, fourth line following Eq. (9.20) reads "simple" -- should be "simply" Page 347, first line of text Equation reference should read "Eq. (7.149)" Page 366, Problem 9.8, last sentence of paragraph (before (a)) should read "The pulse-width modulator circuit obeys Eq. (7.149)." -- not "(7.132)" Page 385, Table 10.2 g2 for the buck-boost converter is equal to 2/MRe Page 419, Figure 11.12(a) ig(t) is the current flowing out of the source vg(t) Page 427, Figure 11.20 Value of ac resistance is inverted. The value should read: -R/D^2 Page 488, second paragraph, fourth sentence Change "greater" to "smaller", so that the sentence reads: "The skin depth d is smaller for high frequencies..." Page 516, Equations (14.14) and (14.15) Summation limits should be j = 1 to k Page 519, Equation (14.28) should contain a factor of 4 in denominator Page 534, Equation (14.84) should read: Bpk = Bmax + LIdc/nAc Page 546, Equation (15.14) Power loss should be (rms voltage)^2 / Rshunt Page 548, last sentence change "displacement factor" to "distortion factor" Page 576, end of first sentence after Eq. (16.12) is should read ir Page 637, Equation (18.27) Corrected version is in attached pdf file. Page 654, Figure 18.20, and Problem 18.2 Flyback transformer turns ratio is 1:1 Appendix 3, page 760, Figure A3.3 Polarity of dependent current source should be reversed.
update 10/25/99
http://ece-www.colorado.edu/~pwrelect/book/errata.html (2 of 2) [25/04/2002 16:42:57]
Fundamentals of Power Electronics Solutions to Selected Problems
Fundamentals of Power Electronics
Fundamentals of Power Electronics Solutions to Selected Problems
Chapter 8 ●
R. W. Erickson
Sample problem 164kB. A solved problem on transfer function asymptotes, similar to problems 8.1 - 8.4.
Chapter 12
Solutions to Selected Problems ●
Solutions to selected problems at the end of chapters in the text Fundamentals of Power Electronics are listed below. These are intended as an educational supplement to the text, for the use of students who have purchased the book. All files copyright R. W. Erickson 1998.
Problem 12.1 116kB. Magnetic circuit analysis.
Update 11/23/98 rwe
These files can be read using the Adobe Acrobat viewer, available free from the Adobe Acrobat web site. Up Chapter 2 ● ●
Problem 2.2 144kB. Analysis of a SEPIC in a voltage regulator application. Problem 2.3 168kB. Inductor current and capacitor voltage ripples in the SEPIC.
Chapter 3 ● ●
Problem 3.3 192kB. Modeling efficiency and losses in a buck converter with input filter. Problem 3.6 148kB. Comparison of boost and buck-boost converters in a voltage step-up application.
Chapter 4 ●
Problem 4.4 196kB. Switch realization in a converter that is probably unfamiliar.
Chapter 5 ● ●
Problem 5.5 81kB. CCM/DCM boundary for the Cuk converter. Problem 5.6 184kB. Solution of the Cuk converter in DCM.
Chapter 6 ●
Problem 6.1 144kB. Analysis of a tapped-inductor boost converter.
http://ece-www.colorado.edu/~pwrelect/book/solutions/solndir.html (1 of 2) [25/04/2002 16:42:59]
http://ece-www.colorado.edu/~pwrelect/book/solutions/solndir.html (2 of 2) [25/04/2002 16:42:59]
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Fundamentals of Power Electronics
1
Robert W. Erickson University of Colorado, Boulder
Chapter 1: Introduction
Fundamentals of Power Electronics
Some applications of power electronics Elements of power electronics
1.2. 1.3.
2
Chapter 1: Introduction
Control input
Switching converter
Power output
Fundamentals of Power Electronics
3
Chapter 1: Introduction
Change and control voltage magnitude Possibly control dc voltage, ac current Produce sinusoid of controllable magnitude and frequency Ac-ac cycloconversion: Change and control voltage magnitude and frequency
Dc-dc conversion: Ac-dc rectification: Dc-ac inversion:
Power input
1.1 Introduction to Power Processing
Fundamentals of Power Electronics
Summary of the course
Introduction to power processing
1.1.
Chapter 1: Introduction
Pout Pin
Fundamentals of Power Electronics
High efficiency leads to low power loss within converter Small size and reliable operation is then feasible Efficiency is a good measure of converter performance
1 –1 Ploss = Pin – Pout = Pout η
η=
4
reference
Controller
Control input
Switching converter
feedback
Power output
5
η
0.2
0.4
0.6
0.8
1
0
0.5
1.5
Chapter 1: Introduction
1
Chapter 1: Introduction
Ploss / Pout
High efficiency is essential
Fundamentals of Power Electronics
feedforward
Power input
Control is invariably required
Converter
Pout
6
Chapter 1: Introduction
Capacitors
Fundamentals of Power Electronics
Resistors
7
Magnetics
DT
Chapter 1: Introduction
s s linearswitched-mode mode Semiconductor devices
T
Devices available to the circuit designer
Fundamentals of Power Electronics
A goal of current converter technology is to construct converters of small size and weight, which process substantial power at high efficiency
Pin
A high-efficiency converter
+ –
Capacitors
Magnetics
8
Chapter 1: Introduction
Capacitors
Magnetics
Fundamentals of Power Electronics
9
Chapter 1: Introduction
Power processing: avoid lossy elements
Resistors
DT
s s linearswitched-mode mode Semiconductor devices
T
Devices available to the circuit designer
Fundamentals of Power Electronics
Signal processing: avoid magnetics
Resistors
DT
s s linearswitched-mode mode Semiconductor devices
T
Devices available to the circuit designer
+ – + –
i(t) = 0 p(t) = v(t) i(t) = 0
Switch open: In either event:
10
–
v(t)
+
Chapter 1: Introduction
i(t)
Dc-dc converter
R 5Ω
–
V 50V
+
11
Chapter 1: Introduction
Input source: 100V Output load: 50V, 10A, 500W How can this converter be realized?
+ –
Fundamentals of Power Electronics
100V
Vg
I 10A
A simple dc-dc converter example
Fundamentals of Power Electronics
Ideal switch consumes zero power
v(t) = 0
Switch closed:
Power loss in an ideal switch
+ –
12
Ploss = 500W
50V –
+ –
Fundamentals of Power Electronics
Pin ≈ 1000W
100V
Vg
Ploss ≈ 500W
13
linear amplifier and base driver
50V –
–+
Vref
Chapter 1: Introduction
Pout = 500W
–
V 50V
+
Chapter 1: Introduction
I 10A
R 5Ω
Series pass regulator: transistor operates in active region +
–
V 50V
+
Pout = 500W
R 5Ω
I 10A
Dissipative realization
Fundamentals of Power Electronics
Pin = 1000W
100V
Vg
+
Resistive voltage divider
Dissipative realization
vs(t)
switch position:
+ –
14
–
vs(t)
1
DTs
Vg
2
0 (1–D) Ts
2
(1 – D) Ts
0
0
Ts
vs(t) dt = DVg
Fundamentals of Power Electronics
Vs = 1 Ts
15
1
t
Vs = DVg
DC component of vs(t) = average value:
switch position:
vs(t)
1
DTs
Vg
2
+
1
t
Vs = DVg
R
I 10A
Chapter 1: Introduction
–
v(t) 50V
+
Chapter 1: Introduction
fs = switching frequency = 1 / Ts
Ts = switching period
D = switch duty cycle 0≤D≤1
The switch changes the dc voltage level
Fundamentals of Power Electronics
100V
Vg
1
Use of a SPDT switch
+ –
–
v(t)
Pout = 500W
R
δ(t)
dTs Ts
δ
transistor gate driver
Fundamentals of Power Electronics
+ –
Power input
vg
16
t
17
i
Load
reference vref input
error signal ve
–
v
+
pulse-width vc G (s) c modulator compensator
Switching converter
Hv
H(s)
Chapter 1: Introduction
sensor gain
Chapter 1: Introduction
Addition of control system for regulation of output voltage
Fundamentals of Power Electronics
This circuit is known as the “buck converter”
C
+
•
Ploss small
–
vs(t)
L
Choose filter cutoff frequency f0 much smaller than switching frequency fs
2
+
i(t)
•
Pin ≈ 500W
100V
Vg
1
Addition of (ideally lossless) L-C low-pass filter, for removal of switching harmonics:
Addition of low pass filter
–+
+ –
V
+ – 2
Fundamentals of Power Electronics
vs(t)
Vg
1
0
Vg
2Vg
3Vg
4Vg
5Vg
0
0.2
18
0.4
1
D
0.6
C
0.8
1
R
+
19
+
vs(t)
load
v(t)
–
–
t
1
2
A single-phase inverter
Fundamentals of Power Electronics
Vg
L
2
The boost converter
–
Chapter 1: Introduction
Modulate switch duty cycles to obtain sinusoidal low-frequency component
“H-bridge”
Chapter 1: Introduction
V
+
20
Chapter 1: Introduction
Fundamentals of Power Electronics
ac line input 85-265Vrms
vac(t)
iac(t) Rectifier
21
dc link
–
+ Dc-dc converter
Chapter 1: Introduction
loads
regulated dc outputs
A computer power supply system
Fundamentals of Power Electronics
• 1000 MW in rectifiers and inverters for utility dc transmission lines
• kW to MW in variable-speed motor drives
• tens, hundreds, or thousands of watts in power supplies for computers or office equipment
• less than 1 W in battery-operated portable equipment
Power levels encountered in high-efficiency converters
1.2 Several applications of power electronics
Batteries
Battery charge/discharge controllers
–
vbus
+
22
Payload
Dc-dc converter
Chapter 1: Introduction
Payload
Dc-dc converter
Fundamentals of Power Electronics
50/60Hz
3øac line
Rectifier
–
vlink
+
23
Dc link
Inverter
Chapter 1: Introduction
Ac machine
variable-frequency variable-voltage ac
A variable-speed ac motor drive system
Fundamentals of Power Electronics
Solar array
Dissipative shunt regulator
A spacecraft power system
Vg – V L
1
DTs
iL(DTs) –V L
2
–V
D'Ts
Ts
t
∆iL
1
t
Fundamentals of Power Electronics
Transformer isolation
Discontinuous conduction mode
0
I iL(0)
iL(t)
switch position:
DTs
Vg – V
Inductor waveforms vL(t)
24
Vg
D Ron
I
D' VD
25
η
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
0
0.1
0.2
0.3
0.4
D' RD
Predicted efficiency
+ –
RL
D
0.5
0.8
0.01
0.9
0.002
–
V
1
R
Chapter 1: Introduction
0.6
0.02
0.7
0.05
D' : 1
RL/R = 0.1
Averaged equivalent circuit +
Chapter 1: Introduction
Part I. Converters in equilibrium
Fundamentals of Power Electronics
analog circuits electronic devices control systems power systems magnetics electric machines numerical simulation
Power electronics incorporates concepts from the fields of
1.3 Elements of power electronics
+ –
p
Gate
n
Collector
p
n-
Emitter
p
n
emitter
collector
26
Switching loss
pA(t)
= vA iA
diode waveforms
transistor waveforms
t1 t2
tr
vB(t)
Fundamentals of Power Electronics
6. Converter circuits
27
5. The discontinuous conduction mode
4. Switch realization
t
t
t
Chapter 1: Introduction
3. Steady-state equivalent circuit modeling, losses, and efficiency
2. Principles of steady state converter analysis
area ~iLVgtr
–Vg
0
0
iL
Chapter 1: Introduction
area ~QrVg
t0
iB(t)
area –Qr
0
iL
0
vA(t)
Vg
iA(t) Qr
Part I. Converters in equilibrium
minority carrier injection
n
gate
Fundamentals of Power Electronics
n
The IGBT
Switch realization: semiconductor devices
δ(t)
dTs Ts
δ(t)
transistor gate driver
vg(t)
t
R
I d(t)
voltage reference vref
–
v(t)
+
Load
v
1:D
28
feedback connection
L
Vg – V d(t)
gate drive
D' : 1
C
–
v(t)
Controller design Ac and dc equivalent circuit modeling of the discontinuous conduction mode Current-programmed control
9. 10.
11.
Chapter 1: Introduction
Converter transfer functions
8.
29
Ac modeling
7.
Fundamentals of Power Electronics
R
t
t
Chapter 1: Introduction
I d(t)
+
averaged waveform
Ts with ripple neglected
actual waveform v(t) including ripple
Averaging the waveforms
Part II. Converter dynamics and control
+ –
Controller
Fundamentals of Power Electronics
t
vc(t)
compensator pulse-width vc Gc(s) modulator
Switching converter
Small-signal averaged equivalent circuit
vg(t) + –
Power input
–+
Closed-loop converter system
Part II. Converter dynamics and control
+ –
Basic magnetics theory Filter inductor design Transformer design
12. 13. 14.
Fundamentals of Power Electronics
250kHz
1811
ik(t)
400kHz
1811
30
Switching frequency
200kHz
2213
Rk
R2
i2(t)
500kHz
2213
1000kHz
2616
0
0.02
0.04
0.06
0.08
0.1
the proximity effect
31
Part III. Magnetics
100kHz
2616
: nk
n1 : n 2
50kHz
3622
iM(t) LM R1
25kHz
4226
i1(t)
Fundamentals of Power Electronics
transformer size vs. switching frequency
transformer design
Pot core size
Part III. Magnetics
Bmax (T)
current density J
i
–i
3i
2i
–2i
d
Φ
2Φ
Chapter 1: Introduction
Chapter 1: Introduction
layer 1
layer 2
layer 3
1
100%
3
91%
5
73%
9
32%
11
13
17
19
32
Model of the ideal rectifier
vcontrol(t)
iac(t)
ac input
–
vac(t)
+
multiplier
va(t)
ig(t) Rs
–
L
vcontrol
Re(vcontrol)
–
v(t)
+
dc output
i(t)
–
v(t)
+ R
Line-commutated rectifiers The ideal rectifier Low harmonic rectifier modeling and control
16. 17. 18.
Chapter 1: Introduction
Power and harmonics in nonsinusoidal systems
15.
33
C
i(t)
Chapter 1: Introduction
p(t) = vac / Re
2
Ideal rectifier (LFR)
controller
and power system harmonics
Fundamentals of Power Electronics
PWM
Q1
D1
boost converter
– verr(t) + Gc(s) vref(t) = kx vg(t) vcontrol(t) compensator
X
vg(t)
vg(t)
+
Part IV. Modern rectifiers,
15
19% 15% 15% 13% 9%
Harmonic number
7
52%
THD = 136% Distortion factor = 59%
Fundamentals of Power Electronics
0%
20%
40%
60%
80%
100%
vac(t)
iac(t)
ig(t)
A low-harmonic rectifier system
and power system harmonics
Part IV. Modern rectifiers,
Pollution of power system by rectifier current harmonics
Harmonic amplitude, percent of fundamental
Q2 D2
D1
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Q4
Q3
0
0.75 1 1.5 2 3.5 5 10 Q = 20
0.5
0.35
0.2
Q = 0.2
D4
D3
0.6
C
0.8
1
F = f s / f0
1.2
34
1.4
35
Resonant conversion Quasi-resonant converters
Fundamentals of Power Electronics
19. 20.
0.4
L
1.6
1.8
R
2
3.5 5 10 Q = 20
2
1.5
1
0.75
0.5
0.35
Q = 0.2
–
V
+
conducting devices:
vds1(t)
commutation interval
X D2 D3
t
Chapter 1: Introduction
turn off Q1, Q4
Q1 Q4
Vg
Zero voltage switching
Chapter 1: Introduction
Part V. Resonant converters
Fundamentals of Power Electronics
Dc characteristics
Vg
+ –
Q1
1:n
Part V. Resonant converters The series resonant converter
M = V / Vg
Fundamentals of Power Electronics
complement D’: D’ = 1 - D
Duty cycle D: 0≤D≤1
Switch output voltage waveform
1
vs(t)
+ –
2
0 switch position:
Vg
1
DTs
Vg
2
DTs
–
vs(t)
2
0
D' Ts Ts
R
1
t
–
v(t)
+
Chapter 2: Principles of steady-state converter analysis
1
+
Chapter 2: Principles of steady-state converter analysis
2.1 Introduction Buck converter
SPDT switch changes dc component
Fundamentals of Power Electronics
2.6. Summary of key points
2.5. Estimating the ripple in converters containing twopole low-pass filters
2.4. Cuk converter example
2.3. Boost converter example
2.2. Inductor volt-second balance, capacitor charge balance, and the small ripple approximation
2.1. Introduction
Chapter 2 Principles of Steady-State Converter Analysis
0
area = DTsVg
Vg
DTs
0
= DVg
Ts
0
Ts
vs(t) dt
3
Chapter 2: Principles of steady-state converter analysis
t
+ –
Fundamentals of Power Electronics
v ≈ vs = DVg
Vg
1
2
4
–
vs(t)
+
0 0
C
R
1
–
v(t)
+
D
Chapter 2: Principles of steady-state converter analysis
Vg
V
L
Insertion of low-pass filter to remove switching harmonics and pass only dc component
Fundamentals of Power Electronics
vs = 1 (DTsVg) = DVg Ts
vs = 1 Ts
Fourier analysis: Dc component = average value
vs(t)
Dc component of switch output voltage
+ –
+ –
+ –
iL(t)
1
L
L
2
iL(t)
2
1
C
iL(t)
5
2
L
R
C
C
–
v
+
–
v
1
-5
-4
-3
-2
-1
0
0
1
2
3
4
5
0
0.2
0.4
0.6
0.8
0.4
0.2
0
0.4
0.4
M(D) = 1––DD
0.2
M(D) = 1 –1 D
0.2
0
0
M(D) = D
D
D
D
0.6
0.6
0.6
0.8
0.8
0.8
1
1
1
Chapter 2: Principles of steady-state converter analysis
–
v
+
R
R
+
6
Chapter 2: Principles of steady-state converter analysis
Develop techniques for easily determining output voltage of an arbitrary converter circuit Derive the principles of inductor volt-second balance and capacitor charge (amp-second) balance Introduce the key small ripple approximation Develop simple methods for selecting filter element values Illustrate via examples
Fundamentals of Power Electronics
●
●
●
●
●
Vg
c)
Vg
b)
Vg
1
Objectives of this chapter
Fundamentals of Power Electronics
Buck-boost
Boost
Buck
a)
Three basic dc-dc converters
M(D) M(D) M(D)
0
V
v(t)
+ –
7
iC(t)
actual waveform v(t) = V + vripple(t)
C
+ vL(t) –
L
0
V Dc component V
actual waveform v(t) = V + vripple(t)
t
Fundamentals of Power Electronics
v(t) ≈ V
8
vripple << V
t
–
v(t)
Chapter 2: Principles of steady-state converter analysis
In a well-designed converter, the output voltage ripple is small. Hence, the waveforms can be easily determined by ignoring the ripple:
v(t) = V + vripple(t)
v(t)
R
+
Chapter 2: Principles of steady-state converter analysis
Dc component V
2
iL(t)
The small ripple approximation
Fundamentals of Power Electronics
v(t) = V + vripple(t)
Actual output voltage waveform
Vg
1
Actual output voltage waveform, buck converter
Inductor volt-second balance, capacitor charge balance, and the small ripple approximation
Buck converter containing practical low-pass filter
2.2.
+ –
+ –
iL(t)
C
+ vL(t) –
L iC(t) R –
v(t)
+
2
9
iL(t)
R –
v(t)
+
+ –
iL(t)
C
+ vL(t) –
L
R
Vg
+ –
iL(t)
C
+ vL(t) –
L iC(t) R
diL(t) dt
Fundamentals of Power Electronics
Solve for the slope: diL(t) vL(t) Vg – V = ≈ L L dt
vL(t) = L
–
v(t)
+
–
v(t)
+
10
Chapter 2: Principles of steady-state converter analysis
⇒ The inductor current changes with an essentially constant slope
Knowing the inductor voltage, we can now find the inductor current via
vL ≈ Vg – V
Small ripple approximation:
vL = Vg – v(t)
Inductor voltage
iC(t)
Chapter 2: Principles of steady-state converter analysis
Vg
iC(t)
switch in position 2
C
+ vL(t) –
L
Inductor voltage and current Subinterval 1: switch in position 1
Fundamentals of Power Electronics
Vg
Vg
switch in position 1
original converter
1
Buck converter analysis: inductor current waveform
Vg
+ – iL(t) C
+ vL(t) –
L iC(t) R
–
v(t)
+
diL(t) dt
11
Chapter 2: Principles of steady-state converter analysis
⇒ The inductor current changes with an essentially constant slope
Fundamentals of Power Electronics
0
I iL(0)
iL(t)
switch position:
vL(t)
Vg – V L
1
DTs
12
DTs
iL(DTs)
Vg – V
Ts
1
t
∆iL
t
vL(t) = L
diL(t) dt
Chapter 2: Principles of steady-state converter analysis
–V L
2
–V
D'Ts
Inductor voltage and current waveforms
Fundamentals of Power Electronics
diL(t) ≈– V L dt
Solve for the slope:
vL(t) = L
Knowing the inductor voltage, we can again find the inductor current via
vL(t) ≈ – V
Small ripple approximation:
vL(t) = – v(t)
Inductor voltage
Inductor voltage and current Subinterval 2: switch in position 2
∆iL =
Ts
Vg – V DTs 2L
13
t
Vg – V DTs 2∆iL
2Ts
Vg – v(t) L – v(t) L
14
(n+1)Ts
t
Chapter 2: Principles of steady-state converter analysis
iL((n + 1)Ts) = iL(nTs)
nTs
iL(nTs)
iL((n+1)Ts)
Chapter 2: Principles of steady-state converter analysis
L=
When the converter operates in equilibrium:
0 DTs Ts
Fundamentals of Power Electronics
iL(Ts) iL(0)=0
iL(t)
DTs
–V L
∆iL
Inductor current waveform during turn-on transient
Fundamentals of Power Electronics
⇒
Vg – V L
iL(DTs)
(change in iL) = (slope)(length of subinterval) Vg – V DTs 2∆iL = L
0
I iL(0)
iL(t)
Determination of inductor current ripple magnitude
0
Ts
vL(t) dt
0
Ts
vL(t) dt
15
vL(t)
DTs
Vg – V
0
Ts
Fundamentals of Power Electronics
16
0 = DVg – (D + D')V = DVg – V
Equate to zero and solve for V: V = DVg
–V
total area λ
Chapter 2: Principles of steady-state converter analysis
⇒
vL(t) dt = (Vg – V)(DTs) + ( – V)(D'Ts) Average voltage is vL = λ = D(Vg – V) + D'( – V) Ts
λ=
Integral of voltage waveform is area of rectangles:
Inductor voltage waveform, previously derived:
t
Chapter 2: Principles of steady-state converter analysis
Inductor volt-second balance: Buck converter example
Fundamentals of Power Electronics
s 0= 1 v (t) dt = vL Ts 0 L The average inductor voltage is zero in steady state.
T
Hence, the total area (or volt-seconds) under the inductor voltage waveform is zero whenever the converter operates in steady state. An equivalent form:
0=
In periodic steady state, the net change in inductor current is zero:
iL(Ts) – iL(0) = 1 L
Integrate over one complete switching period:
Inductor defining relation: di (t) vL(t) = L L dt
The principle of inductor volt-second balance: Derivation
0
Ts
iC(t) dt
0
Ts
iC(t) dt = iC
17
Fundamentals of Power Electronics
Realization using power MOSFET and diode
Boost converter with ideal switch
Vg
Vg
+ –
+ –
iL(t)
iL(t)
18
DTs
Ts
+ –
1
Q1
2
D1
C
iC(t)
C
iC(t)
R
R
–
v
+
–
v
Chapter 2: Principles of steady-state converter analysis
+ vL(t) –
L
+ vL(t) –
L
+
Chapter 2: Principles of steady-state converter analysis
2.3 Boost converter example
Fundamentals of Power Electronics
Hence, the total area (or charge) under the capacitor current waveform is zero whenever the converter operates in steady state. The average capacitor current is then zero.
0= 1 Ts
In periodic steady state, the net change in capacitor voltage is zero:
vC(Ts) – vC(0) = 1 C
Integrate over one complete switching period:
Capacitor defining relation: dv (t) iC(t) = C C dt
The principle of capacitor charge balance: Derivation
+ –
iL(t) + vL(t) –
L
+ –
C
iC(t) R
+ vL(t) –
–
v
+
19
1
R –
v
+
+ –
iL(t)
+ vL(t) –
L
C
iC(t)
switch in position 2
C
iC(t)
Fundamentals of Power Electronics
vL = Vg iC = – V / R
Small ripple approximation:
vL = Vg iC = – v / R
20
Inductor voltage and capacitor current
R
–
v
+
+ –
+ vL(t) –
C
iC(t) R
–
v
Chapter 2: Principles of steady-state converter analysis
Vg
iL(t)
L +
Chapter 2: Principles of steady-state converter analysis
Vg
2
Subinterval 1: switch in position 1
Fundamentals of Power Electronics
Vg
Vg
switch in position 1
original converter
iL(t)
L
Boost converter analysis
21
Vg
+ vL(t) – C
iC(t) R
–
v
+
Chapter 2: Principles of steady-state converter analysis
+ –
iL(t)
L
Fundamentals of Power Electronics
iC(t)
vL(t)
– V/R
DTs
DTs
Vg
22
t
t
Chapter 2: Principles of steady-state converter analysis
D'Ts
I – V/R
Vg – V
D'Ts
Inductor voltage and capacitor current waveforms
Fundamentals of Power Electronics
vL = Vg – V iC = I – V / R
Small ripple approximation:
vL = Vg – v iC = iL – v / R
Inductor voltage and capacitor current
Subinterval 2: switch in position 2
vL(t) dt = (Vg) DTs + (Vg – V) D'Ts
23
Vg – V
D'Ts
0
Fundamentals of Power Electronics
0
1
2
3
4
5
24
0.4
D
0.6
0.8
1
Chapter 2: Principles of steady-state converter analysis
M(D) = 1 = 1 D' 1 – D
0.2
t
Chapter 2: Principles of steady-state converter analysis
DTs
Vg
Conversion ratio M(D) of the boost converter
Fundamentals of Power Electronics
M(D) = V = 1 = 1 Vg D' 1 – D
V =
Vg D' The voltage conversion ratio is therefore
Solve for V:
Vg (D + D') – V D' = 0
Equate to zero and collect terms:
0
Ts
vL(t)
Inductor volt-second balance Net volt-seconds applied to inductor over one switching period:
M(D)
iC(t) dt = ( – V ) DTs + (I – V ) D'Ts R R
25
0
0.2
– V/R
DTs
0.4
D
0.6
D'Ts
I – V/R
0.8
1
I
iL(t)
0
Vg L DTs
Vg – V L Ts
∆iL
t
Vg DTs 2L Fundamentals of Power Electronics
∆iL =
Solve for peak ripple:
26
Chapter 2: Principles of steady-state converter analysis
• Choose L such that desired ripple magnitude is obtained
Change in inductor current during subinterval 1 is (slope) (length of subinterval): Vg 2∆iL = DTs L
Inductor current slope during subinterval 1: diL(t) vL(t) Vg = = L L dt Inductor current slope during subinterval 2: diL(t) vL(t) Vg – V = = L L dt
t
Chapter 2: Principles of steady-state converter analysis
0
2
4
6
8
I (V g / R)
iC(t)
Determination of inductor current ripple
Fundamentals of Power Electronics
Eliminate V to express in terms of Vg: Vg I= 2 D' R
Solve for I: I= V D' R
Collect terms and equate to zero: – V (D + D') + I D' = 0 R
0
Ts
Capacitor charge balance:
Determination of inductor current dc component
v(t)
0
V –V RC DTs
I V C – RC Ts
∆v
t
27
Fundamentals of Power Electronics
Cuk converter: practical realization using MOSFET and diode
Cuk converter, with ideal switch
Vg
Vg
+ –
+ –
i1
i1
28
L1
L1
Q1
+ v1 –
C1
2
D1
L2
L2
C2
i2
C2
–
v2
+
–
v2
+
R
R
Chapter 2: Principles of steady-state converter analysis
1
+ v1 –
C1
i2
Chapter 2: Principles of steady-state converter analysis
• Choose C such that desired voltage ripple magnitude is obtained • In practice, capacitor equivalent series resistance (esr) leads to increased voltage ripple
2.4 Cuk converter example
Fundamentals of Power Electronics
∆v = V DTs 2RC
Solve for peak ripple:
– 2∆v = – V DTs RC
Change in capacitor voltage during subinterval 1 is (slope) (length of subinterval):
Capacitor voltage slope during subinterval 2: dvC(t) iC(t) I = = – V C RC C dt
Capacitor voltage slope during subinterval 1: dvC(t) iC(t) – V = = RC C dt
Determination of capacitor voltage ripple
Vg
Vg
+ –
+ –
i1
29
+ vL1 –
L1
i1 + vL1 –
L1
–
v1
+
C1
+ vL2 –
L2
iC1 + vL2 –
iC2
iC2 C2
i2
C2
i2
Vg
+ –
Fundamentals of Power Electronics
vL1 = Vg vL2 = – V1 – V2 i C1 = I 2 V i C2 = I 2 – 2 R
30
R
R
+
v1
–
C1
iC1 + vL2 –
L2
iC2 C2
i2
–
v2
+ R
Chapter 2: Principles of steady-state converter analysis
Small ripple approximation for subinterval 1:
vL1 = Vg vL2 = – v1 – v2 i C1 = i 2 v i C2 = i 2 – 2 R
i1 + vL1 –
L1
MOSFET conduction interval Inductor voltages and capacitor currents:
–
v2
+
–
v2
+
Chapter 2: Principles of steady-state converter analysis
C1
iC1
+
v1
–
L2
Waveforms during subinterval 1
Fundamentals of Power Electronics
Capacitor C1 is charged from input
Switch in position 2: diode conducts
Capacitor C1 releases energy to output
Switch in position 1: MOSFET conducts
with switch in positions 1 and 2
Cuk converter circuit
Vg
+ –
i1
31
C1
v1
+
–
iC1
+ vL2 –
L2
iC2 C2
i2
DTs
Vg
Fundamentals of Power Electronics
vL1(t)
Inductor voltage vL1(t)
Vg – V1
D'Ts
32
R
Chapter 2: Principles of steady-state converter analysis
t
vL1 = DVg + D'(Vg – V1) = 0
Volt-second balance on L1:
The principles of inductor volt-second and capacitor charge balance state that the average values of the periodic inductor voltage and capacitor current waveforms are zero, when the converter operates in steady state. Hence, to determine the steady-state conditions in the converter, let us sketch the inductor voltage and capacitor current waveforms, and equate their average values to zero.
Waveforms:
–
v2
+
Chapter 2: Principles of steady-state converter analysis
+ vL1 –
L1
Equate average values to zero
Fundamentals of Power Electronics
vL1 = Vg – V1 vL2 = – V2 i C1 = I 1 V i C2 = I 2 – 2 R
Small ripple approximation for subinterval 2:
vL1 = Vg – v1 vL2 = – v2 i C1 = i 1 v i C2 = i 2 – 2 R
Inductor voltages and capacitor currents:
Diode conduction interval
Waveforms during subinterval 2
– V1 – V2
DTs
I2
DTs D'Ts
I1
D'Ts
– V2
i C1 = DI 2 + D'I 1 = 0
vL2 = D( – V1 – V2) + D'( – V2) = 0
Average the waveforms:
Chapter 2: Principles of steady-state converter analysis
D'Ts
t
Fundamentals of Power Electronics
34
i C2 = I 2 –
V2 =0 R
Chapter 2: Principles of steady-state converter analysis
Note: during both subintervals, the capacitor current iC2 is equal to the difference between the inductor current i2 and the load current V2/R. When ripple is neglected, iC2 is constant and equal to zero.
DTs
I2 – V2 / R (= 0)
Capacitor current iC2(t) waveform iC2(t)
33
t
t
Equate average values to zero
Fundamentals of Power Electronics
iC1(t)
Capacitor C1 current
vL2(t)
Inductor L2 voltage
Equate average values to zero
M(D) =
0.2
0.6
35
0.8
Fundamentals of Power Electronics
di 1(t) vL1(t) Vg – V1 = = L1 L1 dt di 2(t) vL2(t) – V2 = = L2 L2 dt
Interval 2 slopes:
di 1(t) vL1(t) Vg = = L1 L1 dt di 2(t) vL2(t) – V1 – V2 = = L2 L2 dt
Interval 1 slopes, using small ripple approximation:
36
i2(t)
I2
I1
i1(t)
1
∆i2
Ts DTs – V2 L2
Ts
DTs
Vg – V1 L1
t
t
Chapter 2: Principles of steady-state converter analysis
– V1 – V2 L2
Vg L1
∆i1
Chapter 2: Principles of steady-state converter analysis
V2 =– D Vg 1–D
0.4
Inductor current waveforms
Fundamentals of Power Electronics
-5
-4
-3
-2
-1
0
0
D
Cuk converter conversion ratio M = V/Vg
M(D)
V1
v1(t)
37
Ts
∆i 1 =
VgDTs 2L 1 VgDTs ∆i 2 = 2L 2 VgD 2Ts ∆v1 = 2D'RC 1
Use dc converter solution to simplify:
38
t
Chapter 2: Principles of steady-state converter analysis
Q: How large is the output voltage ripple?
Fundamentals of Power Electronics
DTs
I1 C1
Chapter 2: Principles of steady-state converter analysis
I2 C1
∆v1
Ripple magnitudes
VgDTs ∆i 1 = 2L 1 V + V2 ∆i 2 = 1 DTs 2L 2 – I DT ∆v1 = 2 s 2C 1
Analysis results
Fundamentals of Power Electronics
dv1(t) i C1(t) I 1 = = C1 C1 dt
Subinterval 2:
dv1(t) i C1(t) I 2 = = C1 C1 dt
Subinterval 1:
Capacitor C1 waveform
+ –
0
I iL(0)
iL(t) Vg – V L
2
39
–V L
C
iC(t)
Ts
–
vC(t)
+
t
∆iL
R
iR(t)
Chapter 2: Principles of steady-state converter analysis
DTs
iL(DTs)
iL(t)
L
Fundamentals of Power Electronics
If the capacitor voltage ripple is small, then essentially all of the ac component of inductor current flows through the capacitor.
Must not neglect inductor current ripple!
V
vC(t)
iC(t)
40
DTs
total charge q
∆v
D'Ts
t
∆v
t
Chapter 2: Principles of steady-state converter analysis
Ts / 2
∆iL
Capacitor current and voltage, buck example
Fundamentals of Power Electronics
What is the capacitor current?
Inductor current waveform.
Vg
1
Buck converter example: Determine output voltage ripple
2.5 Estimating ripple in converters containing two-pole low-pass filters
DTs
total charge q Ts / 2
DTs
total charge q Ts / 2
Fundamentals of Power Electronics
V
vC(t)
iC(t)
∆v
D'Ts
∆iL
41
(change in charge) = C (change in voltage)
q = C (2∆v)
Chapter 2: Principles of steady-state converter analysis
t
∆v
t
Current iC(t) is positive for half of the switching period. This positive current causes the capacitor voltage vC(t) to increase between its minimum and maximum extrema. During this time, the total charge q is deposited on the capacitor plates, where
∆v
D'Ts
∆iL
42
Ts 2
∆iL Ts 8C Note: in practice, capacitor equivalent series resistance (esr) further increases ∆v.
∆v =
Eliminate q and solve for ∆v:
q = 12 ∆iL
Chapter 2: Principles of steady-state converter analysis
t
∆v
t
The total charge q is the area of the triangle, as shown:
Estimating capacitor voltage ripple ∆v
Fundamentals of Power Electronics
V
vC(t)
iC(t)
Estimating capacitor voltage ripple ∆v
DTs
total flux linkage λ Ts / 2
∆i
Vg
D'Ts
+ –
∆v
i1
L1
t
43
∆i
t
C1 –
D1
i2
L2
C2
R
= inductor volt-seconds
λ = inductor flux linkages
can use similar arguments, with λ = L ∆i
Q1
44
Chapter 2: Principles of steady-state converter analysis
1. The dc component of a converter waveform is given by its average value, or the integral over one switching period, divided by the switching period. Solution of a dc-dc converter to find its dc, or steadystate, voltages and currents therefore involves averaging the waveforms. 2. The linear ripple approximation greatly simplifies the analysis. In a welldesigned converter, the switching ripples in the inductor currents and capacitor voltages are small compared to the respective dc components, and can be neglected. 3. The principle of inductor volt-second balance allows determination of the dc voltage components in any switching converter. In steady-state, the average voltage applied to an inductor must be zero.
Fundamentals of Power Electronics
–
v
+
Chapter 2: Principles of steady-state converter analysis
vC1
+
iT
2.6 Summary of Key Points
Fundamentals of Power Electronics
I
iL(t)
vL(t)
Example: problem 2.9
Inductor current ripple in two-pole filters
45
Chapter 2: Principles of steady-state converter analysis
Fundamentals of Power Electronics
1
3.6. Summary of key points
Chapter 3: Steady-state equivalent circuit modeling, ...
3.5. Example: inclusion of semiconductor conduction losses in the boost converter model
3.4. How to obtain the input port of the model
3.3. Construction of equivalent circuit model
3.2. Inclusion of inductor copper loss
3.1. The dc transformer model
Chapter 3. Steady-State Equivalent Circuit Modeling, Losses, and Efficiency
Fundamentals of Power Electronics
4. The principle of capacitor charge balance allows determination of the dc components of the inductor currents in a switching converter. In steadystate, the average current applied to a capacitor must be zero. 5. By knowledge of the slopes of the inductor current and capacitor voltage waveforms, the ac switching ripple magnitudes may be computed. Inductance and capacitance values can then be chosen to obtain desired ripple magnitudes. 6. In converters containing multiple-pole filters, continuous (nonpulsating) voltages and currents are applied to one or more of the inductors or capacitors. Computation of the ac switching ripple in these elements can be done using capacitor charge and/or inductor flux-linkage arguments, without use of the small-ripple approximation. 7. Converters capable of increasing (boost), decreasing (buck), and inverting the voltage polarity (buck-boost and Cuk) have been described. Converter circuits are explored more fully in a later chapter.
Summary of Chapter 2
(ideal conversion ratio)
(η = 100%)
Power input
+ Vg –
Ig
control input
D
Switching dc-dc converter
Pin = Pout
+ Vg –
M(D) I
M(D)Vg
Fundamentals of Power Electronics
Power input
dependent sources Ig
2
+ –
I + V –
Power output
Vg I g = V I
3
I + V –
Power output
+ Vg –
control input
D
+ V –
Power output
Chapter 3: Steady-state equivalent circuit modeling, ...
Power input
Ig
1 : M(D)
I g = M(D) I
Dc transformer
V = M(D) Vg
I
Chapter 3: Steady-state equivalent circuit modeling, ...
Equivalent circuits corresponding to ideal dc-dc converter equations
Fundamentals of Power Electronics
These equations are valid in steady-state. During transients, energy storage within filter elements may cause Pin ≠ Pout
I g = M(D) I
V = M(D) Vg
Vg I g = V I
Pin = Pout
Basic equations of an ideal dc-dc converter:
3.1. The dc transformer model
+ Vg –
control input
D
1 : M(D)
I + V – Power output
• conversion ratio M controllable via duty cycle
• conversion of dc voltages and currents, ideally with 100% efficiency
Models basic properties of ideal dc-dc converter:
+ Vg –
D
Switching dc-dc converter
–
+
+ Vg –
Fundamentals of Power Electronics
V1
R1
1 : M(D)
+ V –
2. Insert dc transformer model
–
+
R1
1. Original system
V1
4
Chapter 3: Steady-state equivalent circuit modeling, ...
+ V –
R
5
R
+ V –
R R + M 2(D) R 1
–
+
R
Chapter 3: Steady-state equivalent circuit modeling, ...
V = M(D) V1
4. Solve circuit
M(D)V1
M 2(D)R1
3. Push source through transformer
Example: use of the dc transformer model
Fundamentals of Power Electronics
• Time-invariant model (no switching) which can be solved to find dc components of converter waveforms
• Solid line denotes ideal transformer model, capable of passing dc voltages and currents
Power input
Ig
The dc transformer model
Vg
+ –
i
+ vL –
L
+ –
i
RL
C
iC
switch in position 1
Vg
Fundamentals of Power Electronics
+ –
i
L
RL
6
1 C
R
–
v
+
L
R
–
v
+
RL
7
Vg
1
–
v
+ vL –
L
RL
C
iC R
–
v
Chapter 3: Steady-state equivalent circuit modeling, ...
+ –
R
+
switch in position 2
C
i
2
+
Chapter 3: Steady-state equivalent circuit modeling, ...
2
Analysis of nonideal boost converter
Fundamentals of Power Electronics
Vg
RL
Insert this inductor model into boost converter circuit:
L
Example: inductor copper loss (resistance of winding):
Dc transformer model can be extended, to include converter nonidealities.
3.2. Inclusion of inductor copper loss
Vg
+ –
i
8
RL
C
iC
+ –
Fundamentals of Power Electronics
Vg
+ vL –
L
RL
C
iC R
9
–
v
–
v
+
Chapter 3: Steady-state equivalent circuit modeling, ...
vL(t) = Vg – i(t) RL – v(t) ≈ Vg – I RL – V iC(t) = i(t) – v(t) / R ≈ I – V / R
i
R
+
Chapter 3: Steady-state equivalent circuit modeling, ...
+ vL –
L
Circuit equations, switch in position 2
Fundamentals of Power Electronics
vL(t) = Vg – I RL iC(t) = –V / R
Small ripple approximation:
iC(t) = –v(t) / R
vL(t) = Vg – i(t) RL
Inductor current and capacitor voltage:
Circuit equations, switch in position 1
0 = Vg – I RL – D'V
We now have two equations and two unknowns:
Fundamentals of Power Electronics
V = 1 1 Vg D' (1 + RL / D' 2R)
Eliminate I and solve for V:
0 = D'I – V / R
10
iC (t)
vL(t)
I – V/R
Vg – IRL – V
D' Ts
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0
11
0.1
t
0.3
0.4
0.5
D
0.6
0.7
0.8
RL / R = 0.1
0.9
1
Chapter 3: Steady-state equivalent circuit modeling, ...
0.2
RL / R = 0.02
RL / R = 0.01
RL / R = 0.05
RL / R = 0
Chapter 3: Steady-state equivalent circuit modeling, ...
-V/R
DTs
Vg – IRL
Solution for output voltage
Fundamentals of Power Electronics
0 = D'I – V / R
Capacitor charge balance:
iC(t) = D ( – V / R) + D' (I – V / R)
Average capacitor current:
0 = Vg – I RL – D'V
Inductor volt-second balance:
vL(t) = 1 v (t)dt Ts 0 L = D(Vg – I RL) + D'(Vg – I RL – V)
Ts
Average inductor voltage:
Inductor voltage and capacitor current waveforms
V / Vg
12
Fundamentals of Power Electronics
• This is a loop equation: the dc components of voltage around a loop containing the inductor sum to zero
• Average inductor voltage then set to zero
• Derived via Kirchoff’s voltage law, to find the inductor voltage during each subinterval
13
Vg
RL
I
D' V
Chapter 3: Steady-state equivalent circuit modeling, ...
• D’V term: for now, leave as dependent source
• IRL term: voltage across resistor of value RL having current I
+ –
L
+ – + IRL – =0
vL = 0 = Vg – I RL – D'V
+ –
Chapter 3: Steady-state equivalent circuit modeling, ...
Inductor voltage equation
Fundamentals of Power Electronics
View these as loop and node equations of the equivalent circuit. Reconstruct an equivalent circuit satisfying these equations
iC = 0 = D'I – V / R
vL = 0 = Vg – I RL – D'V
Results of previous section (derived via inductor volt-sec balance and capacitor charge balance):
3.3. Construction of equivalent circuit model
I
D' V
+ –
D' I
–
V
+
+ –
Fundamentals of Power Electronics
Vg
I
RL
D' : 1
–
V
+
15
The dependent sources are equivalent to a D’ : 1 transformer:
+ –
RL
The two circuits, drawn together:
Vg
14
C
–
V
+
V/R
R
Chapter 3: Steady-state equivalent circuit modeling, ...
• D’I term: for now, leave as dependent source
• V/R term: current through load resistor of value R having voltage V
D' I
=0
node
+ –
n:1
–
V2
+
–
nI1 V2
• reciprocal voltage/current dependence
• sources have same coefficient
I1
nV2
+
Chapter 3: Steady-state equivalent circuit modeling, ...
R
R
I1
Dependent sources and transformers
Complete equivalent circuit
Fundamentals of Power Electronics
• This is a node equation: the dc components of current flowing into a node connected to the capacitor sum to zero
• Average capacitor current then set to zero
• Derived via Kirchoff’s current law, to find the capacitor current during each subinterval
iC = 0 = D'I – V / R
Capacitor current equation
+ –
L
D' : 1
+ – –
V
+ R
16
–
V
+
Vg D'
R R + L2 D'
R
=
Vg D'
+ –
Fundamentals of Power Electronics
Vg
I
I=
2
D' : 1
–
V
+
17
RL D' 2 R
R
Chapter 3: Steady-state equivalent circuit modeling, ...
Vg Vg 1 = 2 D' R + RL D' 1 + RL D' 2 R
RL
1+
1
Chapter 3: Steady-state equivalent circuit modeling, ...
V=
Solution for output voltage using voltage divider formula:
R
Solution for input (inductor) current
Fundamentals of Power Electronics
Vg / D'
D' I
Refer all elements to transformer secondary: R / D' 2
Vg
I
RL
Converter equivalent circuit
Solution of equivalent circuit
1+
RL D' 2 R
1
Pout (V) (D'I) V = = D' Pin Vg (Vg) (I)
1+
RL D' 2 R
1
η
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
Fundamentals of Power Electronics
η=
Vg
+ –
18
I
D' : 1
–
V
+
0
0.1
0.2
19
0.3
R
D
0.5
0.6
0.7
0.8
0.9
1
Chapter 3: Steady-state equivalent circuit modeling, ...
0.4
RL/R = 0.1
0.05
0.02
0.01
0.002
Chapter 3: Steady-state equivalent circuit modeling, ...
RL
Efficiency, for various values of RL
Fundamentals of Power Electronics
η=
η=
Pout = (V) (D'I)
Pin = (Vg) (I)
Solution for converter efficiency
+ –
1 2
iL + vL –
L
RL
C
20
+ –
IL
=0
RL
–
VC
+
Fundamentals of Power Electronics
21
Chapter 3: Steady-state equivalent circuit modeling, ...
R
VC /R
iC = 0 = I L – VC/R
What happened to the transformer? • Need another equation
DVg
+ – =0
vL = 0 = DVg – I LRL – VC
R
Chapter 3: Steady-state equivalent circuit modeling, ...
iC = 0 = I L – VC/R
–
vC
+
Construct equivalent circuit as usual
Fundamentals of Power Electronics
vL = 0 = DVg – I LRL – VC
Average inductor voltage and capacitor current:
Vg
ig
Buck converter example —use procedure of previous section to derive equivalent circuit
3.4. How to obtain the input port of the model
0
area = DTs IL
iL (t) ≈ IL
DTs
0 Ts
0
Ts
22
ig(t) dt = DI L
Vg
ig(t) dt = DI L
Fundamentals of Power Electronics
Ig = 1 Ts
0
Ts
Chapter 3: Steady-state equivalent circuit modeling, ...
t
+ –
23
Ig
Chapter 3: Steady-state equivalent circuit modeling, ...
D IL
Input port equivalent circuit
Fundamentals of Power Electronics
Ig = 1 Ts
Dc component (average value) of ig(t) is
ig(t)
Input current waveform ig(t):
Modeling the converter input port
+ – D IL
+ DV g –
IL
RL
–
VC
+ R
+ –
24
1:D
RL
–
VC
+ R
Chapter 3: Steady-state equivalent circuit modeling, ...
IL
+ – DTs
Ts
L
+ –
C
iC R
–
v
+
Fundamentals of Power Electronics
25
Chapter 3: Steady-state equivalent circuit modeling, ...
Insert these models into subinterval circuits
Diode: constant forward voltage VD plus on-resistance RD
MOSFET: on-resistance Ron
Models of on-state semiconductor devices:
Vg
i
Boost converter example
3.5. Example: inclusion of semiconductor conduction losses in the boost converter model
Fundamentals of Power Electronics
Vg
Ig
Replace dependent sources with equivalent dc transformer:
Vg
Ig
Input and output port equivalent circuits, drawn together:
Complete equivalent circuit, buck converter
+ –
+ vL –
L
Ron
RL
DTs
C
iC
Ts
R
L
+ –
26
–
v
+
Vg
–
v
i
+ vL –
L
RL VD
C
iC
RD
R
-V/R
DTs
Vg – IRL – IRon
I – V/R
Vg – IRL – VD – IRD – V
D' Ts
t
Fundamentals of Power Electronics
27
Chapter 3: Steady-state equivalent circuit modeling, ...
iC = D(–V/R) + D'(I – V/R) = 0
vL = D(Vg – IRL – IRon) + D'(Vg – IRL – VD – IRD – V) = 0
iC (t)
vL(t)
–
v
+
Chapter 3: Steady-state equivalent circuit modeling, ...
+ –
R
+
switch in position 2
C
iC
Average inductor voltage and capacitor current
Fundamentals of Power Electronics
Vg
i
+ –
switch in position 1
Vg
i
Boost converter example: circuits during subintervals 1 and 2
+ –
+ –
Vg
RL
RL
Fundamentals of Power Electronics
+ –
Vg
–
V
+
28
I
D' VD
R
D Ron
D Ron
I
I
D' VD
D' VD
29
+ –
D' V
+ –
D' : 1
D' I
–
V
+
–
V
R
R
Chapter 3: Steady-state equivalent circuit modeling, ...
D' RD
D' V
D' RD +
Chapter 3: Steady-state equivalent circuit modeling, ...
V/R
+ ID'RD -
D' RD
Complete equivalent circuit
Fundamentals of Power Electronics
D' I
+ –
D Ron + IDRon-
+ –
+ –
D'I – V/R = 0
Vg
+ IRL -
RL
Vg – IRL – IDRon – D'VD – ID'RD – D'V = 0
Construction of equivalent circuits
+ –
D'VD Vg
D' VD
1+
D' : 1
30
–
V
+
D'VD Vg
+ –
RL
Fundamentals of Power Electronics
Vg / D' >> VD D' 2R >> RL + DRon + D'RD 31
R
I
D' VD
D' RD
D' : 1
–
V
R
Chapter 3: Steady-state equivalent circuit modeling, ...
D Ron
+
Chapter 3: Steady-state equivalent circuit modeling, ...
1 RL + DRon + D'RD 1+ D' 2R
RL + DRon + D'RD D' 2R
1–
Vg
Conditions for high efficiency:
η = D' V = Vg
Pout = (V) (D'I)
Pin = (Vg) (I)
and
D' RD
D' 2R D' 2R + RL + DRon + D'RD
+ –
Solution for converter efficiency
1–
V = 1 Vg D'
I
Vg – D'VD
D Ron
V= 1 D'
+ –
Fundamentals of Power Electronics
Vg
RL
Solution for output voltage
+ –
I
0
D
32
(1.155) I
D
(1.00167) I
I
DTs
1.1 I
2I
0 Ts
(1.3333) D I2 R on
(1.0033) D I2 R on
D I2 R on
Average power loss in R on
(b) (a)
(c)
t
Chapter 3: Steady-state equivalent circuit modeling, ...
D
MOS FET rms current
i(t)
MOSFET current waveforms, for various ripple magnitudes:
Fundamentals of Power Electronics
33
Chapter 3: Steady-state equivalent circuit modeling, ...
1. The dc transformer model represents the primary functions of any dc-dc converter: transformation of dc voltage and current levels, ideally with 100% efficiency, and control of the conversion ratio M via the duty cycle D. This model can be easily manipulated and solved using familiar techniques of conventional circuit analysis. 2. The model can be refined to account for loss elements such as inductor winding resistance and semiconductor on-resistances and forward voltage drops. The refined model predicts the voltages, currents, and efficiency of practical nonideal converters. 3. In general, the dc equivalent circuit for a converter can be derived from the inductor volt-second balance and capacitor charge balance equations. Equivalent circuits are constructed whose loop and node equations coincide with the volt-second and charge balance equations. In converters having a pulsating input current, an additional equation is needed to model the converter input port; this equation may be obtained by averaging the converter input current.
Summary of chapter 3
Fundamentals of Power Electronics
(c) ∆i = I
(b) ∆i = 0.1 I
(a) ∆i = 0
Inductor current ripple
• Result is the same, provided ripple is small
• To correctly predict power loss in a resistor, use rms values
• Model uses average currents and voltages
Accuracy of the averaged equivalent circuit in prediction of losses
1
0
Fundamentals of Power Electronics
All power semiconductor devices function as SPST switches.
–
v
+
i
1
SPST switch, with voltage and current polarities defined
Vg
Vg
2
L
2
+ –
iA
+
– vB
+ vA –
A
iB
B
C
iL(t)
C
iL(t)
R
R
–
V
+
–
V
Chapter 4: Switch realization
L
with two SPST switches:
+ –
1
with SPDT switch:
Buck converter
+
Chapter 4: Switch realization
SPST (single-pole single-throw) switches
Fundamentals of Power Electronics
4.4. Summary of key points
Transistor switching with clamped inductive load. Diode recovered charge. Stray capacitances and inductances, and ringing. Efficiency vs. switching frequency.
4.3. Switching loss
Power diodes, MOSFETs, BJTs, IGBTs, and thyristors
4.2. A brief survey of power semiconductor devices
Single-, two-, and four-quadrant switches. Synchronous rectifiers
4.1. Switch applications
Chapter 4. Switch Realization
Fundamentals of Power Electronics
0
4
switch off-state voltage
Chapter 4: Switch realization
OFF-state: v > 0
A single-quadrant switch example:
–
switch on-state current
ON-state: i > 0
i
1
v
+
Chapter 4: Switch realization
Quadrants of SPST switch operation
3
A nontrivial step: two SPST switches are not exactly equivalent to one SPDT switch It is possible for both SPST switches to be simultaneously ON or OFF Behavior of converter is then significantly modified —discontinuous conduction modes (ch. 5) Conducting state of SPST switch may depend on applied voltage or current —for example: diode
Fundamentals of Power Electronics
●
●
●
●
Realization of SPDT switch using two SPST switches
switch on-state current
0
i
1
Fundamentals of Power Electronics
–
v
+
switch off-state voltage
switch off-state voltage
5
Fourquadrant switch
Currentbidirectional two-quadrant switch
switch off-state voltage
switch off-state voltage
Chapter 4: Switch realization
switch on-state current
switch on-state current
6
Chapter 4: Switch realization
Single-quadrant switch: on-state i(t) and off-state v(t) are unipolar.
SCR: A special case — turn-on transition is active, while turn-off transition is passive.
Passive switch: Switch state is controlled by the applied current and/or voltage at terminals 1 and 2.
Active switch: Switch state is controlled exclusively by a third terminal (control terminal).
4.1.1. Single-quadrant switches
Fundamentals of Power Electronics
Voltagebidirectional two-quadrant switch
Singlequadrant switch
switch on-state current
Some basic switch applications
0
i
0
–
v
1 i +
0
–
v
1 i +
v
7
i
on off
v
8
instantaneous i-v characteristic
Fundamentals of Power Electronics
C
on
Chapter 4: Switch realization
• provided that the intended on-state and off-state operating points lie on the diode i-v characteristic, then switch can be realized using a diode
• can block negative offstate voltage
• can conduct positive onstate current
• Single-quadrant switch:
• A passive switch
Chapter 4: Switch realization
• provided that the intended on-state and off-state operating points lie on the transistor i-v characteristic, then switch can be realized using a BJT or IGBT
• can block positive off-state voltage
• can conduct positive onstate current
• Single-quadrant switch:
• An active switch, controlled by terminal C
The Bipolar Junction Transistor (BJT) and the Insulated Gate Bipolar Transistor (IGBT)
C
IGBT
BJT
i
instantaneous i-v characteristic
off
Fundamentals of Power Electronics
Symbol
–
v
+
1
The diode
0
–
v
iA
Switch A
on
B
iA
iB
switch A
+
– vB
+ vA –
A
Fundamentals of Power Electronics
SPST switch operating points
Vg
off
v
(reverse conduction)
on
on
9
Chapter 4: Switch realization
• provided that the intended onstate and off-state operating points lie on the MOSFET i-v characteristic, then switch can be realized using a MOSFET
• can block positive off-state voltage
• can conduct positive on-state current (can also conduct negative current in some circumstances)
• Normally operated as singlequadrant switch:
iL
C
10
–Vg
iB
on
vB
Chapter 4: Switch realization
iL
Switch B: diode
Switch A: transistor
switch B
Switch B
off
switch B
–
V
Vg
R
+
off vA
iL(t)
switch A
L
Realization of switch using transistors and diodes Buck converter example
+ –
i
instantaneous i-v characteristic
Fundamentals of Power Electronics
Symbol
C
1 i +
• An active switch, controlled by terminal C
The Metal-Oxide Semiconductor Field Effect Transistor (MOSFET)
iL
0
–
v
+
Fundamentals of Power Electronics
BJT / anti-parallel diode realization
C
1 i
+ –
–Vg
Vg
–
iB
iL(t)
on
switch B
vL(t)
off
11
iB
– vB +
+
L
switch B
–
off vA
vA
switch A
+
iL
(diode conducts)
on
off
12
v
(transistor conducts)
on
instantaneous i-v characteristic
i
vB
Chapter 4: Switch realization
Chapter 4: Switch realization
• provided that the intended onstate and off-state operating points lie on the composite i-v characteristic, then switch can be realized as shown
• can block positive off-state voltage
• can conduct positive or negative on-state current
• Normally operated as twoquadrant switch:
• Usually an active switch, controlled by terminal C
4.1.2. Current-bidirectional two-quadrant switches
Fundamentals of Power Electronics
on
switch A
iA
Vg
iA
Realization of buck converter using single-quadrant switches
0
i
(diode conducts)
on
off
Fundamentals of Power Electronics
13
(diode conducts)
on
off
v
(transistor conducts)
on
v
C
–
v
+
14
Power MOSFET, and its integral body diode
0
1 i
switch off-state voltage
Chapter 4: Switch realization
Use of external diodes to prevent conduction of body diode
Chapter 4: Switch realization
switch on-state current
MOSFET body diode
(transistor conducts)
on
Power MOSFET characteristics
i
Fundamentals of Power Electronics
–
v
+
1
i
Two quadrant switches
+ –
Vg
iB
Q2
Q1
15
–
D2 v B
+
–
D1 vA
+
L
C
iL
R
Chapter 4: Switch realization
–
v0
+
v0(t) = (2D – 1) Vg
1
D
Fundamentals of Power Electronics
–Vg
0.5
16
Vg v0(t) = (2D – 1) R R
Chapter 4: Switch realization
Hence, current-bidirectional two-quadrant switches are required.
iL(t) =
The resulting inductor current variation is also sinusoidal:
D(t) = 0.5 + Dm sin (ωt)
Vg
0
Sinusoidal modulation to produce ac output: v0
v0(t) = (2D – 1) Vg
Inverter: sinusoidal modulation of D
Fundamentals of Power Electronics
+ –
Vg
iA
A simple inverter
+ – ic
ib
Q2 vbus > vbatt
Q1 D2
L
Fundamentals of Power Electronics
18
–
vbatt
Chapter 4: Switch realization
A dc-dc converter with bidirectional power flow.
spacecraft main power bus –
vbus
+
D1 +
Chapter 4: Switch realization
Bidirectional battery charger/discharger
17
Switches must block dc input voltage, and conduct ac load current.
Fundamentals of Power Electronics
Vg
ia
The dc-3øac voltage source inverter (VSI)
0
–
v
+
(transistor blocks voltage)
i
0
1
0
i
–
v
+
20
off (transistor blocks voltage)
on
(diode blocks voltage)
i
off
Fundamentals of Power Electronics
C
–
v
+
1
v
v switch off-state voltage
Chapter 4: Switch realization
switch on-state current
Chapter 4: Switch realization
• The SCR is such a device, without controlled turn-off
• provided that the intended onstate and off-state operating points lie on the composite i-v characteristic, then switch can be realized as shown
• can block positive or negative off-state voltage
• can conduct positive on-state current
• Normally operated as twoquadrant switch:
Two-quadrant switches
19
instantaneous i-v characteristic
off
on
(diode blocks voltage)
i
off
Fundamentals of Power Electronics
BJT / series diode realization
C
i
1
• Usually an active switch, controlled by terminal C
4.1.3. Voltage-bidirectional two-quadrant switches
–
+ vbc(t)
–
+ vab(t)
Fundamentals of Power Electronics
switch on-state current
22
switch off-state voltage
φc
φb
φa
Chapter 4: Switch realization
Chapter 4: Switch realization
• can block positive or negative off-state voltage
• can conduct positive or negative on-state current
• Usually an active switch, controlled by terminal C
4.1.4. Four-quadrant switches
21
Another example: boost-type inverter, or current-source inverter (CSI).
Requires voltage-bidirectional two-quadrant switches.
iL
A dc-3øac buck-boost inverter
Fundamentals of Power Electronics
Vg
+ –
0
i
0
van(t)
vcn(t)
vbn(t)
ic
ib
ia
–
v
+
Chapter 4: Switch realization
0
i
1
3øac output
Fundamentals of Power Electronics
24
• Requires nine four-quadrant switches
Chapter 4: Switch realization
• All voltages and currents are ac; hence, four-quadrant switches are required.
+ –
+ –
3øac input
23
–
–
0
v
v
A 3øac-3øac matrix converter
Fundamentals of Power Electronics
–
v
+
1
+
i
i +
1
1
Three ways to realize a four-quadrant switch
+ –
Q1
iA
vA
C
+
C
–
Q2
Fundamentals of Power Electronics
Vg
+ –
0
–
v
25
MOSFET as synchronous rectifier
C
1 i +
on
iB
– vB +
L
26
iL(t)
v
Chapter 4: Switch realization
instantaneous i-v characteristic
off
on (reverse conduction)
Chapter 4: Switch realization
• Useful in lowvoltage high-current applications
• Semiconductor conduction loss can be made arbitrarily small, by reduction of MOSFET onresistances
• MOSFET Q2 is controlled to turn on when diode would normally conduct
Buck converter with synchronous rectifier
Fundamentals of Power Electronics
ideal switch
0
conventional diode rectifier
–
– 0
v
+
i
i
v
+
1
1
i
Replacement of diode with a backwards-connected MOSFET, to obtain reduced conduction loss
4.1.5. Synchronous rectifiers
Fundamentals of Power Electronics
+
+
+
–
E v
n-
low doping concentration
+
28
depletion region, reverse-biased
{
p
v
A power diode, under reverse-biased conditions:
4.2.1. Power diodes
27
Chapter 4: Switch realization
On resistance vs. breakdown voltage vs. switching times Minority carrier and majority carrier devices
Power diodes Power MOSFETs Bipolar Junction Transistors (BJTs) Insulated Gate Bipolar Transistors (IGBTs) Thyristors (SCR, GTO, MCT)
Fundamentals of Power Electronics
●
●
●
●
●
●
●
4.2. A brief survey of power semiconductor devices
+ – –
Chapter 4: Switch realization
–
–
n
{
+
+
conductivity modulation
+
+ +
n-
–
Chapter 4: Switch realization
–
–
n
(1)
(2)
Fundamentals of Power Electronics
i(t)
v(t)
(3) 30
di dt
(4)
tr
(5)
area –Qr
t
Chapter 4: Switch realization
(6)
0
t
Typical diode switching waveforms
29
minority carrier injection
+
p
v
{ Fundamentals of Power Electronics
i
Forward-biased power diode
– +
31
Chapter 4: Switch realization
R ated max voltage
2500V
SD453N25S20PC
600V 1200V
MUR1560 RHRU100120
45V 150V
444CNQ045 30CPQ150
Fundamentals of Power Electronics
30V
MBR6030L
S chot t ky rect i fi ers
150V
MUR815
Ul t ra-fas t recov ery rect i fi ers
400V
1N3913
Fas t recov ery rect i fi ers
Part number
32
30A
440A
60A
100A
15A
8A
400A
30A
Rated avg current
1.19V
0.69V
0.48V
2.6V
1.2V
0.975V
2.2V
1.1V
60ns
60ns
35ns
2µs
400ns
tr (max)
Chapter 4: Switch realization
V F (typical)
Characteristics of several commercial power rectifier diodes
Fundamentals of Power Electronics
A majority carrier device Essentially no recovered charge Model with equilibrium i-v characteristic, in parallel with depletion region capacitance Restricted to low voltage (few devices can block 100V or more)
Schottky diode
Reverse recovery time and recovered charge specified Intended for converter applications
Fast recovery and ultra-fast recovery
Reverse recovery time not specified, intended for 50/60Hz
Standard recovery
Types of power diodes
p
n
p
n
Fundamentals of Power Electronics
drain
depletion region
n
n
33
p
n
n
n-
+
–
34
n
p
n
MOSFET: Off state
Drain
n
n-
source
Fundamentals of Power Electronics
n
Gate
Source
Chapter 4: Switch realization
• off-state voltage appears across nregion
• p-n- junction is reverse-biased
Chapter 4: Switch realization
• n-channel device is shown
• Vertical current flow
• Consists of many small enhancementmode parallelconnected MOSFET cells, covering the surface of the silicon wafer
• Gate lengths approaching one micron
4.2.2. The Power MOSFET
n
p
drain
n
n-
Body diode
n
source
Fundamentals of Power Electronics
n
p
35
drain current
n
n-
n
n
n
36
p
n
MOSFET body diode
drain
channel
p
Fundamentals of Power Electronics
n
source
MOSFET: on state
Chapter 4: Switch realization
• diode switching speed not optimized —body diode is slow, Qr is large
• diode can conduct the full MOSFET rated current
• negative drain-tosource voltage can forward-bias the body diode
• p-n- junction forms an effective diode, in parallel with the channel
Chapter 4: Switch realization
• on resistance = total resistances of nregion, conducting channel, source and drain contacts, etc.
• drain current flows through n- region and conducting channel
• positive gate voltage induces conducting channel
• p-n- junction is slightly reversebiased
0A
5A
0V
off state 5V
Cgs
Cgd
Cds(vds) =
S
D
Fundamentals of Power Electronics
G
VGS
V
37
10V
on state 1V
V DS = 0.5V
V DS =
V DS
V =2
15V
Chapter 4: Switch realization
• on-resistance has positive temperature coefficient, hence easy to parallel
• MOSFET can conduct peak currents well in excess of average current rating — characteristics are unchanged
• On state: VGS >> Vth
• Off state: VGS < Vth
1+
C0 vds V0
Cds
38
Cds(vds) ≈ C0
Chapter 4: Switch realization
C '0 V0 vds = vds
• switching times determined by rate at which gate driver charges/ discharges Cgs and Cgd
• Cds : intermediate in value, highly nonlinear
• Cgd : small, highly nonlinear
• Cgs : large, essentially constant
A simple MOSFET equivalent circuit
Fundamentals of Power Electronics
ID
10A
V
Typical MOSFET characteristics
00V =2 DS
=1 0V DS
100V 100V 100V 400V 400V 500V 1000V
IRF510 IRF540 APT10M25BNR IRF740 MTM15N40E APT5025BN APT1001RBNR
1.0Ω
0.25Ω
0.3Ω
0.55Ω
0.025Ω
0.077Ω
0.54Ω
0.018Ω
R on
Chapter 4: Switch realization
150nC
83nC
110nC
63nC
171nC
72nC
8.3nC
110nC
Qg (typical)
40
Chapter 4: Switch realization
A majority-carrier device: fast switching speed Typical switching frequencies: tens and hundreds of kHz On-resistance increases rapidly with rated blocking voltage Easy to drive The device of choice for blocking voltages less than 500V 1000V devices are available, but are useful only at low power levels (100W) Part number is selected on the basis of on-resistance rather than current rating
Fundamentals of Power Electronics
●
●
●
●
●
●
●
39
11A
23A
15A
10A
75A
28A
5.6A
50A
Rated avg current
MOSFET: conclusions
60V
IRFZ48
Fundamentals of Power Electronics
Rated max voltage
Part number
Characteristics of several commercial power MOSFETs
p
RB + vBE(t) –
Fundamentals of Power Electronics
vs(t)
+ –
iB(t)
41
n
RL
–
vCE(t)
+
VCC
42
iC(t)
VCC
vCE(t)
iB(t)
–Vs1
vBE(t)
–Vs1
vs(t)
(1) (2) (3)
0
0
Vs2
(4)
IB1
(5)
ICon
IConRon
0.7V
(6)
–IB2
(7)
(9)
t
Chapter 4: Switch realization
(8)
Chapter 4: Switch realization
• on-state: substantial minority charge in p and n- regions, conductivity modulation
• on-state: base-emitter and collector-base junctions are both forward-biased
• minority carrier device
• npn device is shown
• Vertical current flow
• Interdigitated base and emitter contacts
BJT switching times
Collector
n
n-
n
Emitter
iC(t)
Fundamentals of Power Electronics
n
Base
4.2.3. Bipolar Junction Transistor (BJT)
IB1 IBon
43
–IB2
0
Chapter 4: Switch realization
t
–
–
Fundamentals of Power Electronics
p
–IB2
Base
n –
n
n-
+
Collector
+
Emitter
44
–
– –
p
Chapter 4: Switch realization
can lead to formation of hot spots and device failure
Current crowding due to excessive IB2
Fundamentals of Power Electronics
iB(t)
Ideal base current waveform
0V
slope =β
5V
acti
increasing IB
Fundamentals of Power Electronics
IC
45
IB
15V
VCE = 0.2V
VCE = 0.5V
V CE = 5V
CE
V CE = 200V V = 20V
46
BVCBO
open emitter
BVsus BVCEO
IB = 0
VCE
Chapter 4: Switch realization
In most applications, the offstate transistor voltage must not exceed BVCEO.
BVsus: breakdown voltage observed with positive base current
BVCEO: collector-emitter breakdown voltage with zero base current
BVCBO: avalanche breakdown voltage of base-collector junction, with the emitter open-circuited
Chapter 4: Switch realization
• Current gain β decreases rapidly at high current. Device should not be operated at instantaneous currents exceeding the rated value
• On state: IB > IC /β
• Off state: IB = 0
Breakdown voltages
10V
saturation region
qu
tion atura s i s a
n egio ve r
Fundamentals of Power Electronics
0A
cutoff
5A
10A
IC
BJT characteristics
Q2
Conclusions: BJT
47
Chapter 4: Switch realization
• Diode D1 speeds up the turn-off process, by allowing the base driver to actively remove the stored charge of both Q1 and Q2 during the turn-off transition
• In a monolithic Darlington device, transistors Q1 and Q2 are integrated on the same silicon wafer
48
Chapter 4: Switch realization
BJT has been replaced by MOSFET in low-voltage (<500V) applications BJT is being replaced by IGBT in applications at voltages above 500V A minority-carrier device: compared with MOSFET, the BJT exhibits slower switching, but lower on-resistance at high voltages
Fundamentals of Power Electronics
●
●
●
Fundamentals of Power Electronics
D1
Q1
• Increased current gain, for high-voltage applications
Darlington-connected BJT
p
n
i1
E
C
i2
emitter
collector
Fundamentals of Power Electronics
G
Equivalent circuit
gate
Symbol
p
n-
Collector
Fundamentals of Power Electronics
n
Gate
Emitter
p
n
n
50
p i2
n i1
p
n-
p
n
Chapter 4: Switch realization
n
Chapter 4: Switch realization
• compared with MOSFET: slower switching times, lower on-resistance, useful at higher voltages (up to 1700V)
• On-state: minority carriers are injected into n- region, leading to conductivity modulation
Location of equivalent devices
The IGBT
49
minority carrier injection
n
• Similar in construction to MOSFET, except extra p region
• A four-layer device
4.2.4. The Insulated Gate Bipolar Transistor (IGBT)
i1
E
i2
51
pA(t)
= vA iA
diode waveforms
0
–Vg
0
iL
t0
iA(t)
t1
t2
area Woff
vB(t)
iB(t)
Vg iL
vA(t)
il
curr ent t a
0
iL
0
t
t
t
Chapter 4: Switch realization
t3
Vg
Rated max voltage
1200V
HGTG30N120D2
1200V
CM300HA-24E
Fundamentals of Power Electronics
600V
CM400HA-12E
M ul t i pl e-chi p p ow er m o dul es
600V
HGTG32N60E2
S i ngl e-chi p dev i ces
Part number
52
300A
400A
30A
32A
Rated avg current
2.7V
2.7V
3.2A
2.4V
0.3µs
0.3µs
0.58µs
0.62µs
tf (typical)
Chapter 4: Switch realization
V F (typical)
Characteristics of several commercial devices
Fundamentals of Power Electronics
G
C
IGBT waveforms
Current tailing in IGBTs
}
The SCR
Cathode
Q1
Q2
Anode
equiv circuit
Fundamentals of Power Electronics
Cathode (K)
Gate (G)
Anode (A)
symbol
Gate
54
n Q1
K
Q2
n
Chapter 4: Switch realization
A
p
n-
p
G
construction K
Chapter 4: Switch realization
4.2.5. Thyristors (SCR, GTO, MCT)
53
Becoming the device of choice in 500-1700V applications, at power levels of 1-1000kW Positive temperature coefficient at high current —easy to parallel and construct modules Forward voltage drop: diode in series with on-resistance. 2-4V typical Easy to drive —similar to MOSFET Slower than MOSFET, but faster than Darlington, GTO, SCR Typical switching frequencies: 3-30kHz IGBT technology is rapidly advancing —next generation: 2500V
Fundamentals of Power Electronics
●
●
●
●
●
●
●
Conclusions: IGBT
55
reverse breakdown
reverse blocking
iA
Fundamentals of Power Electronics
• Gate-cathode junction becomes reverse-biased only in vicinity of gate contact
• Negative gate current induces lateral voltage drop along gate-cathode junction
• Large feature size
56
+
K
–
n
–
–
A
G
iA
p
n-
p
–
–
–iG
vAK
iG = 0
Chapter 4: Switch realization
forward blocking
increasing iG
forward conducting
–
+
Chapter 4: Switch realization
n
K
Why the conventional SCR cannot be turned off via gate control
Fundamentals of Power Electronics
• 5000-6000V, 1000-2000A devices
• A voltage-bidirectional two-quadrant switch
• Cannot be actively turned off
• Simple construction, with large feature size
• Double injection leads to very low on-resistance, hence low forward voltage drops attainable in very high voltage devices
• A minority carrier device
• Positive feedback —a latching device
The Silicon Controlled Rectifier (SCR)
57
Chapter 4: Switch realization
Fundamentals of Power Electronics
• Small feature size, highly interdigitated, modern fabrication
• A latching SCR, with added built-in MOSFETs to assist the turn-on and turn-off processes
• p-type device
• Still an emerging device, but some devices are commercially available
58
Q4 channel
Q3 channel
Gate
n
n
Chapter 4: Switch realization
Cathode
n
p-
n
p
Anode
The MOS-Controlled Thyristor (MCT)
Fundamentals of Power Electronics
• Maximum controllable on-state current: maximum anode current that can be turned off via gate control. GTO can conduct peak currents well in excess of average current rating, but cannot switch off
• Turn-off current gain: typically 2-5
Turn-off transition:
• Negative gate current is able to completely reverse-bias the gatecathode junction
• Gate and cathode contacts are highly interdigitated
• An SCR fabricated using modern techniques —small feature size
The Gate Turn-Off Thyristor (GTO)
59
Chapter 4: Switch realization
• Maximum current that can be interrupted is limited by the on-resistance of Q4
Summary: Thyristors
Anode
Q2
Q3
• Positive gate-anode voltage turns n-channel MOSFET Q4 on, reversebiasing the base-emitter junction of Q2 and turning off the device
• Negative gate-anode voltage turns p-channel MOSFET Q3 on, causing Q1 and Q2 to latch ON
Fundamentals of Power Electronics
60
Chapter 4: Switch realization
• The MCT: So far, ratings lower than IGBT. Slower than IGBT. Easy to drive. Second breakdown problems? Still an emerging device.
• The GTO: intermediate ratings (less than SCR, somewhat more than IGBT). Slower than IGBT. Slower than MCT. Difficult to drive.
• The SCR: highest voltage and current ratings, low cost, passive turn-off transition
• The thyristor family: double injection yields lowest forward voltage drop in high voltage devices. More difficult to parallel than MOSFETs and IGBTs
Fundamentals of Power Electronics
Gate
Q4
Q1
Cathode
The MCT: equivalent circuit
vA
Ts
–
gate + driver
vB
–
iB
ideal diode
iL(t)
L
vB(t) = vA(t) – Vg i A(t) + iB(t) = iL
W off =
1 2
62
VgiL (t 2 – t 0)
pA(t)
= vA iA
diode waveforms
transistor waveforms
transistor turn-off transition
Buck converter example
DTs
+ physical MOSFET
iA
Fundamentals of Power Electronics
+ –
Chapter 4: Switch realization
0
–Vg
0
iL
t0
t2
iA(t)
0
iL
0
area Woff
Vg
t
t
t
Chapter 4: Switch realization
t1
vB(t)
iB(t)
V g iL
vA(t)
4.3.1. Transistor switching with clamped inductive load
61
• Time required to insert or remove the controlling charge determines switching times
• Semiconductor devices are charge controlled
• Energy stored in device capacitances and parasitic inductances
• Diode stored charge
• Transistor switching times
Fundamentals of Power Electronics
Vg
4.3. Switching loss
• Energy is lost during the semiconductor switching transitions, via several mechanisms:
+ –
Psw = 1 Ts
VgiL (t 2 – t 0)
physical IGBT
iA
DTs
+
vA
Ts
–
vB
– gate + driver
+ –
iB
ideal diode
iL(t)
L
63
64
pA(t) dt = (W on + W off ) fs switching transitions
Fundamentals of Power Electronics
Psw = 1 Ts
pA(t)
= v A iA
diode waveforms
IGBT waveforms
transistor turn-off transition
Example: buck converter with IGBT
+ –
switching transitions
pA(t) dt = (W on + W off ) fs
0
–Vg
0
iL
t0
iA(t)
t1
curr ent t
ail
t3
Vg
0
iL
0
t
t
Chapter 4: Switch realization
t2
area Woff
vB(t)
iB(t)
Vg iL
vA(t)
t
Chapter 4: Switch realization
Switching loss due to current-tailing in IGBT
Fundamentals of Power Electronics
Vg
1 2
Similar result during transistor turn-on transition. Average power loss:
W off =
Energy lost during transistor turn-off transition:
Switching loss induced by transistor turn-off transition
}
–
+
– vB iB
silicon diode
iL(t)
L
• Qr depends on diode on-state forward current, and on the rate-of-change of diode current during diode turn-off transition
• Diode recovered stored charge Qr flows through transistor during transistor turn-on transition, inducing switching loss
+ –
vA
Vg (iL – iB(t)) dt switching transition
Fundamentals of Power Electronics
• Often, this is the largest component of switching loss
= Vg i L t r + Vg Q r
WD ≈
With abrupt-recovery diode:
switching transition
vA(t) i A(t) dt
Energy lost in transistor:
WD =
Qr
area ~QrVg
t0
iB(t)
area –Qr
0
iL
0
vA(t)
Vg
iA(t)
t 1 t2
tr
vB(t)
area ~iLVgtr
–Vg
0
0
iL
pA(t)
= vA iA
diode waveforms
transistor waveforms
66
Qr
area ~QrVg
t0
iB(t)
area –Qr
0
iL
0
vA(t)
Vg
iA(t)
t 1 t2
tr
vB(t)
area ~iLVgtr
–Vg
0
0
iL
t
t
t
Switching loss calculation
65
pA(t)
= vA iA
diode waveforms
transistor waveforms
Chapter 4: Switch realization
(t2 – t1) << (t1 – t0)
Abrupt-recovery diode:
(t2 – t1) >> (t1 – t0)
Soft-recovery diode:
t
Chapter 4: Switch realization
t
t
4.3.2. Diode recovered charge
Fundamentals of Power Electronics
Vg
+
fast transistor
iA
+ –
Σ
capacitive elements
1 2
CiV 2i
67
WL =
Σ
inductive elements
1 2
L jI 2j
Chapter 4: Switch realization
+ –
Cds
Cj
Fundamentals of Power Electronics
68
W C = 12 (Cds + C j) V 2g
Energy lost during MOSFET turn-on transition (assuming linear capacitances):
Vg
Buck converter example
Chapter 4: Switch realization
Example: semiconductor output capacitances
Fundamentals of Power Electronics
WC =
Total energy stored in linear capacitive and inductive elements:
• Inductances that appear effectively in series with switch elements are open-circuited when the switch turns off. Their stored energy is lost during the switch turn-off transition.
• Capacitances that appear effectively in parallel with switch elements are shorted when the switch turns on. Their stored energy is lost during the switch turn-on transition.
4.3.3. Device capacitances, and leakage, package, and stray inductances
+ –
C '0 V0 = vds vds
0
V DS
0
V DS
vds C ds(vds) dvds
C '0(vds) vds dvds = 23 Cds(VDS) V 2DS
vds i C dt =
69
Cds(VDS)
Chapter 4: Switch realization
4 3
Fundamentals of Power Electronics
70
• A significant loss in high current applications
• Transistors
• Diodes
Interconnection and package inductances
Chapter 4: Switch realization
• A significant loss when windings are not tightly coupled
• Effective inductances in series with windings
Transformer leakage inductance
• Significant reverse-biased junction capacitance
• Essentially no stored charge
Schottky diode
Some other sources of this type of switching loss
Fundamentals of Power Electronics
— same energy loss as linear capacitor having value
W Cds =
W Cds =
Energy stored in Cds at vds = VDS :
Cds(vds) ≈ C0
Approximate dependence of incremental Cds on vds :
MOSFET nonlinear Cds
silicon diode
+ – vL(t)
L
–
iB(t) + vB(t) C
71
–V2
0
vB(t)
0
iL(t)
0
vi(t) V1
t1
Qr = – t2
t3
iL(t) dt
Fundamentals of Power Electronics
W L = 12 L i 2L(t 3) = V2 Qr 72
Applied inductor voltage during interval t2 ≤ t ≤ t3 : di (t) vL(t) = L L = – V2 dt Hence, t3 t3 di (t) WL = L L iL(t) dt = ( – V2) iL(t) dt dt t2 t2
t2
Energy stored in inductor during interval t2 ≤ t ≤ t3 : t3 WL = vL(t) iL(t) dt
Recovered charge is
–V2
0
vB(t)
0
iL(t)
0
vi(t)
V1
t1
t2
–V2
t3
t
t
t
t2
t3
t
t
Chapter 4: Switch realization
area – Qr
–V2
t
Chapter 4: Switch realization
area – Qr
Energy associated with ringing
Fundamentals of Power Electronics
• Ringing of L-C network is damped by parasitic losses. Ringing energy is lost.
• When diode becomes reverse-biased, negative inductor current flows through capacitor C.
• Negative inductor current removes diode stored charge Qr
• Diode is forward-biased while iL(t) > 0
vi(t) + –
iL(t)
Ringing induced by diode stored charge
dc asymptote
Ploss = Pcond + Pfixed + W tot fsw
50% 10kHz
60%
70%
80%
90%
100%
73
Chapter 4: Switch realization
Pfixed = fixed losses (independent of load and fsw) Pcond = conduction losses
fsw
100kHz
74
1MHz
Pcond + Pfixed W tot
Chapter 4: Switch realization
This can be taken as a rough upper limit on the switching frequency of a practical converter. For fsw > fcrit, the efficiency decreases rapidly with frequency.
fcrit =
Switching losses are equal to the other converter losses at the critical frequency
Efficiency vs. switching frequency
Fundamentals of Power Electronics
η
Fundamentals of Power Electronics
where
Ploss = Pcond + Pfixed + W tot fsw
Total converter loss can be expressed as
Psw = W tot fsw
Average switching power loss is
W tot = W on + W off + W D + W C + W L + ...
Add up all of the energies lost during the switching transitions of one switching period:
4.3.4. Efficiency vs. switching frequency
fcrit
75
Chapter 4: Switch realization
Fundamentals of Power Electronics
76
Chapter 4: Switch realization
5. Majority carrier devices, including the MOSFET and Schottky diode, exhibit very fast switching times, controlled essentially by the charging of the device capacitances. However, the forward voltage drops of these devices increases quickly with increasing breakdown voltage. 6. Minority carrier devices, including the BJT, IGBT, and thyristor family, can exhibit high breakdown voltages with relatively low forward voltage drop. However, the switching times of these devices are longer, and are controlled by the times needed to insert or remove stored minority charge. 7. Energy is lost during switching transitions, due to a variety of mechanisms. The resulting average power loss, or switching loss, is equal to this energy loss multiplied by the switching frequency. Switching loss imposes an upper limit on the switching frequencies of practical converters.
Summary of chapter 4
Fundamentals of Power Electronics
1. How an SPST ideal switch can be realized using semiconductor devices depends on the polarity of the voltage which the devices must block in the off-state, and on the polarity of the current which the devices must conduct in the on-state. 2. Single-quadrant SPST switches can be realized using a single transistor or a single diode, depending on the relative polarities of the off-state voltage and on-state current. 3. Two-quadrant SPST switches can be realized using a transistor and diode, connected in series (bidirectional-voltage) or in anti-parallel (bidirectionalcurrent). Several four-quadrant schemes are also listed here. 4. A “synchronous rectifier” is a MOSFET connected to conduct reverse current, with gate drive control as necessary. This device can be used where a diode would otherwise be required. If a MOSFET with sufficiently low Ron is used, reduced conduction loss is obtained.
Summary of chapter 4
77
Chapter 4: Switch realization
Fundamentals of Power Electronics
1
Chapter 5: Discontinuous conduction mode
5.4. Summary of results and key points
5.3. Boost converter example
5.2. Analysis of the conversion ratio M(D,K)
5.1. Origin of the discontinuous conduction mode, and mode boundary
Chapter 5. The Discontinuous Conduction Mode
Fundamentals of Power Electronics
8. The diode and inductor present a “clamped inductive load” to the transistor. When a transistor drives such a load, it experiences high instantaneous power loss during the switching transitions. An example where this leads to significant switching loss is the IGBT and the “current tail” observed during its turn-off transition. 9. Other significant sources of switching loss include diode stored charge and energy stored in certain parasitic capacitances and inductances. Parasitic ringing also indicates the presence of switching loss.
Summary of chapter 4
D1 iD(t)
iL(t)
L
C
R
Note that I depends on load, but ∆iL does not.
Minimum diode current is (I – ∆iL) Dc component I = V/R Current ripple is (V – V) Vg DD'Ts ∆iL = g DTs = 2L 2L
+ –
Q1
3
–
V
+
I
0
0
iD(t)
conducting devices:
I
iL(t)
DTs
D1
Ts
Ts Q1
t
∆iL
t
Chapter 5: Discontinuous conduction mode
Q1
DTs
∆iL
continuous conduction mode (CCM)
Buck converter example, with single-quadrant switches
Fundamentals of Power Electronics
Vg
Chapter 5: Discontinuous conduction mode
5.1. Origin of the discontinuous conduction mode, and mode boundary
2
Occurs because switching ripple in inductor current or capacitor voltage causes polarity of applied switch current or voltage to reverse, such that the current- or voltage-unidirectional assumptions made in realizing the switch are violated. Commonly occurs in dc-dc converters and rectifiers, having singlequadrant switches. May also occur in converters having two-quadrant switches. Typical example: dc-dc converter operating at light load (small load current). Sometimes, dc-dc converters and rectifiers are purposely designed to operate in DCM at all loads. Properties of converters change radically when DCM is entered: M becomes load-dependent Output impedance is increased Dynamics are altered Control of output voltage may be lost when load is removed
Fundamentals of Power Electronics
●
●
●
●
Introduction to Discontinuous Conduction Mode (DCM)
D1 iD(t)
C
R
Note that I depends on load, but ∆iL does not. 4
–
V
+
D1 iD(t)
C
R
Note that I depends on load, but ∆iL does not. The load current continues to be positive and non-zero.
Minimum diode current is (I – ∆iL) Dc component I = V/R Current ripple is (V – V) Vg DD'Ts ∆iL = g DTs = 2L 2L
+ –
iL(t)
L
0
0
DTs
5
D1
Ts
Ts Q1
I
iL(t)
iD(t)
0
0
∆iL
∆iL
t
t
DTs
DTs
D2Ts
D2Ts D1
X
D3Ts
Ts
Ts Q1
t
Chapter 5: Discontinuous conduction mode
D1Ts Q1
t
Chapter 5: Discontinuous conduction mode
Q1
DTs
Discontinuous conduction mode
conducting devices:
–
V
+
Increase R some more, such that I < ∆iL Q1
I
iD(t)
conducting devices:
I
iL(t)
CCM-DCM boundary
Further reduce load current
Fundamentals of Power Electronics
Vg
L
Minimum diode current is (I – ∆iL) Dc component I = V/R Current ripple is (V – V) Vg DD'Ts ∆iL = g DTs = 2L 2L
+ –
iL(t)
Fundamentals of Power Electronics
Vg
Q1
Increase R, until I = ∆iL
Reduction of load current
0
Kc ( rit D) = 1
–D
Fundamentals of Power Electronics
0
1
K < Kcrit: DCM
for K < 1:
K = 2L/RTs 1
7
D
0
1
2
0
–D
K = 2L/RTs
1
D
Chapter 5: Discontinuous conduction mode
Kc ( rit D) = 1
for K > 1:
K > Kcrit: CCM
Chapter 5: Discontinuous conduction mode
K and Kcrit vs. D
6
K < K crit(D) for DCM K = 2L and K crit(D) = D' RTs
K > Kcrit: CCM
where
Fundamentals of Power Electronics
2
2L < D' RTs
This expression is of the form
Simplify:
Insert buck converter expressions for I and ∆iL : DVg DD'TsVg < R 2L
I > ∆iL for CCM I < ∆iL for DCM
Mode boundary
8
Chapter 5: Discontinuous conduction mode
or or
R < Rcrit(D) R > Rcrit(D)
for CCM for DCM
max ( K crit )
2 L Ts
Fundamentals of Power Electronics
Chapter 5: Discontinuous conduction mode
9
(1 – D)2 Buck-boost
2L (1 – D) 2 T s
D (1 – D)2 Boost
1
1 4 27
2 L Ts 27 L 2 Ts
min ( Rcrit ) 0≤D≤1
R crit(D) 2L (1 – D)T s 2L D (1 – D) 2 T s 0≤D≤1
(1 – D)
K crit(D) Buck
Converter
Table 5.1. CCM-DCM mode boundaries for the buck, boost, and buck-boost converters
K > K crit(D) K < K crit(D)
Summary: mode boundary
Fundamentals of Power Electronics
where
R < Rcrit(D) for CCM for DCM R > Rcrit(D) Rcrit(D) = 2L D'Ts
Solve Kcrit equation for load resistance R:
Critical load resistance Rcrit
Analysis of the conversion ratio M(D,K)
0
Ts
iC(t) dt = 0
because ∆v << V
Q1
D1 iD(t)
iL(t)
L
C
R
Vg
11
+ –
+ –
+ –
iL(t)
L
C
+ vL(t) –
L
C
+ vL(t) –
iL(t)
C
+ vL(t) –
iC(t)
iC(t)
iC(t)
R
R
R
–
v(t)
+
–
v(t)
+
–
v(t)
+
Chapter 5: Discontinuous conduction mode
Vg
Vg
subinterval 2
subinterval 3
–
V
+
subinterval 1
iL(t)
L
Chapter 5: Discontinuous conduction mode
Example: Analysis of DCM buck converter M(D,K)
10
Converter steady-state equations obtained via charge balance on each capacitor and volt-second balance on each inductor. Use care in applying small ripple approximation.
i(t) ≈ I is a poor approximation when ∆i > I
v(t) ≈ V
Fundamentals of Power Electronics
+ –
vL(t) dt = 0
Small ripple approximation sometimes applies:
iC = 1 Ts
Fundamentals of Power Electronics
Vg
0
Ts
Capacitor charge balance
vL = 1 Ts
Inductor volt-second balance
Analysis techniques for the discontinuous conduction mode:
5.2.
Fundamentals of Power Electronics
vL(t) ≈ – V iC(t) ≈ iL(t) – V / R
12
Vg
+ –
13
Vg
+ –
C
+ vL(t) –
L iC(t) R
–
v(t)
+
iC(t) R
–
v(t)
+
Chapter 5: Discontinuous conduction mode
C
+ vL(t) –
L
Chapter 5: Discontinuous conduction mode
iL(t)
Subinterval 2
Small ripple approximation for v(t) but not for i(t):
vL(t) = – v(t) iC(t) = iL(t) – v(t) / R
Fundamentals of Power Electronics
vL(t) ≈ Vg – V iC(t) ≈ iL(t) – V / R
Small ripple approximation for v(t) (but not for i(t)!):
vL(t) = Vg – v(t) iC(t) = iL(t) – v(t) / R
iL(t)
Subinterval 1
14
Vg
+ –
C
+ vL(t) –
L
D1Ts
Vg – V
–V
D2Ts
0
D3Ts
Ts
Fundamentals of Power Electronics
V = Vg
Solve for V:
D1 D1 + D2
15
t
R
–
v(t)
+
Chapter 5: Discontinuous conduction mode
note that D2 is unknown
vL(t) = D1(Vg – V) + D2( – V) + D3(0) = 0
Volt-second balance:
vL(t)
iC(t)
Chapter 5: Discontinuous conduction mode
iL(t)
Inductor volt-second balance
Fundamentals of Power Electronics
vL(t) = 0 iC(t) = – V / R
Small ripple approximation:
vL = 0, iL = 0 iC(t) = iL(t) – v(t) / R
Subinterval 3
0
Ts
iL(t) dt
iL(t) dt = 1 i pk (D1 + D2)Ts 2
Fundamentals of Power Electronics
D 1T s iL = (Vg – V) (D1 + D2) 2L
Ts
triangle area formula:
iL = 1 Ts
0
Vg – V D 1T s L average current:
iL(D1Ts) = i pk =
peak current:
16
= I
iL(t)
0 D1Ts
Vg – V L
L
= I
iL(t)
D 1T s
DTs D2Ts
ipk –V L
Ts
D 3 Ts
Ts
17
t
t
Chapter 5: Discontinuous conduction mode
V = D1Ts (D + D ) (V – V) 1 2 g R 2L
equate dc component to dc load current:
0
Vg – V L
D3Ts
–
v(t)
+
Chapter 5: Discontinuous conduction mode
DTs
R
v(t)/R iC(t)
D2Ts
ipk –V L
C
iL(t)
Inductor current waveform
Fundamentals of Power Electronics
must compute dc component of inductor current and equate to load current (for this buck converter example)
iL = V / R
hence
iC = 0
capacitor charge balance:
iL(t) = iC(t) + V / R
node equation:
Capacitor charge balance
D1 D1 + D2
0.0
0.2
0.4
0.6
0.8
0.0
0.2
K≥1
0.4
K = 0.5
K = 0.1
K = 0.01
Fundamentals of Power Electronics
M(D,K)
1.0
18
D
0.6
19
0.8
M=
1.0
1+
for K < K crit
Chapter 5: Discontinuous conduction mode
2 1 + 4K / D 2
D
for K > K crit
Chapter 5: Discontinuous conduction mode
(from capacitor charge balance)
(from inductor volt-second balance)
Buck converter M(D,K)
Fundamentals of Power Electronics
V = 2 Vg 1 + 1 + 4K / D 21 where K = 2L / RTs valid for K < K crit
Eliminate D2 , solve for V :
V = D1Ts (D + D ) (V – V) 1 2 g R 2L
V = Vg
Two equations and two unknowns (V and D2):
Solution for V
+ –
Fundamentals of Power Electronics
where
for CCM
for CCM
20
C
iD(t)
21
Kcrit(D)
iC(t) R
I=
Vg D' 2 R
0
0.05
0.1
0.15
0
Vg DTs 2L
–
v(t)
+
0.4
D
0.6
0.8
1
Chapter 5: Discontinuous conduction mode
0.2
Kcrit ( 13 ) = 4 27
Chapter 5: Discontinuous conduction mode
∆iL =
Previous CCM soln:
Q1
D1
Mode boundary
K > K crit(D) for CCM for DCM K < K crit(D) K = 2L and K crit(D) = DD' 2 RTs
2L > DD' 2 RTs
Vg DTsVg > 2 2L D' R
Fundamentals of Power Electronics
L + vL(t) –
I > ∆iL for CCM I < ∆iL for DCM
Mode boundary:
Vg
i(t)
5.3. Boost converter example
+ –
+ vL(t) –
L
Q1
C
D1 i (t) D
Fundamentals of Power Electronics
Vg
i(t)
22
0
0.05
0.1
0
0.2
D
iC(t) R
subinterval 2
23
subinterval 3
–
v(t)
+
subinterval 1
0.6
K
CCM K > Kcrit
0.8
1
+ –
+ –
+ –
i(t)
i(t)
+ vL(t) –
L
+ vL(t) –
L
+ vL(t) –
C
C
C
iC(t) R
R
R
iC(t)
iC(t)
–
–
v(t)
+
–
v(t)
+
v(t)
+
Chapter 5: Discontinuous conduction mode
Vg
Vg
Vg
i(t)
L
Chapter 5: Discontinuous conduction mode
0.4
DCM K < Kcrit
Conversion ratio: DCM boost
Fundamentals of Power Electronics
where
K > K crit(D) for CCM for DCM K < K crit(D) K = 2L and K crit(D) = DD' 2 RTs
0.15
CCM
Mode boundary
(D) K crit
Fundamentals of Power Electronics
vL(t) ≈ Vg – V iC(t) ≈ i(t) – V / R
Vg
24
+ –
i(t)
Vg
25
+ –
i(t)
L iC(t) R
–
v(t)
+
R
–
v(t)
+
Chapter 5: Discontinuous conduction mode
D1Ts < t < (D1 +D2)Ts
C
iC(t)
Chapter 5: Discontinuous conduction mode
+ vL(t) –
L
C
0 < t < D 1T s
+ vL(t) –
Subinterval 2
Small ripple approximation for v(t) but not for i(t):
vL(t) = Vg – v(t) iC(t) = i(t) – v(t) / R
Fundamentals of Power Electronics
vL(t) ≈ Vg iC(t) ≈ – V / R
Small ripple approximation for v(t) (but not for i(t)!):
vL(t) = Vg iC(t) = – v(t) / R
Subinterval 1
26
Vg
+ –
+ vL(t) –
L
C
D1Ts
Vg
Fundamentals of Power Electronics
Solve for V: D + D2 V= 1 Vg D2
27
0
D3Ts
Ts
t
R
–
v(t)
+
Chapter 5: Discontinuous conduction mode
note that D2 is unknown
Vg – V
D2Ts
D1Vg + D2(Vg – V) + D3(0) = 0
Volt-second balance:
vL(t)
iC(t)
Chapter 5: Discontinuous conduction mode
(D1 +D2)Ts < t < Ts
i(t)
Inductor volt-second balance
Fundamentals of Power Electronics
vL(t) = 0 iC(t) = – V / R
Small ripple approximation:
vL = 0, i = 0 iC(t) = – v(t) / R
Subinterval 3
0
iD(t) dt
Fundamentals of Power Electronics
0
iD(t) dt = 1 i pk D2Ts 2
triangle area formula:
iD = 1 Ts
Ts
average diode current:
peak current: Vg i pk = DT L 1 s
Ts
28
iC(t) R
–
v(t)
+
iD(t)
i(t)
0
0
29
D1Ts
D1Ts
L
Vg
DTs
DTs
L
D3Ts
D3Ts
Ts
Ts
t Chapter 5: Discontinuous conduction mode
D2Ts
L
Vg – V
D2Ts
ipk
Vg – V
ipk
t
Chapter 5: Discontinuous conduction mode
C
D1 i (t) D
Inductor and diode current waveforms
Fundamentals of Power Electronics
must compute dc component of diode current and equate to load current (for this boost converter example)
iD = V / R
hence
iC = 0
capacitor charge balance:
iD(t) = iC(t) + v(t) / R
node equation:
Capacitor charge balance
Solution for V
30
Chapter 5: Discontinuous conduction mode
(from capacitor charge balance)
V 2gD 21 =0 K
Fundamentals of Power Electronics
V 2 – VVg –
31
Chapter 5: Discontinuous conduction mode
Substitute into charge balance eqn, rearrange terms:
Eliminate D2 , solve for V. From volt-sec balance eqn: Vg D2 = D1 V – Vg
V g D 1 D 2T s V = R 2L
Two equations and two unknowns (V and D2): D + D2 (from inductor volt-second balance) V= 1 Vg D2
Fundamentals of Power Electronics
V g D 1 D 2T s V = R 2L
equate to dc load current:
V g D 1 D 2T s iD = 1 1 i pk D2Ts = Ts 2 2L
average diode current:
Equate diode current to load current
K = 2L / RTs K < Kcrit(D)
1 + 4D 21 / K 2
0
0.25
K≥
K=
K
4/2
5
D
0.5
7
0.1
=
0.0
0.75
33
1
M =
1+
for K < K crit
Chapter 5: Discontinuous conduction mode
M≈1+ D 2 K
Approximate M in DCM:
1 + 4D 2 / K 2
1 1–D
for K > K crit
Chapter 5: Discontinuous conduction mode
Boost converter characteristics
32
Transistor duty cycle D = interval 1 duty cycle D1
where valid for
V = M(D ,K) = 1 + 1 Vg
1
Fundamentals of Power Electronics
0
1
2
3
4
5
1 + 4D 21 / K 2
Note that one root leads to positive V, while other leads to negative V. Select positive root:
V = 1± Vg
Use quadratic formula:
Fundamentals of Power Electronics
M(D,K)
Solution for V V 2gD 21 V – VVg – =0 K 2
0.0 K=
0
0.2
Fundamentals of Power Electronics
0
1
DCM M(D,K)
0.4
B
D
0.6
35
0.8
1 K
1) ×– t( s oo k-b uc
ost
Buck
Bo
1
D 1 1–D – D 1–D
CCM M(D)
Chapter 5: Discontinuous conduction mode
Chapter 5: Discontinuous conduction mode
• DCM boost characteristic is nearly linear
• CCM and DCM characteristics intersect at mode boundary. Actual M follows characteristic having larger magnitude
• DCM buck-boost characteristic is linear
• DCM buck and boost characteristics are asymptotic to M = 1 and to the DCM buck-boost characteristic
Summary of DCM characteristics
34
K
DCM D2(D,K) K M(D,K) D K M(D,K) D
DCM occurs for K < K crit .
DCM M(D,K) 2 1 + 1 + 4K / D 2 1 + 1 + 4D 2 / K 2 – D K
K = 2L / RT s.
Fundamentals of Power Electronics
with
D (1 – D)2
Boost (1 – D)2
(1 – D)
Buck
Buck-boost
K crit (D)
Converter
Table 5.2. S ummary of CCM-DCM characteristics for the buck, boost, and buck-boost converters
Summary of DCM characteristics
36
Chapter 5: Discontinuous conduction mode
Fundamentals of Power Electronics
37
Chapter 5: Discontinuous conduction mode
4. Extra care is required when applying the small-ripple approximation. Some waveforms, such as the output voltage, should have small ripple which can be neglected. Other waveforms, such as one or more inductor currents, may have large ripple that cannot be ignored. 5. The characteristics of a converter changes significantly when the converter enters DCM. The output voltage becomes loaddependent, resulting in an increase in the converter output impedance.
Summary of key points
Fundamentals of Power Electronics
1. The discontinuous conduction mode occurs in converters containing current- or voltage-unidirectional switches, when the inductor current or capacitor voltage ripple is large enough to cause the switch current or voltage to reverse polarity. 2. Conditions for operation in the discontinuous conduction mode can be found by determining when the inductor current or capacitor voltage ripples and dc components cause the switch on-state current or off-state voltage to reverse polarity. 3. The dc conversion ratio M of converters operating in the discontinuous conduction mode can be found by application of the principles of inductor volt-second and capacitor charge balance.
Summary of key points
+ –
1
2
L
C
R
–
V
+
Chapter 6: Converter circuits
Fundamentals of Power Electronics
• Conversion ratio is M = D
2
Chapter 6: Converter circuits
• Switch changes dc component, low-pass filter removes switching harmonics
Begin with buck converter: derived in chapter 1 from first principles
Vg
1
• For a given application, which converter is best?
• How can we obtain transformer isolation in a converter?
• What converters are possible?
• How can we obtain a converter having given desired properties?
6.1. Circuit manipulations
Fundamentals of Power Electronics
6.5. Summary of key points
6.4. Converter evaluation and design
6.3. Transformer isolation
6.2. A short list of converters
6.1. Circuit manipulations
• Where do the boost, buck-boost, and other converters originate?
Chapter 6. Converter Circuits
Fundamentals of Power Electronics
V2 = DV1
4
V1 = 1 V2 D
Power flow
–
+
–
2
Chapter 6: Converter circuits
+ –
Chapter 6: Converter circuits
port 2
V2
1
L
V1
+
port 1
Interchange power source and load:
Inversion of source and load
3
Power flow
–
+
port 2
–
2
L
V2
1
V2 = DV1
V1
Fundamentals of Power Electronics
+ –
+
port 1
Buck converter example
Interchange power input and output ports of a converter
6.1.1. Inversion of source and load
V1 = M 1(D) Vg
V = M 2(D) V1
V1 = M 1(D) Vg
+ –
Fundamentals of Power Electronics
Vg
converter 1
+ –
5
Chapter 6: Converter circuits
Inversion of buck converter yields boost converter
–
–
Power flow
V2
V1
port 2
+
L
+
port 1
D
V = M (D) 2 V1
6
–
V
+
Chapter 6: Converter circuits
converter 2
V = M(D) = M (D) M (D) 1 2 Vg
–
V1
+
6.1.2. Cascade connection of converters
Fundamentals of Power Electronics
• Interchange of D and D’ V1 = 1 V2 D'
• Transistor conducts when switch is in position 2
• Reversal of power flow requires new realization of switches
Realization of switches as in chapter 4
+ –
2
V = 1 V1 1 – D
V1 =D Vg
1
2
L1
+ –
2
Fundamentals of Power Electronics
Vg
1
L
iL
combine inductors L1 and L2
+ –
C1 –
V1
+
L2 1
2
7
V = D Vg 1 – D
Boost converter
C2
1
8
2
L2 1
–
V
+
2
C2
R
–
V
+
–
V
Chapter 6: Converter circuits
Noninverting buck-boost converter
R
+
Chapter 6: Converter circuits
Buck cascaded by boost: simplification of internal filter remove capacitor C1
Vg
L1
Buck converter
Fundamentals of Power Electronics
Vg
1
Example: buck cascaded by boost
{ {
+ –
+ –
2
subinterval 1
–
V
+
L
9
Vg
iL 1
+ –
–
V
+
Vg
inverting buck-boost
+ –
+ –
Fundamentals of Power Electronics
Vg
noninverting buck-boost
iL
iL
subinterval 1
10
–
V
+
–
V
+
Vg
Vg
+ –
+ –
–
V
+
–
V
+
–
V
Chapter 6: Converter circuits
iL
iL
subinterval 2
+
Chapter 6: Converter circuits
iL
subinterval 2
2
Reversal of output voltage polarity
Fundamentals of Power Electronics
Vg
iL
Vg
1
Noninverting buck-boost converter
+ –
+ – –
V
+ Vg
+ –
iL
subinterval 2
1
iL
11
–
V
+
–
V
+
Chapter 6: Converter circuits
V =– D Vg 1–D
Fundamentals of Power Electronics
12
Cuk converter: boost cascaded by buck
• Other cascade connections are possible
Pulsating output current of boost converter
Pulsating input current of buck converter
Chapter 6: Converter circuits
Equivalent circuit model: buck 1:D transformer cascaded by boost D’:1 transformer
• Properties of buck-boost converter follow from its derivation as buck cascaded by boost
Discussion: cascade connections
2
One side of inductor always connected to ground — hence, only one SPDT switch needed:
Fundamentals of Power Electronics
Vg
Vg
iL
subinterval 1
Reduction of number of switches: inverting buck-boost
Vg
+ –
A
a 1
c C
-t ree th
2
er mi na l c
ell b
B
buck-boost converter
a-A b-C c-B
Vg
+ –
A a
re e
1
c C
2
-ter mi n al c
l b B
Cuk converter
a-A b-C c-B
–
v
+
Chapter 6: Converter circuits
boost converter with L-C output filter
a-C b-A c-B
14
buck converter with L-C input filter
a-A b-B c-C
Fundamentals of Power Electronics
–
v
+
Chapter 6: Converter circuits
Three-terminal cell can be connected between source and load in three nontrivial distinct ways:
A capacitor and SPDT switch as a threeterminal cell:
th
Rotation of a dual three-terminal network
13
boost converter
a-C b-A c-B
Fundamentals of Power Electronics
buck converter
a-A b-B c-C
Three-terminal cell can be connected between source and load in three nontrivial distinct ways:
Treat inductor and SPDT switch as threeterminal cell:
6.1.3. Rotation of three-terminal cell
el
+ –
+ –
Buck converter 1
2
1
1
2
Fundamentals of Power Electronics
Vg
–
V2
+
–
V1
+
15
load
–
V
+
Chapter 6: Converter circuits
The outputs V1 and V2 may both be positive, but the differential output voltage V can be positive or negative.
V = V1 – V2
Differential load voltage is
–
V2
+
–
V1
+
Buck converter 2
16
–
V
+
Chapter 6: Converter circuits
V = (2D – 1) Vg
Simplify:
V = DVg – D'Vg
Differential load voltage is
Converter #2 transistor driven with duty cycle complement D’
Converter #1 transistor driven with duty cycle D
Differential connection using two buck converters
D'
V2 = M(D') Vg
converter 2
D
V1 = M(D) Vg
converter 1
Fundamentals of Power Electronics
Vg
dc source
6.1.4. Differential connection of load to obtain bipolar output voltage
} {
Buck converter 1
2
1
1
2
Buck converter 2
Fundamentals of Power Electronics
+ –
Original circuit
Vg
17
0.5
1
Chapter 6: Converter circuits
D
–
V2
+
–
V1
+
18
Chapter 6: Converter circuits
1
–
–
2
V
V + –
+
2
+
Vg
1
Bypass load directly with capacitor
Simplification of filter circuit, differentially-connected buck converters
Fundamentals of Power Electronics
–1
0
1
M(D)
V = (2D – 1) Vg
Conversion ratio M(D), differentially-connected buck converters
} {
+ – 2 1
2
2
iL
L +
R
V –
C
19
Chapter 6: Converter circuits
Commonly used in single-phase inverter applications and in servo amplifier applications
H-bridge, or bridge inverter
+ –
D3
V3 = M(D 3) Vg
converter 3
D2
V2 = M(D 2) Vg
converter 2
D1
V1 = M(D 1) Vg
converter 1
–
V3
+
–
V2
+
–
V1
+
20
+ vbn –
Vn
3øac load
v cn
Fundamentals of Power Electronics
Vg
dc source
+ –
–
V
+
Vg
1
Re-draw for clarity
Chapter 6: Converter circuits
Control converters such that their output voltages contain the same dc biases. This dc bias will appear at the neutral point Vn. It then cancels out, so phase voltages contain no dc bias.
Van = V1 – Vn Vbn = V2 – Vn Vcn = V3 – Vn
Vn = 1 (V1 + V2 + V3) 3 Phase voltages are
With balanced 3ø load, neutral voltage is
Differential connection to obtain 3ø inverter
Fundamentals of Power Electronics
Vg
1
Combine series-connected inductors
Simplification of filter circuit, differentially-connected buck converters
+ n
va – – +
1
2
+ – + vbn –
Vn
3øac load
Chapter 6: Converter circuits
+
+ –
+ vbn –
3øac load
22
Vn
Chapter 6: Converter circuits
+
Fundamentals of Power Electronics
v cn
“Voltage-source inverter” or buck-derived three-phase inverter
Vg
dc source
va –
Re-draw for clarity:
n
+
3ø differential connection of three buck converters
21
–
V3
+
–
V2
+
–
V1
+
v cn
Fundamentals of Power Electronics
Vg
dc source
3ø differential connection of three buck converters
+ n
va – –
–
23
Chapter 6: Converter circuits
boost
Fundamentals of Power Electronics
current-fed bridge
ac-dc converters
bridge
dc-ac converters
buck
dc-dc converters noninverting buck-boost
24
Chapter 6: Converter circuits
inverse of Watkins-Johnson
Watkins-Johnson
buck-boost
• After elimination of degenerate and redundant cases, eight converters are found:
• There are a limited number of ways to do this, so all possible combinations can be found
• Use switches to connect inductor between source and load, in one manner during first subinterval and in another during second subinterval
Single-input single-output converters containing one inductor
Fundamentals of Power Electronics
• Single-input single-output converters containing two inductors. The switching period is divided into two subintervals. Several of the more interesting members of this class are listed.
• Single-input single-output converters containing a single inductor. The switching period is divided into two subintervals. This class contains eight converters.
Two simple classes of converters are listed here:
An infinite number of converters are possible, which contain switches embedded in a network of inductors and capacitors
6.2. A short list of converters
+ –
+ –
1
2
1
2
1
2
+ –
2
Fundamentals of Power Electronics
Vg
1
4. Noninverting buck-boost
+ –
3. Buck-boost
Vg
M(D) =
–
V
+
–
V
+
1 1–D
M(D) = D
25
0
1
2
3
4
M(D)
0
0.5
1
M(D)
0
0
1
2
M(D) =
–
V
+
D 1–D
–
26
D 1–D
V
+
M(D) = –
0
1
2
3
4
M(D)
M(D)
–4
–3
–2
–1
0
0
0
1
1
D
D
1
D
D
Chapter 6: Converter circuits
0.5
0.5
1
Chapter 6: Converter circuits
0.5
0.5
Converters producing a unipolar output voltage
Fundamentals of Power Electronics
Vg
2. Boost
Vg
1. Buck
Converters producing a unipolar output voltage
2
+ – 1
2
2
1
+
–
V
+
+ – 2
1
1
2
–
V
+
+
Fundamentals of Power Electronics
Vg
2
8. Inverse of Watkins-Johnson
+ –
1
7. Current-fed bridge
Vg
+ –
27
2
1
–
V
+
–3
–2
–1
0
1
M(D)
–1
0
1
M(D)
or Vg
+ –
M(D) =
V –
M(D) =
1
D 2D – 1
1 2D – 1
2
1
2
28
–
V
+
–2
–1
0
1
2
M(D)
–2
–1
0
1
2
M(D)
1
1
D
D
1
D
D
Chapter 6: Converter circuits
0.5
0.5
1
Chapter 6: Converter circuits
0.5
0.5
Converters producing a bipolar output voltage suitable as ac-dc rectifiers
Vg
M(D) = 2D – 1 D
1
2
M(D) = 2D – 1
or
V –
Fundamentals of Power Electronics
Vg
+ –
1
6. Watkins-Johnson
Vg
5. Bridge
Converters producing a bipolar output voltage suitable as dc-ac inverters
+ –
1
1 2
2
M(D) =
–
V
+
–
V
+
D 1–D
D 1–D
M(D) = –
29
0
1
2
3
4
M(D)
M(D)
–4
–3
–2
–1
0
0
0
1
1
D
D
Chapter 6: Converter circuits
0.5
0.5
+ –
+ –
1
2
1
Fundamentals of Power Electronics
Vg
4. Buck 2
Vg
1
3. Inverse of SEPIC
2
D 1–D
2
M(D) = D 2
M(D) =
–
V
+
30
–
V
+
0
0.5
1
M(D)
0
1
2
3
4
M(D)
0
0
1
1
D
D
Chapter 6: Converter circuits
0.5
0.5
Several members of the class of two-inductor converters
Fundamentals of Power Electronics
Vg + –
2. SEPIC
Vg
1. Cuk
Several members of the class of two-inductor converters
–
v3(t)
Fundamentals of Power Electronics
: n3
–
–
+
v2(t)
v1(t)
i3(t)
+
i2(t) +
n1 : n 2
Multiple winding transformer i1(t)
31
iM(t) LM
i1'(t)
32
ideal transformer
: n3
i3(t)
i2(t)
–
v3(t)
+
–
v2(t)
+
Chapter 6: Converter circuits
v1(t) v2(t) v3(t) n 1 = n 2 = n 3 = ... 0 = n 1i1'(t) + n2i2(t) + n 3i3(t) + ...
–
v1(t)
+
i1(t)
Equivalent circuit model n1 : n2
Chapter 6: Converter circuits
A simple transformer model
Fundamentals of Power Electronics
• Obtain multiple output voltages via multiple transformer secondary windings and multiple converter secondary circuits
• Minimization of current and voltage stresses when a large step-up or step-down conversion ratio is needed — use transformer turns ratio
• Reduction of transformer size by incorporating high frequency isolation transformer inside converter
• Isolation of input and output ground connections, to meet safety requirements
Objectives:
6.3. Transformer isolation
33
B(t) ∝
v1(t) dt
Fundamentals of Power Electronics
0
Ts
34
Magnetizing current is determined by integral of the applied winding voltage. The magnetizing current and the winding currents are independent quantities. Volt-second balance applies: in steady-state, iM(Ts) = iM(0), and hence
The magnetizing inductance is a real inductor, obeying di (t) v1(t) = L M M dt integrate: t iM (t) – iM (0) = 1 v (τ) dτ LM 0 1
0= 1 Ts
v1(t) dt
–
v1(t)
+
LM
ideal transformer
: n3
n1 : n2
i3(t)
i2(t)
–
v3(t)
+
–
v2(t)
+
Chapter 6: Converter circuits
iM(t)
i1(t)
i1'(t)
Chapter 6: Converter circuits
H(t) ∝ i M (t)
slope ∝ LM
saturation
Transformer core B-H characteristic
Volt-second balance in LM
Fundamentals of Power Electronics
• Transformer saturates when magnetizing current iM is too large
• At dc: magnetizing inductance tends to short-circuit. Transformers cannot pass dc voltages
• If all secondary windings are disconnected, then primary winding behaves as an inductor, equal to the magnetizing inductance
• Appears effectively in parallel with windings
• Models magnetization of transformer core material
The magnetizing inductance LM
Q2
D2
D1
Fundamentals of Power Electronics
+ –
Q1
Q4
Q3
D4
D3
Full-bridge isolated buck converter
Vg
35
36
–
vT(t)
+
i1(t)
: n
1 : n
D6
D5
iD5(t)
C
R
–
v
+
Chapter 6: Converter circuits
–
vs(t)
+
L
i(t)
Chapter 6: Converter circuits
6.3.1. Full-bridge and half-bridge isolated buck converters
Fundamentals of Power Electronics
• apply volt-second balance to all converter inductors, including magnetizing inductance
• analyze converter as usual, treating magnetizing inductance as any other inductor
• replace transformer by equivalent circuit model containing magnetizing inductance
• To understand operation of transformer-isolated converters:
• The need to reset the transformer volt-seconds to zero by the end of each switching period adds considerable complexity to converters
• “Transformer reset” is the mechanism by which magnetizing inductance volt-second balance is obtained
Transformer reset
+ –
Q2
D2
D1
I
0 Q1 Q4 D5
i
nVg
Vg
Vg LM
DTs
+
i1(t)
–
vT(t)
D4
D3
ideal
: n
1 : n
37
transformer model
LM
iM(t)
i1'(t)
D6
D5
iD6(t)
D5 D6
0.5 i
0
∆i
0
Ts Q2 Q3 D6
0
nVg
–Vg
LM
– Vg
Ts+DTs D5 D6
0.5 i
0
0
38
t 2Ts
+
L
C
R
–
v
+
Chapter 6: Converter circuits
• Effect of nonidealities?
• Transformer volt-second balance is obtained over two switching periods
• During next switching period: transistors Q2 and Q3 conduct for time DTs , applying voltseconds –Vg DTs to primary winding
• During first switching period: transistors Q1 and Q4 conduct for time DTs , applying voltseconds Vg DTs to primary winding
Chapter 6: Converter circuits
–
vs(t)
iD5(t) i(t)
Full-bridge: waveforms
Fundamentals of Power Electronics
conducting devices:
iD5(t)
vs(t)
i(t)
vT(t)
iM(t)
Q3
Q4
Fundamentals of Power Electronics
Vg
Q1
Full-bridge, with transformer equivalent circuit
0 Q1 Q4 D5
i
nVg
DTs
D6
D5
D5 D6
0.5 i
0
iD6(t)
–
Ts Q2 Q3 D6
0
nVg
Ts+DTs
40
D5 D6
0.5 i
0
t 2Ts
–
v
R
vs(t)
C
+
L +
iD5(t) i(t)
Fundamentals of Power Electronics
conducting devices:
iD5(t)
vs(t)
39
Chapter 6: Converter circuits
Chapter 6: Converter circuits
• Essentially no net magnetization of transformer core by secondary winding currents
• Secondary amp-turns add to approximately zero
• Output filter inductor current divides approximately equally between diodes
• During second (D’) subinterval, both secondary-side diodes conduct
Operation of secondary-side diodes
Fundamentals of Power Electronics
Saturation can be prevented by placing a capacitor in series with primary, or by use of current programmed mode (chapter 11)
Magnetizing current slowly increases in magnitude
Net volt-seconds are applied to primary winding
These volt-seconds never add to exactly zero.
– (Vg – (Q2 and Q3 forward voltage drops))( Q2 and Q3 conduction time)
Volt-seconds applied to primary winding during next switching period:
(Vg – (Q1 and Q4 forward voltage drops))( Q1 and Q4 conduction time)
Volt-seconds applied to primary winding during first switching period:
Effect of nonidealities on transformer volt-second balance
M(D) = nD
V = nDVg
V = vs
iD6(t)
–
41
D5 D6
Q2
Q1
D2
D1
Cb
Ca
–
vT(t)
+
i1(t)
: n
1 : n
D4
D3
iD3(t)
Ts Q2 Q3 D6
0
nVg
Ts+DTs
• M = 0.5 nD
42
• vs(t) is reduced by a factor of two
• Voltage at capacitor centerpoint is 0.5Vg
–
D5 D6
0.5 i
0
t 2Ts
C
i(t)
R
–
v
Chapter 6: Converter circuits
L
+
Chapter 6: Converter circuits
vs(t)
+
• Replace transistors Q3 and Q4 with large capacitors
+ –
Q1 Q4 D5
DTs
0
0.5 i
buck converter with turns ratio
0
i
nVg
∆i
Half-bridge isolated buck converter
Fundamentals of Power Electronics
Vg
iD5(t)
vs(t)
I
conducting devices:
–
v
R
vs(t) C
+
L +
iD5(t) i(t)
Fundamentals of Power Electronics
D6
D5
i(t)
Volt-second balance on output filter inductor
+ – Q1
D3
Q1
v1
LM
–
+
iM
–
vQ1
+
i1
i1'
i2
i3 D1
–
v3
v2 +
+
–
n1 : n2 : n3
Fundamentals of Power Electronics
+ –
L
C
44
D2
D3
–
vD3
+
L
–
V
+
R
–
V
Chapter 6: Converter circuits
C
+
Chapter 6: Converter circuits
R
Forward converter with transformer equivalent circuit
43
• Transformer is reset while transistor is off
• Maximum duty cycle is limited
• Single-transistor and two-transistor versions
Fundamentals of Power Electronics
Vg
D1
D2
• Buck-derived transformer-isolated converter
Vg
n1 : n2 : n3
6.3.2. Forward converter
Q1 D2
DTs
n3 n 1 Vg
Vg LM
Vg
+ –
v1
LM
Q1 on
–
+
iM
Fundamentals of Power Electronics
Vg
D1 D3
D 2T s Ts
0
n Vg – n1 2 LM
n – n 1 Vg 2
D3
D3Ts
0
0
0
45
t
i1
i1'
D1 off
i2
i3
–
v3
v2 +
+
–
n1 : n2 : n3
46
+
–
vD3
D2 on
L
C
–
V
Chapter 6: Converter circuits
R
+
Chapter 6: Converter circuits
• Output filter inductor, in conjunction with diode D3, may operate in either CCM or DCM
• Magnetizing current, in conjunction with diode D1, operates in discontinuous conduction mode
Subinterval 1: transistor conducts
Fundamentals of Power Electronics
conducting devices:
vD3
iM
v1
Forward converter: waveforms
+ –
+ –
i1
47
i2
+
v2
–
i3
–
v3
+
48
+
–
D3 on vD3
+
–
D3 on vD3
Subinterval 3
D1 on
i2 = iM n1 /n2
n1 : n2 : n3
i 1'
i3
–
v3
v2 +
+
–
Q1 off D1 off
–
v1
+
Fundamentals of Power Electronics
Vg
iM =0 LM
i1
i1'
Q1 off
–
v1
LM
Fundamentals of Power Electronics
Vg
+
iM
n1 : n2 : n3
L
L
C
R
–
V
+
–
V
Chapter 6: Converter circuits
R
+
Chapter 6: Converter circuits
C
Subinterval 2: transformer reset
Q1 D2
DTs
Vg LM
Vg
D1 D3
D2Ts Ts
n Vg – n1 2 LM
n – n 1 Vg 2
D3
D3Ts
0
0
Transformer reset
49
Fundamentals of Power Electronics
D3 = 1 – D – D2 ≥ 0 n D3 = 1 – D 1 + 2 ≥ 0 n1 Solve for D D ≤ 1n 1+ 2 n1
50
for n1 = n2:
D≤ 1 2
D3 cannot be negative. But D3 = 1 – D – D2. Hence
Solve for D2: n D2 = n 2 D 1
v1 = D ( Vg ) + D2 ( – Vg n 1 / n 2 ) + D3 ( 0 ) = 0
From magnetizing current volt-second balance:
Fundamentals of Power Electronics
t
Chapter 6: Converter circuits
Chapter 6: Converter circuits
v1 = D ( Vg ) + D2 ( – Vg n 1 / n 2 ) + D3 ( 0 ) = 0
conducting devices:
iM
v1
Magnetizing inductance volt-second balance
Q1 D2
DTs
n3 n 1 Vg
D3
Fundamentals of Power Electronics
conducting devices:
vD3
D2
iM(t)
iM(t)
DTs
51
D2Ts
D > 0.5
DTs D2Ts D3Ts
D < 0.5
0
D1 D3
D2Ts Ts
–
vD3
+
L
C
52
D3
D3 Ts
0
R
t
–
V
+
Conversion ratio M(D)
Fundamentals of Power Electronics
magnetizing current waveforms, for n1 = n2
t
t
Chapter 6: Converter circuits
n vD3 = V = n 3 D Vg 1
Chapter 6: Converter circuits
2Ts
What happens when D > 0.5
1 n 1+ 2 n1
and
53
max vQ1 = 2Vg
+ –
Q2
V = nDVg
D2
Fundamentals of Power Electronics
Vg
D1
Q1
D≤ 1 2
1:n
54
D3
D4
C
R
–
V
Chapter 6: Converter circuits
max vQ1 = max vQ2 = Vg
L +
Chapter 6: Converter circuits
The two-transistor forward converter
Fundamentals of Power Electronics
D≤ 1 2
For n1 = n2
n max vQ1 = Vg 1 + n 1 2
which can be increased by increasing the turns ratio n2 / n1. But this increases the peak transistor voltage:
D≤
Maximum duty cycle limited to
Maximum duty cycle vs. transistor voltage stress
Vg
I
0
D1
Q1
i
nVg
Vg
Vg LM
DTs D1 D2
0.5 i
0
∆i
0
Fundamentals of Power Electronics
conducting devices:
iD1(t)
vs(t)
i(t)
vT(t)
iM(t)
Fundamentals of Power Electronics
V = nDVg
Q2
– vT(t) +
– vT(t) +
Q1
55
0≤D≤1
D2
iD1(t)
D1 i(t)
–
vs(t)
+
L
C
Ts
D2
Q2
0
nVg
–Vg
LM
– Vg
Ts+DTs D1 D2
0.5 i
0
0
t
56
2Ts
–
V
Chapter 6: Converter circuits
R
+
Chapter 6: Converter circuits
• Current programmed control can be used to mitigate transformer saturation problems. Duty cycle control not recommended.
• Effect of nonidealities on transformer volt-second balance?
• As in full bridge, transformer volt-second balance is obtained over two switching periods
• Secondary-side circuit identical to full bridge
• Used with low-voltage inputs
Waveforms: push-pull
1 : n
6.3.3. Push-pull isolated buck converter
+ –
Vg
Vg
+ –
+ –
57
Q1
Q1
L
L
1:1
Fundamentals of Power Electronics
Flyback converter having a 1:n turns ratio and positive output:
Isolate inductor windings: the flyback converter
Vg
Vg
+ –
+ –
Q1
58
LM
Q1
1:n
LM
D1
1:1
C
D1
+
V
–
+
V
–
Chapter 6: Converter circuits
D1
D1
Chapter 6: Converter circuits
–
V
+
+
V
–
Derivation of flyback converter, cont.
Fundamentals of Power Electronics
construct inductor winding using two parallel wires:
buck-boost converter:
6.3.4. Flyback converter
●
●
●
●
–
vL
LM
Q1
1:n
+ –
ig
vL LM
–
+
i
1:n
transformer model
Fundamentals of Power Electronics
Vg
D1 C
iC R –
v
+
●
●
●
●
A two-winding inductor Symbol is same as transformer, but function differs significantly from ideal transformer Energy is stored in magnetizing inductance Magnetizing inductance is relatively small
C
60
iC R
Subinterval 1
59
–
v
+
Chapter 6: Converter circuits
vL = Vg iC = – V R ig = I
CCM: small ripple approximation leads to
vL = Vg iC = – v R ig = i
Chapter 6: Converter circuits
Current does not simultaneously flow in primary and secondary windings Instantaneous winding voltages follow turns ratio Instantaneous (and rms) winding currents do not follow turns ratio Model as (small) magnetizing inductance in parallel with ideal transformer
+ –
+
i
Fundamentals of Power Electronics
Vg
ig
transformer model
The “flyback transformer”
+ – –
vL
+
i – v/n +
Q1
DTs
I
–V/R
Vg
Fundamentals of Power Electronics
conducting devices:
ig
iC
vL
1:n
i/n
C
61
iC R –
v
+
Chapter 6: Converter circuits
vL = – V n I iC = n – V R ig = 0
CCM: small ripple approximation leads to
vL = – nv iC = ni – v R ig = 0
Ts D1
D'Ts
0
I/n – V/R
–V/n
t
62
Chapter 6: Converter circuits
I g = ig = D (I) + D' (0)
Conversion ratio is M(D) = V = n D Vg D' Charge balance: iC = D (– V ) + D' ( nI – V ) = 0 R R Dc component of magnetizing current is I = nV D'R Dc component of source current is
vL = D (Vg) + D' (– V n) = 0
Volt-second balance:
CCM Flyback waveforms and solution
Fundamentals of Power Electronics
Vg
ig =0
transformer model
Subinterval 2
+ –
63
I
DI
D' : n
+ DV g –
I
R
+ –
–
V
+
D'I n
64
R
Chapter 6: Converter circuits
D'V n
Chapter 6: Converter circuits
Widely used in low power and/or high voltage applications Low parts count Multiple outputs are easily obtained, with minimum additional parts Cross regulation is inferior to buck-derived isolated converters Often operated in discontinuous conduction mode DCM analysis: DCM buck-boost with turns ratio
Fundamentals of Power Electronics
●
●
●
●
●
●
+ –
Ig
Discussion: Flyback converter
Fundamentals of Power Electronics
Vg
Ig
Vg
1:D
vL = D (Vg) + D' (– V n) = 0 iC = D (– V ) + D' ( nI – V ) = 0 R R I g = ig = D (I) + D' (0)
Equivalent circuit model: CCM Flyback
–
V
+
+ –
+ vT(t) –
L
Q2
Q1
Q4
Q3
–
vT(t)
+
: n
1 : n
D2
D1
C
io(t)
Fundamentals of Power Electronics
66
–
v
Chapter 6: Converter circuits
R
+
Chapter 6: Converter circuits
• Circuit topologies are equivalent to those of nonisolated boost converter • With 1:1 turns ratio, inductor current i(t) and output current io(t) waveforms are identical to nonisolated boost converter
Vg
i(t)
65
Full-bridge transformer-isolated boost-derived converter
Fundamentals of Power Electronics
Of these, the full-bridge and push-pull boost-derived isolated converters are the most popular, and are briefly discussed here.
• push-pull boost-derived converter
• inverse of forward converter: the “reverse” converter
• full-bridge and half-bridge isolated boost converters
• A wide variety of boost-derived isolated dc-dc converters can be derived, by inversion of source and load of buck-derived isolated converters:
6.3.5. Boost-derived isolated converters
Q1 Q2 Q3 Q4
DTs
0
0
Ts Q1 Q4 D1
D'Ts
I/n
V/n
Q1 Q2 Q3 Q4
Vg
Q1 Q4 D1
Vg –V/n
Fundamentals of Power Electronics
conducting devices:
I
i(t)
vL(t)
Q1 Q2 Q3 Q4
0 DTs
0
67
Ts
Q2 Q3 D2
D'Ts
I/n
– V/n
t
Q1 Q2 Q3 Q4
DTs
Vg
68
Ts
Q2 Q3 D2
D'Ts
Vg –V/n
t
Chapter 6: Converter circuits
—boost with turns ratio n
M(D) = V = n Vg D'
Solve for M(D):
vL = D (Vg) + D' (Vg – V / n) = 0
Application of volt-second balance to inductor voltage waveform:
Chapter 6: Converter circuits
• During next switching period: transistors Q2 and Q3 conduct for time DTs , applying volt-seconds –VDTs to secondary winding.
• During first switching period: transistors Q1 and Q4 conduct for time DTs , applying volt-seconds VDTs to secondary winding.
• As in full-bridge buck topology, transformer voltsecond balance is obtained over two switching periods.
Conversion ratio M(D)
Fundamentals of Power Electronics
conducting devices:
io(t)
vT(t)
Transformer reset mechanism
Vg
Q2
+ vL(t) –
L
D2
D1
69
M(D) = V = n Vg D'
– vT(t) +
– vT(t) +
1 : n
C
io(t)
–
V
Chapter 6: Converter circuits
R
+
Vg
+ – Fundamentals of Power Electronics
Q2
Q1
70
1 : n
D2
D1
C
–
V
Chapter 6: Converter circuits
R
+
Push-pull converter based on Watkins-Johnson converter
Fundamentals of Power Electronics
i(t)
Q1
Push-pull boost-derived converter
+ –
Vg
Vg
+ –
Q1
1:n
ideal transformer model
LM = L2
i2
ip
+ –
+ –
M(D) = V = n D D' Vg
C1
Fundamentals of Power Electronics
Vg
i1
L1
Fundamentals of Power Electronics
Isolated SEPIC
Basic nonisolated SEPIC
Q1
71
C1
C1
ip
is
D1
C2
72
R
I2
i2(t)
I1
i1(t)
is(t)
conducting devices:
–
v
+
ip(t)
L2
Q1
DTs
0
– i2
1:n
Isolated SEPIC
i1
L1
Q1
L1
is
D1
C2
R
R
–
v
+
–
v
D1
D'Ts
t
Chapter 6: Converter circuits
Ts
(i1 + i2) / n
i1
Chapter 6: Converter circuits
D1
C2
+
6.3.6. Isolated versions of the SEPIC and Cuk converter
Vg
Vg
+ –
+ –
Q1
73
1:n
L1 D1
D1
C1
C1
C2
C2
Vg
Vg
Fundamentals of Power Electronics
Split capacitor C1 into series capacitors C1a and C1b
Nonisolated Cuk converter
+ –
+ –
Q1
L1
L1
74
C1a
Q1
C1
C1b
D1
D1
R
–
v
+
R
–
v
+
C2
C2
R
R
–
v
+
–
v
Chapter 6: Converter circuits
L2
L2 +
Chapter 6: Converter circuits
L2
L2
Obtaining isolation in the Cuk converter
Fundamentals of Power Electronics
Isolated inverse SEPIC
Nonisolated inverse SEPIC
Q1
Inverse SEPIC
Vg
+ –
Q1
C1a
1:n
C1b
D1
L2
C2
R
–
v
+
75
Chapter 6: Converter circuits
Fundamentals of Power Electronics
Spreadsheet design
76
Chapter 6: Converter circuits
Comparison via switch stress, switch utilization, and semiconductor cost
• An unbiased quantitative comparison of worst-case transistor currents and voltages, transformer size, etc.
• Rough designs of several converter topologies to meet the given specifications
Trade studies
There is no ultimate converter, perfectly suited for all possible applications
For a given application, which converter topology is best?
6.4. Converter evaluation and design
Fundamentals of Power Electronics
• Transformer functions in conventional manner, with small magnetizing current and negligible energy storage within the magnetizing inductance
• Capacitors C1a and C1b ensure that no dc voltage is applied to transformer primary or secondary windings
Discussion
M(D) = V = n D D' Vg
Insert transformer between capacitors C1a and C1b
L1
Isolated Cuk converter
77
Chapter 6: Converter circuits
k
Σ Vj I j j=1
Fundamentals of Power Electronics
78
Chapter 6: Converter circuits
In a good design, the total active switch stress is minimized.
Ij is the rms current applied to switch j (peak current is also sometimes used).
Vj is the peak voltage applied to switch j,
where
S=
In a converter having k active semiconductor devices, the total active switch stress S is defined as
Total active switch stress S
Fundamentals of Power Electronics
Minimization of total switch stresses leads to reduced loss, and to minimization of the total silicon area required to realize the power devices of the converter.
This suggests evaluating candidate converter approaches by comparing the voltage and current stresses imposed on the active semiconductor devices.
• Conduction and switching losses associated with the active semiconductor devices often dominate the other sources of loss
• Largest single cost in a converter is usually the cost of the active semiconductor devices
6.4.1. Switch stress and switch utilization
79
Fundamentals of Power Electronics
S = VQ1,pk I Q1,rms = (Vg + V / n) (I D)
Switch stress S is
Transistor current coincides with ig(t). RMS value is P I Q1,rms = I D = load Vg D
During subinterval 2, the transistor blocks voltage VQ1,pk equal to Vg plus the reflected load voltage: Vg VQ1,pk = Vg + V / n = D'
+ –
ig
conducting devices: 80
Vg
Q1
LM
Q1
DTs
I
1:n
D1
D'Ts
0
C
–
V
+
Chapter 6: Converter circuits
Ts
D1
t
Chapter 6: Converter circuits
CCM flyback example: Determination of S
Fundamentals of Power Electronics
Active switch utilization is a function of converter operating point.
Converters having low switch utilizations require extra active silicon area, and operate with relatively low efficiency.
Active switch utilization is less than 1 in transformer-isolated converters, and is a quantity to be maximized.
The active switch utilization is the converter output power obtained per unit of active switch stress. It is a converter figure-of-merit, which measures how well a converter utilizes its semiconductor devices.
The active switch utilization U is defined as P U = load S
It is desired to minimize the total active switch stress, while maximizing the output power Pload.
Active switch utilization U
Vg
+ –
Pload = D' D S
81
Ig
I
D' : n
Fundamentals of Power Electronics
large D leads to large transistor voltage
small D leads to large transistor current
Single operating point design: choose D = 1/3.
n = V D' Vg D
For given V, Vg, Pload, the designer can arbitrarily choose D. The turns ratio n must then be chosen according to U
82
0
0.1
0.2
0.3
0.4
0
0.2
0.4
R
–
V
+
0.8
Chapter 6: Converter circuits
D
0.6
max U = 0.385 at D = 1/3
1
Chapter 6: Converter circuits
CCM flyback model
1:D
Flyback example: switch utilization U(D)
Fundamentals of Power Electronics
U=
Hence switch utilization U is
S = VQ1,pk I Q1,rms = (Vg + V / n) (I D)
Previously-derived expression for S:
Express load power Pload in terms of V and I: Pload = D' V nI
CCM flyback example: Determination of U
1 2
D' 2 1+D
84
Chapter 6: Converter circuits
Increasing the range of operating points leads to reduced switch utilization Buck converter can operate with high switch utilization (U approaching 1) when D is close to 1 Boost converter can operate with high switch utilization (U approaching ∞) when D is close to 1 Transformer isolation leads to reduced switch utilization Buck-derived transformer-isolated converters U ≤ 0.353 should be designed to operate with D as large as other considerations allow transformer turns ratio can be chosen to optimize design
Fundamentals of Power Electronics
●
●
●
●
●
0
1
Chapter 6: Converter circuits
Switch utilization : Discussion
Fundamentals of Power Electronics
Other isolated buck-derived converters (fullbridge, half-bridge, push-pull) Isolated boost-derived converters (full bridge, push-pull)
83
1 = 0.353 2 2 1 = 0.353 2 2
1 D 2
D 2 2
1 3
2 = 0.385 3 3
D' D
1 2
0
∞
D' D
Boost Buck-boost, flyback, nonisolated SEPIC, isolated SEPIC, nonisolated Cuk, isolated Cuk Forward, n1 = n2
1
1
D
Buck
max U(D) occurs at D =
max U(D)
U(D)
Converter
Table 6.1. Active switch utilizations of some common dc-dc converters, single operating point.
Comparison of switch utilizations of some common converters
Chapter 6: Converter circuits
voltage derating factor
current derating factor
converter switch utilization
Fundamentals of Power Electronics
86
Chapter 6: Converter circuits
Typical cost of active semiconductor devices in an isolated dc-dc converter: $1 - $10 per kW of output power.
(voltage derating factor) and (current derating factor) are required to obtain reliable operation. Typical derating factors are 0.5 - 0.75
(semiconductor device cost per rated kVA) = cost of device, divided by product of rated blocking voltage and rms current, in $/kVA. Typical values are less than $1/kVA
semiconductor cost = per kW output power
semiconductor device cost per rated kVA
Active semiconductor cost vs. switch utilization
85
Nonisolated and isolated versions of buck-boost, SEPIC, and Cuk converters U ≤ 0.385 Single-operating-point optimum occurs at D = 1/3 Nonisolated converters have lower switch utilizations than buck or boost Isolation can be obtained without penalizing switch utilization
Fundamentals of Power Electronics
●
Switch utilization: Discussion
Chapter 6: Converter circuits
260V 15V 200W 20W 100kHz 0.1V
minimum input voltage V g output voltage V maximum load power Pload minimum load power Pload switching frequency fs maximum output ripple ∆v
• Output voltage ripple ≤ 0.1V
• Select switching frequency of 100kHz
• Must operate at 10% load
• Rated load power 200W
• Regulated output of 15V
• Input voltage: rectified 230Vrms ±20%
Fundamentals of Power Electronics
88
Specifications are entered at top of spreadsheet
Chapter 6: Converter circuits
Compare single-transistor forward and flyback converters in this application
390V
maximum input voltage V g
S peci fi cat i o ns
87
Spreadsheet design example
Fundamentals of Power Electronics
• Evaluation of design tradeoffs
• Evaluation of worst-case stresses over a range of operating points
A computer spreadsheet is a very useful tool for this job. The results of the steady-state converter analyses of chapters 1-6 can be entered, and detailed design investigations can be quickly performed:
• Optimize the design of a given converter
• Compare converter topologies and select the best for the given specifications
Given ranges of Vg and Pload , as well as desired value of V and other quantities such as switching frequency, ripple, etc., there are two basic engineering design tasks:
6.4.2. Converter design using computer spreadsheet
+ –
Fundamentals of Power Electronics
90
3A ref to sec
Q1
LM
inductor current ripple ∆i
+ –
0.125
Vg
1:n
turns ratio n2 / n1
Design variables
D3
C
R
D1 C
–
V
+
Chapter 6: Converter circuits
• Design for CCM at full load; may operate in DCM at light load
L
Chapter 6: Converter circuits
• Design for CCM at full load; may operate in DCM at light load
–
V
+
Flyback converter design, CCM
89
2A ref to sec
inductor current ripple ∆i
Fundamentals of Power Electronics
0.125
turns ratio n3 / n1
D1
1
Q1
D2
reset winding turns ratio n2 / n1
Design variables
Vg
n1 : n2 : n3
Forward converter design, CCM
D' V Ts 2 ∆i
D' V Ts 2L
91
Chapter 6: Converter circuits
2 K 2n 3Vg –1 n 1V
2
–1
in DCM
D=
n1 V n 3 Vg
in CCM
Fundamentals of Power Electronics
92
Chapter 6: Converter circuits
at a given operating point, the actual duty cycle is the small of the values calculated by the CCM and DCM equations above. Minimum D occurs at minimum Pload and maximum Vg.
D=
Solve for D:
These equations apply equally well to the forward converter, provided that all quantities are referred to the transformer secondary side.
with K = 2 L / R Ts, and R = V 2 / Pload
Check for DCM at light load. The solution of the buck converter operating in DCM is n 2 V = 3 Vg n1 1 + 4K / D 2
Forward converter example, continued
Fundamentals of Power Electronics
∆i is a design variable. For a given ∆i, the equation above can be used to determine L. To ensure CCM operation at full load, ∆i should be less than the full-load output current. C can be found in a similar manner.
L=
Solve for L:
∆i =
Inductor current ripple is
Maximum duty cycle occurs at minimum Vg and maximum Pload. Converter then operates in CCM, with n D= 1 V n 3 Vg
Enter results of converter analysis into spreadsheet (Forward converter example)
n3 D n1 I 2 + (∆i) 2 / 3 ≈
n3 DI n1
93
Chapter 6: Converter circuits
2A ref to sec
inductor current ripple ∆i
0.308 0.251
minimum D, at full load minimum D, at minimum load
1.13A 0.226 49V 9.1A 49V 11.1A 1.15A
rms transistor current iQ1 transistor utilization U peak diode voltage v D1 rms diode current iD1 peak diode voltage v D2 rms diode current iD2 rms output capacitor current iC
Fundamentals of Power Electronics
780V
peak transistor voltage v Q1
Worst-case stresses
0.462
maximum duty cycle D
Results
0.125
turns ratio n3 / n1
94
rms output capacitor current iC
peak diode current iD1
rms diode current iD1
peak diode voltage v D1
transistor utilization U
rms transistor current iQ1
peak transistor voltage v Q1
Worst-case stresses
9.1A
22.2A
16.3A
64V
0.284
1.38A
510V
0.179
0.235
Chapter 6: Converter circuits
minimum D, at minimum load
minimum D, at full load
maximum duty cycle D
0.316
3A ref to sec
inductor current ripple ∆i
Results
0.125
turns ratio n2 / n1
Design variables 1
Design variables reset winding turns ratio n2 / n1
Fl y back conv ert er desi g n, C C M
Forw ard co nv ert er des i g n, C C M
Results: forward and flyback converter spreadsheets
Fundamentals of Power Electronics
Other component stresses can be found in a similar manner. Magnetics design is left for a later chapter.
(this neglects transformer magnetizing current)
I Q1, rms =
Rms transistor current is
n max vQ1 = Vg 1 + n 1 2
Peak transistor voltage is
Worst-case component stresses can now be evaluated.
More regarding forward converter example
Chapter 6: Converter circuits
Fundamentals of Power Electronics
96
Chapter 6: Converter circuits
CCM flyback exhibits higher peak and rms currents. Currents in DCM flyback are even higher
transistor utilization 0.284
1.38A worst case, 22% higher than forward
Flyback
transistor utilization 0.226
1.13A worst-case
Forward
95
Discussion: rms transistor current
Fundamentals of Power Electronics
A conclusion: reset mechanism of flyback is superior to forward
Fix: use two-transistor forward converter, or change reset winding turns ratio
MOSFETs having voltage rating greater than 1000V are not available (in 1995) —when ringing due to transformer leakage inductance is accounted for, this design will have an inadequate design margin
Ideal peak transistor voltage: 780V, 53% greater than flyback
Forward converter
An 800V or 1000V MOSFET would have an adequate design margin
Actual peak voltage will be higher, due to ringing causes by transformer leakage inductance
Ideal peak transistor voltage: 510V
Flyback converter
Discussion: transistor voltage
97
Chapter 6: Converter circuits
Fundamentals of Power Electronics
98
Chapter 6: Converter circuits
1. The boost converter can be viewed as an inverse buck converter, while the buck-boost and Cuk converters arise from cascade connections of buck and boost converters. The properties of these converters are consistent with their origins. Ac outputs can be obtained by differential connection of the load. An infinite number of converters are possible, and several are listed in this chapter. 2. For understanding the operation of most converters containing transformers, the transformer can be modeled as a magnetizing inductance in parallel with an ideal transformer. The magnetizing inductance must obey all of the usual rules for inductors, including the principle of volt-second balance.
Summary of key points
Fundamentals of Power Electronics
Secondary-side currents, especially capacitor currents, limit the practical application of the flyback converter to situations where the load current is not too great.
rms capacitor current 9.1A
peak diode current 22.2A
rms diode current 16.3A
peak diode voltage 64V
Flyback
rms capacitor current 1.15A
rms diode current 9.1A / 11.1A
peak diode voltage 49V
Forward
Discussion: secondary-side diode and capacitor stresses
99
Chapter 6: Converter circuits
Fundamentals of Power Electronics
100
Chapter 6: Converter circuits
5. In the conventional forward converter, the transformer is reset while the transistor is off. The transformer magnetizing inductance operates in the discontinuous conduction mode, and the maximum duty cycle is limited. 6. The flyback converter is based on the buck-boost converter. The flyback transformer is actually a two-winding inductor, which stores and transfers energy. 7. The transformer turns ratio is an extra degree-of-freedom which the designer can choose to optimize the converter design. Use of a computer spreadsheet is an effective way to determine how the choice of turns ratio affects the component voltage and current stresses. 8. Total active switch stress, and active switch utilization, are two simplified figures-of-merit which can be used to compare the various converter circuits.
Summary of key points
Fundamentals of Power Electronics
3. The steady-state behavior of transformer-isolated converters may be understood by first replacing the transformer with the magnetizing-inductance-plus-ideal-transformer equivalent circuit. The techniques developed in the previous chapters can then be applied, including use of inductor volt-second balance and capacitor charge balance to find dc currents and voltages, use of equivalent circuits to model losses and efficiency, and analysis of the discontinuous conduction mode. 4. In the full-bridge, half-bridge, and push-pull isolated versions of the buck and/or boost converters, the transformer frequency is twice the output ripple frequency. The transformer is reset while it transfers energy: the applied voltage polarity alternates on successive switching periods.
Summary of key points
1
Chapter 7: AC equivalent circuit modeling
Fundamentals of Power Electronics
2
7.8. Summary of key points Chapter 7: AC equivalent circuit modeling
7.7. Modeling the pulse-width modulator
7.6. The canonical circuit model
7.5. Circuit averaging and averaged switch modeling
7.4. State-space averaging
7.3. Example: A nonideal flyback converter
7.2. The basic ac modeling approach
7.1. Introduction
Chapter 7. AC Equivalent Circuit Modeling
Fundamentals of Power Electronics
AC equivalent circuit modeling Converter transfer functions Controller design Ac and dc equivalent circuit modeling of the discontinuous conduction mode 11. Current programmed control
7. 8. 9. 10.
Part II Converter Dynamics and Control
vg(t) + –
Power input
δ(t)
3
dTs Ts
δ(t)
transistor gate driver
t
vc(t)
R
v
feedback connection
Chapter 7: AC equivalent circuit modeling
Controller
t
–
v(t)
+
Load
voltage reference vref
compensator pulse-width vc Gc(s) modulator
Switching converter
A simple dc-dc regulator system, employing a buck converter
Introduction
Fundamentals of Power Electronics
4
Chapter 7: AC equivalent circuit modeling
Control the duty cycle d(t) such that ig (t) accurately follows a reference signal iref (t), and v(t) accurately follows a reference signal vref.
Regulate the ac input current waveform.
Regulate the dc output voltage.
Ac-dc rectifiers
Control the duty cycle d(t) such that v(t) accurately follows a reference signal vref (t).
Regulate an ac output voltage.
Dc-ac inverters
Control the duty cycle d(t) such that v(t) accurately follows a reference signal vref.
Regulate dc output voltage.
Dc-dc converters
Applications of control in power electronics
Fundamentals of Power Electronics
• in R
• in Vg
• in element values
There are uncertainties:
• in R
• in vg(t)
There are disturbances:
Objective: maintain v(t) equal to an accurate, constant value V.
7.1.
+ –
Modeling
5
Chapter 7: AC equivalent circuit modeling
Fundamentals of Power Electronics
6
Chapter 7: AC equivalent circuit modeling
• After basic insight has been gained, model can be refined (if it is judged worthwhile to expend the engineering effort to do so), to account for some of the previously neglected phenomena
• Approximations neglect small but complicating phenomena
• Simplified model yields physical insight, allowing engineer to design system to operate in specified manner
• Model dominant behavior of system, ignore other insignificant phenomena
• Representation of physical behavior by mathematical means
Fundamentals of Power Electronics
• Current-programmed control of converters (Chapter 11)
• Model converters operating in DCM (Chapter 10)
• Design converter control systems (Chapter 9)
• Construct converter small-signal transfer functions (Chapter 8)
• Extend the steady-state converter models of Chapters 2 and 3, to include CCM converter dynamics (Chapter 7)
What are the small-signal transfer functions of the converter?
How do ac variations in vg(t), R, or d(t) affect the output voltage v(t)?
Need dynamic models of converters:
Develop tools for modeling, analysis, and design of converter control systems
Objective of Part II
7
{
ωm
Fundamentals of Power Electronics
8
switching harmonics
ω
Chapter 7: AC equivalent circuit modeling
If ripple is neglected, then only lowfrequency components (modulation frequency and harmonics) remain.
With small switching ripple, highfrequency components (switching harmonics and sidebands) are small.
ωs
switching frequency and sidebands
with sinusoidal modulation of duty cycle modulation frequency and its harmonics
t
t
Chapter 7: AC equivalent circuit modeling
averaged waveform Ts with ripple neglected
actual waveform v(t) including ripple
{ Contains frequency components at: • Modulation frequency and its harmonics • Switching frequency and its harmonics • Sidebands of switching frequency
spectrum of v(t)
gate drive
The resulting variations in transistor gate drive signal and converter output voltage:
Output voltage spectrum
Fundamentals of Power Electronics
where D and Dm are constants, | Dm | << D , and the modulation frequency ωm is much smaller than the converter switching frequency ωs = 2πfs.
d(t) = D + Dm cos ωmt
Suppose the duty cycle is modulated sinusoidally:
Neglecting the switching ripple
{
L
= 1 Ts Ts
Ts
Ts
t
t + Ts
Ts
Ts
x(τ) dτ
= iC(t)
= vL(t)
Fundamentals of Power Electronics
xL(t)
where
dt
dt d vC(t)
d iL(t)
Average over one switching period to remove switching ripple:
C
9
Chapter 7: AC equivalent circuit modeling
10
Ts
Ts
=0
=0
Chapter 7: AC equivalent circuit modeling
by inductor volt-second balance and capacitor charge balance.
iC(t)
vL(t)
Note that, in steady-state,
Averaging to remove switching ripple
Fundamentals of Power Electronics
• Remove switching harmonics by averaging all waveforms over one switching period
Approach:
• Ignore complicated switching harmonics and sidebands
• Ignore the switching ripple
• Predict how low-frequency variations in duty cycle induce lowfrequency variations in the converter voltages and currents
Objective of ac converter modeling
dt
dt d vC(t)
d iL(t) Ts
Ts
= iC(t)
= vL(t) Ts
Ts
βRiB
Fundamentals of Power Electronics
B
βFiB
E
C
Nonlinear Ebers-Moll model
iB
11
Chapter 7: AC equivalent circuit modeling
12
B rE
E
Chapter 7: AC equivalent circuit modeling
iB
βFiB
C
Linearized small-signal model, active region
Small-signal modeling of the BJT
Fundamentals of Power Electronics
Hence, must linearize by constructing a small-signal converter model.
constitute a system of nonlinear differential equations.
C
L
The averaged voltages and currents are, in general, nonlinear functions of the converter duty cycle, voltages, and currents. Hence, the averaged equations
Nonlinear averaged equations
0
13
linearized function
quiescent operating point
1
D = 0.5
Example: linearization at the quiescent operating point
+ –
I d(t)
1:D
L
14
Vg – V d(t)
Small-signal ac equivalent circuit model
I d(t)
C
–
v(t)
+
Chapter 7: AC equivalent circuit modeling
buck-boost example
D' : 1
R
Chapter 7: AC equivalent circuit modeling
D
Result of averaged small-signal ac modeling
Fundamentals of Power Electronics
vg(t)
0.5
actual nonlinear characteristic
Fundamentals of Power Electronics
V
–Vg
0
Buck-boost converter: nonlinear static control-to-output characteristic
+ –
+ – C
dv(t) v(t) =– R dt
vg(t)
+ –
–
v(t)
L
i(t) C
R
Fundamentals of Power Electronics
v(t) dv(t) ≈– R dt iC(t) = C
Ts
di(t) ≈ vg(t) dt vL(t) = L
Ts
16
Chapter 7: AC equivalent circuit modeling
–
v(t)
+
Chapter 7: AC equivalent circuit modeling
R
+
Small ripple approximation: replace waveforms with their low-frequency averaged values:
iC(t) = C
15
i(t)
2
Switch in position 1
L
Inductor voltage and capacitor current are: di(t) vL(t) = L = vg(t) dt
Fundamentals of Power Electronics
vg(t)
1
Buck-boost converter example
7.2. The basic ac modeling approach
vg(t)
+ – L
i(t) C
R
Ts
–
v(t) R
17
Ts
t
t + Ts
x(τ)dτ
t
t + Ts
dt
Ts
= d(t) vg(t) Fundamentals of Power Electronics
L
d i(t)
0
vL(t)
Ts
18
+ d'(t) v(t)
vL(τ)dτ ≈ d(t) vg(t) Insert into Eq. (7.2):
vL(t) T = 1 Ts s
Average the inductor voltage in this manner:
xL(t) T = 1 Ts s
Low-frequency average is found by evaluation of
Inductor voltage waveform
Ts
Ts
dTs Ts
Ts
+ d' v(t)
v(t)
Ts
Ts
t
Chapter 7: AC equivalent circuit modeling
This equation describes how the low-frequency components of the inductor waveforms evolve in time.
Ts
= d vg(t)
+ d'(t) v(t)
Ts
Ts
vL(t)
vg(t)
–
v(t)
+
Chapter 7: AC equivalent circuit modeling
7.2.1 Averaging the inductor waveforms
Fundamentals of Power Electronics
dv(t) ≈ – i(t) dt
i C(t) = C
Ts
di(t) ≈ v(t) dt
vL(t) = L
Small ripple approximation: replace waveforms with their low-frequency averaged values:
v(t) dv(t) i C(t) = C = – i(t) – R dt
di(t) vL(t) = L = v(t) dt
Inductor voltage and capacitor current are:
Switch in position 2
19
vL(t)
vg(t) Ts
Ts
Ts
dTs
L
vg
dTs Ts
L
v
Ts
Ts
Ts
+ d' v(t)
v(t)
Ts
i(dTs)
= d vg(t)
L
di(t) = vL(t) dt
di = 1 L
t
t + Ts
vL(τ)dτ
Ts
Fundamentals of Power Electronics
20
Chapter 7: AC equivalent circuit modeling
So the net change in inductor current over one switching period is exactly equal to the period Ts multiplied by the average slope 〈 vL 〉Ts /L.
i(t + Ts) – i(t) = 1 Ts vL(t) L
Left-hand side is the change in inductor current. Right-hand side can be related to average inductor voltage by multiplying and dividing by Ts as follows:
t
t + Ts
t
i(Ts)
t
Chapter 7: AC equivalent circuit modeling
Divide by L and integrate over one switching period:
Inductor equation:
Ts
Inductor voltage and current waveforms
0
i(0)
i(t)
0
vL(t)
Net change in inductor current is correctly predicted by the average inductor voltage
Fundamentals of Power Electronics
During transients or ac variations, the average inductor voltage is not zero in general, and this leads to net variations in inductor current.
In steady-state, the average inductor voltage is zero (volt-second balance), and hence the inductor current waveform is periodic: i(t + Ts) = i(t). There is no net change in inductor current over one switching period.
Use of the average inductor voltage allows us to determine the net change in inductor current over one switching period, while neglecting the switching ripple.
7.2.2 Discussion of the averaging approximation
d vg(t)
vg(t) L
L
+ d' v(t)
dTs
Ts
v(t) L
Ts
Ts
i(t)
t
i(Ts)
Ts
Averaged waveform
Ts
L
i(t + Ts) – i(t) = vL(t) Ts
Ts
i(t + Ts) – i(t) = 1 Ts vL(t) L
dt
d i(t)
21
Ts
Chapter 7: AC equivalent circuit modeling
L
Ts
dt
d i(t)
Fundamentals of Power Electronics
Hence,
dt
d i(t)
Ts
=
= vL(t)
22
Ts
i(t + Ts) – i(t) Ts
Chapter 7: AC equivalent circuit modeling
Define the derivative of 〈 i 〉Ts as (Euler formula):
Rearrange:
We have
Fundamentals of Power Electronics
The net change in inductor current over one switching period is exactly equal to the period Ts multiplied by the average slope 〈 vL 〉Ts /L.
0
i(0)
i(t)
Actual waveform, including ripple
Average inductor voltage correctly predicts average slope of iL(t)
i(dTs)
=
i(0)
0
i(0)
+
Ts
dTs
L
vg Ts
dTs
L
v
i(dTs)
Ts
t
L
vg(t)
i(Ts)
Ts
=
i(dTs)
+
d'Ts
v(t) L
Ts
Fundamentals of Power Electronics
The final value i(Ts) is equal to the initial value i(0), plus the switching period Ts multiplied by the average slope 〈 vL 〉Ts /L.
The intermediate step of computing i(dTs) is eliminated.
Eliminate i(dTs), to express i(Ts) directly as a function of i(0):
24
0
i(0)
i(t)
L
+ d' v(t)
dTs
Ts
Ts
Ts
t
i(Ts)
Ts
Chapter 7: AC equivalent circuit modeling
d vg(t)
vg(t) L
i(t)
Averaged waveform
Ts
+ d'(t) v(t)
vL(t)
Ts
v(t) L
Ts d(t) vg(t) L
Actual waveform, including ripple
i(Ts) = i(0) +
Ts
Chapter 7: AC equivalent circuit modeling
Net change in inductor current over one switching period
23
(final value) = (initial value) + (length of interval) (average slope)
i(Ts)
(final value) = (initial value) + (length of interval) (average slope)
Fundamentals of Power Electronics
With switch in position 2:
With switch in position 1:
Let’s compute the actual inductor current waveform, using the linear ripple approximation.
i(t)
Computing how the inductor current changes over one switching period
Ts
= d(t) –
v(t) R Ts
+ d'(t) – i(t) Ts
–
v(t) R
dt
d v(t) Ts
= – d'(t) i(t) Ts
–
v(t) R Ts
25
Ts
Ts
during subinterval 2
during subinterval 1
= d(t) i(t)
Ts
Fundamentals of Power Electronics
i g(t)
Average value:
0
i(t)
Buck-boost input current waveform is
26
We found in Chapter 3 that it was sometimes necessary to write an equation for the average converter input current, to derive a complete dc equivalent circuit model. It is likewise necessary to do this for the ac model.
i g(t) =
Ts
v(0)
RC
v(t)
Ts
R
v(t) Ts
–
0
ig(t)
RC
v(t)
Ts
–
Ts
Ts
C
i(t)
t
t
v(Ts)
Ts
Ts
Ts
v(t)
Ts
– i(t)
Ts
i(t)
Ts
dTs
0
i g(t)
Ts
Ts
t
Chapter 7: AC equivalent circuit modeling
Converter input current waveform
0
R
v(t)
v(dTs)
dTs
–
dTs
i C(t)
Capacitor voltage and current waveforms
–
–
Chapter 7: AC equivalent circuit modeling
v(t) 0
0
iC(t)
7.2.4 The average input current
Fundamentals of Power Electronics
C
Collect terms, and equate to C d〈 v 〉Ts /dt:
i C(t)
Average capacitor current:
7.2.3 Averaging the capacitor waveforms
ig(t)
dt
d i(t) dt d v(t)
Ts
Ts
Ts
= d(t) i(t)
–
v(t) R Ts
+ d'(t) v(t)
Ts
Ts
Ts
= – d'(t) i(t)
= d(t) vg(t) Ts
27
Chapter 7: AC equivalent circuit modeling
, Ts
v(t)
, Ts
ig(t)
Ts
Fundamentals of Power Electronics
28
Chapter 7: AC equivalent circuit modeling
reach the quiescent values I, V, and Ig, given by the steady-state analysis as V = – D Vg D' I=– V D' R Ig = D I
i(t)
then, from the analysis of Chapter 2, after transients have subsided the inductor current, capacitor voltage, and input current
s
d(t) = D vg(t) T = Vg
If the converter is driven with some steady-state, or quiescent, inputs
Construct small-signal model: Linearize about quiescent operating point
Fundamentals of Power Electronics
—nonlinear because of multiplication of the time-varying quantity d(t) with other time-varying quantities such as i(t) and v(t).
C
L
Converter averaged equations:
7.2.5. Perturbation and linearization
Ts
= Vg + vg(t)
Ts
Ts
Ts
= I g + i g(t)
= V + v(t)
= I + i(t)
29
Chapter 7: AC equivalent circuit modeling
Fundamentals of Power Electronics
30
Chapter 7: AC equivalent circuit modeling
then the nonlinear converter equations can be linearized.
i g(t) << I g
v(t) << V
i(t) << I
d(t) << D
vg(t) << Vg
If the ac variations are much smaller in magnitude than the respective quiescent values,
The small-signal assumption
Fundamentals of Power Electronics
ig(t)
v(t)
i(t)
In response, and after any transients have subsided, the converter dependent voltages and currents will be equal to the corresponding quiescent values, plus small ac variations:
d(t) = D + d(t)
vg(t)
So let us assume that the input voltage and duty cycle are equal to some given (dc) quiescent values, plus superimposed small ac variations:
Perturbation
dt
0
+ Dc terms
Chapter 7: AC equivalent circuit modeling
2 nd order ac terms (nonlinear)
1 st order ac terms (linear)
2 nd order ac terms (nonlinear)
Fundamentals of Power Electronics
32
Chapter 7: AC equivalent circuit modeling
• Second-order ac terms, containing products of ac quantities. These are nonlinear, because they involve multiplication of ac quantities
• First-order ac terms, containing a single ac quantity, usually multiplied by a constant coefficient such as a dc term. These are linear functions of the ac variations
• Dc terms, containing only dc quantities
The right-hand side contains three types of terms:
Since I is a constant (dc) term, its derivative is zero
Dc terms
0 dI + d i(t) = DV + D'V + Dv (t) + D'v(t) + V – V d(t) + d(t) v (t) – v(t) L ➚ g g g g dt dt
The perturbed inductor equation
31
1 st order ac terms (linear)
d i(t) = DVg+ D'V + Dvg(t) + D'v(t) + Vg – V d(t) + d(t) vg(t) – v(t) dt
Fundamentals of Power Electronics
L
➚ dI
Multiply out and collect terms:
d'(t) = 1 – d(t) = 1 – D + d(t) = D' – d(t)
with D’ = 1 – D
d I + i(t) = D + d(t) Vg + vg(t) + D' – d(t) V + v(t) dt
note that d’(t) is given by
L
Insert the perturbed expressions into the inductor differential equation:
Perturbation of inductor equation
i g(t) << I g
v(t) << V
i(t) << I
d(t) << D
vg(t) << Vg
2 nd order ac terms (nonlinear)
when
d(t) << D
33
Chapter 7: AC equivalent circuit modeling
So neglect second-order terms. Also, dc terms on each side of equation are equal.
d(t) vg(t) << D vg(t)
then the second-order ac terms are much smaller than the first-order terms. For example,
1 st order ac terms (linear)
d i(t) = Dvg(t) + D'v(t) + Vg – V d(t) dt
Fundamentals of Power Electronics
34
Chapter 7: AC equivalent circuit modeling
Note that the quiescent values D, D’, V, Vg, are treated as given constants in the equation.
This is the desired result: a linearized equation which describes smallsignal ac variations.
L
Upon discarding second-order terms, and removing dc terms (which add to zero), we are left with
Linearized inductor equation
Fundamentals of Power Electronics
Provided
Dc terms
0 dI + d i(t) = DV + D'V + Dv (t) + D'v(t) + V – V d(t) + d(t) v (t) – v(t) L ➚ g g g g dt dt
Neglect of second-order terms
d V + v(t) V + v(t) = – D' – d(t) I + i(t) – R dt
d(t)i(t)
dv(t) v(t) = – D'i(t) – + Id(t) R dt
+
1 st order ac term
i g(t)
=
Dc term
DI
+
1 st order ac terms (linear)
Di(t) + Id(t)
I g + i g(t) = D + d(t) I + i(t)
+
2 nd order ac term (nonlinear)
d(t)i(t)
Chapter 7: AC equivalent circuit modeling
Fundamentals of Power Electronics
36
Chapter 7: AC equivalent circuit modeling
This is the linearized small-signal equation which described the converter input port.
i g(t) = Di(t) + Id(t)
Neglect second-order terms. Dc terms on both sides of equation are equal. The following first-order terms remain:
Dc term
Ig
Collect terms:
35
Average input current Perturbation leads to
Fundamentals of Power Electronics
This is the desired small-signal linearized capacitor equation.
C
Dc terms
1 st order ac terms 2 nd order ac term (linear) (nonlinear) Neglect second-order terms. Dc terms on both sides of equation are equal. The following terms remain:
0 ➚ + dv(t) = – D'I – V + – D'i(t) – v(t) + Id(t) + C dV R R dt dt
Collect terms:
C
Perturbation leads to
Capacitor equation
i g(t) = Di(t) + Id(t)
D vg(t)
+ –
38
Chapter 7: AC equivalent circuit modeling
D' v(t)
Chapter 7: AC equivalent circuit modeling
Vg – V d(t)
i(t)
+ d i(t) – L dt
L
d i(t) = Dvg(t) + D'v(t) + Vg – V d(t) dt
Fundamentals of Power Electronics
L
37
Inductor loop equation
Fundamentals of Power Electronics
Reconstruct equivalent circuit corresponding to these equations, in manner similar to the process used in Chapter 3.
L
d i(t) = Dvg(t) + D'v(t) + Vg – V d(t) dt dv(t) v(t) C = – D'i(t) – + Id(t) R dt
The linearized small-signal converter equations:
7.2.6. Construction of small-signal equivalent circuit model
+ –
+ –
39
C
R
v(t)
C
Chapter 7: AC equivalent circuit modeling
–
v(t) R +
dv(t) dt
Fundamentals of Power Electronics
vg(t)
+ –
i g(t) = Di(t) + Id(t)
40
I d(t)
i g(t)
Chapter 7: AC equivalent circuit modeling
D i(t)
Input port node equation
Fundamentals of Power Electronics
I d(t)
dv(t) v(t) = – D'i(t) – + Id(t) R dt
D' i(t)
C
Capacitor node equation
I d(t)
D i(t)
+ – D vg(t)
Vg – V d(t)
D' v(t) D' i(t)
I d(t)
L
Vg – V d(t)
D' : 1
I d(t)
I d(t)
C
i(t)
+ –
L
I d(t)
Fundamentals of Power Electronics
+ –
vg(t)
V d(t)
1:D
Vg d(t)
42
D' : 1
+ –
vg(t)
boost
buck
I d(t)
i(t)
L
–
v(t)
+
C
–
v(t)
+
R
R
C
R
R
–
v(t)
–
v(t)
Chapter 7: AC equivalent circuit modeling
C
+
+
Chapter 7: AC equivalent circuit modeling
7.2.7. Results for several basic converters
41
Small-signal ac equivalent circuit model of the buck-boost converter
+ –
1:D
Replace dependent sources with ideal dc transformers:
+ –
L
+ –
Fundamentals of Power Electronics
vg(t)
vg(t)
i(t)
Collect the three circuits:
Complete equivalent circuit
+ –
+ –
+ –
+ – I d(t)
1:D i(t)
L
43
Vg – V d(t)
I d(t)
C
L
Q1
Fundamentals of Power Electronics
vg(t) + –
ig(t)
1:n
D1
Flyback converter example
C
44
R
–
v(t)
+
–
v(t)
+
R
Chapter 7: AC equivalent circuit modeling
D' : 1
Chapter 7: AC equivalent circuit modeling
• Flyback transformer has magnetizing inductance L, referred to primary
• MOSFET has onresistance Ron
7.3. Example: a nonideal flyback converter
Fundamentals of Power Electronics
vg(t)
buck-boost
Results for several basic converters
+ –
–
vL(t)
L
Q1
ideal
1:n
D1
C
iC(t)
s
Fundamentals of Power Electronics
iC(t) = –
– i(t) Ts
Ts
v(t) R ig(t) = i(t) T
vL(t) = vg(t)
Ts
–
v(t)
+
45
+ –
Ron
vg
46
+ –
Ron
L
i
vg
+ –
ig =0
ig
–
vL
+
i – v/n +
+ vL
i L
C
iC R
R
1:n
C
iC R
–
v
+
–
v
+
–
v
+
Chapter 7: AC equivalent circuit modeling
MOSFET conducts, diode is reverse-biased
Ron
1:n
i/n
C
iC
Chapter 7: AC equivalent circuit modeling
–
vL
+
transformer model
–
1:n
transformer model
Subinterval 2
vg
ig
transformer model
Subinterval 1
Subinterval 1
R
Small ripple approximation:
vL(t) = vg(t) – i(t) Ron v(t) iC(t) = – R ig(t) = i(t)
Circuit equations:
Fundamentals of Power Electronics
vg(t) + –
+
i(t)
ig(t)
Flyback converter, with transformer equivalent circuit
Circuits during subintervals 1 and 2
vg – iRon
dTs – v/n
vL(t)
Ts
= d(t)
d i(t) dt
Ts
Ts
vg(t)
Ts
Ts
= d(t) vg(t) Fundamentals of Power Electronics
L
Hence, we can write:
vL(t)
Ts
vg
47
+ –
ig =0
–
vL
+
i
Ts
Ts
48
0
vg(t)
Ts
C
iC R
–
v
+
–
nL
v(t)
Ts
Ts
i(t)
Ts
t
Chapter 7: AC equivalent circuit modeling
Ts
Ts
v(t) n
– v(t) n
dTs
L
– Ron i(t) Ts
Ts
Chapter 7: AC equivalent circuit modeling
Ron – d'(t)
Ron + d'(t)
– d(t) i(t)
– i(t)
t
i(t)
i/n
MOSFET is off, diode conducts
– v/n +
1:n
transformer model
Inductor waveforms
Average inductor voltage:
0
vL(t)
Fundamentals of Power Electronics
vL(t) = –
v(t) T s n v(t) i(t) T iC(t) = – n s – R ig(t) = 0
Small ripple approximation:
Circuit equations: v(t) vL(t) = – n i(t) v(t) iC(t) = – n – R ig(t) = 0
Subinterval 2
Ts
dTs
Ts
Ts
= d(t)
– v(t) R
Hence, we can write: d v(t) T i(t) s C = d'(t) n dt
iC(t)
Average capacitor current:
– v/R
i C(t)
i v n–R
0
ig(t)
0
= d(t) i(t)
Fundamentals of Power Electronics
Ts
Average input current: ig(t)
Ts
Ts
–
49
v(t) R
+ d'(t)
t
Ts
0 i(t) n
v(t)
Ts
–
–
Ts
Ts
i(t)
Ts
50
dTs
0
i g(t)
Ts
– RC
v(t) Ts
Ts
v(t)
Ts
t
Chapter 7: AC equivalent circuit modeling
Ts
t
Chapter 7: AC equivalent circuit modeling
Ts
dTs
Ts
v(t) R
RC
v(t)
nC
i(t)
Input current waveform
Fundamentals of Power Electronics
0
iC(t)
Capacitor waveforms
ig(t)
dt
d i(t) dt d v(t) i(t) n
= d(t) i(t)
= d'(t)
= d(t) vg(t)
Ts
Ts
Ts
–
v(t) R Ts
– d(t) i(t)
+
Ts
Chapter 7: AC equivalent circuit modeling
= I g + i g(t)
= V + v(t)
v(t) n
Ts
Ron – d'(t)
Ts
V + v(t) – D + d(t) I + i(t) Ron n
v(t) n
Dc terms +
52
Chapter 7: AC equivalent circuit modeling
2 nd order ac terms (nonlinear)
v(t) d(t)vg(t) + d(t) n – d(t)i(t)Ron
1 st order ac terms (linear)
d i(t) v(t) V = DVg– D' V n – DRonI + Dvg(t) – D' n + Vg + n – IRon d(t) – DRon i(t) dt
Fundamentals of Power Electronics
0
Ts
– d(t) i(t)
d I + i(t) = D + d(t) Vg + vg(t) – D' – d(t) dt
= d(t) vg(t)
L
Ts
d i(t) dt
dt
Ts
Ts
Ts
= I + i(t)
Ron – d'(t)
Perturbation of the averaged inductor equation
51
v(t)
d(t) = D + d(t)
ig(t)
i(t)
Ts
= Vg + vg(t)
vg(t)
L
➚ dI
Ts
Next step: perturbation and linearization. Let
Fundamentals of Power Electronics
L
Ts
Ts
Ts
— a system of nonlinear differential equations
C
L
The averaged converter equations
53
dt
d v(t)
Ts
= d'(t)
i(t) n
Ts
–
v(t) R
Ts
d V + v(t) = D' – d(t) dt
V + v(t) I + i(t) – n R
Fundamentals of Power Electronics
Dc terms
54
2 nd order ac term (nonlinear)
d(t)i(t) n
Chapter 7: AC equivalent circuit modeling
1 st order ac terms (linear)
0 D'i(t) v(t) Id(t) ➚ + dv(t) = D'I C dV –V + – n – R – n n R dt dt
Collect terms:
C
Perturb about quiescent operating point:
C
Original averaged equation:
Perturbation of averaged capacitor equation
Fundamentals of Power Electronics
Chapter 7: AC equivalent circuit modeling
d i(t) v(t) = Dvg(t) – D' n + Vg + V n – IRon d(t) – DRon i(t) dt
This is the desired linearized inductor equation.
L
Second-order terms are small when the small-signal assumption is satisfied. The remaining first-order terms are:
0 = DVg– D' V n – DRonI
Dc terms:
Linearization of averaged inductor equation
dv(t) D'i(t) v(t) Id(t) = n – – n R dt
55
Chapter 7: AC equivalent circuit modeling
Ts
= d(t) i(t)
Ts
+
1 st order ac term
i g(t)
Fundamentals of Power Electronics
Dc term
Ig
Collect terms:
=
56
Dc term
DI
I g + i g(t) = D + d(t) I + i(t)
+
Di(t) + Id(t)
+
2 nd order ac term (nonlinear)
d(t)i(t)
Chapter 7: AC equivalent circuit modeling
1 st order ac terms (linear)
Perturb about quiescent operating point:
ig(t)
Original averaged equation:
Perturbation of averaged input current equation
Fundamentals of Power Electronics
This is the desired linearized capacitor equation.
C
Second-order terms are small when the small-signal assumption is satisfied. The remaining first-order terms are:
V 0 = D'I n –R
Dc terms:
Linearization of averaged capacitor equation
57
Chapter 7: AC equivalent circuit modeling
i g(t) = Di(t) + Id(t)
Fundamentals of Power Electronics
58
Next step: construct equivalent circuit models.
L
Chapter 7: AC equivalent circuit modeling
d i(t) v(t) = Dvg(t) – D' n + Vg + V n – IRon d(t) – DRon i(t) dt dv(t) D'i(t) v(t) Id(t) C = n – – n R dt
Small-signal ac equations:
0 = DVg– D' V n – DRonI V 0 = D'I n –R I g = DI
Dc equations:
Summary: dc and small-signal ac converter equations
Fundamentals of Power Electronics
This is the desired linearized input current equation.
i g(t) = Di(t) + Id(t)
Second-order terms are small when the small-signal assumption is satisfied. The remaining first-order terms are:
I g = DI
Dc terms:
Linearization of averaged input current equation
D vg(t)
+ –
Fundamentals of Power Electronics
I d(t) n
dv(t) D'i(t) v(t) Id(t) = n – – n R dt
D' i(t) n
C
59
i(t)
DRon
D' v(t) n
60
C
C
dv(t) dt
R
Chapter 7: AC equivalent circuit modeling
–
v(t)
+
v(t) R
Chapter 7: AC equivalent circuit modeling
+ –
d(t) Vg – IRon + V n
Small-signal ac equivalent circuit: capacitor node
+ d i(t) – L dt
L
d i(t) v(t) = Dvg(t) – D' n + Vg + V n – IRon d(t) – DRon i(t) dt
Fundamentals of Power Electronics
L
Small-signal ac equivalent circuit: inductor loop
+ –
I d(t)
I d(t)
61
D i(t)
+ –
D vg(t)
L i(t)
DRon
D' v(t) n
+ –
d(t) Vg – IRon + V n
D' i(t) n
+ –
I d(t)
i g(t)
1:D
L i(t)
DRon
62
d(t) Vg – IRon + V n
I d(t) C n
–
v(t)
+
–
v(t)
R
R
Chapter 7: AC equivalent circuit modeling
D' : n
I d(t) C n
+
Chapter 7: AC equivalent circuit modeling
D i(t)
Replace dependent sources with ideal transformers:
+ –
Fundamentals of Power Electronics
vg(t)
vg(t)
Combine circuits: i g(t)
+ –
Small-signal ac model, nonideal flyback converter example
Fundamentals of Power Electronics
vg(t)
i g(t)
i g(t) = Di(t) + Id(t)
Small-signal ac equivalent circuit: converter input node
+ –
+ –
63
Chapter 7: AC equivalent circuit modeling
Fundamentals of Power Electronics
64
Chapter 7: AC equivalent circuit modeling
• To solve the differential equations of a system, the initial values of the state variables must be specified
• At a given point in time, the values of the state variables depend on the previous history of the system, rather than the present values of the system inputs
• Other typical physical state variables: position and velocity of a motor shaft
• For a typical converter circuit, the physical state variables are the inductor currents and capacitor voltages
• The physical state variables of a system are usually associated with the storage of energy
• If the system is linear, then the derivatives of the state variables are expressed as linear combinations of the system independent inputs and state variables themselves
• A canonical form for writing the differential equations of a system
7.4.1. The state equations of a network
Fundamentals of Power Electronics
• Computer programs exist which utilize the state-space averaging method
• A general approach: if the state equations of the converter can be written for each subinterval, then the small-signal averaged model can always be derived
• Often cited in the literature
• Uses the state-space matrix description of linear circuits
• Equivalent to the modeling method of the previous sections
• A formal method for deriving the small-signal ac equations of a switching converter
7.4. State Space Averaging
x1(t) x(t) = x2(t) ,
State vector x(t) contains inductor currents, capacitor voltages, etc.:
d x1(t) dt d x2(t) dx(t) = dt dt
iin(t)
C1
iC1(t)
u(t) = iin(t)
Input vector
R1
iR1(t)
–
v1(t)
–
v2(t) R3
R2
y(t) =
vout(t) iR1(t)
Choose output vector as
C2
+
+ vout(t) –
Chapter 7: AC equivalent circuit modeling
L i(t) + + v (t) – L iC2(t)
Example
65
Fundamentals of Power Electronics
66
Chapter 7: AC equivalent circuit modeling
To write the state equations of this circuit, we must express the inductor voltages and capacitor currents as linear combinations of the elements of the x(t) and u( t) vectors.
C1 0 0 K = 0 C2 0 0 0 L
Matrix K
v1(t) x(t) = v2(t) i(t)
State vector
Fundamentals of Power Electronics
Matrix K contains values of capacitance, inductance, and mutual inductance, so that K dx/dt is a vector containing capacitor currents and inductor winding voltages. These quantities are expressed as linear combinations of the independent inputs and state variables. The matrices A, B, C, and E contain the constants of proportionality.
Output vector y(t) contains other dependent quantities to be computed, such as ig(t)
Input vector u(t) contains independent sources such as vg(t)
K
A canonical matrix form:
dx(t) = A x(t) + B u(t) dt y(t) = C x(t) + E u(t)
State equations of a linear system, in matrix form
67
vL(t) = L
–
v2(t)
=
dx(t) dt
C1 0 0 0 C2 0 0 0 L
K
Fundamentals of Power Electronics
=
dv1(t) dt dv2(t) dt di(t) dt
Express in matrix form:
The same equations:
R3
R2 + vout(t) –
68
1
A
–1
0
0 1 – R2 + R3
– 1 R1
0
1
x(t)
+ B
v1(t) 1 v2(t) + 0 0 i(t)
u(t)
Chapter 7: AC equivalent circuit modeling
–1
dv1(t) v (t) = iin(t) – 1 – i(t) R dt dv2(t) v2(t) iC2(t) = C2 = i(t) – R2 + R3 dt di(t) vL(t) = L = v1(t) – v2(t) dt iC1(t) = C1
iin(t)
Chapter 7: AC equivalent circuit modeling
di(t) = v1(t) – v2(t) dt
dv2(t) v (t) = i(t) – 2 R2 + R3 dt
dv1(t) v (t) = iin(t) – 1 – i(t) R dt
C2
+
Equations in matrix form
Fundamentals of Power Electronics
Find vL via loop equation:
iC2(t) = C2
–
v1(t)
Find iC2 via node equation:
C1
+ + v (t) – L iC2(t)
iC1(t) = C1
R1
iC1(t)
L
Find iC1 via node equation:
iin(t)
iR1(t)
i(t)
Circuit equations
vout(t) iR1(t) iin(t)
R1
iR1(t) C1
iC1(t)
–
v1(t) C2
+ + v (t) – L iC2(t)
L
–
v2(t)
+
R3
R2
v1(t) R1
=
Fundamentals of Power Electronics
y(t)
0
1 R1
C
R3 R2 + R3
70
iR1(t) =
0
0
v1(t) R1
vout(t) = v2(t)
0
Express in matrix form:
The same equations:
vout(t) = iR1(t)
69
+ E
u(t)
iin(t)
Chapter 7: AC equivalent circuit modeling
x(t)
v1(t) v2(t) + 0 0 i(t)
R3 R2 + R3
+ vout(t) –
Chapter 7: AC equivalent circuit modeling
Express in matrix form
Fundamentals of Power Electronics
iR1(t) =
Express elements of the vector y as linear combinations of elements of x and u: R3 vout(t) = v2(t) R2 + R3
y(t) =
i(t)
Output (dependent signal) equations
Chapter 7: AC equivalent circuit modeling
Fundamentals of Power Electronics
A = D A 1 + D' A 2 B = D B 1 + D' B 2 C = D C 1 + D' C 2 E = D E 1 + D' E 2
where the averaged matrices are
72
Chapter 7: AC equivalent circuit modeling
X = equilibrium (dc) state vector U = equilibrium (dc) input vector Y = equilibrium (dc) output vector D = equilibrium (dc) duty cycle
and the equilibrium dc components are
0=AX+BU Y=CX+EU
Provided that the natural frequencies of the converter, as well as the frequencies of variations of the converter inputs, are much slower than the switching frequency, then the state-space averaged model that describes the converter in equilibrium is
Equilibrium (dc) state-space averaged model
71
dx(t) = A 2 x(t) + B 2 u(t) dt y(t) = C 2 x(t) + E 2 u(t)
Fundamentals of Power Electronics
K
During subinterval 2, when the switches are in position 2, the converter reduces to another linear circuit, that can be described by the following state equations:
During subinterval 1, when the switches are in position 1, the converter reduces to a linear circuit that can be described by the following state equations: dx(t) K = A 1 x(t) + B 1 u(t) dt y(t) = C 1 x(t) + E 1 u(t)
Given: a PWM converter, operating in continuous conduction mode, with two subintervals during each switching period.
7.4.2. The basic state-space averaged model
73
Chapter 7: AC equivalent circuit modeling
C 1 – C 2 X + E 1 – E 2 U d(t)
y(t) = C x(t) + E u(t) +
x(t) = small – signal (ac) perturbation in state vector u(t) = small – signal (ac) perturbation in input vector y(t) = small – signal (ac) perturbation in output vector d(t) = small – signal (ac) perturbation in duty cycle
A 1 – A 2 X + B 1 – B 2 U d(t)
dx(t) = A x(t) + B u(t) + dt
Fundamentals of Power Electronics
74
Chapter 7: AC equivalent circuit modeling
So if we can write the converter state equations during subintervals 1 and 2, then we can always find the averaged dc and small-signal ac models
where
K
Small-signal ac state-space averaged model
Fundamentals of Power Electronics
Y = – C A– 1 B + E U
X = – A– 1 B U
Solution for X and Y:
0=AX+BU Y=CX+EU
Equilibrium state-space averaged model:
Solution of equilibrium averaged model
= 1 Ts Ts t
t + Ts
x(τ) dτ
dt
d x(t) Ts
x(t)
Ts
u(t)
dx(t) = A 1 x(t) + B 1 u(t) dt y(t) = C 1 x(t) + E 1 u(t)
Fundamentals of Power Electronics
dx(t) = K – 1 A 1 x(t) dt
76
Ts
+ B 1 u(t)
Ts
Chapter 7: AC equivalent circuit modeling
Small ripple assumption: the elements of x(t) and u(t) do not change significantly during the subinterval. Hence the slopes are essentially constant and are equal to
dx(t) = K – 1 A 1 x(t) + B 1 u(t) dt
So the elements of x(t) change with the slope
K
During subinterval 1, we have
Ts
Chapter 7: AC equivalent circuit modeling
+ d(t) B 1 + d'(t) B 2
Change in state vector during first subinterval
75
= d(t) A 1 + d'(t) A 2
Fundamentals of Power Electronics
K
By averaging the inductor voltages and capacitor currents, one obtains:
The low-frequency components of the input and output vectors are modeled in a similar manner.
x(t)
As in Sections 7.1 and 7.2, the low-frequency components of the inductor currents and capacitor voltages are modeled by averaging over an interval of length Ts. Hence, we define the average of the state vector as:
7.4.3. Discussion of the state-space averaging result
Ts
+ B 1 u(t) Ts
initial value
interval length
dTs
77
Ts
0
Ts
Ts
Ts
+ B 2 u(t)
Ts
initial value
Fundamentals of Power Electronics
final value
x(Ts) = x(dTs) +
interval length
d'Ts
78
Ts
+ B 2 u(t)
Ts
Chapter 7: AC equivalent circuit modeling
slope
K – 1 A 2 x(t)
The value of the state vector at the end of the second subinterval is therefore
dx(t) = K – 1 A 2 x(t) dt
State vector now changes with the essentially constant slope
Use similar arguments.
+ dB
Ts
Chapter 7: AC equivalent circuit modeling
Ts
dTs
x
+ B1 u
K –1 dA 1 + d'A 2
K –1 A 1 x
+ B 1 u(t) slope
K – 1 A 1 x(t)
x(0)
x(t)
Change in state vector during second subinterval
Fundamentals of Power Electronics
final value
x(dTs) = x(0) +
Net change in state vector over first subinterval:
dx(t) = K – 1 A 1 x(t) dt
Change in state vector during first subinterval
Ts
0
Ts
Ts
dt
d x(t)
Ts
Ts
80
x(t)
Ts
Ts
Ts
x(Ts)
Ts
u(t)
t
Ts
Chapter 7: AC equivalent circuit modeling
+ d(t) B 1 + d'(t) B 2
u
Ts
+ B2 u
x(t)
K –1 A 2 x
Ts
Ts
Ts
Chapter 7: AC equivalent circuit modeling
u(t)
+ B 2 u(t)
+ TsK – 1 d(t)B 1 + d'(t)B 2
+ dB 1 + d'B 2
Ts
= d(t) A 1 + d'(t) A 2
x(Ts) – x(0) Ts
Ts
≈
Fundamentals of Power Electronics
K
We obtain:
dt
dTs
x
+ B1 u
K –1 dA 1 + d'A 2
K –1 A 1 x
Use Euler approximation:
x(0)
x(t)
d x(t)
79
Ts
+ d'TsK – 1 A 2 x(t)
Approximate derivative of state vector
Fundamentals of Power Electronics
Ts
x(t)
+ B 1 u(t)
x(Ts) = x(0) + TsK – 1 d(t)A 1 + d'(t)A 2
Collect terms:
x(Ts) = x(0) + dTsK – 1 A 1 x(t)
Eliminate x(dTs), to express x(Ts) directly in terms of x(0) :
Ts
Ts
+ B 2 u(t)
+ B 1 u(t)
Ts
Ts
x(Ts) = x(dTs) + d'Ts K – 1 A 2 x(t)
x(dTs) = x(0) + dTs K – 1 A 1 x(t)
We have:
Net change in state vector over one switching period
0
Ts
0
C 2 x(t)
C 1 x(t)
dTs
+ E 2 u(t)
+ E 1 u(t)
Ts
Ts
Ts
Ts
Ts
t
Ts
= d(t) C 1 x(t)
Ts
Ts
Ts
+ d'(t) C 2 x(t)
Ts
u(t)
Ts
+ E 2 u(t)
Ts
Chapter 7: AC equivalent circuit modeling
+ d(t) E 1 + d'(t) E 2
Ts
dt y(t)
d x(t)
Ts
Ts
x(t) x(t)
= d(t) A 1 + d'(t) A 2 = d(t) C 1 + d'(t) C 2
Ts
Ts
+ d(t) E 1 + d'(t) E 2
+ d(t) B 1 + d'(t) B 2
u(t)
u(t)
A = D A 1 + D' A 2 B = D B 1 + D' B 2 C = D C 1 + D' C 2 E = D E 1 + D' E 2 Fundamentals of Power Electronics
where
0=AX+BU Y=CX+EU
82
and
Ts
Ts
Chapter 7: AC equivalent circuit modeling
X = equilibrium (dc) state vector U = equilibrium (dc) input vector Y = equilibrium (dc) output vector D = equilibrium (dc) duty cycle
The converter operates in equilibrium when the derivatives of all elements of < x(t) >Ts are zero. Hence, the converter quiescent operating point is the solution of
K
The averaged (nonlinear) state equations:
Averaged state equations: quiescent operating point
81
x(t)
+ E 1 u(t)
= d(t) C 1 + d'(t) C 2
Fundamentals of Power Electronics
y(t)
Collect terms:
y(t)
Remove switching harmonics by averaging over one switching period:
y(t)
y(t)
Low-frequency components of output vector
Ts
Ts
Ts
Y >> y(t)
X >> x(t)
= Y + y(t)
U >> u(t) D >> d(t)
with
= U + u(t)
= X + x(t)
d(t) = D + d(t) ⇒ d'(t) = D' – d(t)
y(t)
u(t)
x(t)
dx(t) dt
+
dc terms +
A 1 – A 2 X + B 1 – B 2 U d(t)
84
Chapter 7: AC equivalent circuit modeling
second–order nonlinear terms
C 1 – C 2 x(t)d(t) + E 1 – E 2 u(t)d(t)
first–order ac terms
C 1 – C 2 X + E 1 – E 2 U d(t)
second–order nonlinear terms
A 1 – A 2 x(t)d(t) + B 1 – B 2 u(t)d(t)
first–order ac terms
= CX + EU + Cx(t) + Eu(t) +
dc terms
= AX + BU + Ax(t) + Bu(t) +
Fundamentals of Power Electronics
dc + 1st order ac
Y+y(t)
first–order ac
K
Averaged state equations: perturbation and linearization
Chapter 7: AC equivalent circuit modeling
D+d(t) E 1 + D'–d(t) E 2 U+u(t)
+
83
D+d(t) C 1 + D'–d(t) C 2 X+x(t)
Y+y(t) =
Fundamentals of Power Electronics
D+d(t) B 1 + D'–d(t) B 2 U+u(t)
+
Substitute into averaged state equations: d X+x(t) K = D+d(t) A 1 + D'–d(t) A 2 X+x(t) dt
Let
Averaged state equations: perturbation and linearization
x(t) =
i(t) v(t)
state vector
+ –
Q1
Fundamentals of Power Electronics
vg(t)
ig(t)
L
i(t) C
R
u(t) =
86
vg(t) VD
input vector
D1
–
v(t)
+
Chapter 7: AC equivalent circuit modeling
y(t) = ig(t)
output vector
• Diode forward voltage drop VD
• MOSFET onresistance Ron
Model nonidealities:
7.4.4. Example: State-space averaging of a nonideal buck-boost converter
Fundamentals of Power Electronics
Chapter 7: AC equivalent circuit modeling
C 1 – C 2 X + E 1 – E 2 U d(t)
y(t) = C x(t) + E u(t) +
85
A 1 – A 2 X + B 1 – B 2 U d(t)
dx(t) = A x(t) + B u(t) + dt
This is the desired result.
K
Dc terms drop out of equations. Second-order (nonlinear) terms are small when the small-signal assumption is satisfied. We are left with:
Linearized small-signal state equations
y(t)
y(t)
C2
00
ig(t)
=
A2
1 –1 – 1 R
0
dx(t) dt
Fundamentals of Power Electronics
K
d i(t) = dt v(t)
di(t) L = v(t) – VD dt dv(t) v(t) C =– – i(t) R dt ig(t) = 0
L 0 0C
87
vg(t)
Ron
00
i(t) + v(t)
88
E2
B2
R
–
v(t)
L
u(t)
vg(t) VD
u(t)
vg(t) VD
i(t)
VD
C
R
–
v(t)
+
Chapter 7: AC equivalent circuit modeling
0 –1 0 0
x(t)
i(t) + v(t)
+ –
ig(t)
u(t)
C
+
Chapter 7: AC equivalent circuit modeling
E1
vg(t) VD i(t) + 00 v(t) x(t)
u(t)
vg(t) VD
i(t)
B1
x(t)
vg(t)
x(t)
L
i(t) + 10 v(t) 00
+ –
ig(t)
Subinterval 2
C1
10
ig(t)
=
A1
dx(t) dt
– Ron 0 d i(t) = dt v(t) 0 – 1 R
Fundamentals of Power Electronics
K
L 0 0C
di(t) L = vg(t) – i(t) Ron dt dv(t) v(t) C =– R dt ig(t) = i(t)
Subinterval 1
+ –
D – D' 0 0
Fundamentals of Power Electronics
DC solution:
0=AX+BU Y=CX+EU
Fundamentals of Power Electronics
Ig =
I = V
or,
90
1 R 1 + D2 on D' R
D' –1 R
Vg VD
Chapter 7: AC equivalent circuit modeling
Vg VD D2 D D' 2R D'R
00
I + V
Vg VD
D – D' 0 0
I + V
D 1 2 D' R D' R – D 1 D'
D0
– D'
– DRon
1 R D 1 + 2 on D' R
Ig =
0 = 0
Vg VD
Chapter 7: AC equivalent circuit modeling
DC state equations
E = DE 1 + D'E 2 = 0 0
89
– Ron 0 – DRon D' 0 1 + D' = 0 – 1 – D' – 1 –1 – 1 R R R
C = DC 1 + D'C 2 = D 0
B = DB 1 + D'B 2 =
In a similar manner,
A = DA 1 + D'A 2 = D
Evaluate averaged matrices
Ig =
0 = 0
91
I
D'VD
I + V
I + V
–
V
+
Vg VD 00
D' : 1
Vg VD
D – D' 0 0
i(t) + 00 00 v(t)
Fundamentals of Power Electronics
i g(t) = D 0
– DRon D' L 0 d i(t) = 0 C dt v(t) – D' – 1 R
92
vg(t) + 0 d(t) 0 I ➚ vD(t)
i(t) + v(t)
Small-signal ac state equations:
D – D' 0 0
Chapter 7: AC equivalent circuit modeling
vg(t) Vg – V – IRon + VD + d(t) 0 I v➚ D(t)
V – IRon + VD V – V – IRon + VD A 1 – A 2 X + B1 – B2 U = – V + g = g I I 0 C1 – C2 X + E1 – E2 U = I
R
Chapter 7: AC equivalent circuit modeling
Small-signal ac model
+ –
Evaluate matrices in small-signal model:
Fundamentals of Power Electronics
Vg
Ig
1:D
DRon
D' –1 R
D0
– D'
– DRon
Corresponding equivalent circuit:
DC state equations:
Steady-state equivalent circuit
+ –
+ – i(t)
DRon
93
+ –
I d(t)
i g(t)
1:D
L
D' i(t)
i(t)
DRon
94
d(t) Vg – V + V D – IRon
Combine individual circuits to obtain
Fundamentals of Power Electronics
vg(t)
vg(t)
C
I d(t)
C
dv(t) dt
–
v(t)
+
D i(t)
R
v(t) R
I d(t) C
–
v(t)
R
Chapter 7: AC equivalent circuit modeling
D' : 1 +
Chapter 7: AC equivalent circuit modeling
I d(t)
+ –
i g(t)
Complete small-signal ac equivalent circuit
D' v(t)
d(t) Vg – V + V D – IRon
Fundamentals of Power Electronics
D vg(t)
+ d i(t) – L dt
L
inductor equation
input eqn
capacitor eqn
i g(t) = D i(t) + I d(t)
d i(t) = D' v(t) – DRon i(t) + D vg(t) + Vg – V – IRon + VD d(t) dt dv(t) v(t) C = –D' i(t) – + I d(t) R dt L
Corresponding equivalent circuits:
Small-signal ac equations, in scalar form:
Construction of ac equivalent circuit
+ –
+ –
+ –
We will use it to model DCM, CPM, and resonant converters Also useful for incorporating switching loss into ac model of CCM converters Applicable to 3ø PWM inverters and rectifiers Can be applied to phase-controlled rectifiers
Chapter 7: AC equivalent circuit modeling
+ –
Fundamentals of Power Electronics
vg(t)
Power input
–
vC(t)
i1(t)
v1(t)
+
+
C iL(t)
Control input
96
d(t)
Switch network
–
+
R
–
v(t)
+
Chapter 7: AC equivalent circuit modeling
–
v2(t)
i2(t)
L
Time-invariant network containing converter reactive elements
Load
Separate switch network from remainder of converter
95
Rather than averaging and linearizing the converter state equations, the averaging and linearization operations are performed directly on the converter circuit
●
●
●
●
Fundamentals of Power Electronics
●
●
●
●
Historically, circuit averaging was the first method known for modeling the small-signal ac behavior of CCM PWM converters It was originally thought to be difficult to apply in some cases There has been renewed interest in circuit averaging and its corrolary, averaged switch modeling, in the last decade Can be applied to a wide variety of converters
port 1
●
7.5. Circuit Averaging and Averaged Switch Modeling
port 2
–
–
Chapter 7: AC equivalent circuit modeling
v2(t)
v1(t)
–
v(t)
i2(t) +
R
i1(t) +
(b)
C
+
98
Chapter 7: AC equivalent circuit modeling
Definition of the switch network terminal quantities is not unique. Different definitions lead equivalent results having different forms
The switch network terminal waveforms are then the port voltages and currents: v1(t), i1(t), v2(t), and i2(t). Two of these waveforms can be taken as independent inputs to the switch network; the remaining two waveforms are then viewed as dependent outputs of the switch network.
The number of ports in the switch network is less than or equal to the number of SPST switches Simple dc-dc case, in which converter contains two SPST switches: switch network contains two ports
Fundamentals of Power Electronics
●
●
●
Discussion
–
–
97
v2(t)
v1(t)
i2(t) +
vg(t) + –
i(t)
L
i1(t) +
(a)
Fundamentals of Power Electronics
Two ways to define the switch network
Ideal boost converter example
Boost converter example
v2(t) –
v1(t) –
vg(t) + –
i(t)
L
C
99
Fundamentals of Power Electronics
vg(t) + –
i(t)
–
i2(t) v2(t) C
Switch network
v1(t) + –
+
100
Boost converter example
i1(t)
–
v(t)
Chapter 7: AC equivalent circuit modeling
R
+
Replace the switch network with dependent sources, which correctly represent the dependent output waveforms of the switch network
L
–
v(t)
+
Chapter 7: AC equivalent circuit modeling
R
Obtaining a time-invariant network: Modeling the terminal behavior of the switch network
Fundamentals of Power Electronics
Since i1(t) and v2(t) coincide with the converter inductor current and output voltage, it is convenient to define these waveforms as the independent inputs to the switch network. The switch network dependent outputs are then v1(t) and i2(t).
+
i2(t)
i1(t) +
Let’s use definition (a):
Boost converter example
0
0
0
〈i2(t)〉T
0
s
〈v1(t)〉Ts
dTs
dTs i1(t)
v2(t)
Ts
Ts
t
t
101
i1(t)
s
+ –
Fundamentals of Power Electronics
〈vg(t)〉T
Power input
–
s
〈i1(t)〉T
〈vC(t)〉Ts
〈v1(t)〉Ts
+
+
102
Control input
–
v(t)
+
d(t)
–
s
〈v2(t)〉T
+
–
〈v(t)〉T
+
s
Chapter 7: AC equivalent circuit modeling
s
R
Load
Chapter 7: AC equivalent circuit modeling
L
〈i2(t)〉T
〈iL(t)〉Ts
Averaged switch network
–
R
So far, no approximations have been made.
The circuit is therefore electrical identical to the original converter.
Averaged time-invariant network containing converter reactive elements C
–
i2(t) v2(t) C
Switch network
v1(t) + –
Now average all waveforms over one switching period:
port 1
+
The waveforms of the dependent generators are defined to be identical to the actual terminal waveforms of the switch network.
vg(t) + –
i(t)
L
The circuit averaging step
Fundamentals of Power Electronics
0
i2(t)
0
v1(t)
Definition of dependent generator waveforms
port 2
+ –
〈i(t)〉T
s
vg(t) + –
L
i(t)
Fundamentals of Power Electronics
〈vg(t)〉Ts
103
L
+ –
104
–
〈i2(t)〉Ts
Averaged switch network
〈v1(t)〉Ts
s
+ R
C
–
v(t)
+
R
–
〈v(t)〉Ts
Chapter 7: AC equivalent circuit modeling
–
〈v2(t)〉Ts
+
i2(t) v2(t) C
Switch network
v1(t) + –
〈i1(t)〉T
i1(t)
+
Chapter 7: AC equivalent circuit modeling
Averaging step: boost converter example
Fundamentals of Power Electronics
In practice, the only work needed for this step is to average the switch dependent waveforms.
The basic assumption is made that the natural time constants of the converter are much longer than the switching period, so that the converter contains low-pass filtering of the switching harmonics. One may average the waveforms over an interval that is short compared to the system natural time constants, without significantly altering the system response. In particular, averaging over the switching period Ts removes the switching harmonics, while preserving the low-frequency components of the waveforms.
The averaging step
+ –
s
s
0
〈i2(t)〉Ts
0
〈i(t)〉T
0
0
〈v1(t)〉T
L
dTs
dTs i1(t)
v2(t)
+ –
t
t
+ –
L
Ts
= Vg + vg(t)
+ –
i 2(t)
v1(t)
v(t)
Ts
Ts
Ts
s
Ts
= V + v(t)
106
R
s
–
〈v(t)〉T
+
C
R
–
V + v(t)
+
Chapter 7: AC equivalent circuit modeling
D' – d(t) I + i(t)
= I 2 + i 2(t)
= V1 + v1(t)
= v2(t)
s
d(t) = D + d(t) ⇒ d'(t) = D' – d(t) i(t) T = i 1(t) T = I + i(t)
vg(t)
C
Ts
Ts
Chapter 7: AC equivalent circuit modeling
–
〈v2(t)〉T s
+
= d'(t) i 1(t)
= d'(t) v2(t)
Perturb and linearize
105
Ts
Ts
d'(t) 〈i1(t)〉Ts
i 2(t)
v1(t)
Average the waveforms of the dependent sources:
Averaged switch model
d'(t) 〈v2(t)〉Ts
D' – d(t) V + v(t)
Fundamentals of Power Electronics
Vg + vg(t)
I + i(t)
Ts
Ts
〈i1(t)〉Ts
The circuit becomes:
As usual, let:
Fundamentals of Power Electronics
〈vg(t)〉Ts
0
i2(t)
0
v1(t)
Compute average values of dependent sources
107
+ –
Chapter 7: AC equivalent circuit modeling
D' V + v(t)
Fundamentals of Power Electronics
D' I + i(t)
108
I d(t)
Chapter 7: AC equivalent circuit modeling
nonlinear, 2nd order
D' – d(t) I + i(t) = D' I + i(t) – I d(t) – i(t)d(t)
Dependent current source
Fundamentals of Power Electronics
V d(t)
nonlinear, 2nd order
D' – d(t) V + v(t) = D' V + v(t) – V d(t) – v(t)d(t)
Dependent voltage source
+ –
+ –
+ –
I + i(t)
L
V d(t)
V d(t)
D' V + v(t)
+ –
+ –
109
D' : 1
I d(t)
D' I + i(t)
R
R
–
V + v(t)
+
–
V + v(t)
+
Chapter 7: AC equivalent circuit modeling
C
I d(t) C
Fundamentals of Power Electronics
110
Chapter 7: AC equivalent circuit modeling
Perturb and linearize the resulting low-frequency model, to obtain a small-signal equivalent circuit
Average converter waveforms over one switching period, to remove the switching harmonics
Model the switch network with equivalent voltage and current sources, such that an equivalent time-invariant network is obtained
Summary: Circuit averaging method
Fundamentals of Power Electronics
Vg + vg(t)
Vg + vg(t)
I + i(t)
L
Linearized circuit-averaged model
+ –
Switch network
1
–
v(t)
+
111
I + i(t) V d(t) I d(t)
–
V + v(t)
+
Chapter 7: AC equivalent circuit modeling
D' : 1
Switch network
1
2
–
v(t)
+
I + i(t) V d(t)
+ –
D' : 1
I d(t)
–
V + v(t)
+
Fundamentals of Power Electronics
112
Chapter 7: AC equivalent circuit modeling
Circuit averaging modifies only the switch network. Hence, to obtain a smallsignal converter model, we need only replace the switch network with its averaged model. Such a procedure is called averaged switch modeling.
• Introduction of ac voltage and current variations, drive by the control input duty cycle variations
• Transformation of dc and small-signal ac voltage and current levels, according to the D’:1 conversion ratio
For the boost example, we can conclude that the switch network performs two basic functions:
i(t)
Basic functions performed by switch network
Fundamentals of Power Electronics
i(t)
2
Circuit averaging of the boost converter: in essence, the switch network was replaced with an effective ideal transformer and generators:
Averaged switch modeling: CCM
+ –
v2(t) –
+ v1(t) –
Ts
Ts
L
= d(t) v1(t)
= d(t) i 2(t)
Fundamentals of Power Electronics
v2(t)
i 1(t)
–
i2(t) +
– Switch network
iC
v2(t)
+
+ vCE –
v1(t)
i1(t)
113
Ts
Ts
C
i(t)
114
R
–
v(t)
+
0
0 v1
i2
i2
T2
dTs
dTs
0
v2(t)
0
i 1(t)
T2
Ts
Ts
t
t
Chapter 7: AC equivalent circuit modeling
0
v2(t)
0
i1(t)
T2
Chapter 7: AC equivalent circuit modeling
The basic buck-type CCM switch cell
Fundamentals of Power Electronics
vg(t) + –
i2(t) +
i1(t)
2. To derive an averaged switch model, express the average values of two of the terminal quantities, for example 〈 v2 〉Ts and 〈 i1 〉Ts, as functions of the other average terminal quantities 〈 v1 〉Ts and 〈 i1 〉Ts . 〈 v2 〉Ts and 〈 i1 〉Ts may also be functions of the control input d, but they should not be expressed in terms of other converter signals.
1. Define a switch network and its terminal waveforms. For a simple transistor-diode switch network as in the buck, boost, etc., there are two ports and four terminal quantities: v1, i1, v2, i2.The switch network also contains a control input d. Buck example:
Averaged switch modeling: Procedure
–
v1(t)
+ v (t) – 2
i2(t)
Switch network
i1(t)
i(t)
C
Vg + vg + –
L
–
v(t)
V1 d
115
Switch network
I2 d
1:D
–
V1 + v1
+
I1 + i1
L
C
I+i
R
V1 d
–
V +v
+
–
V2 + v2
+
I2 + i2
Chapter 7: AC equivalent circuit modeling
–
V2 + v2
+
I2 + i2
I2 d
1:D
V2 + v2(t) = D V1 + v1(t) + V1 d(t)
v2(t) –
v1(t) –
–
–
v2(t) –
v1(t) –
–
V1 + v1
+
I1 + i1
–
V1 + v1
V2 d
V1 d DD'
+ –
Fundamentals of Power Electronics
+
+
i2(t)
v2(t)
v1(t)
+
I1 + i1
I2 d
+ –
i1(t)
i2(t) + +
–
V1 + v1
+
I1 + i1
116
D' : D
D' : 1
1:D
I2 d DD'
I1 d
V1 d
+ –
i1(t)
i2(t) + +
i1(t)
Appendix 3 Averaged switch modeling of a CCM SEPIC
see also
Chapter 7: AC equivalent circuit modeling
–
V2 + v2
+
I2 + i2
–
V2 + v2
+
I2 + i2
–
V2 + v2
+
I2 + i2
Three basic switch networks, and their CCM dc and small-signal ac averaged switch models
–
V1 + v1
I1 + i1 +
R
+
Perturbation and linearization of switch network: I 1 + i 1(t) = D I 2 + i 2(t) + I 2 d(t)
+ –
Fundamentals of Power Electronics
Resulting averaged switch model: CCM buck converter
vg(t) + –
+
Circuit-averaged model
Replacement of switch network by dependent sources, CCM buck example
+ –
0
v1
tir
tvf
vCE(t)
0
i2
iC(t)
tir
vCE(t)
tvf
Fundamentals of Power Electronics
tvr
tif
L
117
C
i(t)
t2
R
t
–
= i 2(t)
0
t1 Ts
tvr
tif
t2
118
t
Chapter 7: AC equivalent circuit modeling
t 1 + t vf + t vr + 1 t ir + 1 t if 2 2 Ts
i 1(t) dt
Ts
Ts
0
i2
iC(t)
Switch network terminal waveforms: v1, i1, v2, i2. To derive averaged switch model, express 〈 v2 〉Ts and 〈 i1 〉Ts as functions of 〈 v1 〉Ts and 〈 i1 〉Ts . 〈 v2 〉Ts and 〈 i1 〉Ts may also be functions of the control input d, but they should not be expressed in terms of other converter signals.
i 1(t) = i C(t) v2(t) = v1(t) – vCE(t)
Chapter 7: AC equivalent circuit modeling
v(t)
+
Averaging i1(t)
Ts
i 1(t) T = 1 s Ts
0
v1
t1
–
–
+
i2(t)
v2(t)
Switch network
iC
v1(t)
+
+ vCE –
Fundamentals of Power Electronics
vg(t) + –
i1(t)
Example: Averaged switch modeling of CCM buck converter, including switching loss
0
t vf + t vr Ts t ir + t if di = Ts
Ts
= i 2(t)
Ts
d + 1 dv 2
Chapter 7: AC equivalent circuit modeling
i 1(t)
Then we can write
Ts
Ts
Ts
tir
tvf
vCE(t)
= v1(t)
= v1(t)
Ts
Ts
t1
= 1 Ts Ts
0
i2
iC(t)
Ts
0
Ts
d – 1 di 2 120
t 1 + 1 t vf + 1 t vr 2 2 Ts
= v1(t) – vCE(t)
Fundamentals of Power Electronics
v2(t)
v2(t)
v2(t)
0
v1
tif
t2
Ts
t
Chapter 7: AC equivalent circuit modeling
– vCE(t) dt + v1(t)
tvr
Averaging the switch network output voltage v2(t)
Fundamentals of Power Electronics
dv =
119
t 1 + t vf + t vr + 1 t ir + 1 t if 2 2 Ts
i 1(t) dt
Ts
Ts
= i 2(t)
= 1 Ts Ts
t 1 + 1 t vf + 1 t vr + 1 t ir + 1 t if 2 2 2 2 d= Ts
Let
i 1(t)
Given
Expression for 〈 i1(t) 〉
= i 2(t) Ts
1 2 s
di(t) 〈v1(t)〉Ts
= v1(t)
1 2
s
–
〈v2(t)〉T
〈i2(t)〉Ts +
–
〈v2(t)〉Ts
〈i2(t)〉Ts +
d – 1 di 2
Chapter 7: AC equivalent circuit modeling
s
di(t) 〈v1(t)〉T
Ts
1 2
Psw =
1 2
dv(t) 〈i2(t)〉Ts
dv + di
122
i 2(t)
1 : d(t)
Ts
di(t) 〈v1(t)〉Ts
Ts
s
–
〈v2(t)〉Ts
+
〈i2(t)〉T
Chapter 7: AC equivalent circuit modeling
v1(t)
1 2
– +
Fundamentals of Power Electronics
–
〈v1(t)〉Ts
+
s
〈i1(t)〉T
Switching loss predicted by averaged switch model
121
1 2
Ts
+ d(t) 〈v (t)〉 1 Ts –
v2(t)
1 : d(t)
d(t) 〈i2(t)〉Ts
d + 1 dv 2
dv(t) 〈i2(t)〉T
dv(t) 〈i2(t)〉Ts
〈i1(t)〉Ts
1 2
〈i1(t)〉Ts
Ts
– +
Fundamentals of Power Electronics
–
s
〈v1(t)〉T
+
–
〈v1(t)〉Ts
+
i 1(t)
Construction of large-signal averaged-switch model
– +
+ – –
V1
+ 1 2
Di V1
123
L
C
I
R
–
V
+
1–
Chapter 7: AC equivalent circuit modeling
D – 12 Di
Di P 2D η = out = = Pin D D + 12 Dv 1+ v 2D
Pout = VI 2 = V1I 2 D – 12 Di
Pin = VgI 1 = V1I 2 D + 12 Dv
Efficiency calcuation:
–
V2
+
I2
Fundamentals of Power Electronics
124
Chapter 7: AC equivalent circuit modeling
• Plug in parameter values for a given specific converter
• A standard form of equivalent circuit model, which represents the above physical properties
Canonical model:
Hence, we expect their equivalent circuit models to be qualitatively similar.
• Control of waveforms by variation of duty cycle
• Low-pass filtering of waveforms
• Transformation of voltage and current levels, ideally with 100% efficiency
All PWM CCM dc-dc converters perform the same basic functions:
7.6. The canonical circuit model
Fundamentals of Power Electronics
1 2
1 2
Averaged switch network model
D v I2
1:D
D V = D – Di Vg = DVg 1 – i 2D
Output voltage:
Vg
I1
Solution of averaged converter model in steady state
– +
+ –
Power input
Vg
125
–
V
+
Load
R
Chapter 7: AC equivalent circuit modeling
Control input
D
1 : M(D)
Converter model
Fundamentals of Power Electronics
• these variations are also transformed by the conversion ratio M(D)
2. Ac variations in vg(t) induce ac variations in v(t)
Power input
Vg + vg(s)
+ –
126
–
R
Load
V + v(s)
+
Chapter 7: AC equivalent circuit modeling
Control input
D
1 : M(D)
Steps in the development of the canonical circuit model
Fundamentals of Power Electronics
• can refine dc model by addition of effective loss elements, as in Chapter 3
• effective turns ratio M(D)
• modeled as in Chapter 3 with ideal dc transformer
1. Transformation of dc voltage and current levels
7.6.1. Development of the canonical circuit model
Power
Vg + vg(s)
+ –
Control
D
127
Zeo(s)
–
j(s) d(s)
Control input
D + d(s)
Zei(s)
1 : M(D)
filter
low-pass
Effective
He(s)
Zeo(s)
–
R
Load
V + v(s)
+
Fundamentals of Power Electronics
128
Chapter 7: AC equivalent circuit modeling
• Can push all sources to input side as shown. Sources may then become frequency-dependent
• Independent sources represent effects of variations in duty cycle
4. Control input variations also induce ac variations in converter waveforms
Power input
Vg + vg(s) + –
e(s) d(s)
R
Load
V + v(s)
+
Chapter 7: AC equivalent circuit modeling
filter
low-pass
Effective
He(s)
Steps in the development of the canonical circuit model
Fundamentals of Power Electronics
Zei(s)
1 : M(D)
input input • effective filter elements may not coincide with actual element values, but can also depend on operating point
• also filters ac variations
• necessary to filter switching ripple
3. Converter must contain an effective lowpass filter characteristic
Steps in the development of the canonical circuit model
+ –
Load
R
129
+ –
Id
L
Vg – V d
D' : 1
Id
Fundamentals of Power Electronics
• Combine transformers
130
• Push inductor to output side of transformers
C
–
V + v(s)
+
Chapter 7: AC equivalent circuit modeling
• Push independent sources to input side of transformers
Vg + vg(s)
1:D
Small-signal ac model of the buck-boost converter
R
Chapter 7: AC equivalent circuit modeling
7.6.2. Example: manipulation of the buck-boost converter model into canonical form
Fundamentals of Power Electronics
v(s) = e(s) M(D) H e(s) d(s)
–
V + v(s)
Gvd(s) =
Zeo(s)
Control-to-output transfer function:
filter
low-pass
+
v(s) = M(D) H e(s) vg(s)
Control input
D + d(s)
Zei(s)
Effective
He(s)
Gvg(s) =
j(s) d(s)
1 : M(D)
Line-to-output transfer function:
Power input
Vg + vg(s) + –
e(s) d(s)
Transfer functions predicted by canonical model
+ –
+ –
Step 1
+ – Id
1:D
Step 2
131
L
I d D'
C
–
V + v(s)
+
+ –
Id
Fundamentals of Power Electronics
Vg + vg(s)
Vg – V d D
1:D
132
I d D'
node A
I d D'
L
R
C
–
V + v(s)
+
Chapter 7: AC equivalent circuit modeling
D' : 1
R
Chapter 7: AC equivalent circuit modeling
D' : 1
How to move the current source past the inductor: Break ground connection of current source, and connect to node A instead. Connect an identical current source from node A to ground, so that the node equations are unchanged.
Fundamentals of Power Electronics
Vg + vg(s)
Vg – V d D
Push voltage source through 1:D transformer Move current source through D’:1 transformer
+ –
+ –
+ – Id
+ –
1:D
Step 4
133
I d D'
sLI d D'
D' : 1
C
–
V + v(s)
+
Id
DI d D'
Fundamentals of Power Electronics
+ –
DI d D'
+ –
Vg + vg(s)
Vg – V d D
134
1:D
sLI d D'
D' : 1
C
–
V + v(s)
+ R
Chapter 7: AC equivalent circuit modeling
L
Push current source past voltage source, again by: Breaking ground connection of current source, and connecting to node B instead. Connecting an identical current source from node B to ground, so that the node equations are unchanged. Note that the resulting parallel-connected voltage and current sources are equivalent to a single voltage source. node B
R
Chapter 7: AC equivalent circuit modeling
L
Now push current source through 1:D transformer.
Fundamentals of Power Electronics
Vg + vg(s)
Vg – V d D
The parallel-connected current source and inductor can now be replaced by a Thevenin-equivalent network:
Step 3
– + – +
I d(s) D'
135
D' : D
C
–
V + v(s)
+ R
Chapter 7: AC equivalent circuit modeling
Effective low-pass filter
L D' 2
Vg + V s LI – D D D'
Fundamentals of Power Electronics
136
Chapter 7: AC equivalent circuit modeling
Pushing the sources past the inductor causes the generator to become frequency-dependent.
e(s) = – V2 1 – s DL D D' 2 R
Simplification, using dc relations, leads to
e(s) =
Voltage source coefficient is:
Coefficient of control-input voltage generator
Fundamentals of Power Electronics
Vg + vg(s) + –
Vg – V – s LI d(s) D DD'
Combine series-connected transformers.
Push voltage source through 1:D transformer, and combine with existing input-side transformer.
Step 5: final result
+ –
j(s) d(s)
1 : M(D)
Le
C
–
V + v(s)
+ R
L D' 2
D 1 D'
–D D'
Buck
137
– V2 D' R
Fundamentals of Power Electronics
What is the relation between vc(t) and d(t)?
vg(t) + –
Power input
δ(t)
138
dTs Ts
δ(t)
transistor gate driver
t
vc(t)
R
v
feedback connection
Chapter 7: AC equivalent circuit modeling
Controller
t
–
v(t)
+
Load
voltage reference vref
compensator pulse-width vc Gc(s) modulator
Switching converter
–+
Pulse-width modulator converts voltage signal vc(t) into duty cycle signal d(t).
j(s) V R V D' 2 R
Chapter 7: AC equivalent circuit modeling
– V2 1 – s DL D D' 2 R
e(s) V D2 V 1 – s 2L D' R
7.7. Modeling the pulse-width modulator
Fundamentals of Power Electronics
Buck-boost
Boost
L L D' 2
M(D)
Converter
Le
Table 7.1. Canonical model parameters for the ideal buck, boost, and buck-boost converters
Vg + vg(s) + –
e(s) d(s)
7.6.3. Canonical circuit parameters for some common converters + –
PWM waveform
for 0 ≤ vc(t) ≤ VM
Fundamentals of Power Electronics
So d(t) is a linear function of vc(t).
vc(t) VM
For a linear sawtooth waveform:
d(t) =
139
δ(t)
comparator δ(t)
0
dTs
Ts
140
δ(t)
0
VM
0
vsaw(t)
2Ts
t
Ts
2Ts
Chapter 7: AC equivalent circuit modeling
dTs
vc(t)
vsaw(t)
t
Chapter 7: AC equivalent circuit modeling
0
vc(t)
Equation of pulse-width modulator
Fundamentals of Power Electronics
vc(t)
analog input
generator
wave
vsaw(t)
+
Sawtooth
VM
A simple pulse-width modulator
–
vc(t) VM for 0 ≤ vc(t) ≤ VM
Vc + vc(t) VM
141
pulse-width modulator
1 VM
vc
1 VM
Fundamentals of Power Electronics
142
d
Chapter 7: AC equivalent circuit modeling
In practice, this limits the useful frequencies of ac variations to values much less than the switching frequency. Control system bandwidth must be sufficiently less than the Nyquist rate fs/2. Models that do not account for sampling are accurate only at frequencies much less than fs/2.
fs
sampler
Therefore, the pulsepulse-width modulator width modulator samples the control waveform, with sampling rate equal to the switching frequency.
The input voltage is a continuous function of time, but there can be only one discrete value of the duty cycle for each switching period.
D + d(s)
Chapter 7: AC equivalent circuit modeling
Vc VM v (t) d(t) = c VM D=
Dc and ac relations:
Vc + vc(s)
Block diagram:
Sampling in the pulse-width modulator
Fundamentals of Power Electronics
D + d(t) =
Result:
vc(t) = Vc + vc(t) d(t) = D + d(t)
Perturb:
d(t) =
PWM equation:
Perturbed equation of pulse-width modulator
143
Chapter 7: AC equivalent circuit modeling
Fundamentals of Power Electronics
144
Chapter 7: AC equivalent circuit modeling
5. The circuit averaging technique also yields equivalent results, but the derivation involves manipulation of circuits rather than equations. Switching elements are replaced by dependent voltage and current sources, whose waveforms are defined to be identical to the switch waveforms of the actual circuit. This leads to a circuit having a time-invariant topology. The waveforms are then averaged to remove the switching ripple, and perturbed and linearized about a quiescent operating point to obtain a small-signal model.
4. The state-space averaging method of section 7.4 is essentially the same as the basic approach of section 7.2, except that the formality of the state-space network description is used. The general results are listed in section 7.4.2.
Summary of key points
Fundamentals of Power Electronics
3. Ac equivalent circuits can be constructed, in the same manner used in Chapter 3 to construct dc equivalent circuits. If desired, the ac equivalent circuits may be refined to account for the effects of converter losses and other nonidealities.
2. Since switching converters are nonlinear systems, it is desirable to construct small-signal linearized models. This is accomplished by perturbing and linearizing the averaged model about a quiescent operating point.
1. The CCM converter analytical techniques of Chapters 2 and 3 can be extended to predict converter ac behavior. The key step is to average the converter waveforms over one switching period. This removes the switching harmonics, thereby exposing directly the desired dc and low-frequency ac components of the waveforms. In particular, expressions for the averaged inductor voltages, capacitor currents, and converter input current are usually found.
7.8. Summary of key points
145
Chapter 7: AC equivalent circuit modeling
Fundamentals of Power Electronics
146
Chapter 7: AC equivalent circuit modeling
8. The conventional pulse-width modulator circuit has linear gain, dependent on the slope of the sawtooth waveform, or equivalently on its peak-to-peak magnitude.
Summary of key points
Fundamentals of Power Electronics
7. The canonical circuit describes the basic properties shared by all dc-dc PWM converters operating in the continuous conduction mode. At the heart of the model is the ideal 1:M(D) transformer, introduced in Chapter 3 to represent the basic dc-dc conversion function, and generalized here to include ac variations. The converter reactive elements introduce an effective low-pass filter into the network. The model also includes independent sources which represent the effect of duty cycle variations. The parameter values in the canonical models of several basic converters are tabulated for easy reference.
6. When the switches are the only time-varying elements in the converter, then circuit averaging affects only the switch network. The converter model can then be derived by simply replacing the switch network with its averaged model. Dc and small-signal ac models of several common CCM switch networks are listed in section 7.5.4. Switching losses can also be modeled using this approach.
Summary of key points
Single pole response Single zero response Right half-plane zero Frequency inversion Combinations Double pole response: resonance The low-Q approximation Approximate roots of an arbitrary-degree polynomial
1
Chapter 8: Converter Transfer Functions
Series impedances: addition of asymptotes Parallel impedances: inverse addition of asymptotes Another example Voltage divider transfer functions: division of asymptotes
Fundamentals of Power Electronics
2
8.5. Summary of key points
Chapter 8: Converter Transfer Functions
8.4. Measurement of ac transfer functions and impedances
8.3.1. 8.3.2. 8.3.3. 8.3.4.
8.3. Graphical construction of converter transfer functions
Converter Transfer Functions
Fundamentals of Power Electronics
8.2.1. Example: transfer functions of the buck-boost converter 8.2.2. Transfer functions of some basic CCM converters 8.2.3. Physical origins of the right half-plane zero in converters
8.2. Analysis of converter transfer functions
8.1.1. 8.1.2. 8.1.3. 8.1.4. 8.1.5. 8.1.6. 8.1.7. 8.1.8.
8.1. Review of Bode plots
Chapter 8. Converter Transfer Functions
3
Chapter 8: Converter Transfer Functions
Fundamentals of Power Electronics
4
Chapter 8: Converter Transfer Functions
• Graphical construction of Bode plots of transfer functions and polynomials, to avoid algebra mistakes approximate transfer functions obtain insight into origins of salient features
• Analytical approximation of roots of high-order polynomials
• Use of inverted poles and zeroes, to refer transfer function gains to the most important asymptote
• Obtaining simple analytical expressions for asymptotes, corner frequencies, and other salient features, allows element values to be selected such that a given desired behavior is obtained
• Writing transfer functions in normalized form, to directly expose salient features
Some elements of design-oriented analysis, discussed in this chapter
Fundamentals of Power Electronics
Design-oriented analysis is a structured approach to analysis, which attempts to avoid the above problems
Obtain simple equations that can be inverted, so that element values can be chosen to obtain desired behavior. Equations that cannot be inverted are useless for design!
Obtain physical insight which leads engineer to synthesis of a good design
Design objectives:
Algebra mistakes
Long equations
Complicated derivations
Problems:
How to approach a real (and hence, complicated) system
Design-oriented analysis
dB
= 20 log 10 G
dB
= 20 log 10
– 6dB
1/2
20 dB – 6 dB = 14 dB
5 = 10/2
1000 = 10
3 ⋅ 20dB = 60 dB
20dB
6 dB
2
10
0 dB
1
3
Magnitude in dB
Actual magnitude
Table 8.1. Expressing magnitudes in decibels
Bode plot of fn
5
f f0
dB
= 20 log 10
f f0
= 20n log 10
Fundamentals of Power Electronics
• Magnitude is 1, or 0dB, at frequency f = f0
• Slope is 20n dB/decade
G
n
Magnitude in dB is
f f0
6
–60dB
–40dB
–20dB
0dB
20dB
40dB
f0
n
=
n
2
=
–2
n=
1
10f0
–1
n=
–2
–1
f log scale
f f0
f f0
Chapter 8: Converter Transfer Functions
0.1f0
40dB/decade
20 dB/decade
–20dB/decade
–40dB/decade
f f0
Chapter 8: Converter Transfer Functions
Bode plots are effectively log-log plots, which cause functions which vary as fn to become linear plots. Given: f n G = f0 60dB 2
Fundamentals of Power Electronics
60dBµA is a current 60dB greater than a base current of 1µA, or 1mA.
5Ω is equivalent to 14dB with respect to a base impedance of Rbase = 1Ω, also known as 14dBΩ.
Z
Z Rbase
Decibels of quantities having units (impedance example): normalize before taking log
G
Decibels
8.1. Review of Bode plots
+ – C
dB
= – 20 log 10
Fundamentals of Power Electronics
G( jω)
0
7
ω 1+ ω 0
2
ω jω 0 ω 2 ω0
2
8
dB
Re (G( jω)) + Im (G( jω)) 1 ω 2 1+ ω Magnitude in dB:
=
G( jω) =
Magnitude is
–
v2(s)
+
1 1 + sRC
with
2
∠G(jω) Re(G(jω))
Chapter 8: Converter Transfer Functions
Im(G(jω)) G(jω)
Chapter 8: Converter Transfer Functions
ω0 = 1 RC
0
This coincides with the normalized form 1 G(s) = 1 + ωs
G(s) =
Express as rational fraction:
1 v2(s) sC G(s) = = 1 +R v1(s) sC
G(jω) and || G(jω) ||
1– 1 G( jω) = = ω 1+ 1+ j ω 0
Let s = jω:
Fundamentals of Power Electronics
v1(s)
R
Transfer function is
G ||
Simple R-C example
8.1.1. Single pole response
|| (jω )
dB
≈ 0dB
–60dB
–40dB
–20dB
0dB
|| G(jω) ||dB
9
0.1f0
0dB
f0
2
2
2
2
=
f f0
–1
–60dB
–40dB
–20dB
0dB
|| G(jω) ||dB
0.1f0
0dB
G( jω) =
Fundamentals of Power Electronics
10
f f0
f0
2
10f0
–20dB/decade
1 ω 1+ ω 0
f f0
–1
f
–1
f
Chapter 8: Converter Transfer Functions
The high-frequency asymptote of || G(jω) || varies as f-1. Hence, n = -1, and a straight-line asymptote having a slope of -20dB/decade is obtained. The asymptote has a value of 1 at f = f0 .
G( jω) ≈
1 ω ω0
ω ≈ ω 0 Then || G(jω) || becomes
ω 1+ ω 0
ω ω0 >> 1
For high frequency, ω >> ω0 and f >> f0 :
10f0
–20dB/decade
1 ω 1+ ω 0
Chapter 8: Converter Transfer Functions
G( jω) =
Asymptotic behavior: high frequency
Fundamentals of Power Electronics
This is the low-frequency asymptote of || G(jω) ||
G( jω)
G( jω) ≈ 1 = 1 1 Or, in dB,
Then || G(jω) || becomes
ω ω0 << 1
For small frequency, ω << ω0 and f << f0 :
Asymptotic behavior: low frequency
dB
= – 20 log 10
1
ω 1 + ω0 0 2
ω 1 + ω0 0
= 1 2 2
≈ – 3 dB
–30dB
–20dB
–10dB
Fundamentals of Power Electronics
11
1dB
12
0.5f0
f0
Chapter 8: Converter Transfer Functions
f
–20dB/decade
2f0
1dB
3dB
Chapter 8: Converter Transfer Functions
Summary: magnitude
0dB
|| G(jω) ||dB
Fundamentals of Power Electronics
Similar arguments show that the exact curve lies 1dB below the asymptotes.
at f = 0.5f0 and 2f0 :
G( jω0)
G( jω0) =
at f = f0:
Evaluate exact magnitude:
Deviation of exact curve near f = f0
-90˚ 0.01f0
-75˚
-60˚
-45˚
-30˚
-15˚
0.1f0
-45˚
0˚ asymptote
Fundamentals of Power Electronics
∠G(jω)
13
f
f0
f0
14
0
∞
ω
0
ω
ω ω0
–90˚
–45˚
0˚
∠G(jω)
∠G( jω) = – tan – 1
Chapter 8: Converter Transfer Functions
100f0
–90˚ asymptote 10f0
ω ω0
ω jω 0 ω 2 ω0
Chapter 8: Converter Transfer Functions
∠G( jω) = – tan – 1
1– 1 G( jω) = = ω 1+ 1+ j ω 0
Phase of G(jω)
Re G( jω)
Im G( jω)
Fundamentals of Power Electronics
∠G( jω) = tan – 1
0˚
G(jω)
Phase of G(jω)
∠G(jω) Re(G(jω))
|
Im(G(jω)) | |G j( ω )|
Fundamentals of Power Electronics
fa = f0 e – π / 2 ≈ f0 / 4.81 fb = f0 e π / 2 ≈ 4.81 f0
15
-90˚ 0.01f0
-75˚
-60˚
-45˚
-30˚
-15˚
0˚
16
0.1f0
f
f0
fb = 4.81 f0
100f0
Chapter 8: Converter Transfer Functions
-45˚
fa = f0 / 4.81
f0
Chapter 8: Converter Transfer Functions
Phase asymptotes
∠G(jω)
Fundamentals of Power Electronics
fa = f0 e – π / 2 ≈ f0 / 4.81 fb = f0 e π / 2 ≈ 4.81 f0
Try a midfrequency asymptote having slope identical to actual slope at the corner frequency f0. One can show that the asymptotes then intersect at the break frequencies
Hence, need a midfrequency asymptote
Low- and high-frequency asymptotes do not intersect
High frequency: –90˚
Low frequency: 0˚
Phase asymptotes
-90˚ 0.01f0
-75˚
-60˚
-45˚
-30˚
-15˚
17
0.1f0
fa = f0 / 10
-45˚
fb = 10 f0
0˚
Fundamentals of Power Electronics
∠G(jω)
|| G(jω) ||dB
0dB
18
-45˚
5.7˚
0.5f0
-45˚/decade
f0 / 10
1dB
f0
f0
5.7˚
2f0
100f0
-90˚
1 1 + ωs
0
Chapter 8: Converter Transfer Functions
10 f0
–20dB/decade
1dB
3dB
G(s) =
Chapter 8: Converter Transfer Functions
f
f0
f0
Summary: Bode plot of real pole
Fundamentals of Power Electronics
fa = f0 / 10 fb = 10 f0
∠G(jω)
0˚
Phase asymptotes: a simpler choice
ω 1+ ω 0 2
ω ω0
19
Chapter 8: Converter Transfer Functions
0˚
Fundamentals of Power Electronics
∠G(jω)
|| G(jω) ||dB
0dB
G(s) = 1 + ωs 0
f0 / 10
+45˚/decade
1dB
20
5.7˚
45˚
0.5f0
f0
f0
5.7˚
2f0
+90˚
Chapter 8: Converter Transfer Functions
10 f0
3dB
1dB
+20dB/decade
Summary: Bode plot, real zero
Fundamentals of Power Electronics
—with the exception of a missing minus sign, same as simple pole
∠G( jω) = tan – 1
Phase:
+20dB/decade slope at high frequency, ω >> ω0
0dB at low frequency, ω << ω0
Use arguments similar to those used for the simple pole, to derive asymptotes:
G( jω) =
Magnitude:
G(s) = 1 + ωs 0
Normalized form:
8.1.2. Single zero response
ω 1+ ω 0 2
∠G( jω) = – tan – 1
ω ω0
21
0˚
Fundamentals of Power Electronics
∠G(jω)
|| G(jω) ||dB
0dB
G(s) = 1 – ωs 0
-45˚/decade
f0 / 10
1dB
22
-45˚
5.7˚
0.5f0
f0
f0
5.7˚
2f0
Chapter 8: Converter Transfer Functions
10 f0
3dB
1dB
+20dB/decade
-90˚
Chapter 8: Converter Transfer Functions
Summary: Bode plot, RHP zero
Fundamentals of Power Electronics
The RHP zero exhibits the magnitude asymptotes of the LHP zero, and the phase asymptotes of the pole
—same as real pole.
Phase:
—same as conventional (left half-plane) zero. Hence, magnitude asymptotes are identical to those of LHP zero.
G( jω) =
Magnitude:
G(s) = 1 – ωs 0
Normalized form:
8.1.3. Right half-plane zero
1 + ωs 0
∠G(jω)
|| G(jω) ||dB
ω 1 + s0
1
+90˚
Fundamentals of Power Electronics
G(s) =
23
24
+45˚
5.7˚
0.5f0
-45˚/decade
f0 / 10
+20dB/decade
1dB
3dB
f0
f0
5.7˚
2f0
0dB
Chapter 8: Converter Transfer Functions
10 f0
1dB
0˚
Chapter 8: Converter Transfer Functions
Asymptotes, inverted pole
Fundamentals of Power Electronics
The inverted-pole format emphasizes the high-frequency gain.
G(s) =
s ω0
An algebraically equivalent form:
Reversal of frequency axis. A useful form when describing mid- or high-frequency flat asymptotes. Normalized form, inverted pole: 1 G(s) = ω 1 + s0
8.1.4. Frequency inversion
|| G(jω) ||dB
–90˚
Fundamentals of Power Electronics
∠G(jω)
25
Chapter 8: Converter Transfer Functions
f0 / 10 26
5.7˚
–45˚
0.5f0
+45˚/decade
3dB
1dB
–20dB/decade
f0
f0
5.7˚
2f0
0˚
10 f0
Chapter 8: Converter Transfer Functions
0dB
1dB
ω G(s) = 1 + s0
Asymptotes, inverted zero
Fundamentals of Power Electronics
Again, the inverted-zero format emphasizes the high-frequency gain.
1 + ωs 0 G(s) = s ω0
An algebraically equivalent form:
Normalized form, inverted zero: ω G(s) = 1 + s0
Inverted zero
Combinations
27
dB
= R1(ω)
dB
+ R2(ω)
dB
Chapter 8: Converter Transfer Functions
Fundamentals of Power Electronics
28
Chapter 8: Converter Transfer Functions
Composite magnitude, when expressed in dB, is sum of individual magnitudes.
Composite phase is sum of individual phases.
R3(ω)
R3(ω) = R1(ω) R2(ω)
The composite magnitude is
θ 3(ω) = θ 1(ω) + θ 2(ω)
The composite phase is
G3(ω) = R1(ω) R2(ω) e j(θ 1(ω) + θ 2(ω))
Fundamentals of Power Electronics
G3(ω) = R1(ω) R2(ω) e j(θ 1(ω) + θ 2(ω))
G3(ω) = G1(ω) G2(ω) = R1(ω) e jθ 1(ω) R2(ω) e jθ 2(ω)
The product G3(ω) can then be written
G1(ω) = R1(ω) e jθ 1(ω) G2(ω) = R2(ω) e jθ 2(ω) G3(ω) = R3(ω) e jθ 3(ω)
Express the complex-valued functions in polar form:
Suppose that we have constructed the Bode diagrams of two complex-values functions of frequency, G1(ω) and G2(ω). It is desired to construct the Bode diagram of the product, G3(ω) = G1(ω) G2(ω).
8.1.5. Combinations
1 + ωs 1
1 + ωs 2
G0
∠G
0˚
10 Hz
f1/10 10 Hz
0 dB
f1 100 Hz
29
f
1 kHz
10f1 1 kHz
–180˚ 100 kHz
–135˚
–90˚
–45˚
0˚
Chapter 8: Converter Transfer Functions
10 kHz
–45˚/decade
10f2 20 kHz
–40 dB/decade
∠G
0˚
|| A0 ||dB
Fundamentals of Power Electronics
∠A
|| A ||
f1 /10
+45˚/dec
f1
30
f2 /10 –90˚
10f1
Chapter 8: Converter Transfer Functions
0˚
|| A∞ ||dB
10f2
–45˚/dec
+20 dB/dec
f2
Determine the transfer function A(s) corresponding to the following asymptotes:
Example 2
100 Hz
–90˚/decade
f2 2 kHz
–20 dB/decade
f2/10 200 Hz –45˚/decade
G0 = 40 ⇒ 32 dB || G ||
1 Hz
–60 dB
–40 dB
–20 dB
0 dB
20 dB
40 dB
Fundamentals of Power Electronics
|| G ||
with G0 = 40 ⇒ 32 dB, f1 = ω1/2π = 100 Hz, f2 = ω2/2π = 2 kHz
Example 1: G(s) =
A(s) = A 0
1 + ωs 1 1 + ωs 2
A0
1
2
1
1 + ωs ➚ s 1+➚ ω
s 1+➚ ω2
31
s = jω
s = jω
= A0
s ω1 1
ω =A f = A0 ω 0 f1 1
Chapter 8: Converter Transfer Functions
s = jω
= A0 1 = A0 1
1 + ωs ➚ 1 + ωs ➚
2
1
s = jω
= A0
f2 f1
s = jω
s = jω
ω f = A 0 ω2 = A 0 2 f1 1
Fundamentals of Power Electronics
A(s) = A ∞
ω 1 + s1 ω 1 + s2 32
Chapter 8: Converter Transfer Functions
Another way to express A(s): use inverted poles and zeroes, and express A(s) directly in terms of A∞
A∞ = A0
s ω1 s ω2
Example 2, continued
A0
A0
s 1+➚ ω
So the high-frequency asymptote is
For f > f2
Fundamentals of Power Electronics
For f1 < f < f2
For f < f1
Analytical expressions for asymptotes:
One solution:
Example 2, continued
1 1 + a 1s + a 2s 2
v1(s)
33
C
1 1 + a 1s + a 2s 2
1 – ss 1
1 1 – ss 2
with
1–
1–
a1 1– 2a 2 a1 1+ 2a 2
s1 = –
s2 = –
4a 2 a 21
4a 2 a 21
Fundamentals of Power Electronics
34
–
v2(s)
Chapter 8: Converter Transfer Functions
• If 4a2 ≤ a12, then the roots s1 and s2 are real. We can construct Bode diagram as the combination of two real poles. • If 4 a2 > a12, then the roots are complex. In Section 8.1.1, the assumption was made that ω 0 is real; hence, the results of that section cannot be applied and we need to do some additional work.
G(s) =
We might factor the denominator using the quadratic formula, then construct Bode diagram as the combination of two real poles:
G(s) =
R
+
Chapter 8: Converter Transfer Functions
Approach 1: factor denominator
Fundamentals of Power Electronics
L
Two-pole low-pass filter example
+ –
How should we construct the Bode diagram?
with a1 = L/R and a2 = LC
G(s) =
Second-order denominator, of the form
v (s) 1 G(s) = 2 = v1(s) 1 + s L + s 2LC R
Example
8.1.6 Quadratic pole response: resonance
1 s 1 + 2ζ ω + ωs 0 0 2
or
G(s) =
1 s 1+ + ωs Qω0 0
2
The Q-factor
35
Chapter 8: Converter Transfer Functions
(peak stored energy) (energy dissipated per cycle)
Fundamentals of Power Electronics
36
Chapter 8: Converter Transfer Functions
For a second-order passive system, the two equations above are equivalent. We will see that Q has a simple interpretation in the Bode diagrams of second-order transfer functions.
Q = 2π
Q is a measure of the dissipation in the system. A more general definition of Q, for sinusoidal excitation of a passive element or system is
Q= 1 2ζ
In a second-order system, ζ and Q are related according to
Fundamentals of Power Electronics
• In the alternative form, the parameter Q is called the quality factor. Q also controls the shape of the exact curve in the vicinity of f = f0. The roots are complex when Q > 0.5.
• The parameter ζ is called the damping factor. ζ controls the shape of the exact curve in the vicinity of f = f0. The roots are complex when ζ < 1.
• The parameter ω0 is the angular corner frequency, and we can define f0 = ω0/2π
• The parameters ζ, ω0, and Q are found by equating the coefficients of s
• When the coefficients of s are real and positive, then the parameters ζ, ω0, and Q are also real and positive
G(s) =
Approach 2: Define a standard normalized form for the quadratic case
1 s 1+ + ωs Qω0 0
37
f0 =
G(s) =
f f0
38
–60 dB
–40 dB
–20 dB
0 dB
|| G(jω) ||dB
for ω >> ω0
Fundamentals of Power Electronics
G →
–2
G → 1 for ω << ω0
Asymptotes are
2
G( jω) =
1 s 1+ + ωs Qω0 0 let s = jω and find magnitude:
In the form
f0
2 2
10f0
–40 dB/decade
f f0
ω + 12 ω 0 Q
–2
f
Chapter 8: Converter Transfer Functions
0.1f0
0 dB
ω 1– ω 0
1 2
Chapter 8: Converter Transfer Functions
2
v2(s) 1 = v1(s) 1 + s L + s 2LC R
ω0 1 = 2π 2π LC Q=R C L
G(s) =
G(s) =
Magnitude asymptotes, quadratic form
Fundamentals of Power Electronics
Result:
Equate coefficients of like powers of s with the standard form
Two-pole low-pass filter example: we found that
Analytical expressions for f0 and Q
-20dB
-10dB
0.3
0.5
Q = 0.1
Q = 0.2
0.7
Q = 0.5
Q = 0.7
Q=1
Q=2
Q=5
Q=∞
Fundamentals of Power Electronics
|| G ||dB
0dB
10dB
0 dB
dB
dB
–40 dB/decade
f0
= Q
2
1
f / f0
2
40
3
∠G
-180°
-135°
-90°
-45°
0°
| Q |dB
1
f / f0
Chapter 8: Converter Transfer Functions
0.1
Q = 0.1
Q = 0.2
Q=∞ Q = 10 Q =5 Q=2 Q=1 Q = 0.7 Q = 0.5
Chapter 8: Converter Transfer Functions
G( jω0)
ω + 12 ω 0 Q
Two-pole response: exact curves
Fundamentals of Power Electronics
39
or, in dB:
The exact curve has magnitude Q at f = f0. The deviation of the exact curve from the asymptotes is | Q |dB
G( jω0) = Q
1 2 2
|| G ||
ω 1– ω 0
At ω = ω0, the exact magnitude is
G( jω) =
Deviation of exact curve from magnitude asymptotes
10
1 1 + a 1s + a 2s 2
or
G(s) =
1 s + ωs 1+ Qω0 0
2
1 + ωs 1
1 1 + ωs 2
An example
41
v2(s) 1 = L v1(s) 1 + s + s 2LC R
v1(s)
+ –
L
C
R
Fundamentals of Power Electronics
ω1 , ω 2 =
L/R±
42
L / R – 4 LC 2 LC
2
–
v2(s)
Chapter 8: Converter Transfer Functions
Use quadratic formula to factor denominator. Corner frequencies are:
G(s) =
R-L-C network example:
+
Chapter 8: Converter Transfer Functions
A problem with this procedure is the complexity of the quadratic formula used to find the corner frequencies.
Fundamentals of Power Electronics
This is a particularly desirable approach when Q << 0.5, i.e., when the corner frequencies ω1 and ω2 are well separated.
G(s) =
When the roots are real, i.e., when Q < 0.5, then we can factor the denominator, and construct the Bode diagram using the asymptotes for real poles. We would then use the following normalized form:
G(s) =
Given a second-order denominator polynomial, of the form
8.1.7. The low-Q approximation
2
L / R – 4 LC 2 LC
ω2 ≈ 1 RC
43
Chapter 8: Converter Transfer Functions
G(s) =
1 s + ωs 1+ Qω0 0
2
ω0 1 – Q
Fundamentals of Power Electronics
ω1 =
1 – 4Q 2 2
44
ω2 =
1 – 4Q 2 2
Chapter 8: Converter Transfer Functions
ω0 1 + Q
Use quadratic formula to express corner frequencies ω1 and ω2 in terms of Q and ω0 as:
Given
Derivation of the Low-Q Approximation
Fundamentals of Power Electronics
These simpler expressions can be derived via the Low-Q Approximation.
ω1 is then independent of C, and ω2 is independent of L.
ω1 ≈ R , L
When the corner frequencies are well separated in value, it can be shown that they are given by the much simpler (approximate) expressions
This complicated expression yields little insight into how the corner frequencies ω1 and ω2 depend on R, L, and C.
ω1 , ω 2 =
L/R±
Factoring the denominator
ω0 1 + Q 1 – 4Q 2 2
ω0 F(Q) Q
1 – 4Q 2
1 – 4Q 2 2
Q ω0 F(Q)
1 – 4Q 2
for Q << 1 2
Fundamentals of Power Electronics
ω1 ≈ Q ω0
For small Q, F(Q) tends to 1. We then obtain
F(Q) = 1 1 + 2
where
ω1 =
can be written in the form
ω 1– ω1 = 0 Q
0
0.25
0.5
0.75
1
0
0.1
0.2
Q
0.3
0.4
45
0
0.25
0.5
0.75
1
0
0.1
0.2
Q
0.3
0.4
46
0.5
Chapter 8: Converter Transfer Functions
For Q < 0.3, the approximation F(Q) = 1 is within 10% of the exact value.
F(Q)
0.5
Chapter 8: Converter Transfer Functions
For Q < 0.3, the approximation F(Q) = 1 is within 10% of the exact value.
F(Q)
Corner frequency ω1
for Q << 1 2
Fundamentals of Power Electronics
ω ω2 ≈ 0 Q
For small Q, F(Q) tends to 1. We then obtain
F(Q) = 1 1 + 2
where
ω2 =
can be written in the form
ω2 =
Corner frequency ω2
Q f0 F(Q) ≈ Q f0
v2(s) 1 = v1(s) 1 + s L + s 2LC R
Fundamentals of Power Electronics
C 1 =R ω1 ≈ Q ω0 = R L LC L ω0 1 ω2 ≈ = 1 = 1 Q RC C LC R L
48
f0F(Q) Q f ≈ 0 Q f2 =
Chapter 8: Converter Transfer Functions
ω0 1 = 2π 2π LC Q=R C L f0 =
Chapter 8: Converter Transfer Functions
–40dB/decade
R-L-C Example
47
Use of the Low-Q Approximation leads to
G(s) =
f0
–20dB/decade
f1 =
For the previous example:
Fundamentals of Power Electronics
0dB
|| G ||dB
The Low-Q Approximation
49
Chapter 8: Converter Transfer Functions
Fundamentals of Power Electronics
L
50
i g(t) = Di(t) + Id(t)
Chapter 8: Converter Transfer Functions
d i(t) = Dvg(t) + D'v(t) + Vg – V d(t) dt dv(t) v(t) C = – D'i(t) – + Id(t) R dt
Small-signal ac equations of the buck-boost converter, derived in section 7.2:
8.2.1. Example: transfer functions of the buck-boost converter
Fundamentals of Power Electronics
8.2.1. Example: transfer functions of the buck-boost converter 8.2.2. Transfer functions of some basic CCM converters 8.2.3. Physical origins of the right half-plane zero in converters
8.2. Analysis of converter transfer functions
d(s) and
vg(s)
and one
v(s) d(s) vg(s) = 0
51
and Gvg(s) =
d(s) = 0
Chapter 8: Converter Transfer Functions
v(s) vg(s)
v(s) + Id(s) R
i(s) =
52
Dvg(s) + D'v(s) + Vg – V d(s) sL
i(s) , and solve for v(s)
Fundamentals of Power Electronics
Eliminate
sCv(s) = – D'i(s) –
sLi(s) = Dvg(s) + D'v(s) + Vg – V d(s)
Chapter 8: Converter Transfer Functions
Take Laplace transform of converter equations, letting initial conditions be zero:
Algebraic approach
Derivation of transfer functions
Fundamentals of Power Electronics
Gvd(s) =
The control-to-output and line-to-output transfer functions can be defined as
v(s) = Gvd(s) d(s) + Gvg(s) vg(s)
Hence, the ac output voltage variations can be expressed as the superposition of terms arising from the two inputs:
The converter contains two inputs, output, v(s)
Definition of transfer functions
L + s 2 LC D' 2 D' 2 R
1
53
vg(s) –
L + s 2 LC D' 2 D' 2 R
v(s) vg(s)
= – D D' 1 + s d(s) = 0
Fundamentals of Power Electronics
Gvg(s) = Gg0
1 s 1+ + ωs Qω0 0
54
2
which is of the following standard form:
Gvg(s) =
Chapter 8: Converter Transfer Functions
1 L + s 2 LC D' 2 D' 2 R
Hence, the line-to-output transfer function is
d(s)
Chapter 8: Converter Transfer Functions
Vg – V D' 2 1+s
LI Vg – V
Derivation of transfer functions
Fundamentals of Power Electronics
v(s) = – D D' 1 + s
1–s
Vg – V – s LI – DD' vg(s) – 2 d(s) D' + s L + s 2 LC D' + s L + s 2 LC R R 2
write in normalized form:
v(s) =
v(s) sCv(s) = – D' Dvg(s) + D'v(s) + Vg – V d(s) – + Id(s) R sL
Derivation of transfer functions
55
Q = D'R
ω0 = D' LC C L
Chapter 8: Converter Transfer Functions
v(s) d(s)
Fundamentals of Power Electronics
Gvd(s) = Gd0
Standard form:
Gvd(s) =
1 – ωs z 1 + s + ωs Qω0 0
vg(s) = 0
56
2
Vg – V = – D' 2
1+s
LI Vg – V
Chapter 8: Converter Transfer Functions
L + s 2 LC D' 2 D' 2 R
1–s
Control-to-output transfer function
Fundamentals of Power Electronics
1 = L Qω0 D' 2R
1 = LC ω 20 D' 2
Gg0 = – D D'
Equate standard form to derived transfer function, to determine expressions for the salient features:
Salient features of the line-to-output transfer function
(RHP)
57
Chapter 8: Converter Transfer Functions
V = – D Vg D' I=– V D' R
Fundamentals of Power Electronics
D = 0.6 R = 10Ω Vg = 30V L = 160µH C = 160µF
Suppose we are given the following numerical values:
58
Chapter 8: Converter Transfer Functions
Gg0 = D = 1.5 ⇒ 3.5dB D' V = 469V ⇒ 53.4dBV Gd0 = D D' 2 ω f0 = 0 = D' = 400Hz 2π 2π LC Q = D'R C = 4 ⇒ 12dB L ωz D'R = = 6.6kHz fz = 2π 2πDL
Then the salient features have the following numerical values:
Plug in numerical values
Fundamentals of Power Electronics
— Simplified using the dc relations:
Q = D'R
C L
Vg – V D' R = LI DL
Vg – V Vg =– 3 = V 2 2 D' D' D D'
ω0 = D' LC
ωz =
Gd0 = –
Salient features of control-to-output transfer function
|| Gvd ||
∠ Gvd
0˚
f0 400Hz
59
1kHz
f
10kHz
–270˚
–270˚ 1MHz
∠ Gvg
0˚
|| Gvg ||
400Hz
f0
60
1kHz
f
–40dB/dec
–180˚
–270˚ 100kHz
–180˚
–90˚
0˚
∠ Gvg
Chapter 8: Converter Transfer Functions
10kHz
Q = 4 ⇒ 12dB
10 f0 533Hz
1 / 2Q 0
10 –1 / 2Q 0 f0 300Hz
100Hz
Gg0 = 1.5 ⇒ 3.5dB
10Hz
–80dB
–60dB
–40dB
–20dB
0dB
20dB
Fundamentals of Power Electronics
|| Gvg ||
–180˚
–90˚
0˚
∠ Gvd
Chapter 8: Converter Transfer Functions
100kHz
10fz 66kHz
–20dB/dec
Bode plot: line-to-output transfer function
100Hz
10 1 / 2Q 0 f0 533Hz
fz 6.6kHz RHP
–40dB/dec
Q = 4 ⇒ 12dB
fz /10 660Hz
10 –1 / 2Q 0 f0 300Hz
Gd0 = 469V ⇒ 53.4dBV
10Hz
–40dBV
–20dBV
0dBV
20dBV
40dBV
60dBV
80dBV
Fundamentals of Power Electronics
|| Gvd ||
Bode plot: control-to-output transfer function
D' LC
V D' V D D' 2
D
1 D' – D D'
buck boost
C L C L D'R
1 + s + ωs Qω0 0 2
61
1 s 1+ + ωs Qω0 0
2
Fundamentals of Power Electronics
• transient response: output initially tends in wrong direction
• phase reversal at high frequency
G(s) = 1 – ωs 0
uin(s)
62
s ωz
1
D' R DL
D' 2R L2
∞
ωz
Chapter 8: Converter Transfer Functions
Gvg(s) = Gg0
Chapter 8: Converter Transfer Functions
uout(s)
8.2.3. Physical origins of the right half-plane zero
Fundamentals of Power Electronics
Gvd(s) = Gd0
1 – ωs z
C L
D'R
R
Q
where the transfer functions are written in the standard forms
D' LC
1 LC
buck-boost
ω0
Gd0 V D
Gg0
Converter
Table 8.2. Salient features of the small-signal CCM transfer functions of some basic dc-dc converters
8.2.2. Transfer functions of some basic CCM converters
– +
vg
vg
+ –
+ –
Ts
L
L
Ts
63
1
= d' iL
iL(t)
2
iL(t) 1
C
iD(t)
R
–
v
+
R
Fundamentals of Power Electronics
| v(t) |
Ts
• As inductor current increases to its new equilibrium value, average diode current eventually increases
= d' iL
iD(t)
Ts
• Increasing d(t) causes the average diode current to initially decrease
iD
iL(t)
64
d = 0.4
s
t
t
t
Chapter 8: Converter Transfer Functions
T
d = 0.6
–
v
+
Chapter 8: Converter Transfer Functions
C
iD(t)
2
Waveforms, step increase in duty cycle
Fundamentals of Power Electronics
Buck-boost
Boost
iD
Two converters whose CCM control-to-output transfer functions exhibit RHP zeroes
H
0µH
µH 100
1m
H
10m
H
m 100
1H
10H
1µH
100Hz
1F
1kHz
100 mF
10
65
0nH
10kHz
100kHz
µF
100 F 1nH
1m
1mΩ 1MHz
10mΩ
Chapter 8: Converter Transfer Functions
F 10nH
10m
100mΩ
10µ
F
1Ω
10Ω
1µF
nF
100
100Ω
10n
F
1kΩ
10kΩ
1nF
pF
100
+ –
j(s) d(s)
Fundamentals of Power Electronics
vg(s)
66
–
ve(s)
1 : M(D) +
–
v(s)
R
Zout
Chapter 8: Converter Transfer Functions
Z2
C
+
Z1
Zin
Le
{
e(s) d(s)
He(s)
Transfer functions predicted by canonical model
Fundamentals of Power Electronics
1 –60dBΩ 10Hz
–40dBΩ
–20dBΩ
0dBΩ
20dBΩ
40dBΩ
60dBΩ
80dBΩ
Impedance graph paper
{
+ –
67
Zout
Chapter 8: Converter Transfer Functions
R
Fundamentals of Power Electronics
R
1 ωC
68
|| Zout ||
f0
Chapter 8: Converter Transfer Functions
Q = R / R0 R0
|| Z1 || = ωLe
Graphical construction of output impedance
Fundamentals of Power Electronics
Zout = Z1 || Z2
Z2
Z1
C
{
Le
Output impedance Zout: set sources to zero
{
Z1
Z out
f0
69
=
1 ω 2L
eC
Chapter 8: Converter Transfer Functions
ωL e
1 /ωC
Q = R / R0
Fundamentals of Power Electronics
R
1 ωC
70
|| Zout ||
f0
ωL D' 2
Chapter 8: Converter Transfer Functions
Q = R / R0 R0
increasing D
Boost and buck-boost converters: Le = L / D’ 2
Fundamentals of Power Electronics
He =
ωL e =1 ωL e
Graphical construction of filter effective transfer function
vz frequency
vz output
vz magnitude
Injection source
71
vy vx
vy vx
Data bus to computer
Chapter 8: Converter Transfer Functions
– 134.7˚
17.3 dB
Data
Fundamentals of Power Electronics
vy vx
and
• Network analyzer measures
72
x
vy
∠v
Chapter 8: Converter Transfer Functions
Narrowband function is essential: switching harmonics and other noise components are removed
Component of input at injection source frequency is measured
• Signal inputs vx and vy perform function of narrowband tracking voltmeter:
• Injection source produces sinusoid vz of controllable amplitude and frequency
Swept sinusoidal measurements
Fundamentals of Power Electronics
–
+
+
–
vy input
vx input
Measured inputs
Network Analyzer
8.4. Measurement of ac transfer functions and impedances
+ –
+
Z out(s) =
amplifier =0 ac input
G(s)
Device under test
vy(s) i out(s)
input
Data bus to computer
vy(s) = G(s) vx(s)
73
– 162.8˚
–4.7 dB
Data
74
voltage probe
i out
Zs
current probe
DC blocking capacitor R
source
vz
Chapter 8: Converter Transfer Functions
+ – + – vy vx
v(s) i(s)
Zout
Z(s) =
+ –
Chapter 8: Converter Transfer Functions
• Dynamics of blocking capacitor are irrelevant
• Actual device input and output voltages are measured as vx and vy
• Injection sinusoid coupled to device input via dc blocking capacitor
• Potentiometer establishes correct quiescent operating point
Measurement of an output impedance
vy vx
vy vx
output
Fundamentals of Power Electronics
DC bias adjust
VCC
G(s)
–
vy input +
Device under test
–
vx input
Measured inputs
Fundamentals of Power Electronics
DC bias adjust
VCC
DC blocking capacitor
vz frequency
vz output
vz magnitude
+ –
Network Analyzer
input
Injection source
Measurement of an ac transfer function
output
{
=0
75
Impedance under test i out
Fundamentals of Power Electronics
Z probe || Z rz
For an accurate measurement, require
76
Injection current can Z(s) i out k i out return to analyzer via two paths. Injection (1 – k) i out current which returns via voltage probe ground voltage induces voltage drop in probe voltage probe, corrupting the voltage probe measurement. Network return connection analyzer measures Z + (1 – k) Z probe = Z + Z probe || Z rz
Grounding problems cause measurement to fail:
Zprobe
Rsource
vz
vx
– vy
+
–
+
Measured inputs
+ –
Injection source
Network Analyzer
Chapter 8: Converter Transfer Functions
+ – (1 – k) i out Z probe
Zrz
injection source return connection
{
Z >>
Chapter 8: Converter Transfer Functions
Measurement of small impedances
Fundamentals of Power Electronics
• Network analyzer result must be multiplied by appropriate factor, to account for scale factors of current and voltage probes
• Current probe produces voltage proportional to current; this voltage is connected to network analyzer channel vx
• Potentiometer at device input port establishes correct quiescent operating point
ac input
• Treat output impedance as transfer function from output current to output voltage: vy(s) v(s) Z out(s) = Z(s) = i out(s) amplifier i(s)
Measurement of output impedance
{
Z(s)
voltage probe
0
i out
i out
+
Zprobe
77
0V
–
Zrz
injection source return connection Rsource
vz
vx
– vy
+
–
+
Measured inputs
+ –
Injection source
Network Analyzer
Chapter 8: Converter Transfer Functions
1:n
Fundamentals of Power Electronics
78
Chapter 8: Converter Transfer Functions
4. Poles and zeroes can be expressed in frequency-inverted form, when it is desirable to refer the gain to a high-frequency asymptote.
3. The right half-plane zero exhibits the magnitude response of the left half-plane zero, but the phase response of the pole.
2. It is good practice to express transfer functions in normalized polezero form; this form directly exposes expressions for the salient features of the response, i.e., the corner frequencies, reference gain, etc.
1. The magnitude Bode diagrams of functions which vary as (f / f0)n have slopes equal to 20n dB per decade, and pass through 0dB at f = f0.
8.5. Summary of key points
voltage probe return connection
Impedance under test
{
Fundamentals of Power Electronics
Injection current must now return entirely through transformer. No additional voltage is induced in voltage probe ground connection
Improved measurement: add isolation transformer
{
79
Chapter 8: Converter Transfer Functions
Fundamentals of Power Electronics
80
Chapter 8: Converter Transfer Functions
10. Measurement of transfer functions and impedances using a network analyzer is discussed in section 8.4. Careful attention to ground connections is important when measuring small impedances.
9. Approximate magnitude asymptotes of impedances and transfer functions can be easily derived by graphical construction. This approach is a useful supplement to conventional analysis, because it yields physical insight into the circuit behavior, and because it exposes suitable approximations. Several examples, including the impedances of basic series and parallel resonant circuits and the transfer function He(s) of the boost and buck-boost converters, are worked in section 8.3.
8. Salient features of the transfer functions of the buck, boost, and buckboost converters are tabulated in section 8.2.2. The line-to-output transfer functions of these converters contain two poles. Their controlto-output transfer functions contain two poles, and may additionally contain a right half-pland zero.
Summary of key points
Fundamentals of Power Electronics
7. The low- Q approximation can be extended to find approximate roots of an arbitrary degree polynomial. Approximate analytical expressions for the salient features can be derived. Numerical values are used to justify the approximations.
6. When the Q is less than 0.5, the two pole response can be plotted as two real poles. The low- Q approximation predicts that the two poles occur at frequencies f0 / Q and Qf0. These frequencies are within 10% of the exact values for Q ≤ 0.3.
5. A two-pole response can be written in the standard normalized form of Eq. (8-53). When Q > 0.5, the poles are complex conjugates. The magnitude response then exhibits peaking in the vicinity of the corner frequency, with an exact value of Q at f = f0. High Q also causes the phase to change sharply near the corner frequency.
Summary of key points
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Summary of Key Points
10.4.
Vg
Vg
+ –
+ –
?
Fundamentals of Power Electronics
DCM
CCM
1 : M(D)
DC
R
R
–
V
vg(s)
+ –
2
?
C
v(s)
R
R
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
–
v(s)
+
j(s) d(s)
+
+
+ –
Le
–
vg(s)
1 : M(D)
–
V
+
e(s) d(s)
AC
We are missing ac and dc equivalent circuit models for the discontinuous conduction mode
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Generalized Averaged Switch Modeling
10.3.
1
Small-Signal AC Modeling of the DCM Switch Network
10.2.
Fundamentals of Power Electronics
DCM Averaged Switch Model
10.1.
Introduction
Chapter 10 Ac and Dc Equivalent Circuit Modeling of the Discontinuous Conduction Mode
+ –
Steady-state output voltage becomes strongly load-dependent Simpler dynamics: one pole and the RHP zero are moved to very high frequency, and can normally be ignored Traditionally, boost and buck-boost converters are designed to operate in DCM at full load All converters may operate in DCM at light load
3
Fundamentals of Power Electronics
• Let us find the averaged quantities 〈 v1 〉, 〈 i1 〉 , 〈 v2 〉, 〈 i2 〉, for operation in DCM, and determine the relations between them
vg
+ –
4
–
iL
L
C R
–
v
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
+ vL –
+
v2
v1
i2 –
Switch network i1 +
+
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Derivation of DCM averaged switch model: buck-boost example
• Define switch terminal quantities v1, i1, v2, i2, as shown
10.1
Fundamentals of Power Electronics
So we need equivalent circuits that model the steady-state and smallsignal ac models of converters operating in DCM The averaged switch approach will be employed
●
●
●
●
Change in characteristics at the CCM/DCM boundary
– + vL – L iL
Peak inductor current:
Ts
C R
5
–
v
+
= d 1 vg(t)
Ts
Ts
+ d 2 v(t)
Fundamentals of Power Electronics
d 2(t) = – d 1(t)
Ts
Ts
+ d3 ⋅ 0
v(t)
Ts
Ts
6
+ d 2(t) v(t)
vg(t)
= d 1(t) vg(t) Solve for d2:
vL(t)
Ts
ipk
Ts
d1Ts
vg – v
i 2(t)
0
v1(t)
Area q1
Ts
Ts
Ts
d2Ts
0
v2(t)
Ts
Area q2
vg – v
i 1(t)
ipk
d3Ts
–v
vg
t
v2(t)
Ts
ipk
Ts
d1Ts
vg – v
i 2(t)
0
v1(t)
Area q1
Ts
Ts
d2Ts
0
v2(t)
Ts
Area q2
vg – v
i 1(t)
d3Ts
–v
vg
t
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
=0
i2(t)
v1(t)
i1(t) ipk
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
v2(t)
i2(t)
v1(t)
In DCM, the diode switches off when the inductor current reaches zero. Hence, i(0) = i(Ts) = 0, and the average inductor voltage is zero. This is true even during transients.
vL(t)
Average inductor voltage:
vg i pk = d 1Ts L
t
i1(t)
Basic DCM equations
+
v2
i2
v1
0
0
–
Fundamentals of Power Electronics
vg
+ –
v
v L
Switch network
vg
vg L
ipk
+
i1
vL(t)
iL(t)
DCM waveforms
Ts
= d 1(t) ⋅ 0 + d 2(t)
Ts
= vg(t) Ts
vg(t) Ts
– v(t) Ts
+ d 3(t) vg(t)
Ts
= – v(t)
= d 1(t) Ts
vg(t) Ts
– v(t) Ts
7
Ts
v2(t)
i2(t)
v1(t)
Ts
ipk
Ts
d1Ts
vg – v
i 2(t)
0
v1(t)
Area q1
Ts
Ts
d2Ts
0
v2(t)
Ts
Area q2
vg – v
i 1(t)
ipk
t
q i 1(t)dt = 1 Ts
t
Ts
=
d 21(t) Ts v1(t) 2L
Ts
i 1(t)dt = 1 d 1Ts i pk 2
Fundamentals of Power Electronics
Ts
8
t
v2(t)
i2(t)
v1(t)
i1(t)
Ts
ipk
Ts
d1Ts
vg – v
i 2(t)
0
v1(t)
Area q1
Ts
Ts
d2Ts
0
v2(t)
Ts
Area q2
vg – v
i 1(t)
ipk
d3Ts
–v
vg
t
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Similar analysis for i2(t) waveform leads to 2 d 21(t) Ts v1(t) T s i 2(t) T = s 2L v2(t)
Note 〈i1(t)〉Ts is not equal to d 〈iL(t)〉Ts !
i 1(t)
Eliminate ipk:
q1 =
t + Ts
The integral q1 is the area under the i1(t) waveform during first subinterval. Use triangle area formula:
i 1(t) T = 1 Ts s
t + Ts
Average the i1(t) waveform:
d3Ts
–v
vg
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
+ d 2(t) ⋅ 0 + d 3(t) – v(t)
Ts
i1(t)
Average switch network terminal currents
Fundamentals of Power Electronics
v2(t)
Similar analysis for v2(t) waveform leads to
v1(t)
Eliminate d2 and d3:
v1(t)
Average the v1(t) waveform:
Average switch network terminal voltages
Ts
=
Ts
Re(d 1)
v1(t)
Ts
9
Ts
Ts
Re(d1)
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
–
v1(t)
+
i 1(t)
Ts
v2(t)
Ts
=
Fundamentals of Power Electronics
i 2(t)
Ts
2
Ts
Ts
2
Re(d 1)
v1(t)
d 21(t) Ts v1(t) i 2(t) T = s 2L v2(t)
10
= p(t)
Ts
–
v(t)
+
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
p(t)
i(t)
Output port: Averaged equivalent circuit
Fundamentals of Power Electronics
Re(d 1) = 2L d 21 Ts
i 1(t)
d 2(t) Ts i 1(t) T = 1 v1(t) s 2L
Input port: Averaged equivalent circuit
–
v(t)
+
11
v(t)
v(t)i(t) = p(t)
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
i(t)
Fundamentals of Power Electronics
12
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
A power source (or power sink) element is obtained
If the instantaneous power depends only on the external elements connected to one port, then the power is not dependent on the characteristics of the elements connected to the other port. The other port becomes a source of power, equal to the power flowing through the first port
In all but a small number of special cases, the instantaneous power throughput is dependent on the applied external source and load
In a lossless two-port network without internal energy storage: instantaneous input power is equal to instantaneous output power
How the power source arises in lossless two-port networks
Fundamentals of Power Electronics
• Power sink: negative p(t)
• Must avoid open- and short-circuit connections of power sources
p(t)
i(t)
The dependent power source
P1
P2
P1
P3
13
n1 : n 2
Ts
Ts
Re(d1)
p(t)
Ts
Fundamentals of Power Electronics
14
–
v2(t)
+
i 2(t)
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Power entering input port is transferred to output port
Input port obeys Ohm’s Law
A two-port lossless network
–
v1(t)
+
i 1(t)
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
P1
P1 + P2 + P3
The loss-free resistor (LFR)
Fundamentals of Power Electronics
Reflection of power source through a transformer
Series and parallel connection of power sources
Properties of power sources
v2(t) –
–
i2(t) +
v1(t)
+
i1(t)
–
–
i 1(t)
Ts
Re(d1)
p(t)
Ts
1 : d(t)
–
v2(t)
+
i 2(t)
–
v2(t)
+
i 2(t)
Ts
Ts
Ts
Ts
Fundamentals of Power Electronics
Averaged model
Original circuit
vg(t)
Ts
vg
+ –
+ –
–
v1(t)
+
i 1(t)
Ts
Ts
–
16
Re(d)
L
Ts
iL
L
+
v2(t)
–
i 2(t)
C
Ts
Ts
R
C R
–
v(t)
+
–
v
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
p(t)
+ vL –
+
v2
v1
i2 –
Switch network +
i1
+
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Ts
Ts
Averaged switch model: buck-boost example
15
–
v1(t)
+
Ts
–
v1(t)
+
+ v2(t)
i 1(t)
Averaged switch model i2(t)
v1(t)
Fundamentals of Power Electronics
DCM
CCM
i1(t) +
Switch network
Averaged modeling of CCM and DCM switch networks
C → open circuit
L → short circuit
17
+ –
R Re
is a general result, for any system that can be modeled as an LFR.
Fundamentals of Power Electronics
18
–
V
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
which agrees with the previous steady-state solution of Chapter 5.
V =– Vg
R Re
R
+
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
V =± Vg
V 2g V 2 P= = Re R
D 2T s R =– D 2L K
Re(D) = 2L D 2Ts Eliminate Re:
P
Equate and solve:
Re(D)
For the buck-boost converter, we have
V =± Vg
Vg
I1
Steady-state LFR solution
Fundamentals of Power Electronics
2 V P= R
Converter output power:
V 2g P= Re
Converter input power:
Let
Solution of averaged model: steady state
V g,rms Re
Re
V rms R
p(t)
19
R –
v(t)
+
R Re Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Vrms = Vg,rms
Equate and solve:
Note that no average power flows into capacitor
C
i2(t)
v2(t) –
v1(t) –
Fundamentals of Power Electronics
+
i2(t) +
i1(t)
–
Ts
20
v1(t)
+
i 1(t)
Ts Ts
–
v2(t)
+
Ts
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Re(d1)
p(t)
i 2(t)
• Can simply replace transistor and diode with the averaged model as follows:
• In each case, averaged transistor waveforms obey Ohm’s law, while averaged diode waveforms behave as dependent power source
• Determine averaged terminal waveforms of switch network
Averaged models of other DCM converters
Fundamentals of Power Electronics
Pav =
Converter average output power: 2
Pav =
Converter average input power: 2
vg(t) + –
i1(t)
Steady-state LFR solution with ac terminal waveforms
vg(t)
vg(t)
Ts
Ts
+ –
+ –
vg(t)
vg(t)
Ts
Ts
+ –
+ –
Fundamentals of Power Electronics
SEPIC
Cuk
Re(d)
L1
Ts
21
p(t)
L
C
C
C1
p(t)
C1
22
L2
Ts
–
v(t)
+
–
v(t)
+
Ts
Ts
Re = 2L d 2T s
Ts
C2
C2
R
R
–
v(t)
+
–
v(t)
+
Ts
Ts
Re =
d 2Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
p(t)
L2
2 L 1||L 2
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
R
R
DCM Cuk, SEPIC
Re(d)
Ts
Re(d)
L1
L
p(t)
Re(d)
Fundamentals of Power Electronics
Boost
Buck
DCM buck, boost
Vg
Vg
+ –
+ –
23
Re(D)
P
P
R
R
–
V
+
–
V
+
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Re(D)
–D 1–D D 1–D
Buck-boost, Cuk SEPIC
Fundamentals of Power Electronics
24
1 1–D Boost
–
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
R Re
R Re
1 + 1 + 4R/Re 2
2 1 + 1 + 4Re/R
M, DCM
Vg I crit = 1 – D D Re(D)
D
Buck
I > I crit for CCM I < I crit for DCM
M, CCM
Converter
Table 10.1. CCM and DCM conversion ratios of basic converters
Steady-state solution of DCM/LFR models
Fundamentals of Power Electronics
Boost
Buck
C → open circuit
Let L → short circuit
Steady-state solution: DCM buck, boost
Ts
Ts
d(t)
Re(d)
p(t) Ts
Ts
=
d 21(t) Ts v1(t) 2L v2(t) Ts
Ts
Ts
25
Ts
Ts
Ts
Ts
Ts
Ts
= I 2 + i 2(t)
= V2 + v2(t)
= I 1 + i 1(t)
= V1 + v1(t)
=
Re(d(t))
Ts
= f1 v1(t)
, v2(t) Ts
, d(t) Ts
d f1 V1, v2, D dv2
v2 = V 2
+ d(t)
Fundamentals of Power Electronics
26
d f1 V1, V2, d dd d=D
v1 = V 1
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
+ higher–order nonlinear terms
+ v2(t)
d f1 v1, V2, D I 1 + i 1(t) = f1 V1, V2, D + v1(t) dv1
(for simple notation, drop angle brackets)
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
v i 1 = r 1 + j1d + g 1v2 1 v i 2 = – r 2 + j2d + g 2v1 2
i 2(t)
v2(t)
i 1(t)
v1(t)
d(t) = D + d(t)
Expand in three-dimensional Taylor series about the quiescent operating point:
Ts
v1(t)
Given the nonlinear equation
i 1(t)
–
v2(t)
+
i 2(t)
Linearization via Taylor series
Fundamentals of Power Electronics
i 2(t)
2
d 21(t) Ts i 1(t) T = v1(t) s 2L
–
v1(t)
+
i 1(t)
Perturb and linearize: let
Small-signal ac modeling of the DCM switch network
Large-signal averaged model
10.2
d f1 V1, v2, D dv2 v2 = V 2
=0
v1 = V 1
=
1 Re(D)
I 1 = f1 V1, V2, D =
DC
V1 Re(D)
Ts
2
, v2(t) Ts
, d(t) Ts
Fundamentals of Power Electronics
28
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Small-signal ac linearization
Ts
= f2 v1(t)
i 2(t) = v2(t) – r1 + v1(t)g 2 + d(t) j2 2
Re(d(t)) v2(t)
v1(t)
same approach
DC terms
Ts
=
of the discontinuous conduction mode
Output port
27
V 21 I 2 = f2 V1, V2, D = Re(D) V2
i 2(t)
Fundamentals of Power Electronics
j1 =
d f1 V1, V2, d V d Re(d) =– 2 1 dd R e (D) dd d = D d=D 2V1 = DRe(D) Chapter 10: Ac and dc equivalent circuit modeling
g1 =
1 = d f1 v1, V2, D r1 dv1
i 1(t) = v1(t) r1 + v2(t) g 1 + d(t) j1 1
AC
Equate dc and first-order ac terms
Ts
– r1 2
29
Quiescent operating point
Power source characteristic
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
v2(t)
Linearized model
1 R
Load characteristic
=1= 21 R M Re(D) v2 = V 2
j1d
g 1v2
g 2v1
j2d
1 (M – 1) 2 Re
0
Boost, Fig. 10.16(b) Buck-boost, Fig. 10.7(b)
Fundamentals of Power Electronics
1 Re
2V1 DRe
2MV1 D(M – 1)Re
2(1 – M)V1 DRe
j1
30
2V1 DMRe
2V1 D(M – 1)Re
2(1 – M)V1 DMRe
j2
M 2R e
r2
+
v2
M 2R e
(M – 1) 2Re
r2
–
i2
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
2M Re
2M – 1 (M – 1) 2 Re
(M – 1) 2 Re M
Re
2–M MRe
g2
Re
r1
Tabl e 1 0 . 2 . Small-signal DCM switch model parameters
Buck, Fig. 10.16(a)
r1
g1
+ –
Switch type
v1
i1
Small-signal DCM switch model parameters
Fundamentals of Power Electronics
i 2(t)
1 = – d f2 V1, v2, D r2 dv2
Output resistance parameter r2
+ –
–
g 1v2
L
31
iL
j2d
r2
+
C
R
v2(t) –
+ v1(t) –
+
–
v1(t)
i1(t)
–
g 1v2
32
g 2v1
j2d
Fundamentals of Power Electronics
–
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
–
v2 j1d
v1
r1
+
i2
+
i1
r2
v2(t)
+
i2(t)
In any event, a small-signal two-port model is used, of the form
i2(t) +
i1(t)
–
v
+
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
g 2v1
A more convenient way to model the buck and boost small-signal DCM switch networks
Fundamentals of Power Electronics
vg
j1d
v2
r1
v1
i2
–
Switch network small-signal ac model
+
i1
Small-signal ac model, DCM buck-boost example
+ –
+ –
iL
L
–
v1
–
v1
+
i1
r1
r1
j1d
j2d
j1d
g 1v2
33
j2d
r2
–
v2
–
v2
+
i2
L
C
C
iL
R
–
v
+
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
When expressed in terms of R, L, C, and M (not D), the smallsignal transfer functions are the same in DCM as in CCM Hence, DCM boost and buck-boost converters exhibit two poles and one RHP zero in control-to-output transfer functions But, value of L is small in DCM. Hence RHP zero appears at high frequency, usually greater than switching frequency Pole due to inductor dynamics appears at high frequency, near to or greater than switching frequency So DCM buck, boost, and buck-boost converters exhibit essentially a single-pole response A simple approximation: let L → 0 34
R
–
v
+
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
g 2v1
r2
+
i2
DCM small-signal transfer functions
Fundamentals of Power Electronics
●
●
●
●
g 2v1
DCM boost switch network small-signal ac model
g 1v2
DCM buck switch network small-signal ac model
Fundamentals of Power Electronics
vg
vg
+
i1
Small-signal ac models of the DCM buck and boost converters (more convenient forms)
r1
line-to-output
control-to-output
Transfer functions
+ –
Gvg(s) =
v vg
v d
g 1v2
Gvd (s) =
j1d
d=0
p
p
with
j2d
C
M M M
2V 1 – M D 2–M 2V M – 1 D 2M – 1
V D
Buck Boost Buck-boost
2 RC
2M – 1 (M– 1)RC
2–M (1 – M)RC
ωp
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Gg0 Gd0 Converter
36
–
v
Gg0 = g 2 R || r 2 = M
Tabl e 1 0 . 3 . Salient features of DCM converter small-signal transfer functions
Fundamentals of Power Electronics
R
Gd0 = j2 R || r 2 1 ωp = R || r 2 C
r2
+
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Gd0 1 + ωs Gg0 1 + ωs 35
=
vg = 0
=
g 2v1
Transfer function salient features
Fundamentals of Power Electronics
vg
DCM switch network small-signal ac model
Buck, boost, and buck-boost converter models all reduce to
The simple approximation L → 0
37
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
2L = ReTs
ωp 2M – 1 = = 2π 2π (M– 1)RC
Fundamentals of Power Electronics
fp =
(36 V) –1 (24 V)
38
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
= 112 Hz
= 72 V ⇒ 37 dBV
(36 V) 2π – 1 (12 Ω)(470 µF) (24 V)
2
(36 V) 2 –1 (24 V)
(36 V) –1 (24 V)
2(5 µH) = 0.25 (16 Ω)(10 µs) 2(36 V) Gd0 = 2V M – 1 = D 2M – 1 (0.25)
D=
V 2g (24 V) 2 Re = = = 16 Ω P 36 W
P = I V – Vg = 3 A 36 V – 24 V = 36 W
Evaluate simple model parameters
Fundamentals of Power Electronics
operating point where the load current is I = 3 A and the dc input voltage is V g = 24 V.
The output voltage is regulated to be V = 36 V. It is desired to determine Gvd(s) at the
f s = 100 kHz
C = 470 µF
L = 5 µH
R = 12 Ω
DCM boost example
∠ Gvd
0˚
|| Gvd ||
39
f
1 kHz
10 kHz
–270˚ 100 kHz
–180˚
–90˚
0˚
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
–20 dB/decade
∠ Gvd
Fundamentals of Power Electronics
40
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
to the transfer functions of the parent CCM converter, derived in Chapter 7. The models for these other modes, control schemes, and switch implementations are shown to be equivalent to the CCM models of Chapter 7, plus additional effective feedback loops that describe the switch behavior
operating in DCM, and/or with current programmed control, and/or with resonant switches, and/or with other control schemes or switch implementations,
An approach that directly relates the transfer functions of converters
10.3 Generalized Switch Averaging
100 Hz
112 Hz
fp
Gd0 ⇒ 37 dBV
10 Hz
–40 dBV
–20 dBV
0 dBV
20 dBV
40 dBV
60 dBV
Fundamentals of Power Electronics
|| Gvd ||
Control-to-output transfer function, boost example
Switch u (t) inputs s
ys(t) Switch outputs
dx(t) = A Fx(t) + B Fu(t) + B sy s(t) dt y(t) = C Fx(t) + E Fu(t) + E sy s(t) u s(t) = C sx(t) + E uu(t)
41
Control inputs
uc(t)
K
= Cs x
= CF x
= AF x
〈us(t)〉Ts
Ts
Ts
Ts
Ts
Ts
Ts
Ts
Ts
Ts
s
s
42
s
s
Averaged dependent signals 〈y(t)〉T
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Averaged control 〈uc(t)〉T s inputs
s
〈ys(t)〉T = f(〈us〉T , 〈uc〉T )
Ts
Ts
Averaged switch outputs
+ Es ys
+ Bs ys
〈ys(t)〉T
+ Eu u
+ EF u
+ BF u
Averaged switch network
u s(t)
dt y(t)
d x(t)
Converter dependent signals y(t)
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
ys(t) = f'(us(t), uc(t), t)
Switch network
n ports
Time-invariant network containing averaged converter states 〈x(t)〉Ts
Averaged switch inputs
Fundamentals of Power Electronics
s
Averaged independent inputs 〈u(t)〉T
K
Time-invariant network containing converter states x(t)
Averaged system equations
Fundamentals of Power Electronics
Converter independent inputs u(t)
Converter and switch network state equations
{
Ts
= f u s(t)
, u c(t)
Ts
Ts
= µ(t) y s1(t) + µ'(t) y s2(t)
µ’ = 1 – µ
= µ(t) y s1(t) + µ'(t) y s2(t)
Fundamentals of Power Electronics
A proof follows later. 44
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
The switch conversion ratio µ is a generalization of the CCM duty cycle d. In general, µ may depend on the switch independent inputs, that is, converter voltages and currents. So feedback may be built into the switch network.
Small-signal transfer functions are found by replacing d(t) with µ(t)
Steady-state relations are found by replacing D with µ0
then CCM equations can be used directly, simply by replacing the duty cycle d(t) with the switch conversion ratio µ(t):
Ts
If it is true that
y s(t)
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Switch conversion ratio µ
43
µ is called the switch conversion ratio
ys2(t) is the value of ys(t) in the CCM converter during subinterval 2
ys1(t) is the value of ys(t) in the CCM converter during subinterval 1
Ts
Fundamentals of Power Electronics
where
y s(t)
Now attempt to write the converter state equations in the same form used for CCM state-space averaging in Chapter 7. This can be done provided that the above equation can be manipulated into the form
y s(t)
Place switch network dependent outputs in vector ys(t), then average over one switching period, to obtain an equation of the form
Averaging the switch network dependent quantities
Ts
vg
+ –
+ –
Re(d)
p(t) Ts
45
–
Ts
Ts
L
L
C
i L(t)
C
iL
Ts
R
R
v2(t) i 1(t)
Fundamentals of Power Electronics
y s(t) =
Switch output vector
u c(t) = d(t)
Switch control input
v (t) u s(t) = 1 i 2(t)
Switch input vector
vg
+ –
46
i1
–
v1
+
–
v(t)
+
–
v
+
Ts
–
v2
+
i2
L
C
iL
R
–
v
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Switch network
+
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
v2(t)
+
i 2(t)
i2
Defining the switch network inputs and outputs
–
Ts
+
Ts
–
–
+ v2
+
Switch network
v1
v1(t)
i 1(t)
Fundamentals of Power Electronics
vg(t)
DCM large-signal averaged model
Original converter
i1
10.3.1 Buck converter example
v1
Ts
Ts
i 2(t)
dTs
0
0
Ts
i 1(t)
v2(t)
t
Ts
Ts
= µ(t)
y s1(t) =
i 2(t)
v1(t)
v1(t)
Ts
Ts
= µ(t)
=
Ts
Ts
i 2(t)
i 1(t)
i 2(t)
v1(t)
⇒
Ts
Ts
+ (1 – µ(t)) 0 0
= µ(t) y s1(t) + µ'(t) y s2(t)
v2(t)
Ts
Ts
Ts
+ (1 – µ(t)) 0 0
, y s2(t) = 0 0
⇒
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Ts
Ts
Fundamentals of Power Electronics
48
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
This is a general definition of µ, for the switch network as defined previously for the buck converter. It is valid not only in CCM, but also in DCM and ...
µ(t) =
i 1(t)
v2(t)
y s(t)
47
Solve for µ, in general
Fundamentals of Power Electronics
Ts
Ts
i 2(t)
v1(t)
CCM switch outputs during subintervals 1 and 2 are:
For CCM operation, this equation is satisfied with µ = d.
Hence, we should define the switch conversion ratio µ to satisfy
v2(t)
i1(t)
Switch output waveforms, CCM operation
v2(t)
=
Ts
Ts
–
Ts
Ts
+ –
v2(t)
Ts
µ=
i 1(t)
Ts
, i 2(t) Ts
v1(t)
Fundamentals of Power Electronics
µ v1(t)
Ts
49
Ts
v2(t)
Ts
,d = Ts
1
µ=
Ts
Ts
C
i L(t)
1
50
⇒
Ts
Ts
for DCM
v2(t)
i 1(t)
R
Ts
Ts
–
v(t)
+
Ts
Ts
v2(t)
i 1(t)
Ts
Ts
=
v1(t)
i 2(t)
Ts
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
v1(t)
i 2(t)
v2(t)
Ts
1
Ts
1 + Re(d)
L
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Ts
–
v2(t)
+
i 2(t)
⇒
Re(d)
i 1(t)
1 + Re(d)
v2(t)
1 + Re(d)
= i 2(t)
Lossless switch network:
we found that
Hence
Ts
i 1(t)
p(t)
Re(d)
Elimination of dependent quantities
Ts
1– =µ
Ts
Re(d)
–
Ts
+
Ts
v1(t)
i 1(t)
Re(d)
– i 1(t)
i 2(t)
i 1(t)
Ts
Ts
i 1(t)
vg(t)
Fundamentals of Power Electronics
1=
Ts
Ts
Ts
= v1(t)
v1(t)
v2(t)
v1(t)
v2(t)
µ(t) =
Solve averaged model for µ
Evaluation of µ
Ts
, i 2(t)
Ts
,d =
Ts
1 + Re(d)
1 v1(t)
i 2(t) Ts
Ts
51
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
1 1 + Re(D)
Fundamentals of Power Electronics
µ0 =
Buck example:
µ 0 = µ(U s, U c, D)
Steady-state components:
Ts
= U s + u s(t)
I2 V1
52
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
u c(t) = U c + u c(t)
u s(t)
µ(t) = µ 0 + µ(t)
Perturbation and linearization
Fundamentals of Power Electronics
• In DCM, switch conversion ratio is a function of not only the transistor duty cycle d, but also the switch independent terminal waveforms i2 and v1. The switch network contains built-in feedback.
• Replace d of CCM expression with µ to obtain a valid DCM expression
• A general result for DCM
µ v1(t)
DCM switch conversion ratio µ
µ0 =
53
1 + Re(D)
1 I2 V1 Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Can now solve for V to obtain the usual DCM expression for V/Vg.
v1(t) i 2(t) – + k s d(t) Vs Is
Fundamentals of Power Electronics
54
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
µ 20Re(D) 1 = – dµ V1, i 2, D = V1 di 2 Is i2 = I 2 dµ V1, I 2, d 2µ 20I 2Re(D) ks = = DV1 dd d=D
µ 20I 2Re(D) 1 = dµ v1, I 2, D = dv1 Vs V 21 v1 = V 1
The gains are found by evaluation of derivatives at the quiescent operating point
µ(t) =
Express linearized conversion ratio as a function of switch control input and independent terminal inputs:
DCM buck example: small-signal equations
Fundamentals of Power Electronics
with
V = M(µ ) = µ 0 0 Vg I2 = V R
DCM: replace D with µ0:
V = M(D) = D Vg IL = V R
In CCM, we know that
Buck example: steady-state solution
+ –
v1
–
v1
v(s) d(s)
vg(s) = 0
Fundamentals of Power Electronics
ks
µ
Ti(s) Gvd (s) = Gvd∞(s) 1 + Ti(s)
Gvd (s) =
– +
d
µ
d
ks
i2
L
C
i(t)
55
R
1 Is
Ti(s)
Vg µ
i2
56
Vg I sZ ei(s)
+ –
C
R
–
v(t)
+
1 + s L + s 2LC R Z ei(s) = R 1 + sRC
i(t)
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Zei(s)
L
Gvd∞(s) = k sI s R || 1 sC
Ti(s) =
i2
–
v(t)
+
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Small-signal switch network block diagram Eq. (10.72)
1 Is
Vg µ
i2
Control-to-output transfer function
1 Vs
Iµ
– ++
Fundamentals of Power Electronics
vg(t)
+
1 : µ0
CCM buck small-signal model
Result: small-signal model of DCM buck
+ –
0 dB
T0 fz
f0
57
µ0 D
fc
2
2
fs π
Vg 2π fRC I sR 2π f 2LC
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
fc =
f f0
=
Fundamentals of Power Electronics
Converter independent inputs u(t)
K
Switch u (t) inputs s
ys(t) Switch outputs
dx(t) = A Fx(t) + B Fu(t) + B sy s(t) dt y(t) = C Fx(t) + E Fu(t) + E sy s(t) u s(t) = C sx(t) + E uu(t)
Time-invariant network containing converter states x(t)
58
Control inputs
uc(t) Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
ys(t) = f '(us(t), uc(t), t)
Switch network
n ports
Converter dependent signals y(t)
10.3.2 Proof of Generalized Averaged Switch Modeling
Fundamentals of Power Electronics
Ti 1 + Ti
|| Ti ||
T0
f fz
Magnitude of the loop gain Ti(s)
{
K
y s(t)
u s(t)
dt y(t)
d x(t)
Ts
Ts
Ts
Ts
= f u s(t)
= C s x(t)
= C F x(t)
= A F x(t) + E F u(t)
+ B F u(t)
, u c(t)
59
Values of ys(t) during subintervals 1 and 2 are defined as ys1(t) and ys2(t)
Ts
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
= µ(t) y s1(t) + µ'(t) y s2(t)
+ E s y s(t)
+ B s y s(t)
dt y(t)
d x(t)
Ts
Ts
= C F x(t)
= A F x(t)
Ts
Ts
+ E F u(t)
+ B F u(t)
Ts
Ts
+ µE sy s1(t) + µ'E sy s2(t)
+ µB sy s1(t) + µ'B sy s2(t)
dx(t) = A 1x(t) + B 1u(t) dt y(t) = C 1x(t) + E 1u(t)
with similar expressions for subinterval 2
60
dx(t) = A Fx(t) + B Fu(t) + B sy s1(t) dt y(t) = C Fx(t) + E Fu(t) + E sy s1(t) Fundamentals of Power Electronics
K
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Now equate the two expressions
But the time-invariant network equations predict that the converter state equations for the first subinterval are
K
It is desired to relate this to the result of the state-space averaging method, in which the converter state equations for subinterval 1 are written as
K
Ts
Ts
Ts
Ts
System averaged state equations
Fundamentals of Power Electronics
y s(t)
Ts
+ E u u(t)
Ts
Ts
Ts
Ts
y s(t) = f ' u s(t), u c(t), t
dx(t) = A Fx(t) + B Fu(t) + B sy s(t) dt y(t) = C Fx(t) + E Fu(t) + E sy s(t) u s(t) = C sx(t) + E uu(t)
Also suppose that we can write
Average:
K
System state equations
dx(t) = A 1x(t) + B 1u(t) = A Fx(t) + B Fu(t) + B sy s1(t) dt y(t) = C 1x(t) + E 1u(t) = C Fx(t) + E Fu(t) + E sy s1(t)
y(t)
dt
d x(t)
Ts
Ts
Fundamentals of Power Electronics
K
61
Ts
Ts
Ts
62
+ µ' C 2 – C F x(t)
Ts
Ts
Ts
+ B 2 – B F u(t)
+ B 1 – B F u(t)
Ts
Ts
Ts
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
+ E 2 – E F u(t)
+ E 1 – E F u(t)
+ E F u(t) + µ C 1 – C F x(t)
= C F x(t)
+ µ' A 2 – A F x(t)
Ts
+ B F u(t) + µ A 1 – A F x(t)
= A F x(t)
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Averaged state equations
Fundamentals of Power Electronics
Now plug these results back into averaged state equations
E sy s2(t) = C 2 – C F x(t) + E 2 – E F u(t)
B sy s2(t) = A 2 – A F x(t) + B 2 – B F u(t)
Result for subinterval 2:
E sy s1(t) = C 1 – C F x(t) + E 1 – E F u(t)
B sy s1(t) = A 1 – A F x(t) + B 1 – B F u(t)
Solve for Bsys1 and Esys1 :
K
Equate the state equation expressions derived via the two methods
Ts
Ts
+ µE 1 + µ'E 2 u(t)
+ µB 1 + µ'B 2 u(t)
Ts
Ts
63
Ts
Ts
Ts
= Y + y(t)
= U + u(t)
= X + x(t)
u c(t)
Ts
Ts
64
= U c + u c(t)
= U s + u s(t)
µ(t) = µ 0 + µ(t)
y(t)
u(t)
x(t)
= µC 1 + µ'C 2 x(t)
u s(t)
Ts
= µA 1 + µ'A 2 x(t)
Ts
Ts
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
+ µE 1 + µ'E 2 u(t)
+ µB 1 + µ'B 2 u(t)
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Perturb and linearize Ts
Fundamentals of Power Electronics
dt y(t)
d x(t)
Let
K
Fundamentals of Power Electronics
Hence, we can use any result derived via state-space averaging, by simply replacing d with µ.
= µC 1 + µ'C 2 x(t)
= µA 1 + µ'A 2 x(t)
•
Ts
Ts
This is the desired result. It is identical to the large-signal result of the state-space averaging method, except that the duty cycle d has been replaced with the conversion ratio µ.
dt y(t)
d x(t)
•
K
Collect terms
E = µ 0E 1 + µ 0'E 2
C = µ 0C 1 + µ 0'C 2
B = µ 0B 1 + µ 0'B 2
A = µ 0A 1 + µ 0'A 2
where
Result
k =
y(t) = Cx(t) + Eu(t) +
+ –
j(s)µ(s)
µ
e(s)µ(s)
T
u s(t)
u c(t)
u s(t)
u c(t)
Ts
Ts
Ts
Ts
= Uc
= Us
= Uc
= Us
R
–
v(s)
Small-signal switch network block diagram Eq. (10.106)
u s(s) Switch inputs: v(s), vg(s), i(s), etc.
C
+
u c(s) Control input(s): d(s), etc. Chapter 10: Ac and dc equivalent circuit modeling 66 of the discontinuous conduction mode
kc
T
ks
Ts
Ts
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
CCM small-signal canonical model Le 1 : M(µ0)
++
Fundamentals of Power Electronics
vg(s)
65
Ts
, u c(t) Ts
d u c(t)
dµ u s(t)
Ts
, u c(t)
Ts
d u s(t)
dµ u s(t)
A generalized canonical model
Fundamentals of Power Electronics
µ(t) = k Ts u s(t) + k Tc u c(t)
C 1 – C 2 X + E 1 – E 2 U µ(t)
T c
k Ts =
dx(t) = Ax(t) + Bu(t) + A 1 – A 2 X + B 1 – B 2 U µ(t) dt
with the linearized switch gains
K
Small-signal ac model
0 = AX + BU Y = CX + EU
DC model
+ –
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Key points
67
Fundamentals of Power Electronics
68
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
A small-signal ac model for the DCM switch network can be derived by perturbing and linearizing the loss-free resistor network. The result has the form of a two-port y-parameter model. The model describes the small-signal variations in the transistor and diode currents, as functions of variations in the duty cycle and in the transistor and diode ac voltage variations. This model is most convenient for ac analysis of the buck-boost converter. 5. To simplify the ac analysis of the DCM buck and boost converters, it is convenient to define two other forms of the small-signal switch model, corresponding to the switch networks of Figs. 10.16(a) and 10.16(b). These models are also y-parameter twoport models, but have different parameter values.
4.
Fundamentals of Power Electronics
1. In the discontinuous conduction mode, the average transistor voltage and current are proportional, and hence obey Ohm’s law. An averaged equivalent circuit can be obtained by replacing the transistor with an effective resistor Re(d). The average diode voltage and current obey a power source characteristic, with power equal to the power effectively dissipated by R e. In the averaged equivalent circuit, the diode is replaced with a dependent power source. 2. The two-port lossless network consisting of an effective resistor and power source, which results from averaging the transistor and diode waveforms of DCM converters, is called a loss-free resistor. This network models the basic power-processing functions of DCM converters, much in the same way that the ideal dc transformer models the basic functions of CCM converters. 3. The large-signal averaged model can be solved under equilibrium conditions to determine the quiescent values of the converter currents and voltages. Average power arguments can often be used.
10.4 Summary of Key Points
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
Key points
69
Fundamentals of Power Electronics
70
Chapter 10: Ac and dc equivalent circuit modeling of the discontinuous conduction mode
8. The conversion ratio µ of DCM switch networks is a function of the applied voltage and current. As a result, the switch network contains effective feedback. So the small-signal model of a DCM converter can be expressed as the CCM converter model, plus effective feedback representing the behavior of the DCM switch network. Two effects of this feedback are increase of the converter output impedance via current feedback, and decrease of the Q-factor of the transfer function poles. The pole arising from the inductor dynamics occurs at the crossover frequency of the effective current feedback loop.
Fundamentals of Power Electronics
6. Since the inductor value is small when the converter operates in the discontinuous conduction mode, the inductor dynamics of the DCM buck, boost, and buck-boost converters occur at high frequency, above or just below the switching frequency. Hence, in most cases the inductor dynamics can be ignored. In the smallsignal ac model, the inductance L is set to zero, and the remaining model is solved relatively easily for the low-frequency converter dynamics. The DCM buck, boost, and buck-boost converters exhibit transfer functions containing a single lowfrequency dominant pole. 7. It is also possible to adapt the CCM models developed in Chapter 7 to treat converters with switches that operate in DCM, as well as other switches discussed in later chapters. The switch conversion ratio µ is a generalization of the duty cycle d of CCM switch networks; this quantity can be substituted in place of d in any CCM model. The result is a model that is valid for DCM operation. Hence, existing CCM models can be adapted directly.
Key points
Control input
is(t) Rf
ic(t)Rf
is(t)Rf
Measure switch current
0
L
Ts
C
iL(t)
Latch
R
S Q
vref
v(t)
–
v(t)
+ R
Conventional output voltage controller
Compensator
Current-programmed controller
Analog comparator
+ –
Clock
D1
1
0 on
Comparator turns transistor off
dTs off
Switch current is(t)
Ts
Fundamentals of Power Electronics
2
A disadvantage: susceptibility to noise
Chapter 11: Current Programmed Control
• Transformer saturation problems in bridge or push-pull converters can be mitigated
• Transistor failures due to excessive current can be prevented simply by limiting ic(t)
• It is always necessary to sense the transistor current, to protect against overcurrent failures. We may as well use the information during normal operation, to obtain better control
• Simple robust output voltage control, with large phase margin, can be obtained without use of compensator lead networks
• Simpler dynamics —inductor pole is moved to high frequency
Advantages of current programmed control:
t
Chapter 11: Current Programmed Control
Clock turns transistor on
Transistor status:
m1
Control signal ic(t)
The peak transistor current replaces the duty cycle as the converter control input.
Current programmed control vs. duty cycle control
Fundamentals of Power Electronics
vg(t) + –
Q1
is(t)
Buck converter
Chapter 11 Current Programmed Control
–+
Chapter 11: Current Programmed Control
4
Describe artificial ramp scheme
•
Fundamentals of Power Electronics
Stability analysis
•
Chapter 11: Current Programmed Control
Controller can be stabilized by addition of an artificial ramp
•
Objectives of this section:
The current programmed controller is inherently unstable for D > 0.5, regardless of the converter topology
•
11.1 Oscillation for D > 0.5
3
Summary
11.5
Fundamentals of Power Electronics
Discontinuous conduction mode
CPM buck converter example
Current programmed controller model: block diagram
A more accurate model
Averaged switch modeling
11.4
11.3
A simple first-order model
11.2 Simple model via algebraic approach
Oscillation for D > 0.5
11.1
Chapter 11: Outline
0
m1
dTs
– m2
Ts
5
iL(Ts)
t
Fundamentals of Power Electronics
M2 D = M 1 D'
0 = M 1DTs – M 2D'Ts
In steady state:
Second interval: i L(Ts) = i L(dTs) – m 2d'Ts = i L(0) + m 1dTs – m 2d'Ts
Solve for d: i – i (0) d= c L m 1T s
i L(dTs) = i c = i L(0) + m 1dTs
First interval:
iL(0)
ic
iL(t)
0
6
m1
dTs
Ts
iL(Ts)
Chapter 11: Current Programmed Control
– m2
t
Chapter 11: Current Programmed Control
buck converter vg – v m1 = – m2 = – v L L boost converter vg vg – v m1 = – m2 = L L buck–boost converter vg m1 = – m2 = v L L
Steady-state inductor current waveform, CPM
Fundamentals of Power Electronics
iL(0)
ic
iL(t)
Inductor current slopes m1 and –m2
Inductor current waveform, CCM
0
m1
dT s
D + d T s DTs
m1
i L(0)
7
– m2
i L(T s)
ic m1
m1
i L(0)
Fundamentals of Power Electronics
i L(Ts) = i L(0) –
m2 m1
i L(Ts) = m 2 dTs
i L(0) = – m 1 dTs
magnified view
8
dT s
t
Perturbed waveform
Chapter 11: Current Programmed Control
Ts
– m2
Steady-state waveform
– m2
Perturbed waveform
Steady-state waveform
Chapter 11: Current Programmed Control
i L(Ts) = i L(0) – D D'
– m2
i L(T s)
Change in inductor current perturbation over one switching period
Fundamentals of Power Electronics
I L0 + i L(0) IL0
ic
iL(t)
Perturbed inductor current waveform
For stability:
i L(nTs) →
0
i L(0)
Ts
– 1.5i L(0)
Fundamentals of Power Electronics
IL0
ic
∞
0
D < 0.5 9
when – D < 1 D' when – D > 1 D'
n
10
2Ts
2.25i L(0)
t
Chapter 11: Current Programmed Control
3Ts
– 3.375i L(0)
α = – D = – 0.6 = – 1.5 D' 0.4
4Ts
Chapter 11: Current Programmed Control
Example: unstable operation for D = 0.6
Fundamentals of Power Electronics
iL(t)
2
i L(nTs) = i L((n – 1)Ts) – D = i L(0) – D D' D'
i L(2Ts) = i L(Ts) – D = i L(0) – D D' D'
i L(Ts) = i L(0) – D D'
Change in inductor current perturbation over many switching periods
0
i L(0)
Ts
– 1 i L(0) 2
11
2Ts
1 i (0) 4 L
4Ts
t
1 i (0) 16 L
Clock
D1
L
0
Ts
C
iL(t)
Latch
R
S Q
12
Current-programmed controller
Analog comparator
+ –
Artificial ramp
ia(t)Rf
ma
Q1
Fundamentals of Power Electronics
Control input
ic(t)Rf
is(t)Rf
Measure is(t) switch Rf current
vg(t) + –
is(t)
Buck converter
–
v(t)
+ R
0
ma
Ts
2Ts
i L(dTs) = i c – i a(dTs)
Chapter 11: Current Programmed Control
or,
i a(dTs) + i L(dTs) = i c
Now, transistor switches off when
ia(t)
t
Chapter 11: Current Programmed Control
3Ts
– 1 i L(0) 8
Stabilization via addition of an artificial ramp to the measured switch current waveform
Fundamentals of Power Electronics
IL0
ic
iL(t)
α = – D = – 1/3 = – 0.5 D' 2/3
Example: stable operation for D = 1/3
++
0
iL(t) m1
dTs
13
– m2
– ma
t
Chapter 11: Current Programmed Control
Ts
0
dT s
– ma
14
– m2
s)
iL (T
D + d T s DTs
) i L(0 m 1
m1
Fundamentals of Power Electronics
I L0 + i L(0) IL0
ic
(ic – ia(t))
Ts
t
Perturbed waveform
Chapter 11: Current Programmed Control
– m2
Steady-state waveform
Stability analysis: perturbed waveform
Fundamentals of Power Electronics
IL0
ic
(ic – ia(t))
i L(dTs) = i c – i a(dTs)
Steady state waveforms with artificial ramp
15
i L(nTs) →
when α > 1
∞
Buck and buck-boost converters: m2 = – v/L
•
16
Deadbeat control, finite settling time
This leads to α = 0 for 0 ≤ D < 1.
Another common choice: ma = m2
Chapter 11: Current Programmed Control
The minimum α that leads to stability for all D.
This leads to α = –1 at D = 1, and | α | < 1 for 0 ≤ D < 1.
Fundamentals of Power Electronics
•
So if v is well-regulated, then m2 is also well-regulated
For stability, require | α | < 1
A common choice: ma = 0.5 m2
= i L(0) α n
m 1 – ma 2 α=– m D' + a D m2 •
•
n
Chapter 11: Current Programmed Control
when α < 1 0
The characteristic value α
Fundamentals of Power Electronics
m –m α = – m 2 + ma 1 a
Characteristic value:
m –m m –m i L(nTs) = i L((n –1)Ts) – m 2 + ma = i L(0) – m 2 + ma 1 a 1 a
After n switching periods:
m –m i L(Ts) = i L(0) – m 2 + ma 1 a
Net change over one switching period:
i L(Ts) = – dTs m a – m 2
Second subinterval:
i L(0) = – dTs m 1 + m a
First subinterval:
Stability analysis: change in perturbation over complete switching periods
ic(t)
Current programmed controller
d(t)
17
–
v(t)
vref
Ts
= i c(t)
Fundamentals of Power Electronics
18
i L(s) ≈ i c(s)
• Resulting small-signal relation:
Chapter 11: Current Programmed Control
• Accurate when converter operates well into CCM (so that switching ripple is small) and when the magnitude of the artificial ramp is not too large
• Yields physical insight and simple first-order model
• Neglects switching ripple and artificial ramp
i L(t)
v(t)
Chapter 11: Current Programmed Control
Compensator
Converter voltages and currents
R
+
The first-order approximation
Fundamentals of Power Electronics
vg(t) + –
Switching converter
11.2 A Simple First-Order Model
+–
0
+ –
vg L
dTs
L
19
iL(t)
v L
D1
C
Ts
–
v(t)
t Chapter 11: Current Programmed Control
R
+
Fundamentals of Power Electronics
20
i g(t) = Di L + I L d(t) Derived in Chapter 7
L
Chapter 11: Current Programmed Control
d i L(t) = Dvg(t) + D'v(t) + Vg – V d(t) dt dv(t) v(t) = – D'i L – + I L d(t) C R dt
Small-signal equations of CCM buck–boost, duty cycle control
Fundamentals of Power Electronics
ic
iL(t)
vg(t)
Q1
11.2.1 Simple model via algebraic approach: CCM buck-boost example
Chapter 11: Current Programmed Control
i L(s) ≈ i c(s)
The simple approximation
21
22
Vg – V Chapter 11: Current Programmed Control
sLi c(s) – Dvg(s) – D'v(s)
Fundamentals of Power Electronics
d(s) =
Solve for the duty cycle variations:
sLi c(s) ≈ Dvg(s) + D'v(s) + Vg – V d(s)
Eliminate the duty cycle (now an intermediate variable), to express the equations using the new control input iL. The inductor equation becomes:
Now let
Fundamentals of Power Electronics
i g(s) = Di L(s) + I L d(s)
sLi L(s) = Dvg(s) + D'v(s) + Vg – V d(s) v(s) + I L d(s) sCv(s) = – D'i L(s) – R
Take Laplace transform, letting initial conditions be zero:
Transformed equations
Vg – V
sLi c(s) – Dvg(s) – D'v(s)
ig
– D'R D2
24
D 1 + sL i c D'R
D2 v D'R g
Chapter 11: Current Programmed Control
Dv R
2 i g(s) = sLD + D i c(s) – D v(s) – D vg(s) R D'R D'R
Fundamentals of Power Electronics
+ –
Chapter 11: Current Programmed Control
Construct equivalent circuit: input port
23
2 D sLD D i g(s) = + D i c(s) – v(s) – v (s) R D'R D'R g
2 sCv(s) = sLD – D' i c(s) – D + 1 v(s) – D vg(s) R R D'R D'R
Fundamentals of Power Electronics
vg
g
sLi c(s) – Dvg(s) – D'v(s) v(s) + IL R V –V
Collect terms, simplify using steady-state relations:
i g(s) = Di c(s) + I L
sCv(s) = – D'i c(s) –
Substitute this expression to eliminate the duty cycle from the remaining equations:
The simple approximation, continued
D' 1 – sLD ic D' 2R
+ –
ig
r1
f1(s) i c
Fundamentals of Power Electronics
vg
25
R D
C
R
v R
g1 v
26
g 2 vg
r2
C
R
–
v
Chapter 11: Current Programmed Control
f2(s) i c
+
Chapter 11: Current Programmed Control
sCv
CPM Canonical Model, Simple Approximation
Fundamentals of Power Electronics
D2 v D'R g
Dv R
Node
2 sCv(s) = sLD – D' i c(s) – D + 1 v(s) – D vg(s) R R D'R D'R
Construct equivalent circuit: output port
D 1 + sL R
1 D 1 + sL D'R
D R
0 – D R
Buck
Boost
Buck-boost
+ –
– D' 1 – sDL D' 2R
R D
R
∞
r2
28
vg = 0
r2
C
R
–
v
Chapter 11: Current Programmed Control
1 + s RC 1+D
1 – s DL D' 2R
= f2 r 2 || R || 1 sC
f2(s) i c
+
Chapter 11: Current Programmed Control
D' 1+D
v(s) i c(s)
g 2 vg
Gvc(s) = – R
g1 v
Result for buck-boost example
f1(s) i c
Gvc(s) =
r1
Control-to-output transfer function
ig
Fundamentals of Power Electronics
vg
27
2 – D D'R
D' 1 – sL2 D' R
1 D'R
∞ – D'R D2
1
0
– R2 D
f2
g2
r1
Transfer functions predicted by simple model
Fundamentals of Power Electronics
f1
g1
Current programmed mode small-signal equivalent circuit parameters, simple model
Converter
Tabl e 11 . 1
Table of results for basic converters
+ – f1(s) i c
Result for buck-boost example
Line-to-output transfer function
r1
+ –
r1
f1(s) i c
Result for buck-boost example
Output impedance
ig
Fundamentals of Power Electronics
vg
g1 v
v(s) vg(s)
ic
29
r2
C
R
g1 v
f2(s) i c
r2
30
Z out(s) =
R
1 + s RC 1+D
1
C
–
v
+
Chapter 11: Current Programmed Control
R 1+D
Z out(s) = r 2 || R || 1 sC
g 2 vg
–
v
+
Chapter 11: Current Programmed Control
1 + s RC 1+D
1
= g 2 r 2 || R || 1 sC =0
f2(s) i c
2 D Gvg(s) = – 1 – D2
Gvg(s) =
g 2 vg
Transfer functions predicted by simple model
Fundamentals of Power Electronics
vg
ig
Transfer functions predicted by simple model
Ts
Ts
Ts
Ts
= d(t) i 2(t)
= d(t) v1(t)
31
Ts
R
= i c(t)
Ts
=
v2(t)
v1(t)
Ts
Ts
≈ i c(t)
Fundamentals of Power Electronics
32
Ts
Ts
Chapter 11: Current Programmed Control
Ts
Ts
≈ i c(t)
i c(t)
Ts
= p(t)
Ts
Ts
Input port is a dependent current sink
Ts
Ts
•
v1(t)
= d(t) i c(t)
v2(t)
i 2(t)
Output port is a current source
Ts
Ts
Ts
Ts
–
v(t)
+
Chapter 11: Current Programmed Control
i 2(t)
•
So:
i 1(t)
i 1(t)
Eliminate duty cycle:
i 1(t)
v2(t)
= d(t) i 2(t)
= d(t) v1(t)
C
iL(t)
The simple approximation:
L
CPM averaged switch equations
Fundamentals of Power Electronics
i 1(t)
v2(t) Ts
–
– Switch network
v2(t)
+
i2(t)
v1(t)
+
Averaged terminal waveforms, CCM:
vg(t) + –
i1(t)
with the simple approximation
11.2.2 Averaged switch modeling
–
〈v1(t)〉T s
〈vg(t)〉T
s
Fundamentals of Power Electronics
Buck-boost
Boost
s
〈ic(t)〉Ts
33
Averaged switch network
〈 p(t)〉T
–
〈v2(t)〉Ts
+
L
〈vg(t)〉Ts
+ –
s
+ –
〈iL(t)〉T
L
s
〈 p(t)〉Ts
34
〈 p(t)〉Ts
L
R
–
〈v(t)〉Ts
+
C
C
R
–
〈v(t)〉Ts
+
R
–
〈v(t)〉T
+
s
Chapter 11: Current Programmed Control
〈iL(t)〉Ts
〈ic(t)〉Ts
Averaged switch network
Averaged switch network
〈ic(t)〉T
C
〈iL(t)〉Ts
Chapter 11: Current Programmed Control
〈i2(t)〉Ts
Results for other converters
Fundamentals of Power Electronics
〈vg(t)〉Ts
+ –
s
+
〈i1(t)〉T
CPM averaged switch model
i c(t)
i 2(t)
v2(t)
i 1(t)
v1(t)
Ts
Ts
Ts
Ts
Ts
= I c + i c(t)
= I 2 + i 2(t)
= V2 + v2(t)
= I 1 + i 1(t)
= V1 + v1(t)
+ –
–
v1
+
35
î2 = î c
V2 V1
–
V1 I1
v2
Ic V1
ic
i 1(t) = i c(t)
–
v2
i2 +
36
C
R
–
v
+
Chapter 11: Current Programmed Control
L
Chapter 11: Current Programmed Control
V2 I I + v2(t) c – v1(t) 1 V1 V1 V1
Switch network small-signal ac model
ic
Fundamentals of Power Electronics
vg
i1
V2 I I + v2(t) c – v1(t) 1 V1 V1 V1
Output port equation:
i 1(t) = i c(t)
Small-signal result:
V1 + v1(t) I 1 + i 1(t) = I c + i c(t) V2 + v2(t)
Resulting input port equation:
Resulting small-signal model Buck example
Fundamentals of Power Electronics
Let
Perturbation and linearization to construct small-signal model
+ –
ig
37
2 –D R
Dv R
ic
L
C
iL
38
–
v
Chapter 11: Current Programmed Control
R
+
Chapter 11: Current Programmed Control
〈v1(t)〉Ts
1 = – I1 r1 V1
2 i 1(s) = D 1 + s L i c(s) + D v(s) – D vg(s) R R R
i c D 1 + sL R
Fundamentals of Power Electronics
vg
V1
Quiescent operating point
Power source characteristic 〈v1(t)〉Ts 〈i1(t)〉Ts = 〈 p(t)〉Ts
Expressing the equivalent circuit in terms of the converter input and output voltages
Fundamentals of Power Electronics
I1
s
〈i1(t)〉T
Origin of input port negative incremental resistance
+ –
v(s) i c(s)
v(s) vg(s)
Gvg(s) =
=0
R
40
–
v
+
Chapter 11: Current Programmed Control
C
iL
Chapter 11: Current Programmed Control
The simple models of the previous section yield insight into the lowfrequency behavior of CPM converters Unfortunately, they do not always predict everything that we need to know: Line-to-output transfer function of the buck converter Dynamics at frequencies approaching fs More accurate model accounts for nonideal operation of current mode controller built-in feedback loop Converter duty-cycle-controlled model, plus block diagram that accurately models equations of current mode controller
Fundamentals of Power Electronics
●
●
●
●
ic
L
11.3 A More Accurate Model
39
Dv R
= R || 1 sC
ic = 0
vg = 0
2 –D R
Gvc(s) =
i c D 1 + sL R
Fundamentals of Power Electronics
vg
ig
Predicted transfer functions of the CPM buck converter
i L(t)
Ts
Ts
dTs
= i c(t)
= I c + i c(t)
= I L + i L(t) d(t) = D + d(t) m 1(t) = M 1 + m 1(t) m 2(t) = M 2 + m 2(t)
i c(t)
i L(t)
s
– m a dTs – d
s
s
– m2
Ts
iL(t)
ic
Fundamentals of Power Electronics
42
Chapter 11: Current Programmed Control
Chapter 11: Current Programmed Control
buck converter vg – v m1 = m2 = v L L boost converter vg v – vg m1 = m2 = L L buck-boost converter vg m1 = m2 = – v L L
Note that it is necessary to perturb the slopes, since these depend on the applied inductor voltage. For basic converters,
Perturb
41
t
(ic – ia(t))
m 1 dTs m 2 d'Ts – d' Ts Ts 2 2 2 2 d Ts d' Ts = i c(t) T – m a dTs – m 1 – m2 s 2 2
s
It is assumed that the artificial ramp slope does not vary.
Let
〈iL(t)〉d'T
m 2d'T s 2
〈iL(t)〉T = d〈iL(t)〉dT + d'〈iL(t)〉d'T
s
madTs
〈iL(t)〉dT
Fundamentals of Power Electronics
0
m1
– ma
m 1dT s 2
11.3.1Current programmed controller model
2
Ts 2
2
Ts 2
43
Chapter 11: Current Programmed Control
D' 2T s 2L D 2T s 2L
Buck-boost
44
D' 2T s 2L 2D – 1 T s 2L
Boost
Fundamentals of Power Electronics
1 – 2D T s 2L
D 2T s 2L
Buck
Chapter 11: Current Programmed Control
–
Fv
Fg Converter
Table 11.2. Current programmed controller gains for basic converters
Fm = 1/MaTs
d(t) = Fm i c(t) – i L(t) – Fg vg(t) – Fv v(t)
The expression for the duty cycle is of the general form
Equation of the current programmed controller
Fundamentals of Power Electronics
D 2T s D' 2Ts 1 d(t) = i (t) – i L(t) – m 1(t) – m 2(t) M aT s c 2 2
Solve for duty cycle variations:
D 2Ts D' 2Ts i L(t) = i c(t) – M aTs d(t) – m 1(t) – m 2(t) 2 2
Simplify using dc relationships:
D 2T s D' 2Ts i L(t) = i c(t) – M aTs + DM 1Ts – D'M 2Ts d(t) – m 1(t) – m 2(t) 2 2
The first-order ac terms are
– M 2 + m 2(t) D' – d(t)
I L + i L(t) = I c + i c(t) – M aTs D + d(t) – M 1 + m 1(t) D + d(t)
Linearize
Fg
iL
Fv
+ –
vg
I d(t)
Fg
1:D
45
Vg d(t)
d
46
ic
–
– –
Fm
+
Fundamentals of Power Electronics
vg(t)
ic
–
– –
+
v
Append this block diagram to the duty cycle controlled converter model, to obtain a complete CPM system model
•
Chapter 11: Current Programmed Control
Describes the duty cycle chosen by the CPM controller
•
iL
Fv
i L(t)
L
v
C
Chapter 11: Current Programmed Control
Ti
Tv
–
v(t) R
+
CPM buck converter model
Fundamentals of Power Electronics
vg
Fm
d
d(t) = Fm i c(t) – i L(t) – Fg vg(t) – Fv v(t)
Block diagram of the current programmed controller
+ + – +
47
ic
–
– –
Fm
d
D' : 1
iL
Fv
I d(t)
I d(t)
1:D
vg
i L(t)
L
Fg
48
D' : 1
d
ic
–
– –
Fm
Vg – V d(t)
+
Fundamentals of Power Electronics
+ –
Fg
+ –
vg(t)
vg
V d(t)
+
v
Ti
Tv
–
v(t) R
Chapter 11: Current Programmed Control
C
+
iL
Fv
Ti
Tv
–
C v(t) R
Chapter 11: Current Programmed Control
v
I d(t)
+
CPM buck-boost converter model
Fundamentals of Power Electronics
vg(t) + –
i L(t)
L
CPM boost converter model
+
+ – +
+
ic
–
– –
+
Fundamentals of Power Electronics
Fg
Fm
d
V D2
++
vg
vg
I d(t)
49
Fg
d
ic
–
– –
Fm
+
v
C
iL
Fv
Tv
D Z i(s)
v
Ti
Zo(s)
50
iL
–
v(t) R
+
Ti
Tv
Zo(s)
Chapter 11: Current Programmed Control
iL
Fv
Zi(s)
i L(t)
L
Tv(s) 1 Z o(s) Fv 1 + Tv(s)
Chapter 11: Current Programmed Control
i L(s) Ti(s) = i c(s) 1 + Ti(s)
Transfer function from îc to îL:
Fm V2 D D Z i(s) Ti(s) = 1 + Fm V2 D Z o(s) Fv D Z i(s)
Ti(s) =
Current loop gain:
Tv(s) = Fm V2 D Z o(s) Fv D Z i(s)
Voltage loop gain:
Block diagram of entire CPM buck converter
Fundamentals of Power Electronics
+ –
+
i L(s) = D vg(s) + V2 d(s) Z i(s) D v(s) = i L(s) Z o(s)
Z i = sL + R || 1 sC
Z o = R || 1 sC
vg(t)
1:D
{
V d D
11.3.2 Example: analysis of CPM buck converter
+ –
Ti(s) v(s) = Z o(s) 1 + Ti(s) i c(s)
Fundamentals of Power Electronics
CCM for K > D’
K = 2L/RTs
Chapter 11: Current Programmed Control
1 + sRC
52
m 1 – ma 2 α=– m D' + a D m2
Chapter 11: Current Programmed Control
Controller stable for | α | < 1
2DM a 1 – α 2DM a 1 – α 1+s L + s 2 LC R M2 1 + α M2 1 + α
Characteristic value
Ti(s) = K 1 – α 1+α
51
Loop gain Ti(s) Result for the buck converter:
Fundamentals of Power Electronics
which is the simple approximation result, multiplied by the factor Ti/(1 + Ti).
Gvc(s) =
The control-to-output transfer function can be written as
•
i L(s) Ti(s) = i c(s) 1 + Ti(s)
When the loop gain Ti(s) is large in magnitude, then îL is approximately equal to îc. The results then coincide with the simple approximation.
•
Discussion: Transfer function from îc to îL
Ti 1 + Ti
Ti0 fz
f
f0
53
fc
Chapter 11: Current Programmed Control
M 2 fs M a 2πD
Fundamentals of Power Electronics
fc =
Solve for crossover frequency:
54
f 20 Ti( j2π fc) ≈ Ti0 =1 2π fc fz
Chapter 11: Current Programmed Control
Equate magnitude to one at crossover frequency:
Ti(s) ≈ Ti0
s ωz ω 20 = Ti0 s 2 sωz ω0
High frequency asymptote of Ti is
2
Ti0 = K 1 – α 1+α 1 ωz = RC 1 ω0 = 2DM a 1 – α LC M2 1 + α M2 1 + α Q=R C L 2DM a 1 – α
1 + ωs z Ti(s) = Ti0 1 + s + ωs Qω0 0
Crossover frequency fc
Fundamentals of Power Electronics
0dB
Ti
Q
Loop gain
Ti0 1 + Ti0
1 + ωs z
1+
1
1
56
Chapter 11: Current Programmed Control
which agrees with the previous slide
2
Chapter 11: Current Programmed Control
Ti0 1 s 1 + + 1 ω ω z 1 + Ti0 1 + Ti0 0 1 + Ti0 Qω0
sωz Ti0ω 20
1 2 s 1+ ω + s 2 z Ti0 ω 0
Fundamentals of Power Electronics
Gvc(s) ≈ R
55
Exact analysis
1+s
Low Q approximation:
Gvc(s) ≈ R
for | Ti0 | >> 1,
Gvc(s) = R
Fundamentals of Power Electronics
Ti(s) 1 ≈R 1 + Ti(s) 1 + sRC 1 + ωs c
c
—the result of the simple first-order model, with an added pole at the loop crossover frequency
Gvc(s) = Z o(s)
i L(s) Ti(s) 1 = ≈ i c(s) 1 + Ti(s) 1 + ωs
Construction of transfer functions
1+s
v(s) vg(s)
Gg0(s) 1 + Ti(s)
Ti0 1 s 1 + + 1 ω ω 1 + Ti0 0 1 + Ti0 z 1 + Ti0 Qω0
Ma 1 (1 – α) 1 2D 2 – M2 2 1 + Ti0 (1 + α)
i c(s) = 0
=
2
Gvg(0) 1 + ωs 1 + ωs z c Fundamentals of Power Electronics
Gvg(s) ≈
58
Chapter 11: Current Programmed Control
Poles are identical to control-to-output transfer function, and can be factored the same way:
Gvg(s) =
Gvg(s) =
Chapter 11: Current Programmed Control
Line-to-output transfer function
57
1 + V2 FmFvZ o(s) D Z i(s) D
1 – V2 FmFg D
Gg0(s) Ti(s) + vg(s) 1 + Ti(s) 1 + Ti(s)
Gg0(s) = Z o(s) D Z i(s)
Fundamentals of Power Electronics
with
v(s) = i c(s) Z o(s)
Solution of complete block diagram leads to
Line-to-output transfer function
for large Ti0
59
Chapter 11: Current Programmed Control
Result: simply replace
•
DCM CPM buck without artificial ramp is stable for D < 2/3. A small artificial ramp ma ≥ 0.086m2 leads to stability for all D.
•
Chapter 11: Current Programmed Control
DCM CPM boost and buck-boost are stable without artificial ramp
•
60
Small-signal transfer functions contain a single low frequency pole
•
Fundamentals of Power Electronics
Inductor dynamics appear at high frequency, near to or greater than the switching frequency
•
Diode by power source
Transistor by power sink
Again, use averaged switch modeling approach
•
11.4 Discontinuous conduction mode
Fundamentals of Power Electronics
Effective feedforward of input voltage variations in CPM controller then effectively cancels out the vg variations in the direct forward path of the converter.
For any Ti0 , the dc gain goes to zero when Ma/M2 = 0.5
2 M a Gvg(0) ≈ 2D –1 K M2 2
Line-to-output transfer function: dc gain
–
v1(t)
iL(t)
L
Switch network
Ts
m 1 + m a Ts
i c(t)
Fundamentals of Power Electronics
d 1(t) =
= m 1 + m a d 1T s
i c = i pk + m ad 1Ts
v1(t) m1 = L
i pk = m 1d 1Ts
Fundamentals of Power Electronics
vg(t)
+ –
i1(t) +
i2(t)
+
v2(t)
–
–
v(t)
62
Analysis
61
C R
+
vL(t)
ic
iL(t)
vL(t)
ic
iL(t)
Ts
Ts
L
m1
v2(t)
ipk
Ts
m2 =
0
0
L
v2
Ts
t
Ts
Ts
L
m1
v2(t)
ipk
Ts
m2 =
0
0
L
v2
Ts
t
Chapter 11: Current Programmed Control
v1(t)
=
v1
– ma
Chapter 11: Current Programmed Control
v1(t)
=
v1
– ma
DCM CPM buck-boost example
Ts
Ts
Ts
Ts
1
2 1
Ts
v1(t)
v1(t)
=
=
Ts
1 2
2
2
63
= p(t)
q1 Ts
Ts
i2(t)
i1(t)
Ts
Ts
Ts
d2Ts
Area q2
i 1(t)
ipk
d3Ts
During first subinterval, energy is transferred from input voltage source, through transistor, to inductor, equal to
•
1
Averaged transistor waveforms obey a power sink characteristic
•
Ts
1
= W fs = 2 Li 2pk fs
Fundamentals of Power Electronics
64
Chapter 11: Current Programmed Control
which is equal to the power sink expression of the previous slide.
p(t)
This energy transfer process accounts for power flow equal to
W = 2 Li 2pk
t
Chapter 11: Current Programmed Control
d1Ts
i 2(t)
ipk
Area q1
Discussion: switch network input port
ma m1
Li 2c fs
1+
1 2
1+
Li 2c fs
= m 1d (t)Ts
1 2
ma m1
i 1(τ)dτ =
= 2 i pk(t)d 1(t)
t
t + Ts
Fundamentals of Power Electronics
i 1(t)
i 1(t)
i 1(t)
i 1(t)
1 i 1(t) T = s Ts
Averaged switch input port equation
1 2
1 Ts
Ts
Ts
Ts
v2(t)
v2(t)
=
p(t)
=
Ts
Ts
v2(t)
65
= p(t)
Ts
i2(t)
i1(t)
Ts
Ts
Ts
d2Ts
Area q2
i 1(t)
ipk
d3Ts
Hence, in averaged model, diode becomes a power source, having value equal to the power consumed by the transistor power sink element
•
Chapter 11: Current Programmed Control
During second subinterval, all stored energy in inductor is transferred, through diode, to load
•
66
Averaged diode waveforms obey a power sink characteristic
•
Fundamentals of Power Electronics
t
Chapter 11: Current Programmed Control
d1Ts
i 2(t)
ipk
Area q1
Discussion: switch network output port
2
Li (t) fs
2 c
q2 Ts
m 1 + ma 1
1 2
Ts
Ts
i 2(τ)dτ =
v1(t)
t + Ts
Fundamentals of Power Electronics
i 2(t)
i 2(t)
t
i pkd 2Ts
Ts
=
d 2(t) = d 1(t)
q2 =
i 2(t)
Averaged switch output port equation
Ts
+ – –
Ts
Ts
p(t)
67
L
Ts
+
v2(t)
–
R
–
v(t)
+
Ts
M 1+ a M1
LI 2c (t) fs
+ –
Fundamentals of Power Electronics
P=
1 2
V2 = P R
Solution
Vg
2
68
R
–
V
+
RL fs M 2 1+ a M1
Chapter 11: Current Programmed Control
for a resistive load
V= PR = I c
P
2
Chapter 11: Current Programmed Control
Ts
Ts
C
i 2(t)
Steady state model: DCM CPM buck-boost
Fundamentals of Power Electronics
vg(t)
v1(t)
+
i 1(t)
Averaged equivalent circuit
vg(t)
vg(t)
Ts
Ts
+ –
+ –
L
p(t) Ts
69
p(t)
L
Ts
C
C
–
v(t)
+
–
v(t)
+
Pload Pload – P
Depends on load characteristic:
Boost
Buck-boost
Fundamentals of Power Electronics
Pload – P Pload
Buck
for DCM
I < I crit
Chapter 11: Current Programmed Control
2 M –1
for CCM
70
0≤D≤1 Ic –
M m T M –1 a s
0≤M <2 3
Stability range when m a = 0
0≤D≤1
I c – M m a Ts
Icrit
I c – M – 1 m a Ts M 2M
1 2
I > I crit
Pload = P
M
Table 11.3. Steady-state DCM CPM characteristics of basic converters Converter
Ts
Ts
Chapter 11: Current Programmed Control
R
R
Summary of steady-state DCM CPM characteristics
Fundamentals of Power Electronics
Boost
Buck
Models of buck and boost
A
1 2
Vg
+ –
2 3
Vg
Vg
DCM unstable for M > 23
B
71
V
DCM
CCM
with ma = 0 resistive load line I = V/R
–
r1
f1 i c
g 1v2
72
g 2v1
r2
–
v2
L
C
R
–
v
+
Chapter 11: Current Programmed Control
f2 i c
+
+ v1
i2
i1
iL
Chapter 11: Current Programmed Control
• the operating point having V > 0.67Vg can be shown to CPM buck be unstable characteristic
• with a resistive load, there can be two operating points
Linearized small-signal models: Buck
Fundamentals of Power Electronics
vg
Vg V 1– V
P
CCM unstable for M > 12
P +P= Vg – V V
Fundamentals of Power Electronics
4P Vg
1 I 2 c
Ic
I
I=
Buck converter: output characteristic with ma = 0
+ –
L
r1
–
+ –
g 1v2
73
g 2v1
f2 i c
r2
C
–
L
74
iL
–
v
r2
+
C
R
–
v
+
Chapter 11: Current Programmed Control
f2 i c
v2
g 2v1
v1
g 1v2
– + f1 i c
i2
i1
r1
R
+
Chapter 11: Current Programmed Control
–
v2
Linearized small-signal models: Buck-boost
Fundamentals of Power Electronics
vg
f1 i c
+
+ v1
i2
i1
Fundamentals of Power Electronics
vg
iL
Linearized small-signal models: Boost
75
2
I1 Ic
1
Chapter 11: Current Programmed Control
m 1 + ma 1 –R M 2 1 – ma m1
m 2 ma 1 M2 2 – M + M – 1 1 + ma m1
R
m 1 – ma 1
m 1 + ma
–R 1–M M2
I1 Ic
2 I Ic
2
r1
f1
m 1 + ma 1
ma m1 2M R m 1 + ma 1
1 M R M –1
1 M R 1–M
g2
ma m1 2 – M – M
Fundamentals of Power Electronics
Buck-boost
Boost
Buck
Converter
76
2
2
I2 Ic
I2 Ic
2 I Ic
f2
R
R M –1 M
Chapter 11: Current Programmed Control
R
m 1 + ma 1 m 1 – 2M + m a 1 1–M
r2
Table 11.5. Current programmed DCM small-signal equivalent circuit parameters: output port
DCM CPM small-signal parameters: output port
Fundamentals of Power Electronics
0
Buck-boost
m 1 – ma 1 ma 1+ m 1
M – 1 R M –1
1 M2 R 1–M
g1
Boost
Buck
Converter
Table 11.4. Current programmed DCM small-signal equivalent circuit parameters: input port
DCM CPM small-signal parameters: input port
+ –
vg = 0
Gc0 1 + ωs
=
v ic
Gc0 = f2 R || r 2 1 ωp = R || r 2 C
Gvc(s) =
f1 i c
r1
p
g 1v
Buck ωp
77
v vg
ic = 0
f2 i c
=
r2
Gg0 1 + ωs
p
C
–
v
Addition of small artificial ramp stabilizes system. ma > 0.086 leads to stability for all M ≤ 1. Output voltage feedback can also stabilize system, without an artificial ramp
• • •
Chapter 11: Current Programmed Control
ωp then constitutes a RHP pole. Converter is unstable. •
78
For ma = 0, numerator is negative when M > 2/3. •
Fundamentals of Power Electronics
R
+
Chapter 11: Current Programmed Control
Gg0 = g 2 R || r 2
Gvg(s) =
g 2vg
m 2 – 3M 1 – M + m a M 2 – M 2 ωp = 1 RC m 1 – M 1 – M + M ma 2
Plug in parameters:
Fundamentals of Power Electronics
vg
Buck, boost, buck-boost all become
Simplified DCM CPM model, with L = 0
Chapter 11: Current Programmed Control
The more accurate model predicts that the inductor pole occurs at the crossover frequency fc of the effective current feedback loop gain Ti(s). The frequency fc typically occurs in the vicinity of the converter switching frequency fs . The more accurate model also predicts that the line-to-output transfer function Gvg(s) of the buck converter is nulled when ma = 0.5 m2.
6.
80
The more accurate model of Section 11.3 correctly accounts for the difference between the average inductor current 〈 iL(t) 〉Ts and the control input ic(t). This model predicts the nonzero line-to-output transfer function Gvg(s) of the buck converter. The current-programmed controller behavior is modeled by a block diagram, which is appended to the small-signal converter models derived in Chapter 7. Analysis of the resulting multiloop feedback system then leads to the relevant transfer functions.
5.
Fundamentals of Power Electronics
The simple model predicts that one pole is eliminated from the converter line-to-output and control-to-output transfer functions. Current programming does not alter the transfer function zeroes. The dc gains become load-dependent.
4.
Summary of key points
Chapter 11: Current Programmed Control
The behavior of current-programmed converters can be modeled in a simple and intuitive manner by the first-order approximation 〈 iL(t) 〉Ts ≈ ic(t). The averaged terminal waveforms of the switch network can then be modeled simply by a current source of value ic , in conjunction with a power sink or power source element. Perturbation and linearization of these elements leads to the small-signal model. Alternatively, the smallsignal converter equations derived in Chapter 7 can be adapted to cover the current programmed mode, using the simple approximation iL(t) ≈ ic(t).
3.
79
The basic current-programmed controller is unstable when D > 0.5, regardless of the converter topology. The controller can be stabilized by addition of an artificial ramp having slope ma. When ma ≥ 0.5 m2, then the controller is stable for all duty cycle.
2.
Fundamentals of Power Electronics
In current-programmed control, the peak switch current is(t) follows the control input ic(t). This widely used control scheme has the advantage of a simpler control-to-output transfer function. The line-to-output transfer functions of current-programmed buck converters are also reduced.
1.
11.5 Summary of key points
1
12. Basic Magnetics Theory 13. Filter Inductor Design 14. Transformer Design
Fundamentals of Power Electronics
Chapter 11: Current Programmed Control
Chapter 12: Basic Magnetics Theory
Part III. Magnetics
81
Current programmed converters operating in the discontinuous conduction mode are modeled in Section 11.4. The averaged transistor waveforms can be modeled by a power sink, while the averaged diode waveforms are modeled by a power source. The power is controlled by ic(t). Perturbation and linearization of these averaged models, as usual, leads to small-signal equivalent circuits.
Fundamentals of Power Electronics
7.
Summary of key points
12.2.3. Leakage inductances
12.1.2. Magnetic circuits
12.3.2. Low-frequency copper loss
Fundamentals of Power Electronics
i(t)
terminal characteristics
v(t)
3
Ampere's law
Faraday's law
Chapter 12: Basic Magnetics Theory
H(t), F(t)
core characteristics
B(t), Φ(t)
12.1. Review of basic magnetics 12.1.1. Basic relations
Chapter 12: Basic Magnetics Theory
12.4.6. PWM waveform harmonics
12.4.3. MMF diagrams 2
12.4.5. Example: power loss in a transformer winding
12.4.2. The proximity effect
Fundamentals of Power Electronics
12.4.4. Power loss in a layer
12.4.1. The skin effect
12.4. Eddy currents in winding conductors
12.3.1. Core loss
12.3. Loss mechanisms in magnetic devices
12.2.2. The magnetizing inductance
12.2.1. The ideal transformer
12.2. Transformer modeling
12.1.1. Basic relations
12.1. Review of basic magnetics
Chapter 12. Basic Magnetics Theory
{
x2
4
+
–
x2
surface S with area Ac
Chapter 12: Basic Magnetics Theory
{
voltage V = El
total current I current density J
x1
x1
x2
H ⋅ dl
+ MMF F = Hl
Fundamentals of Power Electronics
x1
magnetic field H
length l
–
x2
Example: uniform magnetic field of magnitude H
F=
5
x1
+
–
x2
Chapter 12: Basic Magnetics Theory
voltage V = El
electric field E
length l
Analogous to electric field of strength E, which induces voltage (EMF) V:
Magnetomotive force (MMF) F between points x1 and x2 is related to the magnetic field H according to
Magnetic field H and magnetomotive force F
surface S with area Ac
MMF F = Hl
–
magnetic field H +
length l
electric field E
length l
Fundamentals of Power Electronics
total flux Φ flux density B
x1
Electrical quantities
Magnetic quantities
Basic quantities
{
dΦ(t) dt
dB(t) dt
Fundamentals of Power Electronics
v(t) = A c
For uniform flux distribution, Φ(t) = B(t)Ac and hence
v(t) =
6
surface S
7
flux Φ(t)
{
{
+
– v(t)
Chapter 12: Basic Magnetics Theory
area Ac
Chapter 12: Basic Magnetics Theory
total current I current density J
surface S with area Ac
Analogous to electrical conductor current density of magnitude J, which leads to total conductor current I:
B ⋅ dA
Faraday’s law
surface S with area Ac
Voltage v(t) is induced in a loop of wire by change in the total flux Φ(t) passing through the interior of the loop, according to
Fundamentals of Power Electronics
total flux Φ flux density B
Φ = B Ac
Example: uniform flux density of magnitude B
Φ=
The total magnetic flux Φ passing through a surface of area Ac is related to the flux density B according to
Flux density B and total flux Φ
Fundamentals of Power Electronics
• For uniform magnetic field strength H(t), the integral (MMF) is H(t)lm. So F (t) = H(t) l m = i(t)
• Illustrated path follows magnetic flux lines around interior of core
Example: magnetic core. Wire carrying current i(t) passes through core window. i(t)
9
H magnetic path length lm
Chapter 12: Basic Magnetics Theory
H ⋅ dl = total current passing through interior of path closed path
shorted loop
Chapter 12: Basic Magnetics Theory
induced flux Φ'(t)
Ampere’s law
8
flux Φ(t)
induced current i(t)
The net MMF around a closed path is equal to the total current passing through the interior of the path:
Fundamentals of Power Electronics
• This current induces a flux Φ’(t), which tends to oppose changes in Φ(t)
• This voltage, divided by the impedance of the loop conductor, leads to current i(t)
• Changing flux Φ(t) induces a voltage v(t) around the loop
Example: a shorted loop of wire
The voltage v(t) induced by the changing flux Φ(t) is of the polarity that tends to drive a current through the loop to counteract the flux change.
Lenz’s law
B = µ0 H
Fundamentals of Power Electronics
= 4π · 10 Henries per meter
-7
µ0 = permeability of free space
B
Free space
µ0
10
H
11
µ
Chapter 12: Basic Magnetics Theory
Highly nonlinear, with hysteresis and saturation
B
A magnetic core material
H
Chapter 12: Basic Magnetics Theory
Core material characteristics: the relation between B and H
Fundamentals of Power Electronics
• The total MMF around a closed loop, accounting for winding current MMF’s, is zero
• Previous example: total MMF around core, F(t) = H(t)lm, is equal to the winding current MMF i(t)
• We can view winding currents as sources of MMF
• Relates magnetic field strength H(t) to winding current i(t)
Ampere’s law: discussion
µ = µr µ0
Tesla Ampere / meter Weber
B H Φ
MKS
13
1Wb = 108 Mx 2 1T = 1Wb / m
1A/m = 4π⋅10-3 Oe
4
1T = 10 G
Chapter 12: Basic Magnetics Theory
Maxwell
Oersted
Gauss
B = µr H
unrationalized cgs
conversions
Chapter 12: Basic Magnetics Theory
Table 12.1. Units for magnetic quantities
B = µ 0 µr H
Fundamentals of Power Electronics
12
H
Typical Bsat = 0.3-0.5T, ferrite 0.5-1T, powdered iron 1-2T, iron laminations
– Bsat
µ
Saturation, no hysteresis B Bsat
Units
H
core material equation
quantity
Fundamentals of Power Electronics
Typical µr = 103 - 105
B=µH µ = µr µ0
No hysteresis or saturation B
Piecewise-linear modeling of core material characteristics
+ v(t) –
i(t) n turns
core
Φ
14
i(t) n turns
H magnetic path length lm
Fundamentals of Power Electronics
H(t) lm = n i(t)
From Ampere’s law, we have
15
Chapter 12: Basic Magnetics Theory
Winding contains n turns of wire, each carrying current i(t). The net current passing through the path interior (i.e., through the core window) is ni(t).
For uniform field strength H(t), the core MMF around the path is H lm.
Choose a closed path which follows the average magnetic field line around the interior of the core. Length of this path is called the mean magnetic path length lm.
core permeability µ
core area Ac
Chapter 12: Basic Magnetics Theory
Inductor example: Ampere’s law
Fundamentals of Power Electronics
Express in terms of the average flux density B(t) = Φ(t)/A c dB(t) v(t) = n A c dt
Total winding voltage is dΦ(t) v(t) = n vturn(t) = n dt
dΦ(t) vturn(t) = dt
For each turn of wire, we can write
Faraday’s law:
Example: a simple inductor
– Bsat for H ≤ Bsat / µ
for H < Bsat / µ
µH
H(t) lm = n i(t)
B=
for H < Bsat / µ µH
– Bsat for H ≤ Bsat / µ
for H ≥ Bsat / µ Bsat
17
Chapter 12: Basic Magnetics Theory
For | i | > Isat the flux density is constant and equal to Bsat. Faraday’s law then predicts dBsat —saturation leads to short circuit v(t) = n A c =0 dt Fundamentals of Power Electronics
H
Chapter 12: Basic Magnetics Theory
Eliminate B and H, and solve for relation between v and i. For | i | < Isat, d H(t) µ n 2 A c di(t) v(t) = µ n A c v(t) = dt lm dt which is of the form di(t) µ n2 Ac v(t) = L L = with dt lm —an inductor
dB(t) v(t) = n A c dt
We have:
16
– Bsat
µ
Electrical terminal characteristics
Fundamentals of Power Electronics
Find winding current at onset of saturation: substitute i = Isat and H = Bsat/µ into equation previously derived via Ampere’s law. Result is B l I sat = µsatn m
B=
for H ≥ Bsat / µ
Bsat
B Bsat
Inductor example: core material model
flux Φ
{
+
H
length l MMF F
+ Φ
18
R
F
– = reluctance of element
core permeability µ R = µAl c
area Ac
Chapter 12: Basic Magnetics Theory
R
–
Fundamentals of Power Electronics
19
Chapter 12: Basic Magnetics Theory
• Solve magnetic circuit using Kirchoff’s laws, etc.
• MMF → voltage, flux → current
• Windings are sources of MMF
• Represent each element with reluctance
Magnetic circuits: magnetic structures composed of multiple windings and heterogeneous elements
Fundamentals of Power Electronics
A corresponding model:
Since H = B / µ and Β = Φ / Ac, we can express F as F = µ lA Φ with R = µ lA c c
MMF between ends of element is F =Hl
Uniform flux and magnetic field inside a rectangular element:
12.1.2. Magnetic circuits
20
Φ1
Magnetic circuit
Φ2
Φ3
Φ1 = Φ2 + Φ3
Φ3
Chapter 12: Basic Magnetics Theory
Φ2
node
Φ1
node
Fundamentals of Power Electronics
21
Chapter 12: Basic Magnetics Theory
Total MMF’s around the closed path add up to zero.
Right-hand side: currents in windings are sources of MMF’s. An n-turn winding carrying current i(t) is modeled as an MMF (voltage) source, of value ni(t).
Left-hand side: sum of MMF’s across the reluctances around the closed path
closed path
H ⋅ dl = total current passing through interior of path
Follows from Ampere’s law:
Magnetic analog of Kirchoff’s voltage law
Fundamentals of Power Electronics
Total flux entering a node must be zero
Flux lines are continuous and cannot end
Divergence of B = 0
Physical structure
Magnetic analog of Kirchoff’s current law
n turns
n turns
core permeability µc Φ
Fundamentals of Power Electronics
+ v(t) –
i(t)
22
23
ni=Φ Rc+Rg
Fc + Fg = n i
magnetic path length lm
air gap lg
cross-sectional area Ac
n i(t)
c
l R g = µ gA 0 c
l
R c = µ Ac
Φ(t)
Rc
–
–
R g Fg
+
Chapter 12: Basic Magnetics Theory
+ –
+
Fc
Chapter 12: Basic Magnetics Theory
magnetic path length lm
air gap lg
cross-sectional area Ac
Magnetic circuit model
Fundamentals of Power Electronics
Fc + Fg = n i
Ampere’s law:
+ v(t) –
i(t)
core permeability µc Φ
Example: inductor with air gap
v(t) = n
magnetic path length lm
air gap lg
cross-sectional area Ac
L=
Hence inductance is
Faraday’s law:
Bsat A c Rc+Rg n
Fundamentals of Power Electronics
Effect of air gap: • decrease inductance • increase saturation current • inductance is less dependent on core permeability
I sat =
Φsat = BsatAc
n2 L= Rc+Rg
ni=Φ Rc+Rg
24
25
BsatAc
Φ = BAc
Φ(t)
Rc
Fc –
c
l R g = µ gA 0 c
R c = µ Ac
l
ni=Φ Rc+Rg
+ –
+
–
R g Fg
+
nIsat1
nIsat2
ni ∝ Hc
Chapter 12: Basic Magnetics Theory
– BsatAc
1
Rc+Rg
1
Rc
Chapter 12: Basic Magnetics Theory
n i(t)
Effect of air gap
n2 Rc+Rg
dΦ(t) dt di(t) n2 Substitute for Φ: v(t) = R + R dt c g
n turns
Fundamentals of Power Electronics
+ v(t) –
i(t)
core permeability µc Φ
Solution of model
n1 i1
+ v1(t) –
+ –
26
+
Φ
n1 turns
i1(t)
Fc
Rc –
Fundamentals of Power Electronics
27
Also, by Faraday’s law, v1 = n 1 dΦ dt dΦ v2 = n 2 dt Eliminate Φ : dΦ = v1 = v2 dt n 1 n 2 Ideal transformer equations: v1 v2 = and n 1 i1 + n 2 i2 = 0 n1 n2
MMF Fc = Φ R also approaches zero. We then obtain 0 = n 1 i1 + n 2 i2
In the ideal transformer, the core reluctance R approaches zero.
Chapter 12: Basic Magnetics Theory
Ideal
–
–
i2
n2 i2
v2
–
v1
n1 : n2
Fc
Rc
+
i1
+ –
+
Φ
+
n1 i1
+ n2 v turns 2(t) –
i2(t)
Chapter 12: Basic Magnetics Theory
n2 i2
core
Φ
12.2.1. The ideal transformer
Fundamentals of Power Electronics
Magnetic circuit model:
Φ R = n 1 i1 + n 2 i2
F c = n 1 i1 + n 2 i2
R = µ mA c
l
Two windings, no air gap:
12.2. Transformer modeling
+ –
+ –
imp = i1 +
n2 i2 n1
–
v1
+
28
L mp =
R
n 21
+ –
n2 n 1 i2
Fc
Rc
Ideal
n1 : n2
–
Fundamentals of Power Electronics
29
–
v2
+
Chapter 12: Basic Magnetics Theory
• Magnetizing current causes the ratio of winding currents to differ from the turns ratio
the resulting inductor is the magnetizing inductance, referred to the primary winding
primary winding then behaves as an inductor
we are left with the primary winding on the core
• If the secondary winding is disconnected:
• A real, physical inductor, that exhibits saturation and hysteresis
• Models magnetization of core material
i2
n2 i2
Chapter 12: Basic Magnetics Theory
n i1 + n 2 i2 1
n1 i1
+
Φ
Magnetizing inductance: discussion
Fundamentals of Power Electronics
with
R
d imp dt n2 L mp = 1
v1 = L mp
i1
with v1 = n 1 dΦ dt
This equation is of the form
Eliminate Φ: n2 n v1 = 1 d i1 + 2 i2 R dt n1
Φ R = n 1 i1 + n 2 i2
For nonzero core reluctance, we obtain
12.2.2. The magnetizing inductance
+ –
v1(t) dt
v1(t) dt
Fundamentals of Power Electronics
1 n1 A c
Flux density is proportional:
imp(t) = 1 L mp
Magnetizing current depends on the integral of the applied winding voltage:
B(t) =
30
L mp =
R
n 21
n i1 + n 2 i2 1
n2 n 1 i2
Ideal
n1 : n2
t1
t2
v1(t) dt
31
–
v2
+
Chapter 12: Basic Magnetics Theory
limits of integration chosen to coincide with positive portion of applied voltage waveform
λ1 =
Flux density befcomes large, and core saturates, when the applied volt-seconds λ1 are too large, where
–
v1
+
i1
i2
Chapter 12: Basic Magnetics Theory
Saturation vs. applied volt-seconds
Fundamentals of Power Electronics
• Saturation is caused by excessive applied volt-seconds
then the magnetizing current is zero, and there is no net magnetization of the core.
• Large winding currents i1(t) and i2(t) do not necessarily lead to saturation. If 0 = n 1 i1 + n 2 i2
• When core saturates, the magnetizing current becomes large, the impedance of the magnetizing inductance becomes small, and the windings are effectively shorted out.
• Saturation occurs when core flux density B(t) exceeds saturation flux density Bsat.
Transformer saturation
Φl1
32
ΦM
ΦM +
Φl2
+ v2(t) –
i2(t)
R
n1 n2
n = n 2 L mp 1
Fundamentals of Power Electronics
primary and secondary self-inductances n L 11 = L l1 + n 1 L 12 2 n2 L 22 = L l2 + n L 12 1
L 12 =
mutual inductance
L L v1(t) i (t) = 11 12 d 1 L 12 L 22 dt i2(t) v2(t)
–
v1
+
i1
L 12 L 11 L 22
L 22 L 11
Ll2
Chapter 12: Basic Magnetics Theory
k=
coupling coefficient
33
ne =
Ideal
n1 : n2
effective turns ratio
n L mp = n 1 L 12 2
Ll1
i2
Chapter 12: Basic Magnetics Theory
–
Φl2 v2(t)
i2(t)
Transformer model, including leakage inductance
Fundamentals of Power Electronics
+ v1(t) –
i1(t)
Φl1
+ v1(t) –
i1(t)
12.2.3. Leakage inductances
–
v2
+
one cycle
v(t)i(t)dt
+ v(t) –
i(t) n turns
Fundamentals of Power Electronics
Substitute into integral:
one cycle
35
H dB
dB(t) dt
one cycle
nA c = A cl m
W=
Relate winding voltage and current to core B and H via Faraday’s law and Ampere’s law: dB(t) v(t) = n A c H(t) lm = n i(t) dt
W=
34
core permeability µ
core area Ac
Chapter 12: Basic Magnetics Theory
H(t)l m dt n
core
Φ
Chapter 12: Basic Magnetics Theory
12.3.1. Core loss Energy per cycle W flowing into nturn winding of an inductor, excited by periodic waveforms of frequency f:
Fundamentals of Power Electronics
Effect of PWM waveform harmonics
MMF diagrams, losses in a layer, and losses in basic multilayer windings
Proximity effect: high frequency limit
High frequency copper loss: the proximity effect
Eddy current losses in ferrite cores
Core loss: classical eddy current losses
High-frequency losses: the skin effect
Core loss: hysteresis loss
Dc copper loss
Low-frequency losses:
12.3. Loss mechanisms in magnetic devices
one cycle
H dB
H dB one cycle
one cycle
36
Chapter 12: Basic Magnetics Theory
• Dependence on maximum flux density: how does area of B-H loop depend on maximum flux density (and on applied waveforms)? Emperical equation (Steinmetz equation):
• Hysteresis loss varies directly with applied frequency
Fundamentals of Power Electronics
37
Chapter 12: Basic Magnetics Theory
Dependence of PH on Bmax is predicted by the theory of magnetic domains.
The parameters KH and α are determined experimentally.
PH = K H f B αmax(core volume)
H
H dB
Area
Hysteresis loss is directly proportional to applied frequency
Modeling hysteresis loss
Fundamentals of Power Electronics
PH = f A cl m
(energy lost per cycle) = (core volume) (area of B-H loop)
The term Aclm is the volume of the core, while the integral is the area of the B-H loop.
W = A cl m
B
Core loss: Hysteresis loss
38
core
eddy current i(t)
Chapter 12: Basic Magnetics Theory
Eddy current loss i2(t)R
Fundamentals of Power Electronics
39
Chapter 12: Basic Magnetics Theory
• Ferrite core material impedance is capacitive. This causes eddy current power loss to increase as f4.
• Classical Steinmetz equation for eddy current loss: PE = K E f 2B 2max(core volume)
• Eddy current power loss i2(t)R then varies with square of excitation frequency f.
• If core material impedance Z is purely resistive and independent of frequency, Z = R, then eddy current magnitude is proportional to voltage: i(t) = v(t)/R. Hence magnitude of i(t) is directly proportional to excitation frequency f.
• Ac flux Φ(t) induces voltage v(t) in core, according to Faraday’s law. Induced voltage is proportional to derivative of Φ(t). In consequence, magnitude of induced voltage is directly proportional to excitation frequency f.
Modeling eddy current loss
Fundamentals of Power Electronics
flux Φ(t)
Magnetic core materials are reasonably good conductors of electric current. Hence, according to Lenz’s law, magnetic fields within the core induce currents (“eddy currents”) to flow within the core. The eddy currents flow such that they tend to generate a flux which opposes changes in the core flux Φ(t). The eddy currents tend to prevent flux from penetrating the core.
Core loss: eddy current loss
3
Ferrite core material Power loss density, Watts / cm
0.01
Hz
Hz
40
0.3
1.5 - 2.0 T 0.6 - 0.8 T
0.25 - 0.5 T
Powdered cores powdered iron, molypermalloy Ferrite Manganese-zinc, Nickel-zinc
41
low
medium
high
Chapter 12: Basic Magnetics Theory
20 kHz - 1 MHz transformers, ac inductors
1 kHz transformers, 100 kHz filter inductors
50-60 Hz transformers, inductors
Applications
Chapter 12: Basic Magnetics Theory
Pfe = K fe B βmax A c l m
Empirical equation, at a fixed frequency:
Relative core loss
Core materials
Laminations iron, silicon steel
1M
B sat
Fundamentals of Power Electronics
0.1
Bmax, Tesla
500 k
Core type
Fundamentals of Power Electronics
0.01
0.1
1 200k Hz
Total core loss: manufacturer’s data
z
100k Hz
50kH
Hz 20k
lb Aw
42
Fundamentals of Power Electronics
i(t)
Φ(t)
wire
43
i(t)
eddy currents
R
Chapter 12: Basic Magnetics Theory
i(t)
eddy currents
Chapter 12: Basic Magnetics Theory
wire
current density
12.4. Eddy currents in winding conductors 12.4.1. The skin effect
Fundamentals of Power Electronics
Pcu = I 2rms R
The wire resistance leads to a power loss of
where Aw is the wire bare cross-sectional area, and lb is the length of the wire. The resistivity ρ is equal to 1.724⋅10-6 Ω cm for soft-annealed copper at room temperature. This resistivity increases to 2.3⋅10-6 Ω cm at 100˚C.
R=ρ
DC resistance of wire
12.3.2. Low-frequency copper loss
ρ πµ f penetration depth δ, cm
0.001 10kHz
0.01
0.1
44
Fundamentals of Power Electronics
A multi-layer foil winding, with d >> δ. Each layer carries net current i(t).
Ac current in a conductor induces eddy currents in adjacent conductors by a process called the proximity effect. This causes significant power loss in the windings of high-frequency transformers and ac inductors.
45
1MHz
#40AWG
#30AWG
#20AWG
Wire diameter
i
–i
2i
–2i
d
Φ
2Φ
Chapter 12: Basic Magnetics Theory
layer 1
layer 2
layer 3
3i
Chapter 12: Basic Magnetics Theory
frequency
100kHz
25˚C
˚C
100
12.4.2. The proximity effect
Fundamentals of Power Electronics
δ = 7.5 cm f
For copper at room temperature:
δ=
For sinusoidal currents: current density is an exponentially decaying function of distance into the conductor, with characteristic length δ known as the penetration depth or skin depth.
Penetration depth δ
current density J
46
Rdc d δ
=
M
Σ Pj = M3 j=1
2M 2 + 1 P1
i
d=δ
= M P1
d
= d >> δ
47
Pw d >> δ 1 d = 2M 2 + 1 3 δ Pw,dc Fundamentals of Power Electronics
FR
Φ
2Φ
Chapter 12: Basic Magnetics Theory
For foil thicknesses other than d = δ, the dc resistance and power loss are changed by a factor of d/δ. The total winding dc copper loss is Pw,dc = M P1 δ d So the proximity effect increases the copper loss by a factor of
Pw,dc
2i
–2i
3i
Chapter 12: Basic Magnetics Theory
layer 1
–i
If foil thickness were d = δ, then at dc each layer would produce copper loss P1. The copper loss of the M-layer winding would be
Compare with dc copper loss:
d >> δ
Add up losses in each layer: Pw
2
layer 2
layer 3
Total loss in M-layer winding: high-frequency limit
Fundamentals of Power Electronics
Pm = ((m – 1)2 + m 2) P1
Power loss Pm in layer m is:
Power loss P3 in layer 3 is: 2 P3 = 2I rms Rdc d + 3I rms δ = 13P1
Power loss P2 in layer 2 is: 2 P2 = I 2rms Rdc d + 2I rms Rdc d δ δ = 5P1
P1 = I 2rms Rdc d δ
Let P1 be power loss in layer 1:
Estimating proximity loss: high-frequency limit
current density J
(b)
(c)
48
(d)
lw
Chapter 12: Basic Magnetics Theory
ϕ= η d δ
Effective ratio of conductor thickness to skin depth:
Conductor spacing factor: n η= π d l 4 lw
i
–i 2i
49
–2i 3i
–3i
2i
i
–i
Chapter 12: Basic Magnetics Theory
–2i
secondary layers
primary layers
{ Fundamentals of Power Electronics
core
Two-winding transformer example
12.4.3. Magnetic fields in the vicinity of winding conductors: MMF diagrams
Fundamentals of Power Electronics
d
(a)
Approximating a layer of round conductors as an effective foil conductor
{
i
Fundamentals of Power Electronics
m p – m s i = F (x)
0
i
2i
3i
MMF F(x)
i
+
–
50
i
F(x) –i
H(x) = lw
F (x)
–i
i
i
51
H(x) =
–i
lw
–i
Chapter 12: Basic Magnetics Theory
lw
F (x)
–i
x
Chapter 12: Basic Magnetics Theory
–i
Ampere’s law and MMF diagram
Fundamentals of Power Electronics
x
i
m p – m s i = F (x)
Transformer example: magnetic field lines
i
sec
–i
i
Fundamentals of Power Electronics
0
i
2i
3i
MMF F(x)
52
–2i 3i –3i 2i
i
53
pri
–i
sec
–i
x
–i
x
Chapter 12: Basic Magnetics Theory
i
pri
sec
Chapter 12: Basic Magnetics Theory
–2i i
Interleaved windings
–i 2i
pri
Fundamentals of Power Electronics
0
i
2i
3i
MMF F(x)
MMF diagram for d >> δ
–3i 4
i
54
i
primary
i
Fundamentals of Power Electronics
55
driven by the MMFs F(0) and F(d). Sinusoidal waveforms are assumed, and rms values are used. It is assumed that H(0) and H(d) are in phase.
The magnetic fields H(0) and H(d) are
Assume uniform magnetic fields at surfaces of layer, of strengths H(0) and H(d). Assume that these fields are parallel to layer surface (i.e., neglect fringing and assume field normal component is zero).
Approximate computation of copper loss in one layer
x
d
F(d)
H(d)
Chapter 12: Basic Magnetics Theory
0
F(0)
H(0)
F(x)
–3i 4
Chapter 12: Basic Magnetics Theory
–3i 4
secondary
12.4.4. Power loss in a layer
Fundamentals of Power Electronics
–1.5 i
–i
–0.5 i
0
0.5 i
i
1.5 i
MMF F(x)
–3i 4
secondary
Partially-interleaved windings: fractional layers
layer
ϕ n 2l
π d nl 4 lw 56
Chapter 12: Basic Magnetics Theory
(MLT) = mean-length-per-turn, or circumference, of layer.
Fundamentals of Power Electronics
57
Q'(ϕ, m) = 2m 2 – 2m + 1 G1(ϕ) – 4m m – 1 G2(ϕ)
Power dissipated in the layer can now be written P = I 2 Rdc ϕ Q'(ϕ, m)
F (0) m – 1 = m F (d)
The quantity m is the ratio of the MMF F(d) to the layer ampere-turns nlI. Then,
F (d) = m n l I
Express F(d) in terms of the winding current I, as
F (d) – F (0) = n l I
If winding carries current of rms magnitude I, then
0
F(0) d
F(d)
H(d)
Chapter 12: Basic Magnetics Theory
F(x)
H(0)
Winding carrying current I, with nl turns per layer
Fundamentals of Power Electronics
η=
sinh (ϕ) cos (ϕ) + cosh(ϕ) sin (ϕ) cosh (2ϕ) – cos (2ϕ)
ϕ= η d δ
G2(ϕ) =
sinh (2ϕ) + sin (2ϕ) G1(ϕ) = cosh (2ϕ) – cos (2ϕ)
nl = number of turns in layer, Rdc = dc resistance of layer,
F 2(d) + F 2(0) G1(ϕ) – 4 F (d)F (0) G2(ϕ)
(MLT) n 2l Rdc = ρ lw η d
where
P = Rdc
Solve Maxwell’s equations to find current density distribution within layer. Then integrate to find total copper loss P in layer. Result is
Solution for layer copper loss P
layer
1
10
100
d=δ
= Q'(ϕ, m) Pdc
Fundamentals of Power Electronics
ϕ=1
P
Relative to copper loss when d = δ
Pdc
P
0.1
58
ϕ
1
5 4
0.1
1
10
100
0.1
59
ϕ
10
m = 0.5
1
1.5
2
8
10
m = 0.5
1
1.5
2
3
4
5
6
Chapter 12: Basic Magnetics Theory
m = 15 12 10
1
3
Chapter 12: Basic Magnetics Theory
m = 15 12 10 8 6
Layer copper loss vs. layer thickness
Fundamentals of Power Electronics
2
P I Rdc
P = ϕ Q'(ϕ, m) I Rdc 2
Increased copper loss in layer
1
10
0.1 Fundamentals of Power Electronics
Ppri Ppri,dc
100
Express summation in closed form:
FR =
primary layers
secondary layers
npi
60
npi npi M
npi
npi
61
ϕ
6
5
0.5
1
1.5
2
3
10
4
Chapter 12: Basic Magnetics Theory
number of layers M = 15 12 10 8 7
FR = ϕ G1(ϕ) + 2 M 2 – 1 G1(ϕ) – 2G2(ϕ) 3
1
x
Chapter 12: Basic Magnetics Theory
npi
Increased total winding loss
Fundamentals of Power Electronics
Sum losses over all 0 primary layers: M Ppri FR = = 1 Σ ϕ Q'(ϕ, m) Ppri,dc M m = 1
1
Proximity effect increases copper m= F n pi loss in layer m by the factor P = ϕ Q'(ϕ, m) 2 I 2 Rdc
Each winding consists of M layers
Two winding transformer
12.4.5. Example: Power loss in a transformer winding
{ {
ϕ=1
0.1
1
10
0.1
0
Ipk
DTs
∞
Σ j=1 Fundamentals of Power Electronics
2ϕ 1 Pcu =D+ 2 Dπ 2 D I pk Rdc
Ts
j ϕ 1 G1( j ϕ 1) + 2 M 2 – 1 G1( j ϕ 1) – 2G2( j ϕ 1) 3
i(t)
10
0.5
1
2 1.5
3
10 8 7 6 5 4
63
Chapter 12: Basic Magnetics Theory
sin 2 ( jπD) G1( j ϕ 1) + 2 M 2 – 1 G1( j ϕ 1) – 2G2( j ϕ 1) 3 j j
t
Chapter 12: Basic Magnetics Theory
12
Total, relative to value predicted by low-frequency analysis:
Pj = I 2j Rdc Ac
Pdc = I Rdc
Dc
I 0 = DI pk
2 I j cos ( jωt)
2 0
Copper loss:
Ij =
∞
Σ j=1
2 I pk sin ( jπD) jπ
i(t) = I 0 +
Fourier series:
with
62
ϕ
1
number of layers M = 15
12.4.6. PWM waveform harmonics
Fundamentals of Power Electronics
Ppri,dc
Ppri
100
Total winding loss
P1
64
Chapter 12: Basic Magnetics Theory
1 0.1
Fundamentals of Power Electronics
FH
10 M = 10 8 6
4
5
65
3
ϕ1
1
M = 0.5
1
1.5
2
10
Chapter 12: Basic Magnetics Theory
D = 0.5
Increased proximity losses induced by PWM waveform harmonics: D = 0.5
Fundamentals of Power Electronics
Pcu = I 20 Rdc + FH FR I 21 Rdc
The total winding copper loss can then be written
FH =
Σ Pj j=1
∞
Effect of harmonics: FH = ratio of total ac copper loss to fundamental copper loss
Harmonic loss factor FH
1
10
0.1
M = 10 8 6 4
66
5 3 1.5 1 M = 0.5
ϕ1
1
2
10
Chapter 12: Basic Magnetics Theory
D = 0.3
1
10
0.1
Fundamentals of Power Electronics
FH
100 M = 10 8 6
5
4
67
3 2
ϕ1
1
1.5 1 M = 0.5
10
Chapter 12: Basic Magnetics Theory
D = 0.1
Increased proximity losses induced by PWM waveform harmonics: D = 0.1
Fundamentals of Power Electronics
FH
100
Increased proximity losses induced by PWM waveform harmonics: D = 0.3
Magnetic core materials exhibit hysteresis and saturation. A core material saturates when the flux density B reaches the saturation flux density Bsat.
4.
Chapter 12: Basic Magnetics Theory
The conventional transformer saturates when the applied winding voltseconds are too large. Addition of an air gap has no effect on saturation. Saturation can be prevented by increasing the core cross-sectional area, or by increasing the number of primary turns.
7.
69
Conventional transformers can be modeled using sources representing the MMFs of each winding, and the core MMF. The core reluctance approaches zero in an ideal transformer. Nonzero core reluctance leads to an electrical transformer model containing a magnetizing inductance, effectively in parallel with the ideal transformer. Flux that does not link both windings, or “leakage flux,” can be modeled using series inductors.
6.
Fundamentals of Power Electronics
Air gaps are employed in inductors to prevent saturation when a given maximum current flows in the winding, and to stabilize the value of inductance. The inductor with air gap can be analyzed using a simple magnetic equivalent circuit, containing core and air gap reluctances and a source representing the winding MMF.
5.
Summary of key points
Chapter 12: Basic Magnetics Theory
Ampere’s law relates the total MMF around a loop to the total current passing through the center of the loop. Ampere’s law implies that winding currents are sources of MMF, and that when these sources are included, then the net MMF around a closed path is equal to zero.
3.
68
Faraday’s law relates the voltage induced in a loop of wire to the derivative of flux passing through the interior of the loop.
2.
Fundamentals of Power Electronics
Magnetic devices can be modeled using lumped-element magnetic circuits, in a manner similar to that commonly used to model electrical circuits. The magnetic analogs of electrical voltage V, current I, and resistance R, are magnetomotive force (MMF) F, flux Φ, and reluctance R respectively.
1.
Summary of Key Points
Chapter 12: Basic Magnetics Theory
Fundamentals of Power Electronics
71
Chapter 12: Basic Magnetics Theory
11. An expression for the copper loss in a layer, as a function of the magnetic field strengths or MMFs surrounding the layer, is given in Section 12.4.4. This expression can be used in conjunction with the MMF diagram, to compute the copper loss in each layer of a winding. The results can then be summed, yielding the total winding copper loss. When the effective layer thickness is near to or greater than one skin depth, the copper losses of multiple-layer noninterleaved windings are greatly increased.
10. The magnetic field strengths in the vicinity of the winding conductors can be determined by use of MMF diagrams. These diagrams are constructed by application of Ampere’s law, following the closed paths of the magnetic field lines which pass near the winding conductors. Multiple-layer noninterleaved windings can exhibit high maximum MMFs, with resulting high eddy currents and high copper loss.
Summary of key points
70
The skin and proximity effects lead to eddy currents in winding conductors, which increase the copper loss Pcu in high-current high-frequency magnetic devices. When a conductor has thickness approaching or larger than the penetration depth δ, magnetic fields in the vicinity of the conductor induce eddy currents in the conductor. According to Lenz’s law, these eddy currents flow in paths that tend to oppose the applied magnetic fields.
9.
Fundamentals of Power Electronics
Magnetic materials exhibit core loss, due to hysteresis of the B-H loop and to induced eddy currents flowing in the core material. In available core materials, there is a tradeoff between high saturation flux density Bsat and high core loss Pfe. Laminated iron alloy cores exhibit the highest Bsat but also the highest Pfe, while ferrite cores exhibit the lowest Pfe but also the lowest Bsat. Between these two extremes are powdered iron alloy and amorphous alloy materials.
8.
Summary of key points
72
Chapter 12: Basic Magnetics Theory
Fundamentals of Power Electronics
1
13.5. Summary of key points
Chapter 13: Filter inductor design
13.4. A step-by-step design procedure
13.3. The core geometrical constant Kg
13.2. Filter inductor design constraints
13.1. Several types of magnetic devices, their B-H loops, and core vs. copper loss
Chapter 13. Filter Inductor Design
Fundamentals of Power Electronics
12. Pulse-width-modulated winding currents of contain significant total harmonic distortion, which can lead to a further increase of copper loss. The increase in proximity loss caused by current harmonics is most pronounced in multiple-layer non-interleaved windings, with an effective layer thickness near one skin depth.
Summary of key points
0
DTs
Fundamentals of Power Electronics
I
i(t)
+ –
i(t)
CCM buck example L
Ts
t
3
∆iL
B Bsat
B-H loop, large excitation
Hc0
∆Hc
Hc
Chapter 13: Filter inductor design
minor B-H loop, filter inductor
Filter inductor
Saturable reactor
Magnetic amplifier
Chapter 13: Filter inductor design
SEPIC transformer
Flyback transformer
2
Coupled inductor
Conventional transformer
Fundamentals of Power Electronics
Ac inductor
Filter inductor
Types of magnetic devices:
Different design procedures are employed in the two cases.
• Further reduce Bmax , to reduce core losses
• Choose Bmax to avoid saturation of core, or
A key design decision: the choice of maximum operating flux density Bmax
13.1. Several types of magnetic devices, their B-H loops, and core vs. copper loss
i(t)
∆i
Fundamentals of Power Electronics
i(t)
L
Fundamentals of Power Electronics
4
+ v(t) –
i(t) n turns
∆i
5
t
n i(t)
Φ(t)
Rc
Rg
Rg
air gap reluctance
core B-H loop
∆Hc
Hc
Chapter 13: Filter inductor design
–∆Hc
Bsat
B
Chapter 13: Filter inductor design
+ –
B-H loop, for operation as ac inductor
Ac inductor
• Could use core materials having high saturation flux density (and relatively high core loss), even though converter switching frequency is high
• Air gap is employed
• Flux density chosen simply to avoid saturation
• Loss dominated by dc copper loss
• Negligible core loss, negligible proximity loss
core reluctance Rc Φ
Filter inductor, cont.
Fundamentals of Power Electronics
imp(t)
v1(t)
area λ1
Lmp
v1(t) –
imp(t)
+
i1(t)
6
n1 : n2
+
–
7
t
v2(t)
∆imp
i2(t)
n imp(t) lm
core B-H loop
n 1∆i mp lm
Hc
Chapter 13: Filter inductor design
H(t) =
B-H loop, for operation as conventional transformer
λ1 2n 1A c
B
Chapter 13: Filter inductor design
Conventional transformer
Fundamentals of Power Electronics
• A high-frequency material (ferrite) must be employed
• Flux density is chosen to reduce core loss
• An air gap is employed
• Core loss, copper loss, proximity loss are all significant
Ac inductor, cont.
+ –
Fundamentals of Power Electronics
vg
Two-output forward converter example
Fundamentals of Power Electronics
i1
i2 n2 turns
n1
9
–
v2
+
–
v1
+
I2
i2(t)
I1
i1(t)
B
Hc
Chapter 13: Filter inductor design
B-H loop, large excitation
Hc0
∆Hc
t
∆i2
∆i1
Chapter 13: Filter inductor design
minor B-H loop, coupled inductor
Coupled inductor
8
• A high frequency material (ferrite) must be employed
• Flux density is chosen to reduce core loss
• No air gap is employed
• Core loss, copper loss, and proximity loss are usually significant
Conventional transformer, cont.
Lmp
imp
n1 : n2
i2
B-H loop, for operation in DCM flyback converter
Fundamentals of Power Electronics
vg
+ –
i1
10
B
11
core B-H loop
n 1i 1,pk R c l m R c+R g
–
v
+
Hc
imp(t)
i2(t)
i1(t)
t
t
Chapter 13: Filter inductor design
i1,pk
i1,pk
Chapter 13: Filter inductor design
DCM Flyback transformer
Fundamentals of Power Electronics
• Low-frequency core material can be employed
• Flux density chosen to avoid saturation
• Core loss and proximity loss usually not significant
• Air gap is employed
• A filter inductor having multiple windings
Coupled inductor, cont.
R
L
Chapter 13: Filter inductor design
13
Pcu = Irms2 R
Chapter 13: Filter inductor design
and which has a given winding resistance R, or, equivalently, exhibits a worst-case copper loss of
which carries worst-case current Imax without saturating,
Design inductor having a given inductance L,
Objective:
Fundamentals of Power Electronics
i(t)
12
13.2. Filter inductor design constraints
Fundamentals of Power Electronics
• A high-frequency core material (ferrite) must be used
• Air gap is employed
• Flux density is chosen to reduce core loss
• Core loss, copper loss, proximity loss are significant
DCM flyback transformer, cont.
n i(t)
n turns
+ – Φ(t)
Rc
Rg
14
air gap lg
cross-sectional area Ac
c
>>
R g:
ni ≈ Φ R
Given a peak winding current Imax, it is desired to operate the core flux density at a peak value Bmax. The value of Bmax is chosen to be less than the worst-case saturation flux density of the core material.
g
Fundamentals of Power Electronics
15
Chapter 13: Filter inductor design
This is constraint #1. The turns ratio n and air gap length lg are unknown.
lg nI max = BmaxA cR g = Bmax µ 0
Let I = Imax and B = Bmax :
ni = BA cR
From solution of magnetic circuit:
g
Chapter 13: Filter inductor design
For R
ni = Φ R c + R g
Solve magnetic circuit:
R g = µ gA 0 c
l
l
R c = µ cA c c
13.2.1. Constraint: maximum flux density
Fundamentals of Power Electronics
+ v(t) –
i(t)
Φ
Assumed filter inductor geometry
16
Fundamentals of Power Electronics
K uWA ≥ nA W
Third design constraint:
K uWA
Area available for winding conductors:
nA W
Total area of copper in window:
wire bare area AW
17
Chapter 13: Filter inductor design
core
Wire must fit through core window (i.e., hole in center of core)
core window area WA
Chapter 13: Filter inductor design
13.3.3. Constraint: Winding area
Fundamentals of Power Electronics
This is constraint #2. The turns ratio n, core area Ac, and air gap length lg are unknown.
2 µ0 Ac n2 n L= R g = lg
Must obtain specified inductance L. We know that the inductance is
13.3.2. Constraint: Inductance
18
Chapter 13: Filter inductor design
lb AW
Fundamentals of Power Electronics
R=ρ
n (MLT) AW 19
Chapter 13: Filter inductor design
where (MLT) is the mean-length-per-turn of the winding. The meanlength-per-turn is a function of the core geometry. The above equations can be combined to obtain the fourth constraint:
l b = n (MLT)
where ρ is the resistivity of the conductor material, lb is the length of the wire, and AW is the wire bare area. The resistivity of copper at room temperature is 1.724⋅10-6 Ω-cm. The length of the wire comprising an n-turn winding can be expressed as
R=ρ
The resistance of the winding is
13.2.4 Winding resistance
Fundamentals of Power Electronics
0.65 for low-voltage foil-winding inductor
0.05 to 0.2 for high-voltage transformer (multiple kV)
0.25 to 0.3 for off-line transformer
0.5 for simple low-voltage inductor
Typical values of Ku :
• Additional insulation may be required between windings
• Bobbin uses some window area
• Insulation reduces Ku by a factor of 0.95 to 0.65, depending on wire size and type of insulation
• Round wire does not pack perfectly, which reduces Ku by a factor of 0.7 to 0.55 depending on winding technique
Mechanisms that cause Ku to be less than 1:
Ku is the fraction of the core window area that is filled by copper
also called the “fill factor”
The window utilization factor Ku
R=ρ
20
Chapter 13: Filter inductor design
Fundamentals of Power Electronics
21
A 2c WA Kg = (MLT)
The core geometrical constant Kg is defined as
Chapter 13: Filter inductor design
So we must choose a core whose geometry satisfies the above equation.
• Left-hand side: function of only core geometry
• Right-hand side: specifications or other known quantities
ρL 2I 2max A 2c WA ≥ (MLT) B 2max RK u
Elimination of n, lg, and AW leads to
Core geometrical constant Kg
Fundamentals of Power Electronics
Eliminate the three unknowns, leading to a single equation involving the remaining quantities.
n, lg, and AW, which are unknowns.
Imax, Bmax , µ0, L, Ku, R, and ρ, which are given specifications or other known quantities, and
Ac, WA, and MLT, which are functions of the core geometry,
n (MLT) AW
µ 0 A cn 2 L= lg
These equations involve the quantities
K uWA ≥ nA W
nI max = Bmax
lg µ0
The four constraints:
13.3 The core geometrical constant Kg
22
Chapter 13: Filter inductor design
Fundamentals of Power Electronics
23
Chapter 13: Filter inductor design
The use of centimeters rather than meters requires that appropriate factors be added to the design equations.
The core dimensions are expressed in cm: Core cross-sectional area Ac (cm2) (cm2) Core window area WA Mean length per turn MLT (cm)
The following quantities are specified, using the units noted: Wire resistivity ρ (Ω-cm) Peak winding current Imax (A) Inductance L (H) Winding resistance R (Ω) Winding fill factor Ku Core maximum flux density Bmax (T)
13.4 A step-by-step procedure
Fundamentals of Power Electronics
How the core geometry affects electrical capabilities: A larger Kg can be obtained by increase of Ac ⇒ more iron core material, or WA ⇒ larger window and more copper
How specifications affect the core size: A smaller core can be used by increasing Bmax ⇒ use core material having higher Bsat R ⇒ allow more copper loss
Kg is a figure-of-merit that describes the effective electrical size of magnetic cores, in applications where the following quantities are specified: • Copper loss • Maximum flux density
ρL 2I 2max A 2c WA Kg = ≥ (MLT) B 2max RK u
Discussion
(cm 5)
24
Chapter 13: Filter inductor design
(m)
Fundamentals of Power Electronics
25
Chapter 13: Filter inductor design
The value expressed above is approximate, and neglects fringing flux and other nonidealities.
The air gap length is given in meters.
with Ac expressed in cm2. µ0 = 4π10-7 H/m.
µ 0LI 2max 4 lg = 2 10 B max A c
Determine air gap length
Fundamentals of Power Electronics
Note the values of Ac, WA, and MLT for this core.
Choose a core which is large enough to satisfy this inequality (see Appendix 2 for magnetics design tables).
ρL 2I 2max Kg ≥ 2 10 8 B max RK u
Determine core size
(Henries)
26
(mH/1000 turns)
Chapter 13: Filter inductor design
Units: Ac cm2, L Henries, Bmax Tesla.
Fundamentals of Power Electronics
n=
27
LI max 10 4 Bmax A c
Chapter 13: Filter inductor design
Determine number of turns n
Fundamentals of Power Electronics
L = A L n 2 10 – 9
10B 2max A 2c AL = LI 2max
The required AL is given by:
When AL is specified, it is the core manufacturer’s responsibility to obtain the correct gap length.
AL is equal to the inductance, in mH, obtained with a winding of 1000 turns.
Core manufacturers sell gapped cores. Rather than specifying the air gap length, the equivalent quantity AL is used.
AL
K uW A n (cm 2)
28
ρn (MLT) Aw
(Ω)
Chapter 13: Filter inductor design
29
Chapter 13: Filter inductor design
The core geometrical constant Kg is a measure of the magnetic size of a core, for applications in which copper loss is dominant. In the Kg design method, flux density and total copper loss are specified.
2.
Fundamentals of Power Electronics
A variety of magnetic devices are commonly used in switching converters. These devices differ in their core flux density variations, as well as in the magnitudes of the ac winding currents. When the flux density variations are small, core loss can be neglected. Alternatively, a low-frequency material can be used, having higher saturation flux density.
1.
13.5 Summary of key points
Fundamentals of Power Electronics
R=
As a check, the winding resistance can be computed:
Select wire with bare copper area AW less than or equal to this value. An American Wire Gauge table is included in Appendix 2.
AW ≤
Evaluate wire size
Selection of operating flux density to optimize total loss Multiple winding design: how to allocate the available window area among several windings A transformer design procedure How switching frequency affects transformer size
• •
• •
i1(t)
R1
: nk
–
–
ik(t)
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14.6. Summary
14.5. Ac inductor design
14.4. Examples
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14.3. A step-by-step transformer design procedure
14.2. Transformer design: Basic constraints
14.1. Winding area optimization
Rk
R2
i2(t)
&KDSWHU7UDQVIRUPHUGHVLJQ
–
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+
v2(t)
+
v1(t)
+
n1 : n2
&KDSWHU7UDQVIRUPHU'HVLJQ
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Inclusion of core loss
•
Some more advanced design issues, not considered in previous chapter:
&KDSWHU7UDQVIRUPHU'HVLJQ
v (t) = nk k
Fill factor Ku
Wire resistivity ρ
Core mean length per turn (MLT)
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Q: how should the window area WA be allocated among the windings?
: nk
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etc.
Winding 1 allocation α1WA Winding 2 allocation α2WA
0 < αj < 1 α1 + α2 +
{
{
Ik
rms current
I2
I1
Window area WA
rms current
rms current
n1 : n2
+ αk = 1
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Total window area WA
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Given: application with k windings having known rms currents and desired turns ratios
:LQGLQJDUHDRSWLPL]DWLRQ
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j=1
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&KDSWHU7UDQVIRUPHUGHVLJQ
Need to select values for α1, α2, …, αk such that the total copper loss is minimized
Pcu,tot = Pcu,1 + Pcu,2 +
ρ (MLT) + Pcu,k = WAK u
Sum previous expression over all windings:
7RWDOFRSSHUORVVRIWUDQVIRUPHU
&KDSWHU7UDQVIRUPHUGHVLJQ
wire area, winding j
length of wire, winding j
n 2j i 2j ρ (MLT) Pcu, j = WAK uα j
A W, j =
l j = n j (MLT)
lj A W, j
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Hence
with
Rj = ρ
Resistance of winding j is
Pcu, j = I 2j R j
Copper loss (not accounting for proximity loss) is
&RSSHUORVVLQZLQGLQJM
0
Pcu,tot
1
α1
+ αk = 1
f (α 1, α 2,
, α k) = 1 –
Σα j=1
j
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
and ξ is the Lagrange multiplier
is the constraint that must equal zero
g(α 1, α 2,
k
, α k, ξ) = Pcu,tot(α 1, α 2,
Define the function
α1 + α2 +
subject to the constraint
Pcu,tot = Pcu,1 + Pcu,2 +
Σ
j=1
k
n 2j I 2j αj
&KDSWHU7UDQVIRUPHUGHVLJQ
, α k) + ξ g(α 1, α 2,
ρ (MLT) + Pcu,k = WAK u
, α k)
&KDSWHU7UDQVIRUPHUGHVLJQ
There is a choice of α1 that minimizes the total copper loss
For α1 = 1: wires of remaining windings have zero area. Their copper losses tend to infinity
For α1 = 0: wire of winding 1 has zero area. Pcu,1 tends to infinity
WRPLQLPL]HWRWDOFRSSHUORVV Minimize the function
where
Pc
cu,k
0HWKRGRI/DJUDQJHPXOWLSOLHUV
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Copper loss
u, 2
+
P
cu, 3
+.. .+ P
9DULDWLRQRIFRSSHUORVVHVZLWKα
P cu,1
αm =
αm =
V mI m n=1
V mI m n=1
j j
Σ VI
∞
Ij is the rms current
= Pcu,tot
&KDSWHU7UDQVIRUPHUGHVLJQ
j j
Σ VI
∞
2
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
&KDSWHU7UDQVIRUPHUGHVLJQ
Window area should be allocated according to the apparent powers of the windings
where
j j
Σ nI
n=1
Vj is the rms or peak applied voltage
Apparent power in winding j is V j Ij
n mI m
∞
j j
ΣnI
k
j=1
An alternate form:
αm =
ρ (MLT) ξ= WAK u
Result:
,QWHUSUHWDWLRQRIUHVXOW
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
∂ f (α 1, α 2, , α k,ξ) =0 ∂α k ∂ f (α 1, α 2, , α k,ξ) =0 ∂ξ
∂ f (α 1, α 2, , α k,ξ) =0 ∂α 1 ∂ f (α 1, α 2, , α k,ξ) =0 ∂α 2
Optimum point is solution of the system of equations
FRQWLQXHG
/DJUDQJHPXOWLSOLHUV
{ n2 turns
n2 turns
i3(t)
} }
i2(t)
0
1 2Ts
1 2Ts
2T s
0
2T s
i3(t)
i2(t)
i1(t)
n2 I D n1
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
i1(t)
i3(t)
i2(t)
i 22(t)dt = 12 I 1 + D
i 21(t)dt =
see Appendix 1
I2 = I3 =
I1 =
I
0
0
I
n2 I n1
DTs
0.5I
0.5I
0
I
0
Ts+DTs
0.5I
0.5I
0
0
0
I
n2 I n1
DTs
0.5I
0.5I
0
2Ts
I
0
Ts+DTs
0.5I
0.5I
0
&KDSWHU7UDQVIRUPHUGHVLJQ
Ts
–
n2 I n1
2Ts
&KDSWHU7UDQVIRUPHUGHVLJQ
Ts
–
n2 I n1
([SUHVVLRQVIRU506ZLQGLQJFXUUHQWV
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
• Treat as a threewinding transformer
• Note that waveshapes (and hence rms values) of the primary and secondary currents are different
n1 turns
i1(t)
3:0IXOOEULGJHWUDQVIRUPHU
([DPSOH
t
t
1+
D 1+D
1
j j
&KDSWHU7UDQVIRUPHUGHVLJQ
Fraction of window area allocated to each secondary winding
Fraction of window area allocated to primary winding
1XPHULFDOH[DPSOH
1+
1 1+D D
n=1
Σ VI
∞
V mI m
2
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
&KDSWHU7UDQVIRUPHUGHVLJQ
ρ(MLT) 3 Pcu,tot = n jI j WAK u jΣ =1 ρ(MLT)n 22 I 2 = 1 + 2D + 2 D(1 + D) WAK u
The total copper loss is then given by
α 1 = 0.396 α 2 = 0.302 α 3 = 0.302
Suppose that we decide to optimize the transformer design at the worst-case operating point D = 0.75. Then we obtain
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
α 2 = α 3 = 12
α1 =
Plug in rms current expressions. Result:
$OORFDWLRQRIZLQGRZDUHD
αm =
t1
t2
v1(t)dt
λ1 2n 1A c )XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Bmax =
v1(t)
&KDSWHU7UDQVIRUPHUGHVLJQ
n1 =
λ1 2Bmax A c
t2
area λ1
&KDSWHU7UDQVIRUPHUGHVLJQ
To attain a given flux density, the primary turns should be chosen according to
t1
&RQVWUDLQW
This causes the flux to change from its negative peak to its positive peak. From Faraday’s law, the peak value of the ac component of flux density is
λ1 =
)OX[GHQVLW\ Flux density B(t) is related to the applied winding voltage according to Faraday’s Law. Denote the voltseconds applied to the primary winding during the positive portion of v1(t) as λ1:
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
This is the first constraint
So increasing Bmax causes core loss to increase rapidly
Bmax is the peak value of the ac component of B(t)
Typical value of β for ferrite materials: 2.6 or 2.7
Pfe = K feB βmax A cl m
Core loss
%DVLFFRQVWUDLQWV
7UDQVIRUPHUGHVLJQ
t
ρ(MLT)n I WAK u
(MLT) WAA 2c
1 B 2max
I tot =
with j=1
k
Σ
nj n1 I j
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
(MLT) WAA 2c
1 B 2max
Power loss Ptot
&KDSWHU7UDQVIRUPHUGHVLJQ
Bmax
&KDSWHU7UDQVIRUPHUGHVLJQ
Optimum Bmax
u
ρ λ 21 I 2tot Pcu = Ku
Pfe = K feB βmax A cl m
Ptot = Pfe + Pcu
There is a value of Bmax that minimizes the total power loss
7RWDOSRZHUORVV 3WRW 3FX3IH
Co r e l oss P
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Note that copper loss decreases rapidly as Bmax is increased
ρ λ 21 I 2tot Pcu = Ku
Eliminate n1, using result of previous slide:
Pcu =
2 2 1 tot
fe
• Total copper loss is then equal to
• Allocate window area between windings in optimum manner, as described in previous section
&RQVWUDLQW
&RSSHUORVV
s Pc los r e opp
C
&KDSWHU7UDQVIRUPHUGHVLJQ
A cl m
2 2 1 tot
ρλ I 2K u
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Bmax =
Now, substitute into
(MLT) 1 WAA 3c l m βK fe
1 β+2
&KDSWHU7UDQVIRUPHUGHVLJQ
Optimum Bmax for a given core and application
and solve for Bmax:
1 B 2max (MLT) – 3 B max WAA 2c
(MLT) WAA 2c ρλ 21 I 2tot dPcu =–2 4K u d Bmax
ρ λ 21 I 2tot Pcu = Ku
dPfe dPcu =– dBmax dBmax
dPfe β–1 = βK feB max A cl m dBmax
Pfe = K feB
β max
7DNHGHULYDWLYHVRIFRUHDQGFRSSHUORVV
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
dPfe dPcu =– dBmax dBmax
Note: optimum does not necessarily occur where Pfe = Pcu. Rather, it occurs where
dPfe dPtot dPcu = + =0 d Bmax d Bmax d Bmax
Then, at the Bmax that minimizes Ptot, we can write
Ptot = Pfe + Pcu
Given that
)LQGRSWLPXPIOX[GHQVLW\%PD[
β 2
ρλ I 4K u
β – β+2
β+2 β
=
β β+2
β + 2 /β
2/β
β + 2
ρλ 21I 2totK fe
–
4K u Ptot
β 2
K gfe =
(MLT) l m2/β
β 2
β – β+2
&KDSWHU7UDQVIRUPHUGHVLJQ
4K u Ptot
β + 2 /β
β + 2
2 β+2
–
β+2 β
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
&KDSWHU7UDQVIRUPHUGHVLJQ
• Kgfe is used when Bmax is to be chosen to minimize total loss
• Kg is used when Bmax is specified
Kgfe is similar to the Kg geometrical constant used in Chapter 13:
Appendix 2 lists the values of Kgfe for common ferrite cores
K gfe ≥
2/β
ρλ 21I 2totK fe
Design procedure: select a core that satisfies
Define
WA A c
2(β – 1)/β
2 β+2
Right side: terms depend on specifications of the application
–
β β+2
7KHFRUHJHRPHWULFDOFRQVWDQW.JIH
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
2 β+2
(MLT) WAA 2c
β + 2
2 2 1 tot
Left side: terms depend on core geometry
(MLT) l m2/β
WA A c
2(β – 1)/β
Rearrange as follows:
Ptot = A cl mK fe
2 β+2
Substitute optimum Bmax into expressions for Pcu and Pfe. The total loss is:
7RWDOORVV
4K u Ptot
β + 2 /β
2/β
10 8
'HWHUPLQHFRUHVL]H
&KDSWHU7UDQVIRUPHUGHVLJQ
(cm2) (cm2) (cm) (cm) (cm2) (T)
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
&KDSWHU7UDQVIRUPHUGHVLJQ
It may be possible to reduce the core size by choosing a core material that has lower loss, i.e., lower Kfe.
Select a core from Appendix 2 that satisfies this inequality.
K gfe ≥
3URFHGXUH
ρλ 21I 2totK fe
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Other quantities and their dimensions: Core cross-sectional area Ac Core window area WA Mean length per turn MLT Magnetic path length le Wire areas Aw1, … Peak ac flux density Bmax
The following quantities are specified, using the units noted: Wire effective resistivity ρ (Ω-cm) Total rms winding current, ref to pri Itot (A) Desired turns ratios n2/n1, n3/n1, etc. (V-sec) Applied pri volt-sec λ1 Allowed total power dissipation Ptot (W) Winding fill factor Ku Core loss exponent β Core loss coefficient Kfe (W/cm3Tβ)
6WHSE\VWHS WUDQVIRUPHUGHVLJQSURFHGXUH
ρλ I 2K u (MLT) 1 WAA 3c l m βK fe
n1 =
λ1 10 4 2Bmax A c
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
n2 n1 n3 n1
n2 = n1 n3 = n1
&KDSWHU7UDQVIRUPHUGHVLJQ
&KDSWHU7UDQVIRUPHUGHVLJQ
(YDOXDWHWXUQV
Choose secondary turns according to desired turns ratios:
Primary turns:
DQG
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
• Choose a core material having greater core loss, then repeat steps 1 and 2
• Specify Bmax using the Kg method of Chapter 13, or
If the core will saturate, then there are two choices:
At this point, one should check whether the saturation flux densityis exceeded. If the core operates with a flux dc bias Bdc, then Bmax + Bdc should be less than the saturation flux density.
2 2 1 tot
1 β+2
(YDOXDWHSHDNDFIOX[GHQVLW\
Bmax = 10 8
n kI k n 1I tot
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
R1 =
ρn 1(MLT) A w1 ρn (MLT) R2 = 2 A w2
Predicted winding resistances:
λ i M, pk = 1 2L M
Peak magnetizing current:
2 1
µn A c LM = lm
Predicted magnetizing inductance, referred to primary:
i1(t)
R1
LM
iM(t)
Rk
R2 ik(t)
&KDSWHU7UDQVIRUPHUGHVLJQ
: nk
n1 : n2 i2(t)
&KDSWHU7UDQVIRUPHUGHVLJQ
α 1K uWA n1 αKW A w2 ≤ 2 u A n2 A w1 ≤
Choose wire sizes according to:
&KRRVHZLUHVL]HV
&KHFNFRPSXWHGWUDQVIRUPHUPRGHO
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
αk =
α1 =
n 1I 1 n 1I tot nI α2 = 2 2 n 1I tot
Fraction of window area assigned to each winding:
DQG
Kfe = 24.7
I
I/n
Area λ1
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
i2(t)
i1(t)
DTs
VC1
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
v1(t)
20 A
I
– nIg
– Ig
– nVC2
D'Ts
:DYHIRUPV
β = 2.6
–
5V
V
+
D nI
2
+ D' I g
2
=4A
&KDSWHU7UDQVIRUPHUGHVLJQ
Total rms winding current: I tot = I 1 + 1n I 2 = 8 A
Applied secondary rms current: I 2 = nI 1 = 20 A
I1 =
Applied primary rms current:
λ 1 = DTsVc1 = (0.5) (5 µsec ) (25 V) = 62.5 V–µsec
Applied primary voltseconds:
&KDSWHU7UDQVIRUPHUGHVLJQ
Use a ferrite pot core, with Magnetics Inc. P material. Loss parameters at 200 kHz are
Allow Ptot = 0.25 W
Ku = 0.5
i2(t)
n=5
n:1
–
v2(t)
+
– vC2(t) +
D = 0.5
i1(t)
+
v1(t)
–
+ vC1(t) –
fs = 200 kHz
+ –
4A
Ig
([DPSOH6LQJOHRXWSXWLVRODWHG &XNFRQYHUWHU
100 W
25 V
Vg
–6 2
2
(1.724⋅10 )(62.5⋅10 ) (8) (4.42) 1 2 (0.5) (0.297)(0.635) 3(3.15) (2.6)(24.7)
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
&KDSWHU7UDQVIRUPHUGHVLJQ
This is much less than the saturation flux density of approximately 0.35 T. Values of Bmax in the vicinity of 0.1 T are typical for ferrite designs that operate at frequencies in the vicinity of 100 kHz.
= 0.0858 Tesla
Bmax = 10 8
–6
1/4.6
&KDSWHU7UDQVIRUPHUGHVLJQ
(YDOXDWHSHDNDFIOX[GHQVLW\
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Next smaller pot core is not large enough.
Kgfe = 0.0049
Pot core data of Appendix 2 lists 2213 pot core with
K gfe ≥
(1.724⋅10 – 6)(62.5⋅10 – 6) 2(8) 2(24.7) 2/2.6 10 8 4.6/2.6 4 (0.5) (0.25) = 0.00295
&KRRVHFRUHVL]H
and n2 = 1
&KDSWHU7UDQVIRUPHUGHVLJQ
1 5
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
A w1 =
&KDSWHU7UDQVIRUPHUGHVLJQ
AWG #9
AWG #16
From wire table, Appendix 2:
(Since, in this example, the ratio of winding rms currents is equal to the turns ratio, equal areas are allocated to each winding)
(0.5)(0.5)(0.297) = 14.8⋅10 – 3 cm 2 (5) (0.5)(0.5)(0.297) A w2 = = 74.2⋅10 – 3 cm 2 (1)
8A
= 0.5
= 0.5 20 A
8A
4A
Wire areas:
α2 =
α1 =
Fraction of window area allocated to each winding:
'HWHUPLQHZLUHVL]HV
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
This would lead to a slightly higher flux density and slightly higher loss.
n1 = 5
In practice, we might select
n n 2 = n1 = 1.15 turns
n 1 = 10 4
(62.5⋅10 – 6) 2(0.0858)(0.635) = 5.74 turns
(YDOXDWHWXUQV
25 kHz
4226
50 kHz
3622
100 kHz
2616
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
250 kHz
1811
400 kHz
1811
500 kHz
2213
1000 kHz
0
0.02
0.04
0.06
0.08
0.1
&KDSWHU7UDQVIRUPHUGHVLJQ
• As switching frequency is increased from 400 kHz to 1 MHz, core size increases
Switching frequency
200 kHz
2213
2616
IRUWKLV3PDWHULDO&XNFRQYHUWHUH[DPSOH
• As switching frequency is increased from 25 kHz to 250 kHz, core size is dramatically reduced
Pot core size
(IIHFWRIVZLWFKLQJIUHTXHQF\RQWUDQVIRUPHUVL]H
Use Litz wire or parallel strands of wire
•
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Use foil windings
•
&KDSWHU7UDQVIRUPHUGHVLJQ
One turn of #9 AWG is not a practical solution
Very large conductors!
Some alternatives
•
•
1 turn #9 AWG
Secondary
5 turns #16 AWG
Primary
:LUHVL]HVGLVFXVVLRQ
Bmax , Tesla
D = 0.75
Optimize transformer at
–
n1 :
T1
: n3
: n3
: n2
: n2
D8
D7
D6
D5
i2b(t)
Use Magnetics, Inc. ferrite P material. Loss parameters at 75 kHz:
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
ρ = 1.724·10-6 Ω-cm
&KDSWHU7UDQVIRUPHUGHVLJQ
(approximately 0.5% of total output power) Use copper wire, with
Ptot = 4 W
–
15 V
+
–
5V
+
(reduced fill factor accounts for added insulation required in multiple-output off-line application) Allow transformer total power loss of
Ku = 0.25
Assume fill factor of
Use E-E core shape
β = 2.6
Kfe = 7.6 W/Tβcm3
15 A
I15V
100 A
I5V
&KDSWHU7UDQVIRUPHUGHVLJQ
i3a(t)
i2b(t)
i2a(t)
2WKHUWUDQVIRUPHUGHVLJQGHWDLOV
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
110:5:15
D4
Turns ratio
Q4
75 kHz
D2
+ i1(t) v1(t)
D3
Transformer frequency
Q2
Q3
150 kHz
+ –
D1
0XOWLSOH2XWSXW)XOO%ULGJH%XFN&RQYHUWHU
([DPSOH
Switching frequency
160 V
Vg
Q1
+
n1 :
T1
: n3
: n3
: n2
: n2
D8
D7
D6
D5
i2b(t)
i3a(t)
i2b(t)
i2a(t)
0
0
DTs
0
0
– Vg
Vg
0
Area λ1 = Vg DTs
– Vg
Ts+DTs 2Ts
0
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
t
&KDSWHU7UDQVIRUPHUGHVLJQ
Ts
0.5I15V
0.5I5V
n2 n I + 3I n 1 5V n 1 15V
I15V
I5V
–
n n2 I + 3I n 1 5V n 1 15V
0
Area λ1 = Vg DTs
0
&KDSWHU7UDQVIRUPHUGHVLJQ
λ 1 = DTsVg = (0.75) (6.67 µsec ) (160 V) = 800 V–µsec
v1(t)
i3a(t)
i2a(t)
i1(t)
Vg
$SSOLHGSULPDU\YROWVHFRQGV
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
D4
–
i1(t) v (t) 1
D3
v1(t)
$SSOLHGWUDQVIRUPHUZDYHIRUPV
n2 n I 5V + 3 I 15V n1 n1
0
0
DTs
Ts
0.5I15V
0.5I5V
0
0
I 3 = 12 I 15V 1 + D = 9.9 A
I 2 = 12 I 5V 1 + D = 66.1 A
I15V
I5V
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
i3a(t)
i2a(t)
&KDSWHU7UDQVIRUPHUGHVLJQ
Ts+DTs 2Ts
t
&KDSWHU7UDQVIRUPHUGHVLJQ
D = 5.7 A
$SSOLHGUPVFXUUHQWVHFRQGDU\ZLQGLQJV
n n I 1 = n 2 I 5V + n 3 I 15V 1 1
–
n n2 I 5V + 3 I 15V n1 n1
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
i1(t)
$SSOLHGSULPDU\UPVFXUUHQW
Σ
all 5 windings
K gfe ≥
1.27 2.26
6.7·10-3 11.8·10-3 28.4·10-3
85.7·10-3 0.209 0.909
EE30 EE40 EE50
1.09
1.8·10-3
8.26·10-3 EE22
0.41
Kgfe cmx
Kg cm5 (A) (mm)
Crosssectional area Ac (cm2)
Geometrical constant
Geometrical constant
EE core data
Core type
A2.2
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
From Appendix 2
6HOHFWFRUHVL]H
1.78
1.10
0.476
0.196
Bobbin winding area WA (cm2)
9.58
7.70
5.77
3.96
Magnetic path length lm (cm)
116
50.3
32.4
8.81
(g)
Core weight
&KDSWHU7UDQVIRUPHUGHVLJQ
10.0
8.50
6.60
3.99
Mean length per turn MLT (cm)
A
&KDSWHU7UDQVIRUPHUGHVLJQ
(1.724⋅10 – 6)(800⋅10 – 6) 2(14.4) 2(7.6) 2/2.6 10 8 4.6/2.6 4 (0.25) (4) = 0.00937
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
= 5.7 A + 5 66.1 A + 15 9.9 A 110 110 = 14.4 A
I tot =
nj n2 n3 I = I + 2 I + 2 j 1 2 n1 n1 n1 I 3
RMS currents, summed over all windings and referred to primary
,WRW
–6
–6 2
Bmax = 10 8
ρλ I 2K u
2
2 2 1 tot
(MLT) 1 WAA 3c l m βK fe
1 β+2
(1.724⋅10 )(800⋅10 ) (14.4) (8.5) 1 3 2 (0.25) (1.1)(1.27) (7.7) (2.6)(7.6)
15 n = 1.87 turns 110 1 )XQGDPHQWDOVRI3RZHU(OHFWURQLFV
n3 =
5 n2 = n = 0.62 turns 110 1
Choose secondary turns according to desired turns ratios:
n 1 = 10 4
(800⋅10 – 6) 2(0.23)(1.27) = 13.7 turns
λ1 n1 = 10 4 2Bmax A c
1/4.6
&KDSWHU7UDQVIRUPHUGHVLJQ
&KDSWHU7UDQVIRUPHUGHVLJQ
• Increased total loss
• More copper loss
• Less core loss
Increased n1 would lead to
22:1:3
we might round the actual turns to
110:5:15
To obtain desired turns ratio of
Rounding the number of turns
(YDOXDWHWXUQV
Choose n1 according to Eq. (14.42):
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
This is less than the saturation flux density of approximately 0.35 T
= 0.23 Tesla
Bmax = 10 8
Plug in values:
Eq. (14.41):
(YDOXDWHDFIOX[GHQVLW\%PD[
&KDSWHU7UDQVIRUPHUGHVLJQ
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Which is close enough to 4 W.
&KDSWHU7UDQVIRUPHUGHVLJQ
Bmax = 0.08 T, Pcu = 3.89 W, Pfe = 0.23 W, Ptot = 4.12 W
Again round n1 to 22. Then
Bmax = 0.14 T, n1 = 12, Ptot = 2.3 W
Repeat previous calculations for EE50 core size. Results:
&DOFXODWLRQVZLWK((
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Which exceeds design goal of 4 W by 50%. So use next larger core size: EE50.
Ptot = Pfe + Pcu = 5.9 W
Pcu =
(8.5) (1.724⋅10 – 6)(800⋅10 – 6) 2(14.4) 2 8 1 2 2 10 4 (0.25) (1.1)(1.27) (0.143) = 5.4 W
Pfe = (7.6)(0.143) 2.6(1.27)(7.7) = 0.47 W
The resulting losses will be
(800⋅10 – 6) Bmax = 10 4 = 0.143 Tesla 2 (22) (1.27)
With n1 = 22, the flux density will be reduced to
ZLWKURXQGHGWXUQV
/RVVFDOFXODWLRQ
⇒ AWG #8 αKW (0.094)(0.25)(1.78) A w3 = 3 u A = = 13.9⋅10 – 3 cm 2 (3) n3
n 2I 2 = 5 66.1 = 0.209 n 1I tot 110 14.4
n 3I 3 = 15 9.9 = 0.094 n 1I tot 110 14.4
α2 =
α3 =
Part IV
&KDSWHU7UDQVIRUPHUGHVLJQ
Low Harmonic Rectifier Modeling and Control
Chapter 18
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
The Ideal Rectifier
Chapter 17
1
Line-Commutated Rectifiers
Chapter 16
Fundamentals of Power Electronics
Power and Harmonics in Nonsinusoidal Systems
Chapter 15
Modern Rectifiers and Power System Harmonics
)XQGDPHQWDOVRI3RZHU(OHFWURQLFV
Might actually use foil or Litz wire for secondary windings
⇒ AWG #16
α 1K uWA (0.396)(0.25)(1.78) = = 8.0⋅10 – 3 cm 2 (22) n1 ⇒ AWG #19 αKW (0.209)(0.25)(1.78) = 93.0⋅10 – 3 cm 2 A w2 = 2 u A = (1) n2
A w1 =
I1 = 5.7 = 0.396 I tot 14.4
Wire gauges
α1 =
Window allocations
:LUHVL]HVIRU((GHVLJQ
Fundamentals of Power Electronics
Express voltage and current as Fourier series:
Source
i(t) = I0
∞
n
n
Surface S
–
v(t)
3
n=1
n=1 ∞
n
n
Load
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Σ V cos nωt – ϕ + Σ I cos nωt – θ v(t) = V0 +
+ –
+
i(t)
Observe transmission of energy through surface S
relate energy transmission to harmonics
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
15.1. Average power
2
Average power in terms of Fourier series RMS value of a waveform Power factor THD Distortion and Displacement factors Power phasors in sinusoidal systems Harmonic currents in three-phase systems AC line current harmonic standards
Fundamentals of Power Electronics
15.4. 15.5. 15.6.
15.2. 15.3.
15.1.
Power And Harmonics in Nonsinusoidal Systems
Chapter 15
0
v(t)i(t)dt
0
T
v(t)i(t)dt
0
n=1
∞
ΣV
4
I0 +
n=1
∞
ΣI
n
cos nωt – θ n dt
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
n cos nωt – ϕ n
Vn cos nωt – ϕ n
I m cos mωt – θ m dt =
n=1
Σ
V nI n cos ϕ n – θ n 2
if n ≠ m if n = m
0 V nI n cos ϕ n – θ n 2
Fundamentals of Power Electronics
2
3 3
5
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
cos ϕ 3 – θ 3
So net energy is transmitted to the load only when the Fourier series of v(t) and i(t) contain terms at the same frequency. For example, if the voltage and current both contain third harmonic, then they lead to the average power VI
Pav = V0I 0 +
∞
Expression for average power becomes
0
T
Orthogonality of harmonics: Integrals of cross-product terms are zero
Evaluation of integral
V0 +
Fundamentals of Power Electronics
Pav = 1 T
T
Investigate influence of harmonics on average power: substitute Fourier series
Wcycle 1 Pav = = T T
This is related to average power as follows:
Wcycle =
T
Energy transmittted to load, per cycle
Fundamentals of Power Electronics
Power: nonzero average
Current: third harmonic only, in phase with voltage
Voltage: third harmonic only
Fundamentals of Power Electronics
Power: zero average
Current: third harmonic only
Voltage: fundamental only
6
7
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
i(t)
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
p(t) = v(t) i(t)
v(t), i(t)
Pav = 0.5
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Pav = 0
p(t) = v(t) i(t)
v(t)
Example 2
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
Example 1
Fundamentals of Power Electronics
Power: net energy is transmitted at fundamental and fifth harmonic frequencies
Current: 1st, 5th, 7th
Voltage: 1st, 3rd, 5th
p(t) = v(t) i(t)
i(t)
v(t)
9 -0.2 Chapter 15: Power and Harmonics in Nonsinusoidal Systems
0.0
0.2
0.4
0.6
-1.0
-0.5
0.0
0.5
1.0
Pav = 0.32
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Example 3
8
(1.2)(0.6) (0.2)(0.1) cos (30°) + cos (45°) = 0.32 2 2
Fundamentals of Power Electronics
Pav =
Average power calculation:
v(t) = 1.2 cos (ωt) + 0.33 cos (3ωt) + 0.2 cos (5ωt) i(t) = 0.6 cos (ωt + 30°) + 0.1 cos (5ωt + 45°) + 0.1 cos (7ωt + 60°)
Fourier series:
Example 3
0
T
v 2(t)dt
Σ
(average power) (rms voltage) (rms current)
Fundamentals of Power Electronics
11
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Power factor always lies between zero and one.
power factor =
For efficient transmission of energy from a source to a load, it is desired to maximize average power, while minimizing rms current and voltage (and hence minimizing losses). Power factor is a figure of merit that measures how efficiently energy is transmitted. It is defined as
15.3. Power factor
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Increased rms values mean increased losses
•
10
Harmonics do not necessarily increase average power
•
Fundamentals of Power Electronics
Harmonics always increase rms value
•
n=1
V 2n 2
Similar expression for current
V +
∞
•
(rms value) =
2 0
Insert Fourier series. Again, cross-multiplication terms have zero average. Result is
(rms value) =
1 T
15.2. Root-mean-square (RMS) value of a waveform, in terms of Fourier series
Vn R
I +
2 0
Σ
∞
Σ
n=1
V 2n 2
I 2n = 2
n=1
∞
V +
2 0
θn = ϕn
12
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
∞ V 20 V 2n +Σ R 2 n = 1 2R 2
so cos (θ n – ϕ n) = 1
Fundamentals of Power Electronics
n=1
I 20 + ∑
∞
I 2n 2
I 2n 2 cos (ϕ 1 – θ 1)
13
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
= (distortion factor) (displacement factor)
(power factor) =
Σ
∞
n=1
I1 2
I +
2 0
V 1I 1 cos (ϕ 1 – θ 1) 2
(rms current) =
Pav =
With a sinusoidal voltage, current harmonics do not lead to average power. However, current harmonics do increase the rms current, and hence they decrease the power factor.
15.3.2. Nonlinear dynamical load, sinusoidal voltage
Fundamentals of Power Electronics
= 1 (rms voltage) R ∞ VI Pav = V0I 0 + Σ n n cos (ϕ n – θ n) 2 n=1
(rms current) =
(rms voltage) =
In =
Then current harmonics are in phase with, and proportional to, voltage harmonics. All harmonics result in transmission of energy to load, and unity power factor occurs.
15.3.1. Linear resistive load, nonsinusoidal voltage
∞
n=1
I +∑ 2 0
I 2n 2
I1
n=2
2 n
70%
80%
90%
100%
14
(rms fundamental current) (rms current)
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
1 1 + (THD) 2
=
15
80%
60%
40%
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
THD
Distortion factor vs. THD
0% Fundamentals of Power Electronics
Distortion factor
Fundamentals of Power Electronics
(distortion factor) =
(THD) =
ΣI
∞
Related to Total Harmonic Distortion (THD):
(distortion factor) =
I1 2
Defined only for sinusoidal voltage.
Distortion factor
100%
20%
0%
20%
40%
60%
80%
100%
16
1
100%
5
73%
9
32%
11
13
15
17
19
19% 15% 15% 13% 9%
Harmonic number
7
52%
THD = 136% Distortion factor = 59%
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
3
91%
Fundamentals of Power Electronics
17
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
(ac voltage) (derated breaker current) (power factor) (rectifier efficiency) = (120 V) (80% of 15 A) (0.99) (0.93) = 1325 W
at unity power factor
(ac voltage) (derated breaker current) (power factor) (rectifier efficiency) = (120 V) (80% of 15 A) (0.55) (0.98) = 776 W
with peak detection rectifier
Maximum power obtainable from 120V 15A wall outlet
Fundamentals of Power Electronics
Typical ac line current spectrum
Conventional singlephase peak detection rectifier
Peak detection rectifier example
Harmonic amplitude, percent of fundamental
18
Fundamentals of Power Electronics
19
|
I
ϕ1 – θ1
ϕ1 – θ1 θ ϕ1 1
||S|
I ms s r rm V =
V
P
Real axis
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Imaginary axis Q
In this purely sinusoidal case, the distortion factor is unity, and the power factor coincides with the displacement factor.
power factor = P = cos ϕ 1 – θ 1 S
The phase angle between the voltage and current, or (ϕ1 – θ1), coincides with the angle of S. The power factor is
S = VI*
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Example: power phasor diagram
Fundamentals of Power Electronics
with I* = complex conjugate of I, j = square root of –1. The magnitude of S is the apparent power (VA). The real part of S is the average power P (watts). The imaginary part of S is the reactive power Q (reactive volt-amperes, or VARs).
S = VI * = P + jQ
Apparent power is the product of the rms voltage and rms current It is easily measured —simply the product of voltmeter and ammeter readings Unit of apparent power is the volt-ampere, or VA Many elements, such as transformers, are rated according to the VA that they can supply So power factor is the ratio of average power to apparent power With sinusoidal waveforms (no harmonics), we can also define the real power P reactive power Q complex power S If the voltage and current are represented by phasors V and I, then
15.4. Power phasors in sinusoidal systems
20
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Fundamentals of Power Electronics
21
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
At the fundamental frequency, phase-controlled rectifiers can be viewed as consumers of reactive power Q, similar to inductive loads.
This lagging current does not arise from energy storage, but it does nonetheless lead to a reduced displacement factor, and to rms current and apparent power that are greater than the minimum amount necessary to transmit the average power.
It will be seen in the next chapter that phase-controlled rectifiers produce a nonsinusoidal current waveform whose fundamental component lags the voltage.
Lagging fundamental current of phasecontrolled rectifiers
Fundamentals of Power Electronics
They are often placed in the utility power distribution system near inductive loads. If Q supplied by capacitor is equal to Q consumed by inductor, then the net current (flowing from the source into the capacitorinductive-load combination) isin phase with the voltage, leading to unity power factor and minimum rms current magnitude.
Capacitors supply reactive power Q.
Capacitor: current leads voltage by 90˚, hence displacement factor is zero.
Just as resistors consume real (average) power P, inductors can be viewed as consumers of reactive power Q.
The alternate storing and releasing of energy in an inductor leads to current flow and nonzero apparent power, but P = 0.
Inductor: current lags voltage by 90˚, hence displacement factor is zero.
The reactive power Q does not lead to net transmission of energy between the source and load. When Q ≠ 0, the rms current and apparent power are greater than the minimum amount necessary to transmit the average power P.
Reactive power Q
22
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
b
vbn(t)
+ –
i a(t) = I a0 +
n
vcn(t)
∞
ib(t)
neutral connection
in(t)
c ic(t)
ia(t)
Fundamentals of Power Electronics
k=1
23
van(t) = Vm cos (ωt) vbn(t) = Vm cos (ωt – 120˚) vcn(t) = Vm cos (ωt + 120˚)
nonlinear loads
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Σ I ak cos (kωt – θ ak) Fourier series of k=1 ∞ line currents and i b(t) = I b0 + Σ I bk cos (k(ωt – 120˚) – θ bk) voltages: k=1 ∞ i c(t) = I c0 + Σ I ck cos (k(ωt + 120˚) – θ ck)
ideal 3ø source
van(t)
+ –
a
15.5.1. Harmonic currents in three-phase four-wire networks
Fundamentals of Power Electronics
The presence of harmonic currents can also lead to some special problems in three-phase systems: • In a four-wire three-phase system, harmonic currents can lead to large currents in the neutral conductors, which may easily exceed the conductor rms current rating • Power factor correction capacitors may experience significantly increased rms currents, causing them to fail In this section, these problems are examined, and the properties of harmonic current flow in three-phase systems are derived: • Harmonic neutral currents in 3ø four-wire networks • Harmonic neutral voltages in 3ø three-wire wye-connected loads
15.5. Harmonic currents in three phase systems
+ –
I ak cos (kωt – θ ak) + I bk cos (k(ωt – 120˚) – θ bk) + I ck cos (k(ωt + 120˚) – θ ck)
24
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
3I k cos (kωt – θ k)
Fundamentals of Power Electronics
i n, rms = 3
25
Σ
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
k = 3,6,9,
I +
I 2k 2
Rms neutral current is
•
∞
Triplen (triple-n, or 0, 3, 6, 9, ...) harmonics do not cancel out, but add. Dc components also add.
•
2 0
Fundamental and most harmonics cancel out
3I k cos (kωt – θ k) •
k = 3,6,9,
Σ
∞
Neutral currents
k = 3,6,9,
Σ
∞
i n(t) = 3I 0 +
Fundamentals of Power Electronics
i n(t) = 3I 0 +
In the balanced case, Iak = Ibk = Ick = Ik and θak = θbk = θck = θk , for all k; i.e., the harmonics of the three phases all have equal amplitudes and phase shifts. The neutral current is then
If the load is unbalanced, then there is nothing more to say. The neutral connection may contain currents having spectrum similar to the line currents.
k=1
Σ
∞
i n(t) = I a0 + I b0 + I c0 +
Neutral current
(0.2I 1) 2 0.6 I 1 i n, rms = 3 = 2 2 2 2 I 1 + (0.2I 1) I I i 1, rms = = 1 1 + 0.04 ≈ 1 2 2 2
26
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
b
vbn(t)
Fundamentals of Power Electronics
ideal 3ø source
van(t)
+ –
+ –
a
n
vcn(t)
27
in(t) = 0
+ vn'n –
n'
nonlinear loads
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
ib(t)
c ic(t)
ia(t)
Wye-connected nonlinear load, no neutral connection:
15.5.2. Harmonic currents in three-phase three-wire networks
Fundamentals of Power Electronics
• Yet the presence of the third harmonic has very little effect on the rms value of the line current. Significant unexpected neutral current flows.
• The triplen harmonics in the three phases add, such that 20% third harmonic leads to 60% third harmonic neutral current.
• The neutral current magnitude is 60% of the line current magnitude!
Solution
A balanced nonlinear load produces line currents containing fundamental and 20% third harmonic: ian(t) = I1 cos(ωt – θ 1 ) + 0.2 I1 cos(3ωt – θ 3). Find the rms neutral current, and compare its amplitude to the rms line current amplitude.
Example
+ –
i n(t) = 3I 0 +
Σ
k = 3,6,9,
∞
3I k cos (kωt – θ k)
b
vbn(t)
+ – n
vcn(t)
in(t) = 0 ib(t)
c ic(t)
ia(t)
deltaconnected nonlinear loads
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Delta-connected load
28
Fundamentals of Power Electronics
29
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
• The load currents may contain dc and triplen harmonics: with a balanced nonlinear load, these circulate around the delta.
• There is again no neutral connection, so the ac line currents contain no dc or triplen harmonics
ideal 3ø source
van(t)
+ –
a
Fundamentals of Power Electronics
With an unbalanced load, the line currents can still contain dc and triplen harmonics.
The load neutral point voltage contains dc and triplen harmonics.
A voltage is induced at the load neutral point, that causes the line current dc and triplen harmonics to become zero.
What happens:
So the ac line currents cannot contain dc or triplen harmonics.
But in(t) = 0, since there is no neutral connection.
If the load is balanced, then it is still true that
No neutral connection
+ –
I rms 2π fC
30
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
C
esr
Fundamentals of Power Electronics
31
15.6.3. IEEE/ANSI Standard 519
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
15.6.2. International Electrotechnical Commission Standard 555
15.6.1. US MIL-STD-461B
15.6. AC line current harmonic standards
Fundamentals of Power Electronics
I 2rms rated reactive power = 2π fC
rated rms voltage Vrms =
Heating in capacitors is a function of capacitor equivalent series resistance (esr) and rms current. The maximum allowable rms current then leads to the capacitor rating:
PFC capacitors are usually not intended to conduct significant harmonic currents.
Harmonic currents in power factor correction capacitors
The shipboard application is a good example of the problems faced in a relatively small stand-alone power system having a large fraction of electronic loads.
•
33
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
First draft of their IEC-555 standard:1982. It has since undergone a number of revisions. Enforcement of IEC-555 is the prerogative of each individual country, and hence it has been sometimes difficult to predict whether and where this standard will actually be given the force of law. Nonetheless, IEC-555 is now enforced in Europe, making it a de facto standard for commercial equipment intended to be sold worldwide. IEC-555 covers a number of different types of low power equipment, with differing harmonic limits. Harmonics for equipment having an input current of up to 16A, connected to 50 or 60 Hz, 220V to 240V single phase circuits (two or three wire), as well as 380V to 415V three phase (three or four wire) circuits, are limited.
Fundamentals of Power Electronics
•
•
•
•
15.6.2. International Electrotechnical Commission Standard 555
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Harmonic limits are now employed by all of the US armed forces. The specific limits are often tailored to the specific application.
•
32
For the nth harmonic with n > 33, the harmonic magnitude may not exceed (1/n) times the fundamental magnitude.
•
Fundamentals of Power Electronics
For loads of 1kW or greater, no current harmonic magnitude may be greater than 3% of the fundamental magnitude.
•
15.6.1. US MIL-STD-461B
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Fundamentals of Power Electronics
35
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Class B: Portable tools, and similar devices. The limits are equal to the Table 15.1 limits, multiplied by 1.5.
Class A: Balanced three-phase equipment, and any equipment which does not fit into the other categories. This class includes low harmonic rectifiers for computer and other office equipment. These limits are given in Table 15.1, and are absolute ampere limits.
IEC-555: Class A and B
34
In a city environment such as a large building, a large fraction of the total power system load can be nonlinear Example: a major portion of the electrical load in a building is comprised of fluorescent lights, which present a very nonlinear characteristic to the utility system. A modern office may also contain a large number of personal computers, printers, copiers, etc., each of which may employ peak detection rectifiers. Although each individual load is a negligible fraction of the total local load, these loads can collectively become significant.
Fundamentals of Power Electronics
•
•
•
•
Low-power harmonic limits
1.08A 0.43A 0.30A 0.23A · (8/n)
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
2 4 6 8 ≤ n ≤ 40
IEC-555: Class C
36
2.30A 1.14A 0.77A 0.40A 0.33A 0.21A 0.15A · (15/n)
Even harmonics Harmonic number Maximum curre
Fundamentals of Power Electronics
37
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
Class C: Lighting equipment, including dimmers and gas discharge lamps. The input current harmonics of ballasted lamps with an input power of more than 25W must meet the limits of Table 15.2, expressed as a percent of the fundamental current. If the input power is less than 25W, then Table 15.3 applies. Incandescent lamp fixtures containing phasecontrol dimmers, rated at greater than 600W, must meet the limits of Table 15.1. When testing for compliance, the dimmer must drive a rated-power lamp, with the phase control set to a firing angle of 90˚±5˚. Incandescent lamp dimmers rated at less than 600W are not covered by the standard. Discharge lamps containing dimmers must meet the limits of both Tables 15.2 and 15.3 at maximum load. Harmonic currents at any dimming position may not exceed the maximum load harmonic currents.
Fundamentals of Power Electronics
3 5 7 9 11 13 15 ≤ n ≤ 39
Odd harmonics Harmonic number Maximum current
Table 15.1. IEC-555 Harmonic current limits, Class A and certain Class C
Class A limits
Fundamentals of Power Electronics
39
θpk – 30˚
θpk + 30˚
180˚
ωt
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
0˚
Class D: Equipment not covered by Class B or C, which has a special waveshape as defined below. This class is directed at diode-capacitor peak-detection rectifiers. Equipment is placed in class D if the ac input current waveshape lies within the shaded area for at least 95% of the duration of each half-cycle. The center line of the shaded area is set to coincide with the peak of the current waveform. A Ipk sinusoid, and a typical peak detection rectifier waveform, are shown for reference; the sinusoid is not Class D but the peak detection rectifier 0.35 Ipk waveform is. The limits for Class D equipment are given in Table 15.3. θpk
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
IEC-555: Class D
38
2% (30%) · (power factor) 10% 7% 5% 3%
2 3 5 7 9 11 ≤ n ≤ 39
Fundamentals of Power Electronics
Maximum current, percent of fundamental
Harmonic number
Table 15.2. IEC-555 Harmonic current limits, certain Class C
Class C limits
3.6 2.0 1.5 1.0 0.6 · (11/n)
3 5 7 9 11 ≤ n ≤ 39
1.0 0.5
Relative limit (mA/W)
0.3A 0.15A
Absolute limit (A)
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
2 4
Harmonic number
41
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
In 1993, the IEEE published a revised draft standard limiting the amplitudes of current harmonics, IEEE Guide for Harmonic Control and Reactive Compensation of Static Power Converters. Harmonic limits are based on the ratio of the fundamental component of the load current IL to the short circuit current at the point of common (PCC) coupling at the utility Isc. Stricter limits are imposed on large loads than on small loads. The limits are similar in magnitude to IEC-555, and cover high voltage loads (of much higher power) not addressed by IEC-555. Enforcement of this standard is presently up to the local utility company. The odd harmonic limits are listed in Tables 15.4 and 15.5. The limits for even harmonics are 25% of the odd harmonic limits. Dc current components and half-wave rectifiers are not allowed.
Fundamentals of Power Electronics
•
•
•
•
40
0.33A
2.30A 1.14A 0.77A 0.40A
Absolute limit (A)
Even harmonics
15.6.3. IEEE/ANSI Standard 519
Fundamentals of Power Electronics
Relative limit (mA/W)
Harmonic number
Odd harmonics
Table 15.3. IEC-555 Harmonic current limits, Class D and certain Class C
Class D limits
4.0% 7.0% 10.0% 12.0% 15.0%
<20 20–50 50–100 100–1000 >1000
7.0%
5.5%
3.5% 4.5%
2.0%
11≤n<17
42
2.5%
2.0%
1.0% 1.5%
0.6%
23≤n<35
1.4%
1.0%
0.5% 0.7%
0.3%
35≤n
20.0%
15.0%
8.0% 12.0%
5.0%
THD
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
6.0%
5.0%
2.5% 4.0%
1.5%
17≤n<23
2.0% 3.5% 5.0% 6.0% 7.5%
<20 20–50 50–100 100–1000 >1000
Fundamentals of Power Electronics
n < 11
Isc/IL
3.5%
1.0% 1.75% 2.25% 2.75%
11≤n<17
43
1.25%
0.3% 0.5% 0.75% 1.0%
23≤n<35
0.7%
0.15% 0.25% 0.35% 0.5%
35≤n
10.0%
2.5% 4.0% 6.0% 7.5%
THD
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
3.0%
0.75% 1.25% 2.0% 2.5%
17≤n<23
Table 15.5. IEEE-519 Maximum odd harmonic current limits for general distribution systems, 69.001kV through 161kV
IEEE-519 current limits, high voltage systems
Fundamentals of Power Electronics
n < 11
Isc/IL
Table 15.4. IEEE-519 Maximum odd harmonic current limits for general distribution systems, 120V through 69kV
IEEE-519 current limits, low voltage systems
3.0% 1.5% 1.0%
69kV and below 69.001kV–161kV above 161kV
Fundamentals of Power Electronics
1
16.1 The single-phase full-wave rectifier 16.1.1 Continuous conduction mode 16.1.2 Discontinuous conduction mode 16.1.3 Behavior when C is large 16.1.4 Minimizing THD when C is small 16.2 The three-phase bridge rectifier 16.2.1 Continuous conduction mode 16.2.2 Discontinuous conduction mode
1.5%
5.0% 2.5%
THD
Chapter 15: Power and Harmonics in Nonsinusoidal Systems
&KDSWHU
44
Chapter 16: Line-commutated rectifiers
Phase control 16.3.1 Inverter mode 16.3.2 Harmonics and power factor 16.3.3 Commutation 16.4 Harmonic trap filters 16.5 Transformer connections 16.6 Summary
16.3
/LQH&RPPXWDWHG5HFWLILHUV
Fundamentals of Power Electronics
It is the responsibility of the utility to meet these limits.
Individual harmonics
Bus voltage at PCC
Table 15.6. IEEE-519 voltage distortion limits
IEEE-519 voltage limits
D3
D4
D2
D1
L
C
+ R
10 ms
ig(t)
vg(t)
20 ms
THD =
3
40 ms
– 1 = 48.3%
Chapter 16: Line-commutated rectifiers
2
I 1, rms = 4 = 90.0% I rms π 2
30 ms
THD = 29%
t
Chapter 16: Line-commutated rectifiers
1 distortion factor
distortion factor =
CCM results, for L →∞ :
Fundamentals of Power Electronics
As L →∞, ac line current approaches a square wave
Typical ac line waveforms for CCM :
Large L
2
&RQWLQXRXVFRQGXFWLRQPRGH
Fundamentals of Power Electronics
• Filter conducted EMI generated by dc load (small L and C)
• Obtain good dc output voltage (large C) and acceptable ac line current waveform (large L)
–
v(t)
Two common reasons for including the dc-side L-C filter:
Zi
iL(t)
Full-wave rectifier with dc-side L-C filter
vg(t)
ig(t)
7KHVLQJOHSKDVHIXOOZDYHUHFWLILHU
Solution of the full-wave rectifier circuit for infinite C:
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.0001
cos (ϕ1 − θ1), PF, M
Fundamentals of Power Electronics
M= V Vm
K L = 2L RTL
Define
30 ms
40 ms
t
4
5
KL
DCM
β
180˚
135˚
90˚
45˚
0˚
200%
150%
100%
50%
0
β
Chapter 16: Line-commutated rectifiers
CCM
THD
M
PF
cos (ϕ1 − θ1)
THD
Chapter 16: Line-commutated rectifiers
Typical distortion factor of a full-wave rectifier with no inductor is in the range 55% to 65%, and is governed by ac system inductance.
0.001
%HKDYLRUZKHQ&LVODUJH
Fundamentals of Power Electronics
20 ms
THD = 145%
As the inductance is reduced, the THD rapidly increases, and the distortion factor decreases.
10 ms
0.01
As L →0, ac line current approaches impulse functions (peak detection)
ig(t)
0.1
Typical ac line waveforms for DCM :
vg(t)
1
Small L
'LVFRQWLQXRXVFRQGXFWLRQPRGH
10
1
Fundamentals of Power Electronics
2π LC L R0 = C Q= R R0 f fp = 1 = 0 2πRC Q
f0 =
Fundamentals of Power Electronics
vg(t)
0.1
1
10
50
D2
D3 6
D1
D4
L
Zi
1
7
10
f0 / fL
–
v(t)
+ R
100
THD=0.5%
THD=1%
THD=3%
THD=10%
Chapter 16: Line-commutated rectifiers
THD=30%
Chapter 16: Line-commutated rectifiers
C
$SSUR[LPDWH7+'
g
An approximate argument: the L-C filter has negligible effect on the ac line current waveform provided that the filter input impedance Zi has zero phase shift at the second harmonic of the ac line frequency, 2 fL. i (t) i (t) L
Sometimes the L-C filter is present only to remove high-frequency conducted EMI generated by the dc load, and is not intended to modify the ac line current waveform. If L and C are both zero, then the load resistor is connected directly to the output of the diode bridge, and the ac line current waveform is purely sinusoidal.
0LQLPL]LQJ7+'ZKHQ&LVVPDOO
Q
20 ms
30 ms
40 ms
THD = 3.6%
t
øc
øb
øa
ia(t)
Fundamentals of Power Electronics
Line current waveform for infinite L
3ø ac
8
Chapter 16: Line-commutated rectifiers
ia(ωt)
D4
D1
0
D5
D2
9
90˚
iL
D6
D3
iL(t)
180˚
L
C
360˚
ωt
dc load R
Chapter 16: Line-commutated rectifiers
–iL
270˚
–
V
+
7KH7KUHH3KDVH%ULGJH5HFWLILHU
Fundamentals of Power Electronics
and discontinuous modes and with small dc filter capacitor. f0/fL = 10, Q = 1
Typical ac line current and voltage waveforms, near the boundary between continuous
10 ms
ig(t)
vg(t)
([DPSOH
Σ
THD = 31% Distortion factor = 3/π = 95.5% In comparison with single phase case:
• • •
ia(ωt) 0 90˚
iL
180˚
–iL
270˚
10
vcn(t)
20 ms
vbn(t)
30 ms
40 ms
THD = 31.9%
Fundamentals of Power Electronics
11
Chapter 16: Line-commutated rectifiers
Inductor current contains sixth harmonic ripple (360 Hz for a 60 Hz ac system). This ripple is superimposed on the ac line current waveform, and influences the fifth and seventh harmonic content of ia(t).
10 ms
van(t)
ia(t)
t
Chapter 16: Line-commutated rectifiers
$W\SLFDO&&0ZDYHIRUP
Fundamentals of Power Electronics
ωt 360˚
the missing 60˚ of current improves the distortion factor from 90% to 95%, because the triplen harmonics are removed
Similar to square wave, but missing triplen harmonics
n = 1,5,7,11,...
nπ nπ 4 nπ I L sin 2 sin 3 sin nωt
•
i a(t) =
∞
Fourier series:
&RQWLQXRXVFRQGXFWLRQPRGH
20 ms
vbn(t)
vcn(t)
30 ms
40 ms
THD = 99.3%
t
Q4
Q1
Q5
Q2
Q6
Q3
iL(t)
–
vd(t)
+
L
C
30˚+α
90˚+α
3 Vm sin(θ + 30˚)dθ
Fundamentals of Power Electronics
= 3 π2 VL-L, rms cos α
3 V=π
Average (dc) output voltage:
øc
øb
ia(t)
Replace diodes with SCRs:
øa
–
V
+
13
0˚
0
Lower thyristor:
vab
Q5
Q1
– vca
90˚
iL
Q6
Q1
vbc
Q6
Q2
180˚
– vab
Q4
Q2
270˚
vca vd (t)
Q4
Q3
– iL
van(t) = Vm sin (ωt)
Chapter 16: Line-commutated rectifiers
Q5
Q3
– vbc
Upper thyristor:
dc load R
ia(t)
α
Phase control waveforms:
ωt
Chapter 16: Line-commutated rectifiers
3KDVHFRQWURO
12
Phase a current contains pulses at the positive and negative peaks of the line-to-line voltages vab(t) and vac(t). Distortion factor and THD are increased. Distortion factor of the typical waveform illustrated above is 71%.
Fundamentals of Power Electronics
3ø ac
van(t)
10 ms
ia(t)
'LVFRQWLQXRXVFRQGXFWLRQPRGH
0
30
øc
øb
90
14
120
Inversion
180
α, degrees
150
IL
L
30˚+α
90˚+α
3 Vm sin(θ + 30˚)dθ
–
V
+ + –
Chapter 16: Line-commutated rectifiers
= 3 π2 VL-L, rms cos α
3 V=π
,QYHUWHUPRGH
60
Rectification
Fundamentals of Power Electronics
15
Chapter 16: Line-commutated rectifiers
If the load is capable of supplying power, then the direction of power flow can be reversed by reversal of the dc output voltage V. The delay angle α must be greater than 90˚. The current direction is unchanged.
3ø ac
øa
Fundamentals of Power Electronics
–1.5
–1
–0.5
0
0.5
1
1.5 V V L–L, rms
'FRXWSXWYROWDJHYVGHOD\DQJOHα
∞
n = 1,5,7,11,...
Σ
nπ nπ 4 nπ I L sin 2 sin 3 sin (nωt – nα)
16
Chapter 16: Line-commutated rectifiers
S
IL
α
=
s rm
|| S || cos α
2 V 3 π
L L–
P
Fundamentals of Power Electronics
17
Chapter 16: Line-commutated rectifiers
Q = 3 I a, rmsVL-L, rms sin α = I L 3 π2 VL-L, rms sin α
P = I L 3 π2 VL-L, rms cos α
|| S || sin α
Q
5HDODQGUHDFWLYHSRZHULQFRQWUROOHGUHFWLILHU DWIXQGDPHQWDOIUHTXHQF\
Fundamentals of Power Electronics
Q = 3 I a, rmsVL-L, rms sin α = I L 3 π2 VL-L, rms sin α
When the dc output voltage is small, then the delay angle α is close to 90˚ and the power factor becomes quite small. The rectifier apparently consumes reactive power, as follows:
power factor = 0.955 cos (α)
Same as uncontrolled rectifier case, except that waveform is delayed by the angle α. This causes the current to lag, and decreases the displacement factor. The power factor becomes:
i a(t) =
Fourier series of ac line current waveform, for large dc-side inductance:
+DUPRQLFVDQGSRZHUIDFWRU
Z2
Z3
Zs
19
Z2
Rectifier model
ir
Z3
Rectifier model
ir
Chapter 16: Line-commutated rectifiers
...
Chapter 16: Line-commutated rectifiers
...
Harmonic traps (series resonant networks)
Z1
+DUPRQLFWUDS
18
Harmonic traps (series resonant networks)
Z1
ac source model
is
Fundamentals of Power Electronics
Z s(s) = Z s'(s) + sL s'
Ac source: model with Thevenin-equiv voltage source and impedance Zs’(s). Filter often contains series inductor sLs’. Lump into effective impedance Zs(s):
Fundamentals of Power Electronics
ac source model
is
Zs
A passive filter, having resonant zeroes tuned to the harmonic frequencies
+DUPRQLFWUDSILOWHUV
ac source model
is
Zs
Z 1 || Z 2 || i s(s) = i R(s) Z s + Z 1 || Z 2 ||
Fifth-harmonic trap Z1
C1
L1
R1
Fundamentals of Power Electronics
is
Ls
Fundamentals of Power Electronics
H(s) =
Z2
Z3
H(s) =
ir
21
ωL
s
Zs
Z1 || Zs fp ≈
Rectifier model
ir
1
ωC 1 f1 =
1 2π L 1C 1
Ls C1
Q1 =
R1
R01 R1
R01 =
1 ωL
L1 C1
Chapter 16: Line-commutated rectifiers
Z1
1 2π L sC 1
R0p R1 R0p ≈
Qp ≈
Chapter 16: Line-commutated rectifiers
...
i s(s) Z s || Z 1 || Z 2 || = i R(s) Zs
6LPSOHH[DPSOH
20
Harmonic traps (series resonant networks)
Z1
or
)LOWHUWUDQVIHUIXQFWLRQ
– 40 dB/decade
fp
C2 7th harmonic trap Z2
C1 5th harmonic trap Z1
23
L2
L1
([DPSOH
22
L1 L1 + Ls
R2
Q1
f1
R1
Fundamentals of Power Electronics
is
Ls
Fundamentals of Power Electronics
1
Qp
ir
Chapter 16: Line-commutated rectifiers
11th harmonic trap Z3
C3
L3
R3
Chapter 16: Line-commutated rectifiers
• Q of parallel resonance is larger than Q of series resonance
• Parallel resonance tends to increase amplitude of third harmonic
• Parallel resonance: C1 and Ls
• Series resonance: fifth harmonic trap
6LPSOHH[DPSOHWUDQVIHUIXQFWLRQ
s
Q1
Fundamentals of Power Electronics
1
f1 Q1 R1
1 ωL
24
2
ωC 1 f2 Q2
2
R2
ωL
f1
25
Q2
f2
f3 Q3
R3
ωL
3
Chapter 16: Line-commutated rectifiers
Q3
f3
Chapter 16: Line-commutated rectifiers
3
ωC 1
7UDQVIHUIXQFWLRQDV\PSWRWHV
Fundamentals of Power Electronics
ωL
1
ωC 1
Zs || Z1 || Z2 || Z3
$SSUR[LPDWHLPSHGDQFHDV\PSWRWHV
Cn
Cn
Cb
Rbp
26
1 fp
Z1
1
f1
f1
ωL
1
fbp
fbp
Fundamentals of Power Electronics
Ls
L7 C7
L5 C5
Cb
Rbp
27
Chapter 16: Line-commutated rectifiers
Fifth-harmonic Seventh-harmonic trap trap with highfrequency rolloff
R7
R5
– 20 dB/decade
Rbp
Chapter 16: Line-commutated rectifiers
– 40 dB/decade
Zs
fp
s
ωC 1
ωL
+DUPRQLFWUDSILOWHUZLWKKLJKIUHTXHQF\UROORII
Fundamentals of Power Electronics
Ln
Rbp
Rn
Ln
Rn
Z1 || Zs
%\SDVVUHVLVWRU
T1 T3
T2
c
b
ia(t)
ia2(t)
ia1(t)
T4
T1
Fundamentals of Power Electronics
3øac source
a
n'
T2
T1
c'
28
b'
a'
c
T3
T2
T6
T3
T5
T2
1:n
29
T6
3 :n
T3 n'
n'
T5
T4
T2
T1
b
c'
n'
b'
T2
T1
Secondary voltages
T3
30˚
a'
–
vd (t)
+
L
dc load
Chapter 16: Line-commutated rectifiers
IL
Chapter 16: Line-commutated rectifiers
Primary voltages
T1
a
7ZHOYHSXOVHUHFWLILHU
T3
3 :n
Fundamentals of Power Electronics
c
b
a
ωt
Three-phase transformer connections can be used to shift the phase of the voltages and currents This shifted phase can be used to cancel out the low-order harmonics Three-phase delta-wye transformer connection shifts phase by 30˚:
7UDQVIRUPHUFRQQHFWLRQV
nI L 3
nI L 1 + 2 3 3
90˚
nIL
180˚
nI L 1 + 3 3
270˚
– nIL
30
360˚
ωt
Chapter 16: Line-commutated rectifiers
• 5th, 7th, 17th, 19th, etc. harmonics are eliminated
• Ac line current contains 1st, 11th, 13th, 23rd, 25th, etc. These harmonic amplitudes vary as 1/n
Fundamentals of Power Electronics
31
Chapter 16: Line-commutated rectifiers
If p is pulse number, then rectifier produces line current harmonics of number n = pk ± 1, with k = 0, 1, 2, ...
• No 5th, 7th, 11th, 13th, 17th, or 19th harmonics
• Transformer connections shift phase by 0˚, 15˚, –15˚, and 30˚
• Use four six-pulse rectifiers
Twenty-four-pulse rectifier
• No 5th, 7th, 11th, 13th harmonics
• Transformer connections shift phase by 0˚, +20˚, and –20˚
• Use three six-pulse rectifiers
Eighteen-pulse rectifier:
5HFWLILHUVZLWKKLJKSXOVHQXPEHU
Fundamentals of Power Electronics
ia(t)
ia2(t)
ia1(t)
:DYHIRUPVRISXOVHUHFWLILHU
RMS values of rectifier waveforms Ideal three-phase rectifiers
17.4 17.5
vac(t) Re
Fundamentals of Power Electronics
2
Re is called the emulated resistance
i ac(t) =
–
Re
Chapter 17: The Ideal Rectifier
vac(t)
+
iac(t)
• ac line current has same waveshape as voltage
• unity power factor
It is desired that the rectifier present a resistive load to the ac power system. This leads to
3URSHUWLHVRIWKHLGHDOUHFWLILHU
Chapter 17: The Ideal Rectifier
Single-phase converter systems employing ideal rectifiers
17.3
1
Realization of a near-ideal rectifier
17.2
Fundamentals of Power Electronics
Properties of the ideal rectifier
17.1
7KH,GHDO5HFWLILHU
&KDSWHU
Fundamentals of Power Electronics
3
–
vac(t)
+
iac(t)
4
p(t) =
ac input
–
vac(t)
+
iac(t)
–
v(t)
+
Chapter 17: The Ideal Rectifier
v 2ac(t) Re
dc output
v(t)i(t) = p(t) =
2
i(t)
Chapter 17: The Ideal Rectifier
p(t) = vac /Re
v 2ac(t) Re(vcontrol(t))
vcontrol
Re(vcontrol)
Re(vcontrol)
vcontrol
Ideal rectifier (LFR)
2XWSXWSRUWPRGHO
The ideal rectifier is lossless and contains no internal energy storage. Hence, the instantaneous input power equals the instantaneous output power. Since the instantaneous power is independent of the dc load characteristics, the output port obeys a power source characteristic.
Fundamentals of Power Electronics
Power apparently “consumed” by Re is actually transferred to rectifier dc output port. To control the amount of output power, it must be possible to adjust the value of Re.
V 2ac,rms Pav = Re(vcontrol)
&RQWURORISRZHUWKURXJKSXW
power source
–
–
i(t)
5
p(t)
Chapter 17: The Ideal Rectifier
i-v characteristic
vac(t) Re(vcontrol)
v 2ac(t) Re(vcontrol(t))
Fundamentals of Power Electronics
p(t) =
v(t)i(t) = p(t)
i ac(t) =
Defining equations of the ideal rectifier:
6
v(t)
v(t)i(t) = p(t)
R Re
R Re
Chapter 17: The Ideal Rectifier
I ac,rms = I rms
Vrms = Vac,rms
When connected to a resistive load of value R, the input and output rms voltages and currents are related as follows:
(TXDWLRQVRIWKHLGHDOUHFWLILHU/)5
power sink
v(t)
+
v(t)
+
Fundamentals of Power Electronics
p(t)
i(t)
i(t)
7KHGHSHQGHQWSRZHUVRXUFH
VM
VM /Re
VM
Fundamentals of Power Electronics
vg(t)
iac(t)
vac(t)
Fundamentals of Power Electronics
vac(t)
iac(t)
t
t
vg
ig
R
v(t) V = vg(t) VM sin (ωt)
Chapter 17: The Ideal Rectifier
M min = V VM
vg(t) = VM sin (ωt) 8
M(d(t)) =
Mmin
V
vac(t) = VM sin (ωt)
M(t)
v(t)
ig(t)
C
Chapter 17: The Ideal Rectifier
Controller
:DYHIRUPV
7
d(t)
– –
i(t) + v(t)
+
1 : M(d(t))
dc–dc converter
vg(t)
ig(t)
Control the duty cycle of a dc-dc converter, such that the input current is proportional to the input voltage:
5HDOL]DWLRQRIDQHDULGHDOUHFWLILHU
v(t) V = vg(t) VM sin (ωt) M(t)
Mmin
vcontrol(t)
iac(t)
Multiplier
Fundamentals of Power Electronics
vac(t)
%RRVWFRQYHUWHU
9
vg(t)
10
L
PWM
Q1
D1
Controller
–
v(t)
+
Chapter 17: The Ideal Rectifier
C
i(t)
Compensator
v (t) +– err Gc(s)
va(t)
ig(t) Rs
–
vref(t) = kx vg(t) vcontrol(t)
X
+ vg(t)
ig(t)
Boost converter
R
Chapter 17: The Ideal Rectifier
ZLWKFRQWUROOHUWRFDXVHLQSXWFXUUHQWWRIROORZLQSXWYROWDJH
Fundamentals of Power Electronics
• We will see that the boost converter exhibits lowest transistor stresses. For this reason, it is most often chosen
• Boost, buck-boost, Cuk, SEPIC, and other converters with similar conversion ratios are suitable
• Above expression neglects converter dynamics
• To avoid distortion near line voltage zero crossings, converter should be capable of producing M(d(t)) approaching infinity
M(d(t)) =
&KRLFHRIFRQYHUWHU
v(t) V = vg(t) VM sin (ωt)
1 1 – d(t)
11
vg(t) V
in CCM
Chapter 17: The Ideal Rectifier
vg(t)d(t)Ts 2L
Ts
=
vg(t) Re
Ts
> ∆i g(t) ⇒ d(t) < 2L ReTs 2L vg(t) Ts 1 – V Fundamentals of Power Electronics
Re <
12
for CCM
Substitute CCM expression for d(t):
i g(t)
The converter operates in CCM when
i g(t)
Chapter 17: The Ideal Rectifier
Low-frequency (average) component of inductor current waveform is
∆i g(t) =
Inductor current ripple is
&&0'&0ERXQGDU\ERRVWUHFWLILHU
Fundamentals of Power Electronics
d(t) = 1 –
The duty ratio should therefore follow
M(d(t)) =
If the converter operates in CCM, then
Since M ≥ 1 in the boost converter, it is required that V ≥ VM
M(d(t)) =
9DULDWLRQRIGXW\F\FOHLQERRVWUHFWLILHU
for CCM
Re < 2L Ts
Ts 1 –
2L VM V
Chapter 17: The Ideal Rectifier
Re(d(t))
p
R
–
V
+
Fundamentals of Power Electronics
14
Beware! This DCM Re(d) from Chapter 10 is not the same as the rectifier emulated resistance Re = vg/ig
+ –
ig(t)
i g(t) =
Chapter 17: The Ideal Rectifier
vg(t) p(t) + Re(d(t)) V – vg(t) v 2g(t) with p(t) = Re(d(t)) Re(d(t)) = 22L d (t)Ts
Solve for input current:
The input characteristic in DCM is found by solution of the averaged DCM model (Fig. 10.12(b)):
vg(t) = 1 – d(t) V
A plot of input current ig(t) vs input voltage vg(t), for various duty cycles d(t). In CCM, the boost converter equilibrium equation is
6WDWLFLQSXWFKDUDFWHULVWLFV RIWKHERRVWFRQYHUWHU
13
For Re between these limits, the converter operates in DCM when vg(t) is near zero, and in CCM when vg(t) approaches VM.
Re >
The converter always operates in DCM provided that
Fundamentals of Power Electronics
vg(t)
2L vg(t) Ts 1 – V
Note that vg(t) varies with time, between 0 and VM. Hence, this equation may be satisfied at some points on the ac line cycle, and not at others. The converter always operates in CCM provided that
Re <
&&0'&0ERXQGDU\
0
0.25
0.5
0.75
0
1–
15
vg(t) V
Chapter 17: The Ideal Rectifier
0.25
d = 0.6
m g(t) =
0.5
) i g(t
16
vg(t) V
DCM
( = vg
d = 0.4 0.75
CCM
Re t) /
1
1–
vg(t) V
Chapter 17: The Ideal Rectifier
2L i (t) > vg(t) VTs g V
CCM when
2L i (t) 1 – vg(t) = d 2(t) vg(t) VTs g V V
DCM:
vg(t) = 1 – d(t) V
CCM:
ZLWKVXSHULPSRVHGUHVLVWLYHFKDUDFWHULVWLF
%RRVWLQSXWFKDUDFWHULVWLFV
d = 0.8 Fundamentals of Power Electronics
jg(t) = 2L i g(t) VT s
1
d=1
Fundamentals of Power Electronics
2L i (t) > vg(t) V VTs g
CCM/DCM mode boundary, in terms of vg(t) and ig(t):
2L i (t) 1 – vg(t) = d 2(t) vg(t) V V VTs g
Now simplify DCM current expression, to obtain
6WDWLFLQSXWFKDUDFWHULVWLFV RIWKHERRVWFRQYHUWHU
d=0
d = 0.2
vcontrol(t) Multiplier
vg(t)
Controller
17
Compensator
C
i(t)
–
v(t)
+ R
vg(t) = i g(t)
Fundamentals of Power Electronics
k xvcontrol(t)
Rs
–
〈vg(t)〉Ts
+
〈ig(t)〉Ts
18
Re(t)
Re
Rs k x vcontrol(t)
vcontrol(t)
C
〈i(t)〉Ts
–
〈v(t)〉Ts
R
Chapter 17: The Ideal Rectifier
This model also applies to other converters that are controlled in the same manner, including buck-boost, Cuk, and SEPIC.
Re(t) =
〈p(t)〉Ts
Ideal rectifier (LFR)
+
Chapter 17: The Ideal Rectifier
k xvcontrol(t)
Rs
va(t) Rs
vref (t) k xvcontrol(t)
Re(vcontrol(t)) =
simplify:
Re =
then Re is
vref (t) = k xvg(t)vcontrol(t)
va(t) ≈ vref (t) multiplier equation
va(t) = i g(t)Rs when the error signal is small,
Current sensor gain
/RZIUHTXHQF\V\VWHPPRGHO
Re(vcontrol(t)) =
vac(t)
iac(t)
PWM
Q1
D1
– verr(t) + Gc(s)
va(t)
ig(t) Rs
–
L
Boost converter
vref(t) = kx vg(t) vcontrol(t)
X
+ vg(t)
Fundamentals of Power Electronics
vac(t)
iac(t)
ig(t)
Solve circuit to find Re:
5HRIWKHPXOWLSO\LQJDYHUDJHFXUUHQW FRQWUROOHU
g
19
Chapter 17: The Ideal Rectifier
so
with
i g(t) =
vg(t) Re
20
Chapter 17: The Ideal Rectifier
V 2M V 2M 2 pac(t) = sin ωt = 1 – cos 2ωt Re 2Re
vg(t) = VM sin (ωt)
pac(t) = vg(t)i g(t)
The instantaneous input power of a single-phase ideal rectifier is not constant:
Fundamentals of Power Electronics
•
For a given constant load characteristic, the instantaneous load current and power are then also constant:
•
pload (t) = v(t)i(t) = VI
It is usually desired that the output voltage v(t) be regulated with high accuracy, using a wide-bandwidth feedback loop
•
6LQJOHSKDVHFRQYHUWHUV\VWHPV FRQWDLQLQJLGHDOUHFWLILHUV
Fundamentals of Power Electronics
Input current contains harmonics. If v is sufficiently greater than vg, then harmonics are small.
e
The boost converter can also be operated in DCM as a low harmonic rectifier. Input characteristic is vg(t) v 2g(t) i g(t) T = + Re s R v(t) – v (t)
Like other DCM applications, this approach is usually restricted to low power (< 200W).
Disadvantages: higher peak currents, larger input current EMI
Advantage: simple control
We found in Chapter 10 that the buck-boost, SEPIC, and Cuk converters, when operated open-loop in DCM, inherently behave as loss-free resistors. This suggests that they could also be used as near-ideal rectifiers, without need for a multiplying controller.
2SHQORRS'&0DSSURDFK
Pload
pac(t)
Fundamentals of Power Electronics
vc(t)
21
Chapter 17: The Ideal Rectifier
22
t
=
d
1 2
= pac(t) – pload(t)
Chapter 17: The Ideal Rectifier
dt
Cv 2C(t)
&DSDFLWRUHQHUJ\VWRUDJHLQ¡V\VWHP
Fundamentals of Power Electronics
Energy storage capacitor voltage must be allowed to vary, in accordance with this equation
dt
= pac(t) – pload(t)
Capacitor energy storage: instantaneous power flowing into capacitor is equal to difference between input and output powers:
•
Cv 2C(t)
An energy storage element must be added
•
1 2
Hence instantaneous input and output powers must be equal
•
d EC(t) d pC(t) = = dt
Ideal rectifier is lossless, and contains no internal energy storage.
•
3RZHUIORZLQVLQJOHSKDVHLGHDOUHFWLILHUV\VWHP
–
vg(t)
+ Re
d
1 2
dt
Cv 2C(t)
s
–
vC(t)
+ Dc–dc converter
Wide-bandwidth control of input current waveform Internal independent energy storage
• •
Fundamentals of Power Electronics
24
Chapter 17: The Ideal Rectifier
During the hold-up time, the load power is supplied entirely by the energy storage capacitor
A typical hold-up time requirement: supply load for one complete missing ac line cycle, or 20msec in a 50Hz system
Hold up time: the duration which the dc output voltage v(t) remains regulated after vac(t) has become zero
A common example: continue to supply load power in spite of failure of ac line for short periods of time.
Internal energy storage allows the system to function in other situations where the instantaneous input and output powers differ.
Chapter 17: The Ideal Rectifier
Wide-bandwidth control of output voltage
load
i(t)
•
–
v(t)
+
pload(t) = VI = Pload
This system is capable of
Energy storage capacitor
C
+ROGXSWLPH
= pac(t) – pload(t) 23
〈 pac(t)〉T
Ideal rectifier (LFR) i (t) 2
Energy storage capacitor voltage vC(t) must be independent of input and output voltage waveforms, so that it can vary according to
Fundamentals of Power Electronics
=
vac(t)
iac(t)
ig(t)
6LQJOHSKDVHV\VWHPZLWKLQWHUQDOHQHUJ\VWRUDJH
,QUXVKFXUUHQW
25
Chapter 17: The Ideal Rectifier
Fundamentals of Power Electronics
26
Chapter 17: The Ideal Rectifier
Converters having buck-boost characteristics are capable of controlling the inrush current. Unfortunately, these converters exhibit higher transistor stresses.
Additional circuitry is needed to limit the magnitude of this inrush current.
Even with d = 0, a large current flows through the boost converter diode to the capacitor, as long as v(t) < vg(t).
Boost converter is not capable of controlling this inrush current.
A problem caused by the large energy storage capacitor: the large inrush current observed during system startup, necessary to charge the capacitor to its equilibrium value.
Fundamentals of Power Electronics
So a single big capacitor is the best solution
But the capacitor is considerably smaller, lighter, and less expensive
each store 1 Joule of energy
100A 100µH inductor
100V 100µF capacitor
Example
But, inductors are not good energy-storage elements
Instead of a capacitor, and inductor or higher-order LC network could store the necessary energy.
(QHUJ\VWRUDJHHOHPHQW
27
Chapter 17: The Ideal Rectifier
Fundamentals of Power Electronics
2
–
vC(t)
28
Energy storage capacitor
C
+
Dc-dc converter
i(t)
load
–
v(t) load
+
Chapter 17: The Ideal Rectifier
Pload V + –
load
Dc-dc converter input port behaves as a power sink. A low frequency converter model is p (t) = VI = P i (t)
To the extent that dc-dc converter losses can be neglected, then dc-dc converter input power is Pload , regardless of capacitor voltage vc(t).
Load power is therefore the constant value Pload = VI.
Load therefore draws constant current I.
Dc-dc converter produces well-regulated dc load voltage V.
/RZIUHTXHQF\PRGHORIGFGFFRQYHUWHU
Fundamentals of Power Electronics
Regardless of the input voltage and frequency, the near-ideal rectifier produces a constant nominal dc output voltage. With a boost converter, this voltage is 380 or 400V.
100V, 110V, 115V, 120V, 132V, 200V, 220V, 230V, 240V, 260V
Nominal rms line voltages of 100V to 260V:
50Hz and 60Hz
The capability to operate from the ac line voltages and frequencies found everywhere in the world:
8QLYHUVDOLQSXW
–
vC(t)
–
vg(t)
+
Ideal rectifier (LFR)
Re
〈 pac(t)〉Ts C
–
vC(t)
+
Energy storage capacitor
i2(t)
Dc-dc converter
Pload V + –
There is a net increase in capacitor charge when the reverse is true.
30
–
Chapter 17: The Ideal Rectifier
This suggests that rectifier and load powers can be balanced by regulating the energy storage capacitor voltage.
i(t) v(t) load
+
pload(t) = VI = Pload
If the load power exceeds the average rectifier power, then there is a net discharge in capacitor energy and voltage over one ac line cycle.
iac(t)
ig(t)
2EWDLQLQJDYHUDJHSRZHUEDODQFH
Chapter 17: The Ideal Rectifier
An additional feeback loop is necessary, to adjust Re such that the rectifier average power is equal to the load power
•
29
But the system contains no mechanism to accomplish this
–
v(t) load
i(t)
•
Dc-dc converter
Pload V + –
+
In equilibrium, average rectifier and load powers must be equal
Energy storage capacitor
C
+
•
Fundamentals of Power Electronics
vac(t)
s
Ideal rectifier (LFR)
Re
〈 pac(t)〉T
pload(t) = VI = Pload
Difference between rectifier output power and dc-dc converter input power flows into capacitor
–
vg(t)
+
i2(t)
•
Fundamentals of Power Electronics
vac(t)
iac(t)
ig(t)
A complete low-frequency system model:
/RZIUHTXHQF\HQHUJ\VWRUDJHSURFHVV¡V\VWHP
vcontrol(t) Rs
L
PWM
Q1
D1
i2(t) +
v(t)
–
Chapter 17: The Ideal Rectifier
Average rectifier power is controlled by variation of Re. Re must not vary too quickly; otherwise, ac line current harmonics are generated Extreme limit: loop has infinite bandwidth, and vc(t) is perfectly regulated to be equal to a constant reference value
• • •
Fundamentals of Power Electronics
i ac(t) =
32
pac(t) p (t) Pload = load = vac(t) vac(t) VM sin ωt Chapter 17: The Ideal Rectifier
• Instantaneous load and ac line powers are then equal
• Energy storage capacitor voltage then does not change, and this capacitor does not store or release energy
The energy-storage-capacitor voltage feedback loop causes the dc component of vc(t) to be equal to some reference value
•
• Input current becomes
–+ vref3 Wide-bandwidth output voltage controller
%DQGZLGWKRIFDSDFLWRUYROWDJHORRS
31
+
Load v(t)
i(t)
Compensator and modulator
d(t)
DC–DC Converter
–+ vref2
vC(t)
C
Low-bandwidth energy-storage capacitor voltage controller
Compensator
–
vC(t)
Compensator
v (t) +– err Gc(s)
va(t)
ig(t)
vref1(t) = kxvg(t)vcontrol(t)
X
vg(t)
–
vg(t)
+
Boost converter
Wide-bandwidth input current controller
Multiplier
Fundamentals of Power Electronics
vac(t)
iac(t)
ig(t)
FRQWDLQLQJWKUHHIHHGEDFNORRSV
$FRPSOHWH¡V\VWHP
t
Chapter 17: The Ideal Rectifier
So bandwidth of capacitor voltage loop must be limited, and THD increases rapidly with increasing bandwidth
Power factor → 0
THD → ∞
t
Fundamentals of Power Electronics
34
Chapter 17: The Ideal Rectifier
Generate tables of component rms and average currents for various rectifier converter topologies, and compare
Approximate here using double integral
Computation of rms value of this waveform is complex and tedious
iQ(t)
Doubly-modulated transistor current waveform, boost rectifier:
506YDOXHVRIUHFWLILHUZDYHIRUPV
33
pac(t) p (t) Pload = load = vac(t) vac(t) VM sin ωt
Fundamentals of Power Electronics
iac(t)
vac(t)
i ac(t) =
,QSXWFXUUHQWZDYHIRUPH[WUHPHOLPLW
1 Tac 0
T ac
i 2Q(t)dt
1 T Tac s n=1
∑
T ac/T s
1 Ts (n-1)T s
nT s
iQ(t)
i 2Q(t)dt
35
Chapter 17: The Ideal Rectifier
1 T Tac s
n=1
∑
T ac/T s
1 Ts
(n-1)T s
nT s
i 2Q(t)dt
Fundamentals of Power Electronics
=
Ts
T ac
36
i 2Q(t)
=
0
1 Tac
I Qrms ≈
T ac
1 Ts
1 lim T Tac T s→0 s
t
t+T s
n=1
∑
T ac/T s
(n-1)T s
nT s
i 2Q(τ)dτ dt
1 Ts
Chapter 17: The Ideal Rectifier
i 2Q(τ)dτ
When Ts << Tac, then the summation can be approximated by an integral, which leads to the double-average:
I Qrms =
$SSUR[LPDWLRQRI506H[SUHVVLRQ
Fundamentals of Power Electronics
Quantity in parentheses is the value of iQ2, averaged over the nth switching period.
I Qrms =
Express as sum of integrals over all switching periods contained in one ac line period:
I Qrms =
RMS transistor current is
506WUDQVLVWRUFXUUHQW
t
t+T
vac(t) = VM sin ωt
s = 1 i 2Q(t)dt Ts t = d(t)i 2ac(t)
VM sin ωt Re
37
Ts
=
Fundamentals of Power Electronics
I Qrms =
=
I Qrms =
0
0
T ac
38
2 2 VM Tac R 2e
1 Tac
1 Tac
0
T ac
dt
sin 2 ωt
T ac/2
sin 2 ωt –
Chapter 17: The Ideal Rectifier
VM sin 3 ωt dt V
V 2M V 1 – M sin ωt 2 V Re
i 2Q
Ts
V 2M V 1 – M sin ωt 2 V Re
VM sin ωt V
Now plug this into the RMS formula:
i 2Q
Evaluate the first integral:
d(t) = 1 –
Duty cycle is therefore
sin 2 ωt dt
Chapter 17: The Ideal Rectifier
(this neglects converter dynamics)
%RRVWUHFWLILHUH[DPSOH
Fundamentals of Power Electronics
V = 1 vac(t) 1 – d(t)
and the duty cycle will ideally be
i ac(t) =
then the input current will be given by
If the input voltage is
i
2 Q T s
For the boost converter, the transistor current iQ(t) is equal to the input current when the transistor conducts, and is zero when the transistor is off. The average over one switching period of iQ2(t) is therefore
%RRVWUHFWLILHUH[DPSOH
0
sin n (θ)dθ =
3 8 16 15π
15 48
4 5 6
Chapter 17: The Ideal Rectifier
4 3π 3
2 π
sin n (θ)dθ
1 2
0
π
2
1 π
VM 2 Re
V 1– 8 M 3π V
= I ac rms
V 1– 8 M 3π V
Fundamentals of Power Electronics
40
Chapter 17: The Ideal Rectifier
When the dc output voltage is not too much greater than the peak ac input voltage, the boost rectifier exhibits very low transistor current. Efficiency of the boost rectifier is then quite high, and 95% is typical in a 1kW application.
I Qrms = 0.39I ac rms
Transistor RMS current is minimized by choosing V as small as possible: V = VM. This leads to
I Qrms =
%RRVWH[DPSOH7UDQVLVWRU506FXUUHQW
39
2 2⋅4⋅6 (n – 1) if n is odd π 1⋅3⋅5 n 1⋅3⋅5 (n – 1) if n is even 2⋅4⋅6 n
Fundamentals of Power Electronics
1 π
π
1
n
,QWHJUDWLRQRISRZHUVRIVLQθRYHUFRPSOHWHKDOIF\FOH
I dc
I ac rms
I ac rms
16 V 3π V M
VM 1 – 8 3π V
rms
Transistor
CCM SEPIC, nonisolated
I dc
3 + 16 nV 2 3π V M
transistor
I dc
I ac rms V M 2 V I dc
VM 3 V 2 3 + 16 V 2 3π V M
I dc
Diode, xfmr secondary
0
I dc
3 + 16 nV 2 3π V M
I ac rms 2 π2 I ac rms 8 VM 3π nV
I ac rms 2 π2 V 1+ 8 M 3π nV
Fundamentals of Power Electronics
with, in all cases,
42
I ac rms = 2 V , ac input voltage = V M sin(ω t) VM I dc dc output voltage = V
I ac rms
I ac rms
0
I ac rms 2 π2 I ac rms 8 VM 3π V
I ac rms 2 π2
FRQWLQXHG
V 1+ 8 M 3π V
I ac rms
I ac rms
I ac rms
C1 , xfmr primary
L1
41
0
I ac rms 2 π2
I ac rms 8 VM 3π nV
I ac rms 2 π2
V 1+ 8 M 3π nV
CCM SEPIC, with n:1 isolation transformer
Diode
L2
C1
L1
I dc
I ac rms
I ac rms
VM nV
V n
VM 2 V
VM V
Chapter 17: The Ideal Rectifier
2I dc 1 + nV VM
n
VM nV
I ac rms 2 max 1,
I ac rms 2
I ac rms 2 1 +
2I dc 1 + V VM
I ac rms
I ac rms max 1,
I ac rms 2
I ac rms 2 1 +
VM V
Chapter 17: The Ideal Rectifier
2I dc 1 + nV VM
I ac rms 2 max 1,
I ac rms 2
I ac rms 2 1 +
I ac rms 2
I ac rms 2 π2
7DEOHRIUHFWLILHUFXUUHQWVWUHVVHV
Fundamentals of Power Electronics
Diode, xfmr secondary
C1
L1
Transistor, xfmr primary
2 I dc V VM
I ac rms 2
Peak
I dc
V I ac rms 2 π2 1 – π M 8 V
Average
Summary of rectifier current stresses for several converter topologies
CCM flyback, with n:1 isolation transformer and input filter
Inductor
Diode
Transistor
CCM boost
Tabl e 17. 2
7DEOHRIUHFWLILHUFXUUHQWVWUHVVHVIRUYDULRXVWRSRORJLHV
Isolated topologies are possible, with higher transistor stress No limiting of inrush current Output voltage must be greater than peak input voltage
• • •
Output voltage can be greater than or less than peak input voltage
•
2A 5.5 A 5.5 A
Boost Nonisolated SEPIC Isolated SEPIC
719 V
719 V
380 V
Transistor voltage
36.4 A
4.85 A
3.6 A
Diode rms current
11.4 A
9.8 A
6.6 A
Transistor rms current, 120V
42.5 A
6.1 A
5.1 A
Diode rms current, 120V
Fundamentals of Power Electronics
44
Chapter 17: The Ideal Rectifier
Isolated SEPIC example has 4:1 turns ratio, with 42V 23.8A dc load
Transistor rms current
Converter
1kW, 240Vrms example. Output voltage: 380Vdc. Input current: 4.2Arms
&RPSDULVRQRIUHFWLILHUWRSRORJLHV
Chapter 17: The Ideal Rectifier
Inrush current limiting is possible
•
43
Isolated topologies are possible, without increased transistor stress
•
Fundamentals of Power Electronics
Higher transistor rms current, lower efficiency
•
Buck-boost, SEPIC, and Cuk converters
Lowest transistor rms current, highest efficiency
•
Boost converter
&RPSDULVRQRIUHFWLILHUWRSRORJLHV
ic
ib
ia Re
Re
Re
45
pc(t)
pb(t)
pa(t) R
dc output
–
v
+
Chapter 17: The Ideal Rectifier
ic
ib
ia Re
Fundamentals of Power Electronics
øc
øb
øa
3øac input
Re
46
ptot = pa + pb + pc
Re
–
v
+
Chapter 17: The Ideal Rectifier
R
dc output
Combine parallel-connected power sources into a single source ptot(t):
,GHDO¡UHFWLILHUPRGHO
Fundamentals of Power Electronics
øc
øb
øa
3øac input
Ideal 3ø rectifier, modeled as three 1ø ideal rectifiers:
,GHDOWKUHHSKDVHUHFWLILHUV
øc
øb
øa
3øac input
47
ic
ib
ia Re pc(t)
pb(t)
pa(t)
Fundamentals of Power Electronics
• a constant power load can be supplied, without need for lowfrequency internal energy storage
• In a balanced system, the ideal 3ø rectifier supplies constant power to its dc output
V2 ptot(t) = pa(t) + pb(t) + pc(t) = 3 M 2 Re
48
øc
øb
øa
3øac input
ic
ib
ia
R
dc output
–
v
+
Re
Re
R
dc output
–
v
Chapter 17: The Ideal Rectifier
ptot = pa + pb + pc
Re
+
Chapter 17: The Ideal Rectifier
• total 3ø power ptot(t) is constant
• 2nd harmonic terms add to zero
V 2M 3 ptot(t) = pa(t) + pb(t) + pc(t) = 2 Re
Total 3ø instantaneous power:
Re
Re
,QVWDQWDQHRXVSRZHULQLGHDO¡UHFWLILHU
Fundamentals of Power Electronics
pa(t) =
v 2an(t) V 2M = 1 – cos 2ωt Re 2Re v 2bn(t) V 2M pb(t) = = 1 – cos 2ωt – 240° Re 2Re v 2 (t) V 2 pc(t) = cn = M 1 – cos 2ωt – 120° Re 2Re
Instantaneous phase powers:
vcn(t) = VM sin ωt – 240°
vbn(t) = VM sin ωt – 120°
van(t) = VM sin ωt
Ac input voltages:
9DOXHRISWRWW
L
L
3
– 2 v10(t)
– 0
v20(t)
+
1 + + v12(t)
Q4
Q1
D4
D1
i1(t)
Q5
Q2
Q6
Q3
D6
D3
i3(t)
C
+
Chapter 17: The Ideal Rectifier
–
Load v(t)
dc output
• Operation of each individual phase is similar to the 1ø boost rectifier
D5
D2
i2(t)
Converter is capable of bidirectional power flow Dc output voltage V must be greater than peak ac line-line voltage VL,pk. Ac input currents are nonpulsating. In CCM, input EMI filtering is relatively easy Very low RMS transistor currents and conduction loss The leading candidate to replace uncontrolled 3ø rectifiers Requires six active devices Cannot regulate output voltage down to zero:
• • • • • • •
Fundamentals of Power Electronics
50
Chapter 17: The Ideal Rectifier
cannot replace traditional buck-type controlled rectifiers
no current limiting
Voltage-source inverter, operated backwards as a rectifier
•
7KH¡DF²GFERRVWUHFWLILHU
49
• Uses six current-bidirectional switches
ic(t)
ib(t)
L
Fundamentals of Power Electronics
øc
øb
øa
ia(t)
3øac input
3øac–dc boost rectifier
7KUHHSKDVHUHFWLILHUVRSHUDWLQJLQ&&0
– 0
v10(t) Q4 v20(t)
+
Q1
D4
D1
Q5
Q2
Q6
Q3
Ts
Ts
Ts
= d 3(t) v(t)
= d 2(t) v(t)
= d 1(t) v(t)
Ts
Ts
Ts
= v30(t)
= v20(t)
Ts
Ts
Ts
51
D6
D3
i3(t)
– v10(t)
– v30(t)
– v20(t)
Ts
Ts
Ts
= d 3(t) – d 1(t)
= d 2(t) – d 3(t)
= d 1(t) – d 2(t)
v(t)
v(t)
v(t)
i 3(t)
i 2(t)
i 1(t)
Ts
Ts
Ts
= d 3(t) i c(t)
= d 2(t) i b(t)
= d 1(t) i a(t)
Ts
Ts
Ts
52
Average switch output-side currents:
Ts
Ts
Ts
= v10(t)
Average line-line voltages:
v30(t)
v20(t)
v10(t)
Average the switch voltages:
Fundamentals of Power Electronics
v31(t)
v23(t)
v12(t)
D5
D2
i2(t)
Conducting devices:
v30(t)
v20(t)
v10(t)
0
0
0
s
s
s
Ts
Ts
Ts
Conducting devices:
v30(t)
v20(t)
v10(t)
0
0
0
0
Q6 / D6
Q3 / D3
v
s
s
s
0
t
t
t
Q6 / D6
Q3 / D3
Ts
Ts
Ts
t
t
t
Chapter 17: The Ideal Rectifier
Q5 / D5
Q2 / D2
0
Q4 / D4
0
s
s
s
Q1 / D1
d3Ts
〈 v30(t)〉T = d3 〈 v(t)〉T
v
d2Ts
〈 v20(t)〉T = d2 〈 v(t)〉T
d1Ts
〈 v10(t)〉T = d1 〈 v(t)〉T
v
Ts
Ts
Ts
Chapter 17: The Ideal Rectifier
Q5 / D5
Q2 / D2
0
Q4 / D4
0
s
s
s
Q1 / D1
d3Ts
〈 v30(t)〉T = d3 〈 v(t)〉T
v
d2Ts
〈 v20(t)〉T = d2 〈 v(t)〉T
v
d1Ts
〈 v10(t)〉T = d1 〈 v(t)〉T
v
$YHUDJHVZLWFKZDYHIRUPV
Fundamentals of Power Electronics
3
– 2
1 + + v12(t)
i1(t)
Drive lower transistors (Q4 – Q6) with complements of duty cycles of respective upper transistors (Q1 – Q3). Each phase operates independently, with its own duty cycle.
Pulse-width modulation:
&RQWURORIVZLWFKHVLQ&&0¡DFGFERRVWUHFWLILHU
Ts
Ts
s
Ts
Ts
Ts
– v10(t)
– v30(t)
– v20(t)
Ts
Ts
Ts
= d 3(t) – d 1(t)
= d 2(t) – d 3(t)
= d 1(t) – d 2(t)
s
s
d1 〈 ia 〉T (d3 – d1) 〈 v 〉T
v(t)
v(t)
v(t)
Ts
Ts
Ts
s
d2 〈 ib 〉T s
i 3(t)
i 2(t)
i 1(t)
d3 〈 ic 〉T
Ts
Ts
Ts
= d 3(t) i c(t)
= d 2(t) i b(t)
Ts
Ts
Ts
–
Fundamentals of Power Electronics
54
Chapter 17: The Ideal Rectifier
The modulation index is defined as one-half of the peak amplitude of the fundamental component of the duty cycle modulation. In some other modulation schemes, it is possible that Dm > 1.
For D0 = 0.5, Dm in the above equations must be less than 1.
Dm is the modulation index
D0 is a dc bias
ω is the ac line frequency
where
d 3(t) = D0 + 12 Dm sin ωt – ϕ – 240°
d 2(t) = D0 + 12 Dm sin ωt – ϕ – 120°
s
Chapter 17: The Ideal Rectifier
Vary duty cycles sinusoidally, in synchronism with ac line d 1(t) = D0 + 12 Dm sin ωt – ϕ
A simple modulation scheme: Sinusoidal PWM
6LQXVRLGDO3:0
53
+ Load 〈 v 〉T
= d 1(t) i a(t)
C
How to vary d(t) such that the desired ac and dc waveforms are obtained?
= v30(t)
= v20(t)
= v10(t)
〈 ic 〉Ts
Ts
s
(d2 – d3) 〈 v 〉T
〈 ib 〉T
Solution is not unique.
Q:
v31(t)
v23(t)
v12(t)
L
L
s
(d1 – d2) 〈 v 〉T
〈 ia 〉Ts
Fundamentals of Power Electronics
øc
øb
L
+ –
øa
$YHUDJHGFLUFXLWPRGHO
+ –
+ –
Ts
= d 1(t) – d 2(t)
v(t) Ts
≈ vab(t)
Dm sin ωt – ϕ – sin ωt – ϕ – 120° v(t) Ts
= VM sin ωt – sin ωt – 120°
D mV = V M
2VM Dm
55
Chapter 17: The Ideal Rectifier
VL,pk VL,pk V= 2 = 1.15 3 Dm Dm
Fundamentals of Power Electronics
56
Chapter 17: The Ideal Rectifier
With sinusoidal PWM, the dc output voltage must be greater than 1.15 times the peak line-line input voltage. Hence, the boost rectifier increases the voltage magnitude.
VL,pk VL,pk V= 2 = 1.15 3 Dm Dm
%RRVWUHFWLILHUZLWKVLQXVRLGDO3:0
Fundamentals of Power Electronics
V=
Solve for the output voltage:
1 2
For small L, ϕ tends to zero. The expression then becomes
1 2
Substitute expressions for duty cycle and ac line voltage variations:
v12(t)
If the switching frequency is high, then the inductors can be small and have negligible effect at the ac line frequency. The averaged switch voltage and ac line voltage are then equal:
6ROXWLRQOLQHDUVLQXVRLGDO3:0
0
0.5
1
d1(t) d2(t)
d3(t)
Input filter
ic(t)
ib(t)
ia(t)
Q4
Q1
D4
D1
Q5
Q2
D5
D2
Q6
Q3
D6
D3
iL(t)
L
C
Load
dc output
Fundamentals of Power Electronics
58
Chapter 17: The Ideal Rectifier
• Can operate in inverter mode by reversal of output voltage polarity
• Exhibits greater active semiconductor stress than boost topology
• Requires two-quadrant voltage-bidirectional switches
–
v(t)
+
Chapter 17: The Ideal Rectifier
• Can produce controlled dc output voltages in the range 0 ≤ V ≤ VL,pk
øc
øb
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3øac input
57
%XFNW\SH¡DF²GFUHFWLILHU
Fundamentals of Power Electronics
〈 v12(t) 〉Ts /V • Overmodulation, in -0.5 which the ωt -1 modulation index 0˚ 60˚ 120˚ 180˚ 240˚ 300˚ 360˚ Dm is increased beyond 1, also leads to undistorted line-line voltages provided that Dm ≤ 1.15. The pulse width modulator saturates, but the duty ratio variations contain only triplen harmonics. V = VL,pk is obtained at Dm = 1.15. Further increases in Dm cause distorted ac line waveforms.
• Triplen harmonics can be added to the duty ratio modulation, without appearing in the line-line voltages.
1RQOLQHDUPRGXODWLRQ
Input filter
ic(t)
ib(t)
ia(t)
Q4
Q1
L1 i (t) c
L1 ib(t)
L1 ia(t)
Q4
Q1
Fundamentals of Power Electronics
øc
øb
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3øac input
Fundamentals of Power Electronics
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3øac input
D4
D1
D4
D1
D5
D2
59
Q6
Q3
D6
D3
iL(t)
Q5
Q2
60
D5
D2
Q6
Q3
D6
D3
&XNWRSRORJ\
Q5
Q2
C1
L
D7
%XFN²ERRVWWRSRORJ\
D7
C
Load
–
v(t)
+
C2
–
Load v(t)
+
Chapter 17: The Ideal Rectifier
Q7
L2
dc output
Chapter 17: The Ideal Rectifier
Q7
dc output
Dc–dc converter with isolation
Dc–dc converter with isolation
Dc–dc converter with isolation
61
i(t)
C
dc output
–
v(t)
+
Chapter 17: The Ideal Rectifier
• Because of the isolation requirement, semiconductor stresses are greater than in 3ø boost rectifier
• Outputs can be connected in series or parallel
• Isolation transformers must be rated to carry the pulsating single-phase ac power pac(t)
• Each rectifier must include isolation between input and output
Fundamentals of Power Electronics
62
Chapter 17: The Ideal Rectifier
So let’s search for approaches that use just one active switch, and only high-frequency reactive elements
• To avoid low-frequency reactive elements, at least one highfrequency switch is needed
• When control of the output voltage is needed, then there must be at least one active device
• No active devices are needed: diodes and harmonic traps will do the job, but these require low-frequency reactive elements
What is the minimum active silicon required to perform the function of 3ø low-harmonic rectification?
Low-harmonic rectification requires active semiconductor devices that are much more expensive than simple peak-detection diode rectifiers.
6RPHRWKHUDSSURDFKHV WRWKUHHSKDVHUHFWLILFDWLRQ
Fundamentals of Power Electronics
øc
øb
øa
3øac input
8VHRIWKUHHVLQJOHSKDVHUHFWLILHUV
D4
D1
D5
D2
D6
D3 Q1
D7
C
Input filter
63
L3
L2
L1
ic(t)
ib(t)
ia(t)
D4
D1
D5
D2
D6
D3 Q1
D7
64
C
dc output
–
v(t)
+
ia(t)
van(t)
–
v(t)
+
dT s L
t
Chapter 17: The Ideal Rectifier
van(t)
VM t
Chapter 17: The Ideal Rectifier
7KHVLQJOHVZLWFK'&0ERRVW¡UHFWLILHU
Fundamentals of Power Electronics
øc
øb
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3øac input
ic(t)
ib(t)
ia(t)
dc output
Inductors L1 to L3 operate in discontinuous conduction mode, in conjunction with diodes D1 to D6. Average input currents 〈ia(t)〉Ts, 〈ib(t)〉Ts, and 〈ic(t)〉Ts are approximately proportional to the instantaneous input line-neutral voltages. Transistor is operated with constant duty cycle; slow variation of the duty cycle allows control of output power.
Input filter
L3
L2
L1
Fundamentals of Power Electronics
øc
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3øac input
7KHVLQJOHVZLWFK'&0ERRVW¡UHFWLILHU
Input filter
T3
T2
T1
ic(t)
ib(t)
ia(t)
D4
D1
D5
D2
D6
D3
65
Q1
T3
T2
T1
D11
D8
D12
D9
C
Fundamentals of Power Electronics
66
High peak currents, needs an input EMI filter
Inrush current limiting and isolation are obtained easily.
Basic converter has a boost characteristic, but buck-boost characteristic is possible (next slide).
Since the open-loop DCM flyback converter can be modeled as a LossFree Resistor, three-phase low-harmonic rectification is obtained naturally.
This converter is effectively three independent single-phase DCM flyback converters that share a common switch.
ia(t)
van(t)
dT s L
t
t
Chapter 17: The Ideal Rectifier
van(t)
VM
–
v(t)
+
Chapter 17: The Ideal Rectifier
D10
D7
dc output
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Fundamentals of Power Electronics
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7KHVLQJOHVZLWFK¡'&0IO\EDFNUHFWLILHU
input filter
ic(t)
ib(t)
ia(t)
T2
D5
D2
67
T3
D6
D3
T3
Q1
T1
D7
T3
D9
C
ic(t)
ib(t)
ia(t)
D4
D1
D5
D2
D6
D3
D7
Q1
Cr
L
C
dc output
–
v(t)
+
Chapter 17: The Ideal Rectifier
When the resonant current pulses return to zero, diodes D1 to D6 are reverse-biased. Transistor Q1 can then be turned off at zero current.
Turning on Q1 initiates resonant current pulses, whose amplitudes depend on the instantaneous input line-neutral voltages.
68
–
v(t)
+
Chapter 17: The Ideal Rectifier
T2
D8
dc output
Inductors Lr and capacitor Cr form resonant tank circuits, having resonant frequency slightly greater than the switching frequency.
Input filter
Lr
Lr
Lr
Fundamentals of Power Electronics
øc
øb
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3øac input
T1
D4
D1
T2
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Fundamentals of Power Electronics
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ic(t)
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La
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Cr1
Cr1
Cr1
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D4
D1
D5
D2
D6
D3
D7
Q1
Lr
Cr2
Ld
C
dc output
Fundamentals of Power Electronics
70
Chapter 17: The Ideal Rectifier
All diodes switch off when their respective tank voltages reach zero. Transistor Q1 is turned off at zero current.
Turning on Q1 initiates resonant voltage pulses in vcra(t), whose amplitudes depend on the instantaneous input line-neutral currents ia(t) to ic(t).
–
v(t)
+
Chapter 17: The Ideal Rectifier
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69
Inductors Lr and capacitors Cr1 and Cr2 form resonant tank circuits, having resonant frequency slightly greater than the switching frequency.
øc
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t
Lowest total active semiconductor stress of all buck-type 3ø low harmonic rectifiers Fundamentals of Power Electronics
3øac input
van(t) R0
Input line currents are approximately sinusoidal pulses, whose amplitudes follow the input line-neutral voltages.
ia(t)
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~ i a(t)R 0
t
Fundamentals of Power Electronics
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Fundamentals of Power Electronics
72
Harmonic corrector
Nonlinear load
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Chapter 17: The Ideal Rectifier
Chapter 17: The Ideal Rectifier
Higher total active semiconductor stress than previous approach, but less EMI filtering is needed. Low THD: < 4% THD can be obtained.
Input-side resonant voltages are approximately sinusoidal pulses, whose amplitudes follow the input currents. Input filter inductors operate in CCM.
vcra(t)
0XOWLUHVRQDQWVLQJOHVZLWFK]HURFXUUHQW VZLWFKLQJ¡EXFNUHFWLILHU
73
Chapter 17: The Ideal Rectifier
Fundamentals of Power Electronics
74
Chapter 17: The Ideal Rectifier
2. The dc-dc boost converter, as well as other converters capable of increasing the voltage, can be adapted to the ideal rectifier application. A control system causes the input current to be proportional to the input voltage. The converter may operate in CCM, DCM, or in both modes. The mode boundary is expressed as a function of Re , 2L/Ts, and the instantaneous voltage ratio vg(t)/V. A well-designed average current controller leads to resistor emulation regardless of the operating mode; however, other schemes discussed in the next chapter may lead to distorted current waveforms when the mode boundary is crossed.
1. The ideal rectifier presents an effective resistive load, the emulated resistance Re, to the ac power system. The power apparently “consumed” by Re is transferred to the dc output port. In a three-phase ideal rectifier, input resistor emulation is obtained in each phase. In both the single-phase and three-phase cases, the output port follows a power source characteristic, dependent on the instantaneous ac input power. Ideal rectifiers can perform the function of low-harmonic rectification, without need for low-frequency reactive elements.
6XPPDU\RINH\SRLQWV
Fundamentals of Power Electronics
• In the majority of applications, this approach exhibits greater total active semiconductor stress than the simple 3ø CCM boost rectifier.
• Total active semiconductor stress is high when the nonlinear load generates large harmonic currents having high THD.
• Does not need to conduct the average load power.
• An active filter that is controlled to cancel the harmonic currents created by a nonlinear load.
+DUPRQLFFRUUHFWLRQ
75
Chapter 17: The Ideal Rectifier
Fundamentals of Power Electronics
76
Chapter 17: The Ideal Rectifier
6. Other three-phase rectifier topologies are known, including six-switch rectifiers having buck and buck-boost characteristics. In addition, threephase low-harmonic rectifiers having a reduced number of active switches, as few as one, are discussed here.
5. In the three-phase case, a boost-type rectifier based on the PWM voltagesource inverter also exhibits low rms transistor currents. This approach requires six active switching elements, and its dc output voltage must be greater than the peak input line-to-line voltage. Average current control can be used to obtain input resistor emulation. An equivalent circuit can be derived by averaging the switch waveforms. The converter operation can be understood by assuming that the switch duty cycles vary sinusoidally; expressions for the average converter waveforms can then be derived.
6XPPDU\RINH\SRLQWV
Fundamentals of Power Electronics
4. RMS values of rectifiers waveforms can be computed by double integration. In the case of the boost converter, the rms transistor current can be as low as 39% of the rms ac input current, when V is close in value to VM. Other converter topologies such as the buck-boost, SEPIC, and Cuk converters exhibit significantly higher rms transistor currents but are capable of limiting the converter inrush current.
3. In a single-phase system, the instantaneous ac input power is pulsating, while the dc load power is constant. Whenever the instantaneous input and output powers are not equal, the ideal rectifier system must contain energy storage. A large capacitor is commonly employed; the voltage of this capacitor must be allowed to vary independently, as necessary to store and release energy. A slow feedback loop regulates the dc component of the capacitor voltage, to ensure that the average ac input power and dc load power are balanced.
6XPPDU\RINH\SRLQWV
L
Q1
D1
C
R
i(t)
Dc-dc boost converter circuit
+ –
RL
ECEN5807 Power Electronics 2
vg(t)
ig(t)
Boost example
2
–
+ –
DRon
D' : 1
VF
Averaged dc model
RL
R
–
v(t)
+
Chapter 18: Low harmonic rectifier modeling and control
v(t) vg(t)
+
ig(t)
Approach: Use the models developed in Chapter 3. Integrate over one ac line cycle to determine steady-state waveforms and average power.
Objective: extend procedure of Chapter 3, to predict the output voltage, duty cycle variations, and efficiency, of PWM CCM low harmonic rectifiers.
i(t)
Chapter 18: Low harmonic rectifier modeling and control
0RGHOLQJORVVHVDQGHIILFLHQF\ LQ&&0KLJKTXDOLW\UHFWLILHUV
1
Control system modeling Modeling the outer low-bandwidth control system Modeling the inner wide-bandwidth average current controller
18.3
ECEN5807 Power Electronics 2
Controller schemes Average current control Feedforward Current programmed control Hysteretic control Nonlinear carrier control
Design example
Modeling losses and efficiency in CCM high-quality rectifiers Expression for controller duty cycle d(t) Expression for the dc load current Solution for converter efficiency η
18.2
18.1
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0
0.2
0°
30°
60°
90°
120°
120°
150°
150°
4
180°
180°
0
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C
id(t)
R
i(t) = I
C
–
v(t) = V
+
–
v(t)
+
i(t)
R
0°
30°
ωt
90°
i(t) = I
vg(t) Re
60°
ig(t) =
120°
id(t)
150°
(low frequency components)
Typical waveforms
180°
Chapter 18: Low harmonic rectifier modeling and control
0
1
2
3
4
5
90°
2
4
6
8
0.8
0.4
VF
Q1
D1
id(t)
Chapter 18: Low harmonic rectifier modeling and control
6
0.6
L
controller
d'(t) : 1
1
60°
ECEN5807 Power Electronics 2
d(t)
0
100
ig(t)
10
200
ig(t)
30°
3
d(t) Ron
300
0°
RL
vg(t) vg(t)
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–
vg(t)
RL
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vg(t)
iac(t)
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Boost rectifier circuit
ig(t) +
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+ –
d(t) Ron
d'(t) : 1
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v(t) = V
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5
Chapter 18: Low harmonic rectifier modeling and control
vg(t) = VM sin ωt
vg(t)
+ –
ig(t)
d(t) =
(large)
C
id(t)
v – vg(t) R v – vg(t) on Re
solve for d(t):
d(t) Ron
d'(t) : 1
R
i(t) = I
ECEN5807 Power Electronics 2
6
–
v(t) = V
+
Chapter 18: Low harmonic rectifier modeling and control
Again, these expressions neglect converter dynamics, and assume that the converter always operates in CCM.
vg(t) d(t)Ron = vg(t) – d'(t)v Re
eliminate ig(t):
with
vg(t) ig(t) = Re
i g(t)d(t)Ron = vg(t) – d'(t)v
Solve input side of model:
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Inductor dynamics are neglected, a good approximation when the ac line variations are slow compared to the converter natural frequencies
Averaged model
vg(t)
ig(t)
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vg(t) Re
vg(t)
d(t) Ron
d'(t) : 1
(large)
C
id(t)
1–
Ron Re v (t) i d (t) = Re R v – vg(t) on Re 2 g
R
i(t) = I
hence id(t) can be expressed as
+ –
ig(t)
T ac
= 2 Tac 0
V Re
2 M
a=
ECEN5807 Power Electronics 2
with
VM V
Ron Re
R V 2M I= 2 1 – on Tac VRe Re
0
8
T ac/2
Ron sin 2 ωt Re
dt
Chapter 18: Low harmonic rectifier modeling and control
1 – a sin ωt
sin 2 ωt
dt
V R v – M on sin ωt Re
1–
This can be written in the normalized form
I = id
T ac/2
–
v(t) = V
+
Chapter 18: Low harmonic rectifier modeling and control
'FORDGFXUUHQW,
7
Now substitute vg (t) = VM sin ωt, and integrate to find 〈id(t)〉Tac:
ECEN5807 Power Electronics 2
Next, average id(t) over an ac line period, to find the dc load current I.
vg(t) 1 –
Ron Re d'(t) = R v – vg(t) on Re
Butd’(t) is:
i d (t) = d'(t)i g(t) = d'(t)
ac
Solve output side of model, using charge balance on capacitor C: I = id T
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0
π/2
ECEN5807 Power Electronics 2
For | a | ≤ 0.15, this approximate expression is within 0.1% of the exact value. If the a2 term is omitted, then the accuracy drops to ±2% for | a | ≤ 0.15. The accuracy of F(a) coincides with the accuracy of the rectifier efficiency η.
F(a) ≈ 1 + 0.862a + 0.78a 2
10
F(a)
dθ
1 – a2
4 sin – 1 a + 2 cos – 1 a
–0.05
a
0.00
0.05
1 – a2
0.10
Chapter 18: Low harmonic rectifier modeling and control
–0.10
– 2a – π +
4 sin – 1 a + 2 cos – 1 a
0.15
Chapter 18: Low harmonic rectifier modeling and control
• a is typically much smaller than unity
– 2a – π +
0.85 –0.15
0.9
0.95
1
1.05
1.1
1.15
dθ = F(a) = 22 aπ 1 – a sin θ
sin 2 θ
Approximation via polynomial:
4 π
1 – a sin θ
sin 2 θ
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• a is a measure of the loss resistance relative to Re 9
dθ = F(a) = 22 aπ 1 – a sin θ
sin 2 θ
• Result is in closed form
0
π/2
ECEN5807 Power Electronics 2
4 π
0
π/2
This integral is obtained not only in the boost rectifier, but also in the buck-boost and other rectifier topologies. The solution is
V 2M R 2 I= 1 – on π VRe Re
By waveform symmetry, we need only integrate from 0 to Tac/4. Also, make the substitution θ = ωt:
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0.0
0.85
0.2
Ro
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1 + 0.862
11
VM /V
0.6
12
2
a=
VM V
Ron Re
1.0
• Losses other than MOSFET on-resistance are not included here
Chapter 18: Low harmonic rectifier modeling and control
0.8
• Efficiencies in the range 90% to 95% can then be obtained, even with Ron as high as 0.2Re
• To obtain high efficiency, choose V slightly larger than VM
η=
Pout R = 1 – on F(a) Pin Re
Chapter 18: Low harmonic rectifier modeling and control
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R on /R e = 0
0.9
0.95
1
ECEN5807 Power Electronics 2
η
ECEN5807 Power Electronics 2
η≈ 1–
with
VM Ron V R + 0.78 M on V Re V Re
Pout R = 1 – on F(a) Pin Re
V 2M R F(a) 1 – on VRe Re 2
Polynomial approximation:
η=
So the efficiency is
Pout = VI = V
Average load power is
V 2M Pin = pin(t) T = ac 2Re
Converter average input power is
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R on/R e
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0.4
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0.9
0.95
1
VM /V
0.6
14
1.0
Chapter 18: Low harmonic rectifier modeling and control
0.8
= (0.075) (27.4 Ω) = 2 Ω
Ron ≤ (0.075) Re
So we require a MOSFET with on resistance of
95% efficiency with VM/V = 0.435 occurs with Ron/Re ≈ 0.075.
Chapter 18: Low harmonic rectifier modeling and control
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13
VM 120 2 V = = 0.435 V 390 V
ECEN5807 Power Electronics 2
η
Also,
ECEN5807 Power Electronics 2
V 2g, rms (120 V) 2 Re = = = 27.4 Ω Pin 526 W
Then the emulated resistance is
P Pin = ηout = 500 W = 526 W 0.95
Average input power is
Assume that losses other than the MOSFET conduction loss are negligible.
Let us design for a given efficiency. Consider the following specifications: Output voltage 390 V Output power 500 W rms input voltage 120 V Efficiency 95%
'HVLJQH[DPSOH
ECEN5807 Power Electronics 2
Low frequency (average) component of input current is controlled to follow input voltage
Boost example
15
+ –
16
Current v (t) reference r
s
–
L
Gc(s) Compensator
Pulse width modulator
Gate driver
–
v(t)
Chapter 18: Low harmonic rectifier modeling and control
+
va(t) ≈ Rs 〈 ig(t)〉T
vg(t)
ig(t) +
Chapter 18: Low harmonic rectifier modeling and control
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Nonlinear carrier control
Hysteretic control
Current programmed control
Feedforward
Average current control
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x y
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verr(t)
x z y
vg(t)
xy z2
18
kv
multiplier
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v(t) +
–
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–
v(t)
–
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Gate driver
– +
vref2(t) Voltage reference
–
v(t)
+
Chapter 18: Low harmonic rectifier modeling and control
+
va(t) vref1(t)
+ –
vcontrol(t)
vg(t)
Gcv(s)
Gc(s) Compensator
Pulse width modulator
Gate driver
C
+
Chapter 18: Low harmonic rectifier modeling and control
vcontrol(t)
+
va(t) vref1(t)
+ –
+ –
Peak detector V M
Pavvg(t)Rs V 2g,rms
To maintain a given power throughput Pav, the reference voltage vref1(t) should be
vref 1(t) =
17
kx xy
Multiplier
vg(t)
ig(t)
)HHGIRUZDUG Feedforward is sometimes used to cancel out disturbances in the input voltage vg(t).
ECEN5807 Power Electronics 2
An analog multiplier introduces the dependence of Re on v(t).
As discussed in Chapter 17, an output voltage feedback loop adjusts the emulated resistance Re such that the rectifier power equals the dc load power: V 2g,rms Pav = = Pload vg(t) Re
8VHRIPXOWLSOLHUWRFRQWURODYHUDJHSRZHU
+ –
k vvcontrol(t)vg(t) V 2M
k vvcontrol(t) 2Rs Peak detector V M
x z y
vg(t)
19
kv
xy z2
multiplier
–
Gcv(s)
Gc(s) Compensator
Pulse width modulator
Gate driver
vcontrol(t)
20
Multiplier
ma
+ –
Clock
Q1
D1
0
Current-programmed controller
ic(t) Comparator = kx vg(t) vcontrol(t)
ia(t)
is(t)
L
Boost converter
Ts
C
i2(t)
Latch
R
S Q
–
v(t)
+ R
Chapter 18: Low harmonic rectifier modeling and control
X
vg(t)
vg(t) + –
ig(t)
++
ECEN5807 Power Electronics 2
Peak transistor current differs from average inductor current, because of inductor current ripple and artificial ramp. This leads to significant input current waveform distortion.
Peak transistor current is programmed to follow input voltage.
Current programmed control is a natural approach to obtain input resistor emulation:
– +
vref2(t) Voltage reference
–
v(t)
+
Chapter 18: Low harmonic rectifier modeling and control
+
va(t) vref1(t)
+ –
vcontrol(t)
vg(t)
ig(t)
&XUUHQWSURJUDPPHGFRQWURO
ECEN5807 Power Electronics 2
Pav =
The average power is then given by
vref 1(t) =
Controller with feedforward produces the following reference:
)HHGIRUZDUGFRQWLQXHG
+ –
i c(t) >
Ts
vg(t) TsV vg(t) 1– V 2L V
base
0.2
0.4
base
vg(t) V
vac(t)
vcontrol(t)
iac(t)
Inner wide-bandwidth controller
ECEN5807 Power Electronics 2
PWM
Q1
D1
+
i2(t)
Re = 0 .2R ba
se
22
0.6
ma = V 2L Rbase = 2L Ts
0.8
+
1.0
Re =
–+ vref3 Wide-bandwidth output voltage controller
v(t)
–
Load v(t)
i(t)
Compensator and modulator
d(t)
DC–DC Converter
–+ vref2
vC(t)
C
Chapter 18: Low harmonic rectifier modeling and control
Low-bandwidth energy-storage capacitor voltage controller
Compensator
–
vC(t)
Compensator
v (t) +– err Gc(s)
va(t)
ig(t) Rs
–
vref1(t) = kxvg(t)vcontrol(t)
X
vg(t)
L
Boost converter
Wide-bandwidth input current controller
Multiplier
+ vg(t)
ig(t)
base
CCM
R = 0.5R
DCM
e
Chapter 18: Low harmonic rectifier modeling and control
0.0
Re = 0.1R
RIKLJKTXDOLW\UHFWLILHUV
Outer low-bandwidth controller
Two loops:
21
0
0.2
0.4
0.6
0.8
Re = 0.33R
&RQWUROV\VWHPPRGHOLQJ
ECEN5807 Power Electronics 2
Minimum slope compensation:
vg(t) V
ma = V 2L
1–
vg(t) i c(t) = Re
TsV m a L vg(t) + L V V
>
It is desired that
or,
i g(t)
Mode boundary: CCM occurs when
1
e
=
bas
Ts
R = e R
i g(t)
Rbase V Ts
jg(t) = i g(t)
Static input characteristics of CPM boost, with minimum slope compensation:
R
R
vg(t)
e
=2 R
e
=4 R
Li 2c (t) fs in DCM vg(t) m a L V – vg(t) + V V v 2g(t)Ts vg(t) –1 + in CCM i c(t) + m a T s V 2LV
&30ERRVWFRQYHUWHU6WDWLFLQSXWFKDUDFWHULVWLFV
bas e bas e b as e
10 R
s
+ –
ac input vcontrol
Re (vcontrol )
〈 p(t)〉T
s
Ideal rectifier (LFR)
dc output
C
s
〈 i2(t)〉T
DYHUDJHGRYHUVZLWFKLQJSHULRG7V
s
–
〈 v(t)〉T
+
ECEN5807 Power Electronics 2
Nonlinear and time-varying 24
Chapter 18: Low harmonic rectifier modeling and control
Ac line-frequency harmonics are included in model
High-frequency switching harmonics are removed via averaging
Load
Chapter 18: Low harmonic rectifier modeling and control
/DUJHVLJQDOPRGHO
23
Ideal rectifier model, assuming that inner wide-bandwidth loop operates ideally
〈 vg(t)〉T
〈 ig(t)〉Ts
ECEN5807 Power Electronics 2
Objective of our modeling efforts: low-frequency small-signal model that predicts transfer functions at frequencies below the ac line frequency
The loop must be slow, to avoid introducing variations in Re at the harmonics of the ac line frequency
This loop maintains power balance, stabilizing the rectifier output voltage against variations in load power, ac line voltage, and component values
0RGHOLQJWKHRXWHUORZEDQGZLGWK FRQWUROV\VWHP
Ts
= Re(vcontrol(t))
Ts
2
=
s
+ –
ac input vcontrol
Re (vcontrol )
v 2g,rms 1 – cos 2ωt Re(vcontrol(t))
〈 vg(t)〉T
25
dc output
C
s
–
〈 v(t)〉T
+
s
V 2g,rms Re
C
s
〈 i2(t)〉T
–
〈 v(t)〉T
+
s
Load
ECEN5807 Power Electronics 2
26
Chapter 18: Low harmonic rectifier modeling and control
The second-harmonic variation in power leads to second-harmonic variations in the output voltage and current
Rectifier output port
V 2g,rms – cos 2 2ωt Re
Load
Chapter 18: Low harmonic rectifier modeling and control
s
〈 i2(t)〉T
6HSDUDWLRQRISRZHUVRXUFHLQWRLWVFRQVWDQWDQG WLPHYDU\LQJFRPSRQHQWV
ECEN5807 Power Electronics 2
〈 p(t)〉T
Ideal rectifier (LFR)
which contains a constant term plus a secondharmonic term
p(t)
vg(t)
Then the instantaneous power is:
vg(t) = 2 vg,rms sin ωt
If the input voltage is
〈 ig(t)〉Ts
3UHGLFWLRQVRIODUJHVLJQDOPRGHO
27
1 2
V 2g,rms Re
C
〈 i2(t)〉T2L
28
t
–
〈 v(t)〉T2L
+ Load
Chapter 18: Low harmonic rectifier modeling and control
Power source is nonlinear
Time invariant model
ECEN5807 Power Electronics 2
2L
Chapter 18: Low harmonic rectifier modeling and control
2π = π ω ω
〈 v(t)〉T
5HVXOWLQJDYHUDJHGPRGHO
T2L =
〈 v(t)〉Ts
Rectifier output port
ECEN5807 Power Electronics 2
v(t)
5HPRYDORIHYHQKDUPRQLFVYLDDYHUDJLQJ
T 2L
=
29
dvcontrol ECEN5807 Power Electronics 2
j2 =
v g,rms = V g,rms
T 2L
30
I2 V
v control = V control
Chapter 18: Low harmonic rectifier modeling and control
V 2g,rms dRe(vcontrol) =– 2 VR e (Vcontrol) dvcontrol
=–
Vg,rms 2 Re(Vcontrol) V
vcontrol(t) r2
v T =V 2L
=
v control = V control
, Vcontrol) T 2L
df Vg,rms, V, vcontrol)
d v
df Vg,rms, v
dvg,rms
df vg,rms, V, Vcontrol)
– 1 = r2
g2 =
where
I 2 + i 2(t) = g 2vg,rms(t) + j2v(t) –
/LQHDUL]HGUHVXOW
Chapter 18: Low harmonic rectifier modeling and control
Vcontrol >> vcontrol(t)
Vg,rms >> vg,rms(t)
V >> v(t)
vg,rms = Vg,rms + vg,rms(t) vcontrol(t) = Vcontrol + vcontrol(t)
= V + v(t)
T 2L
with
, vcontrol(t))
T 2L
Re(vcontrol(t)) v(t)
v 2g,rms(t)
I 2 >> i 2(t)
T 2L
T 2L
v(t)
= f vg,rms(t), v(t)
=
T 2L
= I 2 + i 2(t)
i 2(t)
v(t)
T 2L
ECEN5807 Power Electronics 2
Let
i 2(t)
p(t)
The averaged model predicts that the rectifier output current is
3HUWXUEDWLRQDQGOLQHDUL]DWLRQ
Rectifier output port
j2 vcontrol
R
Chapter 18: Low harmonic rectifier modeling and control
Pav VVcontrol
Pav VVcontrol Pav VT on 2Pav VD
2Pav VVg,rms
2Pav VVg,rms 2Pav VVg,rms 2Pav VVg,rms
Current-programmed control, Fig. 18.10 Nonlinear-carrier charge control of boost rectifier, Fig. 18.14 Boost with hysteretic control, Fig. 18.13(b) DCM buck–boost, flyback, SEPIC, or Cuk converters
ECEN5807 Power Electronics 2
32
Chapter 18: Low harmonic rectifier modeling and control
V2 Pav
V2 Pav
V2 2Pav
V2 Pav
V2 Pav Pav VVcontrol
Average current control with feedforward, Fig. 18.9
0
g2
Controller type
r2
Small-signal model parameters for several types of rectifier control schemes
0RGHOSDUDPHWHUV
31
j2
Tabl e 18. 1
ECEN5807 Power Electronics 2
v(s) 1 = g 2 R||r 2 vg,rms(s) 1 + sC R||r 2
–
v
Line-to-output
C
v(s) 1 = j2 R||r 2 vcontrol(s) 1 + sC R||r 2
r2
+
Control-to-output
Predicted transfer functions
g 2 vg,rms
i2
6PDOOVLJQDOHTXLYDOHQWFLUFXLW
–
vg(t)
+
Ideal rectifier (LFR)
Re
s
〈 pac(t)〉T C –
vC(t)
+
Energy storage capacitor
i2(t)
Dc-dc converter
Pload
V + –
Incremental resistance R of constant power load is negative, and is
33
Chapter 18: Low harmonic rectifier modeling and control
–
ECEN5807 Power Electronics 2
g v(s) = 2 vg,rms(s) sC
j v(s) = 2 vcontrol(s) sC
34
Chapter 18: Low harmonic rectifier modeling and control
When r2 = –R, the parallel combination r2 || R becomes equal to zero. The small-signal transfer functions then reduce to
7UDQVIHUIXQFWLRQVZLWKFRQVWDQWSRZHUORDG
ECEN5807 Power Electronics 2
which is equal in magnitude and opposite in polarity to rectifier incremental output resistance r2 for all controllers except NLC
2 R=– V Pav
i(t) v(t) load
+
pload(t) = VI = Pload
Rectifier and dc-dc converter operate with same average power
vac(t)
iac(t)
ig(t)
&RQVWDQWSRZHUORDG
+ –
Ts
〈i2(t)〉Ts
= Vg + vg(t)
–
〈v2(t)〉Ts
+
i 2(t)
v1(t)
v(t)
i(t)
Ts
Ts
Ts
Ts Ts
Ts
= I 2 + i 2(t) 35
R
s
–
〈v(t)〉T
So we are faced with the design of a control system that exhibits significant nonlinear time-varying behavior.
Problem: variations in vg, i1 , and d are not small.
C
+
Chapter 18: Low harmonic rectifier modeling and control
= V + v(t)
= I + i(t)
= V1 + v1(t)
= v2(t)
= i 1(t)
d(t) = D + d(t) ⇒ d'(t) = D' – d(t)
vg(t)
+ –
Averaged switch network
〈v1(t)〉Ts
s
〈i1(t)〉T
= V + v(t)
v(t) << V
Ts
dt
d i g(t)
dt
d i g(t)
ECEN5807 Power Electronics 2
L
substitute:
L
Ts
Ts
= vg(t)
= vg(t)
36
Ts
Ts
Chapter 18: Low harmonic rectifier modeling and control
– d'(t)V – d'(t)v(t)
– d'(t)V – d'(t)v(t)
The boost converter average inductor voltage is
In the special case of the boost rectifier, this is sufficient to linearize the equations of the average current controller.
with
v(t)
When the rectifier operates near steady-state, it is true that
/LQHDUL]LQJWKHHTXDWLRQVRIWKHERRVWUHFWLILHU
ECEN5807 Power Electronics 2
In Chapter 7, we perturbed and linearized using the assumptions
〈vg(t)〉Ts
s
〈i(t)〉T
L
Averaged (but not linearized) boost converter model, Fig. 7.42:
0RGHOLQJWKHLQQHUZLGHEDQGZLGWK DYHUDJHFXUUHQWFRQWUROOHU
dt
Ts
= vg(t) Ts
– d'(t)V – d'(t)v(t)
Ts
37
+ –
= vg(t)
vg(t)
Ts Ts
Ts
+ –
d'(t)V
Chapter 18: Low harmonic rectifier modeling and control
i g(t)
L
– d'(t)V
ECEN5807 Power Electronics 2
38
Chapter 18: Low harmonic rectifier modeling and control
The converter small-signal transfer functions derived in Chapters 7 and 8 are evaluated, using the time-varying operating point. The poles, zeroes, and gains vary slowly as the operating point varies. An average current controller is designed, that has a positive phase margin at each operating point.
Assume that the ac line variations are much slower than the converter dynamics, so that the rectifier always operates near equilibrium. The quiescent operating point changes slowly along the input sinusoid, and we can find the slowly-varying “equilibrium” duty ratio as in Section 18.1.
An approximate approach that is sometimes used in these cases: the quasi-static approximation
The above approach is not sufficient to linearize the equations needed to design the rectifier averaged current controllers of buck-boost, Cuk, SEPIC, and other converter topologies. These are truly nonlinear timevarying systems.
7KHTXDVLVWDWLFDSSUR[LPDWLRQ
ECEN5807 Power Electronics 2
i g(s) V = d(s) sL
dt
d i g(t)
Equivalent circuit:
L
The nonlinear term is much smaller than the linear ac term. Hence, it can be discarded to obtain
L
d i g(t)
/LQHDUL]HGERRVWUHFWLILHUPRGHO
Chapter 18: Low harmonic rectifier modeling and control
Soft switching Zero current switching Zero voltage switching The zero voltage transition converter Load-dependent properties of resonant converters
19.4
19.5
Chapter 19: Resonant Conversion
Exact characteristics of the series and parallel resonant converters
19.3
1
Examples Series resonant converter Parallel resonant converter
19.2
Fundamentals of Power Electronics
Sinusoidal analysis of resonant converters
5HVRQDQW&RQYHUVLRQ
&KDSWHU
39
19.1
Introduction
ECEN5807 Power Electronics 2
It is well-understood in the field of control systems that, when the converter dynamics are not sufficiently fast, then the quasi-static approximation yields neither necessary nor sufficient conditions for stability. Such behavior can be observed in rectifier systems. Worstcase analysis to prove stability should employ simulations.
No good condition on system parameters, which can justify the approximation, is presently known for the basic converter topologies
Should be valid provided that the converter dynamics are suffieiently fast, such that the converter always operates near the assumed operating points
In the literature, several authors have reported success using this method
4XDVLVWDWLFDSSUR[LPDWLRQGLVFXVVLRQ
2
dc source vg(t)
+ –
Cs
Fundamentals of Power Electronics
Series tank network
L
L
Cp
Switch network
3
–
vs(t)
+
is(t)
Parallel tank network
Several resonant tank networks
Basic circuit
NS
Cp
Cs
Cp
Cs
–
v(t)
+
Resistive load R
Chapter 19: Resonant Conversion
LCC tank network
L
Resonant tank network
L
NT i(t)
Chapter 19: Resonant Conversion
$EDVLFFODVVRIUHVRQDQWLQYHUWHUV
Fundamentals of Power Electronics
• Resonant inverters or rectifiers producing line-frequency ac
• Resonant dc-dc converters
• Dc-to-high-frequency-ac inverters
Some types of resonant converters:
Resonant power converters contain resonant L-C networks whose voltage and current waveforms vary sinusoidally during one or more subintervals of each switching period. These sinusoidal variations are large in magnitude, and the small ripple approximation does not apply.
,QWURGXFWLRQWR5HVRQDQW&RQYHUVLRQ
fs
fs
fs
3fs
3fs
3fs
4
5fs
5fs
5fs
f
f
f
NS Switch network
–
Cs
Fundamentals of Power Electronics
5
–
vR(t)
+
iR(t)
NT Resonant tank network
vs(t)
L
The series resonant dc-dc converter
dc source + – vg(t)
+
is(t)
Transfer function H(s)
NF
–
v(t)
+
i(t)
Chapter 19: Resonant Conversion
Rectifier network Low-pass dc filter load network
NR
Rectify and filter the output of a dc-high-frequency-ac inverter
R
Chapter 19: Resonant Conversion
Output can be controlled by variation of switching frequency, closer to or away from the tank resonant frequency
Tank current and output voltage are essentially sinusoids at the switching frequency fs.
'HULYDWLRQRIDUHVRQDQWGFGFFRQYHUWHU
Fundamentals of Power Electronics
Tank current spectrum
Resonant tank response
Switch output voltage spectrum
7DQNQHWZRUNUHVSRQGVRQO\WRIXQGDPHQWDO FRPSRQHQWRIVZLWFKHGZDYHIRUPV
Switch network
v2(t) –
v1(t) –
7
+
i2(t) +
Fundamentals of Power Electronics
Two switch networks:
+
–
v1(t)
i1(t)
–
v1(t)
i1(t) +
Switch network
L
C
i(t)
–
v(t)
–
v2(t)
i2(t) +
R
+
Chapter 19: Resonant Conversion
Cr
Lr
ZCS quasi-resonant switch network
–
v2(t)
+
i2(t)
R
Low-pass ac filter load network
–
v(t)
+
Chapter 19: Resonant Conversion
Switch network
Buck converter example
vg(t) + –
PWM switch network
In a conventional PWM converter, replace the PWM switch network with a switch network containing resonant elements.
i1(t)
6
Resonant tank network
Cs
4XDVLUHVRQDQWFRQYHUWHUV
Fundamentals of Power Electronics
dc source + – vg(t)
L
i(t)
Same as dc-dc series resonant converter, except output rectifiers are replaced with four-quadrant switches:
$VHULHVUHVRQDQWOLQNLQYHUWHU
8
Chapter 19: Resonant Conversion
Fundamentals of Power Electronics
Complexity of analysis
9
Chapter 19: Resonant Conversion
Resonant converters are usually controlled by variation of switching frequency. In some schemes, the range of switching frequencies can be very large
These considerations lead to increased conduction losses, which can offset the reduction in switching loss
Quasi-sinusoidal waveforms exhibit higher peak values than equivalent rectangular waveforms
Significant currents may circulate through the tank elements, even when the load is disconnected, leading to poor efficiency at light load
Can optimize performance at one operating point, but not with wide range of input voltage and load power variations
5HVRQDQWFRQYHUVLRQGLVDGYDQWDJHV
Fundamentals of Power Electronics
High voltage converters: significant transformer leakage inductance and winding capacitance leads to resonant network
In specialized applications, resonant networks may be unavoidable
Zero-current switching can be used to commutate SCRs
Zero-voltage switching also reduces converter-generated EMI
Turn-on or turn-off transitions of semiconductor devices can occur at zero crossings of tank voltage or current waveforms, thereby reducing or eliminating some of the switching loss mechanisms. Hence resonant converters can operate at higher switching frequencies than comparable PWM converters
Zero-voltage switching
Zero-current switching
The chief advantage of resonant converters: reduced switching loss
5HVRQDQWFRQYHUVLRQDGYDQWDJHV
10
Chapter 19: Resonant Conversion
NS Switch network
–
–
vR(t)
NT Resonant tank network
vs(t)
+
Cs
+
L
iR(t)
is(t)
Transfer function H(s)
NF
–
11
Chapter 19: Resonant Conversion
Let us model all ac waveforms by their fundamental components.
Fundamentals of Power Electronics
R
Rectifier network Low-pass dc filter load network
NR
v(t)
+
i(t)
If tank responds primarily to fundamental component of switch network output voltage waveform, then harmonics can be neglected.
dc source + – vg(t)
A resonant dc-dc converter:
6LQXVRLGDODQDO\VLVRIUHVRQDQWFRQYHUWHUV
Fundamentals of Power Electronics
• Ac modeling of quasi-resonant converters via averaged switch modeling
• Steady-state analysis of quasi-resonant converters
• Quasi-resonant converter topologies
• Circulating currents, and the dependence (or lack thereof) of conduction loss on load power
• Mechanisms of soft switching
• Simple and exact results for the series and parallel resonant converters
• Simple steady-state analysis via sinusoidal approximation
5HVRQDQWFRQYHUVLRQ2XWOLQHRIGLVFXVVLRQ
fs
fs
fs
+ –
1
2
Switch network
2
1
NS
–
vs(t)
+
is(t)
Fundamentals of Power Electronics
n = 1, 3, 5,...
If the switch network produces a square wave, then its output voltage has the following Fourier series: 4Vg vs(t) = π Σ 1n sin (nωst)
vg
3fs
3fs
3fs
12
5fs
5fs
5fs
f
f
f
Chapter 19: Resonant Conversion
Remaining ac waveforms can be found via phasor analysis.
Neglect harmonics of switch output voltage waveform, and model only the fundamental component.
Tank current and output voltage are essentially sinusoids at the switching frequency fs.
13
vs(t)
Chapter 19: Resonant Conversion
So model switch network output port with voltage source of value vs1(t)
4Vg vs1(t) = π sin (ωst) = Vs1 sin (ωst)
t
Fundamental component vs1(t)
The fundamental component is
– Vg
Vg
4 π Vg
&RQWUROOHGVZLWFKQHWZRUNPRGHO
Fundamentals of Power Electronics
Tank current spectrum
Resonant tank response
Switch output voltage spectrum
7KHVLQXVRLGDODSSUR[LPDWLRQ
+ – 1
2
Switch network
2
–
vs(t)
+
is(t)
2I s1 π cos (ϕ s)
0 T /2
T s/2
i g(τ)dτ
is1(t) = Is1 sin (ωst – ϕs)
15
Chapter 19: Resonant Conversion
• Model is power conservative: predicted average input and output powers are equal
• Fundamental ac components of output port waveforms are modeled
Fundamentals of Power Electronics
ωst
Chapter 19: Resonant Conversion
+ 4Vg – π sin (ωst)
vs1(t) =
is(t)
ig(t)
s ≈ 2 I s1 sin (ωsτ – ϕ s)dτ Ts 0 2 I cos (ϕ ) =π s1 s
= 2 Ts Ts
• Dc components of input port waveforms are modeled
• Switch network converts dc to ac
–
vg
+
14
i g(t)
ϕs
Is1
6ZLWFKQHWZRUNHTXLYDOHQWFLUFXLW
Fundamentals of Power Electronics
It is desired to model the dc component (average value) of the switch network input current.
i s(t) ≈ I s1 sin (ωst – ϕ s)
Assume that switch network output current is
vg
1
NS
0RGHORIVZLWFKQHWZRUNLQSXWSRUW
16
Chapter 19: Resonant Conversion
Fundamentals of Power Electronics
ϕR
iR(t)
V
Actual waveforms
–V
vR(t) ωst
17
ϕR
vR1(t) fundamental ωst
Chapter 19: Resonant Conversion
vR1(t) Re 8 Re = 2 R π
iR1(t) i R1(t) =
4 πV
with harmonics ignored
vR1(t) = 4V π sin (ωst – ϕ R) = V R1 sin (ωst – ϕ R)
Again, since tank responds only to fundamental components of applied waveforms, harmonics in vR(t) can be neglected. vR(t) becomes
6LQXVRLGDODSSUR[LPDWLRQUHFWLILHU
Fundamentals of Power Electronics
i R(t) = I R1 sin (ωst – ϕ R)
Then vR(t) has the following Fourier series: ∞ 1 vR(t) = 4V Σ π n = 1, 3, 5, n sin (nωst – ϕ R)
–V
ωst
vR(t) is a square wave, having zero crossings in phase with tank output current iR(t).
ϕR
iR(t)
vR(t)
If iR(t) is a sinusoid:
dc load
R
V
Assume large output filter capacitor, having small ripple.
Low-pass filter network
Rectifier network
– NF
NR
–
i(t) + v(t)
| iR(t) |
vR(t)
+
iR(t)
0RGHOLQJWKHUHFWLILHUDQGFDSDFLWLYH ILOWHUQHWZRUNV
–V
ωst
=I
I R1 sin (ωst – ϕ R) dt
Ts
Fundamentals of Power Electronics
Re = 82 R = 0.8106R π 19
–
vR1(t)
+ 2 π I R1
–
V
I
R
Chapter 19: Resonant Conversion
Rectifier equivalent circuit
Re = 82 R π
Re
iR1(t) +
Chapter 19: Resonant Conversion
T s/2
i R(t)
I= 2 TS 0 2I =π R1
Hence
With a resistive load R, this becomes
vR1(t) 8 V = i R(t) π 2 I
Effective resistance Re is
Hence rectifier presents a resistive load to tank network
Fundamental components of current and voltage are sinusoids that are in phase
Rectifier input port:
Re =
18
dc load
R
Output capacitor charge balance: dc load current is equal to average rectified tank output current
(TXLYDOHQWFLUFXLWRIUHFWLILHU
Fundamentals of Power Electronics
ϕR
iR(t)
vR(t)
Low-pass filter network
Rectifier network
V
NF
NR
–
–
i(t) + v(t)
| iR(t) |
vR(t)
+
iR(t)
5HFWLILHUGFRXWSXWSRUWPRGHO
+ –
Zi
Resonant network –
vR1(t)
+
iR1(t)
Re
20
s = jω s
Fundamentals of Power Electronics
21
vR1(s) H(s) = v (s) Re Re s1 which has peak magnitude H(s) s = jω s I R1 = Vs1 Re
i R(s) =
Solution for tank output current:
VR1 = H(s) Vs1
Ratio of peak values of input and output voltages:
vR1(s) = H(s) vs1(s)
Transfer function:
vs1(t)
+ –
Zi
is1(t)
–
vR1(t)
Re
Chapter 19: Resonant Conversion
Resonant network
Transfer function H(s) iR1(t) +
Chapter 19: Resonant Conversion
6ROXWLRQRIWDQNQHWZRUNZDYHIRUPV
Fundamentals of Power Electronics
Can solve for transfer function via conventional linear circuit analysis.
Model of ac waveforms is now reduced to a linear circuit. Tank network is excited by effective sinusoidal voltage (switch network output port), and is load by effective resistive load (rectifier input port).
vs1(t)
is1(t)
Transfer function H(s)
5HVRQDQWWDQNQHWZRUN
Zi
s = jω s
22
VR1 Vs1
H(s)
Resonant network
Vs1 Vg
4 π
–
V = H(s) Vg
s = jω s
Re
2 π I R1
–
V
+
I
s = jω s
Fundamentals of Power Electronics
23
R
Chapter 19: Resonant Conversion
V = H(s) Vg
Eliminate Re:
Re = 82 R π
vR1(t)
&RQYHUVLRQUDWLR0
I R1 VR1
1 Re
4Vg π sin (ωst)
vs1(t) =
+ –
iR1(t) +
Chapter 19: Resonant Conversion
So we have shown that the conversion ratio of a resonant converter, having switch and rectifier networks as in previous slides, is equal to the magnitude of the tank network transfer function. This transfer function is evaluated with the tank loaded by the effective rectifier input resistance Re.
Fundamentals of Power Electronics
I I R1
V I
2I s1 π cos (ϕ s)
2 π
+ –
M= V = R Vg
Vg
is1(t)
Transfer function H(s)
6ROXWLRQRIFRQYHUWHU YROWDJHFRQYHUVLRQUDWLR0 99J
NS switch network
+ –
L
Cs
24
+
–
vR(t)
NT resonant tank network
2I s1 π cos (ϕ s)
4Vg π sin (ωst)
Fundamentals of Power Electronics
C
25
NF
–
v(t)
i(t) + R
–
Re
Re = 82 R π
vR1(t)
iR1(t) + 2 π I R1
I
R
2
Chapter 19: Resonant Conversion
1 + Q 2e 1 – F F
1
–
V
+
Chapter 19: Resonant Conversion
rectifier network low-pass dc filter load network
NR
1 = 2π f 0 LC L R0 = M = H( jωs) = C R Qe = 0 Re
series tank network
L
transfer function H(s)
ω0 =
Zi
is1(t)
vs1(t) =
+ –
Re Re = Z i(s) R + sL + 1 e sC s Q eω 0 = 2 s 1+ + ωs Q eω 0 0 H(s) =
Vg
–
vs(t)
+
iR(t)
0RGHOVHULHVUHVRQDQWFRQYHUWHU
Fundamentals of Power Electronics
dc source + – vg(t)
is(t)
transfer function H(s)
([DPSOHV 6HULHVUHVRQDQWFRQYHUWHU
Re
1 ωC
C ωR e
Fundamentals of Power Electronics
|| H ||
Fundamentals of Power Electronics
|| Zi ||
26
Re / R0
1
27
f0
&RQVWUXFWLRQRI+
R0
f0
ωL
Chapter 19: Resonant Conversion
Qe = R0 / Re
Chapter 19: Resonant Conversion
R / e ω L
Qe = Re / R0
&RQVWUXFWLRQRI=L
fs
fs
fs
etc. 1 f 5 0
Fundamentals of Power Electronics
1 3 1 5
1
M
3fs
3fs
3fs
28
5fs
5fs
5fs
f
f
f
29
1 f 3 0
f0
Chapter 19: Resonant Conversion
fs
Chapter 19: Resonant Conversion
H( jnωs) M= V = n Vg
Result of analysis:
4Vg vs(t) ≈ vsn(t) = nπ sin (nωst)
Can now approximate vs(t) by its third harmonic:
Example: excitation of tank by third harmonic of switching frequency
6XEKDUPRQLFPRGHVRI65&
Fundamentals of Power Electronics
tank current spectrum
resonant tank response
switch output voltage spectrum
6XEKDUPRQLFPRGHVRIWKH65&
NS switch network
–
vs(t)
Cp
+
–
vR(t)
NT resonant tank network
L
iR(t)
rectifier network
NR
Fundamentals of Power Electronics
ϕR
–I
iR(t)
iR1(t) fundamental
vR1(t) Re 2 Re = π R 8
vR1(t) i R1(t) =
4 πI
vR(t)
I
30
ωst
ωst
31
NR rectifier network
NF low-pass filter network
R
dc load
–
v(t)
+
i(t)
R
Chapter 19: Resonant Conversion
i R1(t) = 4I π sin (ωst – ϕ R)
Fundamental component of iR(t):
k
–
vR(t)
+
iR(t)
dc load
–
v(t)
+
i(t)
Chapter 19: Resonant Conversion
0RGHORIXQFRQWUROOHGUHFWLILHU ZLWKLQGXFWLYHILOWHUQHWZRUN
Fundamentals of Power Electronics
Need a new model for rectifier and filter networks
ϕR
NF low-pass filter network
Rectifier is driven by sinusoidal voltage, and is connected to inductive-input low-pass filter
Different tank network
Differs from series resonant converter as follows:
dc source + – vg(t)
+
is(t)
3DUDOOHOUHVRQDQWGFGFFRQYHUWHU
vR1(t) πVR1 = 4I i R1(t)
0
T s/2
2V VR1 sin (ωst – ϕ R) dt = π R1
32
Chapter 19: Resonant Conversion
Re
2 Re = π R 8
Fundamentals of Power Electronics
–
vR1(t)
+
iR1(t)
33
2 π V R1
+ –
R
Chapter 19: Resonant Conversion
–
V
+
I
(TXLYDOHQWFLUFXLWPRGHORIXQFRQWUROOHGUHFWLILHU ZLWKLQGXFWLYHILOWHUQHWZRUN
Fundamentals of Power Electronics
2 π Re = R = 1.2337R 8
For a resistive load, V = IR. The effective resistance Re can then be expressed
V= 2 TS
In steady state, the dc output voltage V is equal to the average value of | vR |:
Re =
Again define
(IIHFWLYHUHVLVWDQFH5H
+ –
vs1(t) = 4Vg π sin (ωst)
Re
Fundamentals of Power Electronics
|| Zo ||
C
34
parallel tank network
s = jω s
Zi
L
iR1(t)
ωL
R0
35
f0
2 π V R1
+ –
–
V
+
I
R
Chapter 19: Resonant Conversion
1 ωC
Qe = Re / R0
Chapter 19: Resonant Conversion
Z o(s) = sL || 1 || Re sC
Z o(s) sL
Re = π R 8 2
Re
H(s) =
–
vR1(t)
+
&RQVWUXFWLRQRI=R
M = V = 82 H(s) Vg π
2I s1 π cos (ϕ s)
+ –
Fundamentals of Power Electronics
Vg
is1(t)
transfer function H(s)
(TXLYDOHQWFLUFXLWPRGHO 3DUDOOHOUHVRQDQWGFGFFRQYHUWHU
1
Re / R0
36
f0
1–F
2
R M = 82 e = R π R0 R0
+ F Qe
1 s + s ω0 Q eω 0
s = jω s
Fundamentals of Power Electronics
37
Chapter 19: Resonant Conversion
• PRC can produce M approaching infinity, provided output current is limited to value less than Vg / R0
• PRC can step up the voltage, provided R > R0
2 2
1
= 82 π 1+ s = jω s
At resonance, this becomes
= 82 π
Z o(s) M = 82 sL π
2
Chapter 19: Resonant Conversion
1 ω LC 2
Qe = Re / R0
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Fundamentals of Power Electronics
|| H ||
&RQVWUXFWLRQRI+
f0 / 3
k=2
+ –
38
k=1
ξ=1
f0
mode index k
Q2
Q1
D2
D1
Q4
Q3
D4
D3
L
Fundamentals of Power Electronics
M= V nVg
39
J=
C
InR0 Vg
1:n
fs
–
V
Chapter 19: Resonant Conversion
R
+
Chapter 19: Resonant Conversion
k=0
subharmonic index ξ
1
Normalized load voltage and current:
Vg
f0 / 2
or
([DFWFKDUDFWHULVWLFVRIWKH VHULHVUHVRQDQWFRQYHUWHU
Fundamentals of Power Electronics
etc. k = 3
ξ=3
1 + ( – 1) k ξ=k+ 2
f0 f < fs < 0 k+1 k
Define
([DFWFKDUDFWHULVWLFVRIWKH VHULHVDQGSDUDOOHOUHVRQDQWGFGFFRQYHUWHUV
Q1 π
Q1
40
(k – 1) complete half-cycles γ
D1
π
Vg
D1
π
Q1
Fundamentals of Power Electronics
iL(t)
D1
π
41
k complete half-cycles γ
Vg
Waveforms for type k CCM, even k : vs(t)
π
Q1
D1
π
D1
Q2
– Vg
ωst
D2 Q2
ωst
Chapter 19: Resonant Conversion
Q1
– Vg
Chapter 19: Resonant Conversion
6HULHVUHVRQDQWFRQYHUWHU
Fundamentals of Power Electronics
iL(t)
vs(t)
Tank current rings continuously for entire length of switching period Waveforms for type k CCM, odd k :
&RQWLQXRXVFRQGXFWLRQPRGH65&
γ=
ω0Ts π = F 2
J=
InR0 Vg
γ Jγ + 12 + (– 1) k 2 2 ξ
M= V nVg
2
2
cos 2
γ =1 2
42
0≤M ≤ 1 ξ
ξ tan 4
2
2
(–1) k+1 +
1+
Fundamentals of Power Electronics
43
Exact, closed-form, valid for any CCM
M=
Qγ 2 γ Qγ + 2 2
ξ 2 – cos 2
2
cos 2
ξ 4tan 2
γ 2
2
Chapter 19: Resonant Conversion
Qγ 2
γ 2
For a resistive load, eliminate J and solve for M vs. γ
Qγ γ + 2 2
Chapter 19: Resonant Conversion
&RQWUROSODQHFKDUDFWHULVWLFV
Fundamentals of Power Electronics
• M is restricted to the range
• Output characteristic, i.e., the relation between M and J, is elliptical
where
M ξ sin 2 2
6HULHVUHVRQDQWFRQYHUWHU
([DFWVWHDG\VWDWHVROXWLRQ&&0
π
Q1
44
k complete half-cycles γ
D1
π
Vg
π
Q1
X
ωst
Chapter 19: Resonant Conversion
Q2
– Vg
Fundamentals of Power Electronics
f0 k
45
2(k – 1) 2(k + 1) > J > γ γ
fs <
Conditions for operation in type k DCM, odd k :
M=1 k
Chapter 19: Resonant Conversion
6WHDG\VWDWHVROXWLRQW\SHN'&0RGGN
Fundamentals of Power Electronics
iL(t)
vs(t)
Type k DCM: during each half-switching-period, the tank rings for k complete half-cycles. The output diodes then become reverse-biased for the remainder of the half-switching-period.
'LVFRQWLQXRXVFRQGXFWLRQPRGH
J = 2k γ
0
0.75 1 1.5 2 3.5 5 10 Q = 20
0.5
0.35
0.2
Q = 0.2
Fundamentals of Power Electronics
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
– 46
Vg
+
Ig = gV
g = 2k γR0
g
0.4
0.6
1
F = f s / f0 47
0.8
1.2
1.4
1.8
2
Chapter 19: Resonant Conversion
1.6
3.5 5 10 Q = 20
2
1.5
1
0.75
0.5
0.35
Q = 0.2
Chapter 19: Resonant Conversion
–
V
+
&RQWUROSODQHFKDUDFWHULVWLFV65&
Fundamentals of Power Electronics
gyrator model, SRC operating in an even DCM:
f0 k
1 >M> 1 k–1 k+1
fs <
Ig = gVg
6WHDG\VWDWHVROXWLRQW\SHN'&0HYHQN
Conditions for operation in type k DCM, even k :
M = V / Vg
0
0.2
0
1
2
3
4
5
6
0
F = 1.15
0.2
F = 1.30
Fundamentals of Power Electronics
J
0.4
0.6
48
0.8
k = 1 CCM
1
F
1.2
1.4
1.8
2
Chapter 19: Resonant Conversion
1.6
k = 0 CCM
49
0.4
F = 1.10
M
0.6
F = 1.07
F = 1.01
1
Chapter 19: Resonant Conversion
0.8
F = 1.05
2XWSXWFKDUDFWHULVWLFV65&DERYHUHVRQDQFH
Fundamentals of Power Electronics
0
0.1
etc.
k=4 DCM
0.3 0.2
k = 3 DCM
k = 2 DCM
k = 1 DCM
0.4
0.5
0.6
0.7
0.8
0.9
1
k = 3 CCM
M
0RGHERXQGDULHV65&
k = 2 CCM
0 0.2
F = .25 F = .1
F = .5
+ –
0.4
50
M
0.6
k = 2 DCM
k = 1 CCM
D2
Q4
Q3
D4
D3
L
C
M= V nVg 51
Normalized load voltage and current:
Q2
D1
Fundamentals of Power Electronics
Vg
Q1
F = .75
F = .85
F = .90
J=
1:n
InR0 Vg
D8
D7
F = .96
F = 1.0
1
4 π
R
–
V
Chapter 19: Resonant Conversion
D6
D5
+
Chapter 19: Resonant Conversion
0.8
F = .93
([DFWFKDUDFWHULVWLFVRI WKHSDUDOOHOUHVRQDQWFRQYHUWHU
Fundamentals of Power Electronics
0
2 π
1
1.5
2
2.5
3
J
2XWSXWFKDUDFWHULVWLFV65&EHORZUHVRQDQFH
k = 1 DCM
γ γ + J sin 2 2
+ cos – 1 cos
Mode boundary
γ + 1 sin 2 γ 2 4
vC(t)
Fundamentals of Power Electronics
(require iteration)
M C0 = 1 – cos (β) J L0 = J + sin (β) cos (α + β) – 2 cos (α) = –1 iL(t) – sin (α + β) + 2 sin (α) + (δ – α) = 2J β+δ=γ M = 1 + 2γ (J – δ)
sin 2
for DCM for CCM
DCM equations
J crit(γ) = – 1 sin (γ) + 2
J > J crit(γ) J < J crit(γ)
52
for π < γ < 2π
for 0 < γ < π (
vC(t)
vC(t)
iL(t)
vs(t) γ
Vg
53
α
D6 D7 δ
γ
I
D5 D8 D6 D7
β
D5 D8
V=
vC(t)
ω0t
Ts
–I
D5 D8 D6 D7
D 6 D7
ω0t
Chapter 19: Resonant Conversion
D5 D8
ω 0t
Chapter 19: Resonant Conversion
– Vg
γ
3DUDOOHOUHVRQDQWFRQYHUWHULQ'&0
Fundamentals of Power Electronics
ϕ=
γ γ + J sin 2 2
– cos – 1 cos
sin (ϕ) M = 2γ ϕ – γ cos 2
CCM closed-form solution
3DUDOOHOUHVRQDQWFRQYHUWHULQ&&0
0.0
F=2 0.5
1.5
0.8 0.9 1.0
0.7
Solid curves: CCM
0.0
0.5
1.0
1.5
0.6
1.0
M
1.5
54
Shaded curves: DCM
1.1
0.5
Q=2
Q = 0.2
1.0
Q=1
Q = 0.5
Fundamentals of Power Electronics
0.0
0.5
1.0
1.5
2.0
2.5
3.0 Q=5
55
1.5
fs /f0
2.0
ZLWKUHVLVWLYHORDG
2.5
Chapter 19: Resonant Conversion
2.5
3.0
Chapter 19: Resonant Conversion
2.0
&RQWUROFKDUDFWHULVWLFVRIWKH35&
1.3
1.2
F = 0.51
2XWSXWFKDUDFWHULVWLFVRIWKH35&
Fundamentals of Power Electronics
J
2.0
2.5
3.0
M = V/Vg
56
Q2
D2
D1
Q4
Q3
D4
D3
–
vs(t)
+
L
is(t)
C
Fundamentals of Power Electronics
57
Zero-current switching (ZCS) occurs
Chapter 19: Resonant Conversion
Operation below resonance: input tank current leads voltage
+ –
vds1(t)
+
iQ1(t) –
Q1
Series resonant converter example
Vg
Chapter 19: Resonant Conversion
2SHUDWLRQRIWKHIXOOEULGJHEHORZ UHVRQDQFH=HURFXUUHQWVZLWFKLQJ
Fundamentals of Power Electronics
Soft switching can mitigate some of the mechanisms of switching loss and possibly reduce the generation of EMI Semiconductor devices are switched on or off at the zero crossing of their voltage or current waveforms: Zero-current switching: transistor turn-off transition occurs at zero current. Zero-current switching eliminates the switching loss caused by IGBT current tailing and by stray inductances. It can also be used to commutate SCR’s. Zero-voltage switching: transistor turn-on transition occurs at zero voltage. Diodes may also operate with zero-voltage switching. Zero-voltage switching eliminates the switching loss induced by diode stored charge and device output capacitances. Zero-voltage switching is usually preferred in modern converters. Zero-voltage transition converters are modified PWM converters, in which an inductor charges and discharges the device capacitances. Zero-voltage switching is then obtained.
6RIWVZLWFKLQJ
|| Zi ||
58
Re
1 ωC
R0
Qe = R0 /Re
Q1 Q4
tβ
D1 D4
Ts 2
vs1(t)
Q2 Q3
Fundamentals of Power Electronics
D2 D3
t
t
“Soft” turn-off of Q2, Q3
Ts + tβ 2
vs(t)
“Soft” “Hard” “Hard” turn-on of turn-off of turn-on of Q 1, Q 4 Q2, Q3 Q1, Q4
Conducting devices:
is(t)
– Vg
Vg
59
D1
Q4
Q3
D4
D3
–
vs(t)
+
L
is(t)
Chapter 19: Resonant Conversion
Q1 is turned off during D1 conduction interval, without loss
Conduction sequence: Q1–D1–Q2–D2
Q2
D2
vds1(t)
+
iQ1(t) –
Q1
=HURFXUUHQWVZLWFKLQJ
C
Chapter 19: Resonant Conversion
f0
ωL
6ZLWFKQHWZRUNZDYHIRUPVEHORZUHVRQDQFH
Fundamentals of Power Electronics
Zero crossing of the tank input current waveform is(t) occurs before the zero crossing of the voltage vs(t).
∠Zi is positive, and tank input current leads tank input voltage.
Operation below resonance: tank input impedance Zi is dominated by tank capacitor.
7DQNLQSXWLPSHGDQFH
Q1 Q4
tβ D1 D4
Ts 2 Q2 Q3
Ts + tβ 2
Vg
D2 D3
t
t
60
Q4
Q3
D4
D3
–
vs(t)
+
L
Q2
D2
D1
Q4
Q3
D4
D3
L
–
vs(t)
+
61
Zero-voltage switching (ZVS) occurs
Fundamentals of Power Electronics
is(t)
C
Chapter 19: Resonant Conversion
is(t)
C
Chapter 19: Resonant Conversion
Operation above resonance: input tank current lags voltage
+ –
vds1(t)
+
iQ1(t) –
Q1
Series resonant converter example
Vg
D1
Q1 turns on while D2 is conducting. Stored charge of D2 and of semiconductor output capacitances must be removed. Transistor turn-on transition is identical to hardswitched PWM, and switching loss occurs.
Q2 D2
vds1(t)
+
iQ1(t) –
Q1
2SHUDWLRQRIWKHIXOOEULGJHEHORZ UHVRQDQFH=HURYROWDJHVZLWFKLQJ
Fundamentals of Power Electronics
“Soft” “Hard” turn-on of turn-off of Q1, Q4 Q 1, Q 4
Conducting devices:
ids(t)
vds1(t)
=&6WXUQRQWUDQVLWLRQKDUGVZLWFKLQJ
|| Zi ||
62
Re
1 ωC
R0
Qe = R0 /Re
tα
D2 D3
Q2 Q3
vs(t)
t
t
“Hard” “Soft” “Hard” turn-off of turn-on of turn-off of Q1, Q4 Q2, Q3 Q2, Q3
Q1 Q4
Ts 2
vs1(t)
Fundamentals of Power Electronics
“Soft” turn-on of Q 1, Q 4
Conducting D1 devices: D 4
is(t)
– Vg
Vg
63
D1
Q4
Q3
D4
D3
–
vs(t)
+
L
is(t)
Chapter 19: Resonant Conversion
Q1 is turned on during D1 conduction interval, without loss
Conduction sequence: D1–Q1–D2–Q2
Q2
D2
vds1(t)
+
iQ1(t) –
Q1
=HURYROWDJHVZLWFKLQJ
C
Chapter 19: Resonant Conversion
f0
ωL
6ZLWFKQHWZRUNZDYHIRUPVDERYHUHVRQDQFH
Fundamentals of Power Electronics
Zero crossing of the tank input current waveform is(t) occurs after the zero crossing of the voltage vs(t).
∠Zi is negative, and tank input current lags tank input voltage.
Operation above resonance: tank input impedance Zi is dominated by tank inductor.
7DQNLQSXWLPSHGDQFH
tα
“Hard” turn-off of Q1, Q4
Q1 Q4
Ts 2 D2 D3
+ –
Conducting devices:
vds1(t)
Q2
Q1
D2
Turn off Q1, Q4
Q1 Q4
Cleg
D1 C leg
Vg
Cleg
Cleg
Commutation interval
X D2 D3
–
vds1(t)
+
Fundamentals of Power Electronics
Vg
Q2 Q3
Vg
t
t
D1
Q4
Q3
D4
D3
–
vs(t)
+
L
is(t)
64
t
65
D4
D3
C
Chapter 19: Resonant Conversion
When Q1 turns off, D2 must begin conducting. Voltage across Q1 must increase to Vg. Transistor turn-off transition is identical to hard-switched PWM. Switching loss may occur (but see next slide).
Q2 D2
vds1(t)
+
iQ1(t) –
Q1
–
vs(t)
+
is(t)
to remainder of converter
L
• Introduce delay between turn-off of Q1 and turn-on of Q2.
• Introduce small capacitors Cleg across each device (or use device output capacitances).
Chapter 19: Resonant Conversion
So zero-voltage switching exhibits low switching loss: losses due to diode stored charge and device output capacitances are eliminated.
Tank current is(t) charges and discharges Cleg. Turn-off transition becomes lossless. During commutation interval, no devices conduct.
Q4
Q3
6RIWVZLWFKLQJDWWKH=96WXUQRIIWUDQVLWLRQ
Fundamentals of Power Electronics
“Soft” turn-on of Q1, Q4
Conducting D1 devices: D 4
ids(t)
vds1(t)
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Commutation interval
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t
Fundamentals of Power Electronics
67
Chapter 19: Resonant Conversion
• To obtain good efficiency at light load, the transistor current should scale proportionally to load current (in resonant converters, it often doesn’t!)
3. Minimize transistor currents and conduction losses
• Preferably, obtain these properties at all loads • Could allow ZVS property to be lost at light load, if necessary
2. Obtain zero-voltage switching or zero-current switching
• With a nonlinear load, must properly match inverter output characteristic to load characteristic
1. Operate with a specified load characteristic and range of operating points
Resonant inverter design objectives:
Vg
Chapter 19: Resonant Conversion
Turn off Q2
Conducting devices:
v2(t)
ic(t)
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Fundamentals of Power Electronics
• Phase-shift control
• Secondary-side diodes switch at zero-current, with loss
• Can obtain ZVS of all primaryside MOSFETs and diodes
Vg
Q1
Basic version based on full-bridge PWM buck converter
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= H ∞( jω)
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Fundamentals of Power Electronics
with
vo
2
2
2
/ R2
vi
The output voltage magnitude is:
69
and let Zo0 be the output impedance (with vi 6short-circuit). Then, R vo( jω) = H ∞( jω) vi( jω) R + Z o0( jω)
R→∞
Let H∞ be the open-circuit (R6∞) transfer function: vo( jω) vi( jω)
68
Zi
ii(t)
purely reactive
resonant network
–
+ Zo v (t) o
resistive load R
2
+ io
2
Z o0
2
= H∞
2
vi
2
Chapter 19: Resonant Conversion
Hence, at a given frequency, the output characteristic (i.e., the relation between ||vo|| and ||io||) of any resonant inverter of this class is elliptical.
vo
This result can be rearranged to obtain
sinusoidal source + vi(t) –
transfer function H(s) io(t)
Chapter 19: Resonant Conversion
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Fundamentals of Power Electronics
• Series • Parallel • LCC
Examples and interpretation
• Dependence of transistor current on load current • Dependence of zero-voltage/zero-current switching on load resistance • Simple, intuitive frequency-domain approach to design of resonant converter
Two theorems
Inverter output i-v characteristics
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vi
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vi
|| vo ||
Fundamentals of Power Electronics
lamp characteristic
inverter characteristic
Electronic ballast
|| io ||
70
|| vo ||
71
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Chapter 19: Resonant Conversion
hed tc ma
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inverter characteristic
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Chapter 19: Resonant Conversion
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Fundamentals of Power Electronics
This result is valid provided that (i) the resonant network is purely reactive, and (ii) the load is purely resistive.
with
2
vo io + 2 =1 2 V oc I sc
2
General resonant inverter output characteristics are elliptical, of the form
,QYHUWHURXWSXWFKDUDFWHULVWLFV
50Ω
R→0
Z i0 Z o0 = Z i∞ Z o∞
H∞
2
= Z o0
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H ∞(s) 1+ R Z o0
Fundamentals of Power Electronics
where
H(s) =
Tank transfer function
Reciprocity
vi ii R→∞
72
Z o0 =
1 – 1 Z i0 Z i∞ 73
vo – io
Zo
Effective resistive load R
vi → open circuit
–
v(t)
+
i(t)
i0 * i∞ * o0 * o∞ * ∞
2
2
Chapter 19: Resonant Conversion
Hence, the input impedance magnitude is 2 1+ R Z o0 2 2 Z i = Z iZ *i = Z i0 2 1+ R Z o∞
Z i∞ = – Z Z o0 = – Z Z o∞ = – Z H∞ = – H
If the tank network is purely reactive, then each of its impedances and transfer functions have zero real parts: Z = – Z* i0
Chapter 19: Resonant Conversion
Z o∞ =
Purely reactive
Resonant network
vi → short circuit
Zi
is(t)
vo – io
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2WKHUUHODWLRQV
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Z i∞ =
Fundamentals of Power Electronics
where v Z i0 = i ii
Z (s) 1+ R 1 + o0 R Z o0(s) Z i(s) = Z i0(s) = Z i∞(s) Z (s) 1+ R 1 + o∞ Z o∞(s) R
Transfer function H(s)
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L
L
Cp
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p
1 ωC
|| Zi0 ||
1 ωC + s
|| Zi0 ||
p
1 ωC
s
1 ωC
74
|| Zi∞ ||
f
f
f
|| Zi∞ ||
ωL
|| Zi∞ ||
ωL
|| Zi0 ||
ωL
Chapter 19: Resonant Conversion
Z i∞(s) = sL + 1 + 1 sC p sC s
Z i0(s) = sL + 1 sC s
Z i∞(s) = sL + 1 sC p
Z i0(s) = sL
Z i∞(s) = ∞
Z i0(s) = sL + 1 sC s
75
Chapter 19: Resonant Conversion
So as the load resistance R varies from 0 to ∞, the resonant network input impedance || Zi || varies monotonically from the short-circuit value || Zi0 || to the open-circuit value || Zi∞ ||. The impedances || Zi∞ || and || Zi0 || are easy to construct. If you want to minimize the circulating tank currents at light load, maximize || Zi∞ ||. Note: for many inverters, || Zi∞ || < || Zi0 || ! The no-load transistor current is therefore greater than the short-circuit transistor current.
Fundamentals of Power Electronics
O
O
O
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Theorem 1: If the tank network is purely reactive, then its input impedance || Zi || is a monotonic function of the load resistance R.
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Fundamentals of Power Electronics
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L
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76
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77
|| Zi ||
s
–
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f
Chapter 19: Resonant Conversion
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Chapter 19: Resonant Conversion
In the special case || Zi0 || = || Zi∞ ||, || Zi || is independent of R.
1 ωC + s
1 ωC
1+
2
gR sin rea
Fundamentals of Power Electronics
• for f < f m, || Zi || increases with increasing R . • for f > f m, || Zi || decreases with increasing R . • at a given frequency f, || Zi || is a monotonic function of R. • It’s not necessary to draw the entire plot: just construct || Zi0 || and || Zi∞ ||.
2
1 Z o0
So the resonant network input impedance is a monotonic function of R, over the range 0 < R < ∞.
d Zi dR
2
; Differentiate:
([DPSOH__=L__RI/&&
Fundamentals of Power Electronics
(i) R = 0 (ii) R = ∞ (iii) Z o0 = Z o∞ , or Z i0 = Z i∞
; Derivative has roots at:
2
1+
R ing rea s inc
Zi
Previously shown:
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L
L
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s
1 ωC
1 ωC + s
1 ωC
Fundamentals of Power Electronics
Zi
LCC
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L
Parallel
Zi
Series
78
|| Zi ||
Zi∞
L
p
1 ωC
|| Zi0 ||
1 ωC s
1 ωC + s
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fm
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|| Zi∞ ||
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79
f
f
|| Zi∞ ||
ωL
|| Zi∞ ||
ωL
|| Zi0 ||
ωL
L
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Chapter 19: Resonant Conversion
f0
Chapter 19: Resonant Conversion
• Below resonance: no-load transistor current is less than short-circuit current (for f
• Above resonance: no-load transistor current is greater than short-circuit transistor current. ZVS.
• ZCS below resonance, ZVS above resonance
• No-load transistor current = 0, both above and below resonance.
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Fundamentals of Power Electronics
then || Zi0 || > || Zi∞ || ; hence transistor current increases as load current decreases, and transistor current is greater than or equal to short-circuit current for all R
For f > fm
then || Zi0 || < || Zi∞ || ; hence transistor current decreases as load current decreases
For f < fm
|| Zi0 || and || Zi∞ || both represent series resonant impedances, whose Bode diagrams are easily constructed. || Zi0 || and || Zi∞ || intersect at frequency fm.
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Fundamentals of Power Electronics
Im Z i(Rcrit) = 0
81
Chapter 19: Resonant Conversion
– Z i∞ Z i0
1+
R 2crit
Z o∞
2
Z o0Z o∞ R 2crit
Chapter 19: Resonant Conversion
Rcrit = Z o0
Solution for Rcrit yields
= Im Z i∞ Re
1–
Note that Zi∞, Zo0, and Zo∞ have zero real parts. Hence, Z 1 + o0 Rcrit Im Z i(Rcrit) = Im Z i∞ Re Z 1 + o∞ Rcrit
3URRIRI7KHRUHP
80
If ZCS occurs when Zi is capacitive, while ZVS occurs when Zi is inductive, then the boundary is determined by ∠Zi = 0. Hence, the critical load Rcrit is the resistance which causes the imaginary part of Zi to be zero:
Previously shown: Z 1 + o0 R Z i = Z i∞ Z o∞ 1+ R
Fundamentals of Power Electronics
It is assumed that zero-current switching (ZCS) occurs when the tank input impedance is capacitive in nature, while zero-voltage switching (ZVS) occurs when the tank is inductive in nature. This assumption gives a necessary but not sufficient condition for ZVS when significant semiconductor output capacitance is present.
Rcrit = Z o0
Theorem 2: If the tank network is purely reactive, then the boundary between zero-current switching and zero-voltage switching occurs when the load resistance R is equal to the critical value Rcrit, given by
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|| Zi ||
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Chapter 19: Resonant Conversion
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/&&H[DPSOH For f > f∞, ZVS occurs for all R. For f < f0, ZCS occurs for all R. For f0 < f < f∞, ZVS occurs for R< Rcrit, and ZCS occurs for R> Rcrit. Note that R = || Zo0 || corresponds to operation at matched load with maximum output power. The boundary is expressed in terms of this matched load impedance, and the ratio Zi∞ / Zi0.
Fundamentals of Power Electronics
O
O
O
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– Z i∞ Z i0
Again, Zi∞, Zi0, and Zo0 are pure imaginary quantities. If Zi∞ and Zi0 have the same phase (both inductive or both capacitive), then there is no real solution for Rcrit. Hence, if at a given frequency Zi∞ and Zi0 are both capacitive, then ZCS occurs for all loads. If Zi∞ and Zi0 are both inductive, then ZVS occurs for all loads. If Zi∞ and Zi0 have opposite phase (one is capacitive and the other is inductive), then there is a real solution for Rcrit. The boundary between ZVS and ZCS operation is then given by R = Rcrit. Note that R = || Zo0 || corresponds to operation at matched load with maximum output power. The boundary is expressed in terms of this matched load impedance, and the ratio Zi∞ / Zi0.
Fundamentals of Power Electronics
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Rcrit = Z o0
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Chapter 19: Resonant Conversion
Chapter 19: Resonant Conversion
The sinusoidal approximation allows a great deal of insight to be gained into the operation of resonant inverters and dc–dc converters. The voltage conversion ratio of dc–dc resonant converters can be directly related to the tank network transfer function. Other important converter properties, such as the output characteristics, dependence (or lack thereof) of transistor current on load current, and zero-voltageand zero-current-switching transitions, can also be understood using this approximation. The approximation is accurate provided that the effective Q–factor is sufficiently large, and provided that the switching frequency is sufficiently close to resonance. Simple equivalent circuits are derived, which represent the fundamental components of the tank network waveforms, and the dc components of the dc terminal waveforms.
Fundamentals of Power Electronics
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Fundamentals of Power Electronics
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Chapter 19: Resonant Conversion
Fundamentals of Power Electronics
87
Chapter 19: Resonant Conversion
6. The input impedance magnitude || Zi ||, and hence also the transistor current magnitude, are monotonic functions of the load resistance R. The dependence of the transistor conduction loss on the load current can be easily understood by simply plotting || Zi || in the limiting cases as R 6 ∞ and as R 6 0, or || Zi∞ || and || Zi0 ||. 7. The ZVS/ZCS boundary is also a simple function of Zi∞ and Zi0. If ZVS occurs at open-circuit and at short-circuit, then ZVS occurs for all loads. If ZVS occurs at short-circuit, and ZCS occurs at open-circuit, then ZVS is obtained at matched load provided that || Zi∞ || > || Zi0 ||. 8. The output characteristics of all resonant inverters considered here are elliptical, and are described completely by the open-circuit transfer function magnitude || H∞ ||, and the output impedance || Zo0 ||. These quantities can be chosen to match the output characteristics to the application requirements.
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Exact solutions of the ideal dc–dc series and parallel resonant converters are listed here as well. These solutions correctly predict the conversion ratios, for operation not only in the fundamental continuous conduction mode, but in discontinuous and subharmonic modes as well. Zero-voltage switching mitigates the switching loss caused by diode recovered charge and semiconductor device output capacitances. When the objective is to minimize switching loss and EMI, it is preferable to operate each MOSFET and diode with zero-voltage switching. Zero-current switching leads to natural commutation of SCRs, and can also mitigate the switching loss due to current tailing in IGBTs.
Fundamentals of Power Electronics
5.
4.
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Waveforms of the half-wave ZCS quasi-resonant switch cell The average terminal waveforms The full-wave ZCS quasi-resonant switch cell
1
Chapter 20: Quasi-Resonant Converters
–
v1(t)
+ Switch cell
Fundamentals of Power Electronics
vg(t) + –
i1(t)
–
v2(t)
+
i2(t)
L
2
C
i(t)
R
–
–
–
Chapter 20: Quasi-Resonant Converters
v2(t)
+
i2(t)
v1(t)
+
PWM switch cell
v(t)
+
i1(t)
Example: realization of the switch cell in the buck converter
A quite general idea: 1. PWM switch network is replaced by a resonant switch network 2. This leads to a quasi-resonant version of the original PWM converter
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Fundamentals of Power Electronics
Summary of key points
20.4
The zero-voltage-switching quasi-resonant switch The zero-voltage-switching multiresonant switch Quasi-square-wave resonant switches
Ac modeling of quasi-resonant converters
20.2.1 20.2.2 20.2.3
Resonant switch topologies
20.1.1 20.1.2 20.1.3
The zero-current-switching quasi-resonant switch cell
20.3
20.2
20.1
Introduction
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Cr
+ –
–
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–
–
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Lr
v1(t)
i1(t) + Q1
D1
D2
Cr
i2r(t)
2π L rC r
=
2π 3
–
v1(t)
i1(t) Switch cell
L
Fundamentals of Power Electronics
4
Performances of half-wave and full-wave cells differ significantly.
Full-wave: Q1 and D1 in parallel, transistor turns off at second zero crossing of current waveform
Half-wave: Q1 and D1 in series, transistor turns off at first zero crossing of current waveform
Two-quadrant switch is required:
Tank capacitor Cr in parallel with diode D2 : diode switches at zero crossings of capacitor voltage waveform
Tank inductor Lr in series with transistor: transistor switches at zero crossings of inductor current waveform
R
–
v(t)
+
–
v1r(t)
+
Q1
D2
Switch network
D1
Cr
i2r(t)
D2
Switch network
Q1
D1
Cr
i2r(t)
–
v2(t)
i2(t) +
–
v2(t)
i2(t) +
Chapter 20: Quasi-Resonant Converters
Full-wave ZCS quasi-resonant switch cell
–
–
+ v1r(t)
Lr
Half-wave ZCS quasi-resonant switch cell
Lr
v1(t)
i1(t) +
–
v1(t)
i1(t) +
C
i(t)
–
v2(t)
i2(t) +
Chapter 20: Quasi-Resonant Converters
–
v2(t)
+
i2(t)
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Fundamentals of Power Electronics
f0 =
Insert either of the above switch cells into the buck converter, to obtain a ZCS quasi-resonant version of the buck converter. Lr vg(t) and Cr are small in value, and their resonant frequency f0 is greater than the switching frequency fs. ω0 1
Full-wave ZCS quasi-resonant switch cell
D2
i2r(t)
Half-wave ZCS quasi-resonant switch cell
Q1
Switch network
D1
Switch network
–
–
+ v1r(t)
Lr
v1(t)
i1(t) +
7ZRTXDVLUHVRQDQWVZLWFKFHOOV
Ts
Ts
5
Fundamentals of Power Electronics
Generalized switch averaging, and µ, are defined and discussed in Section 10.3.
If V/Vg = M(d) for a PWM CCM converter, then V/Vg = M(µ) for the same converter with a switch network having conversion ratio µ.
The switch conversion ratio µ is the ratio of the average terminal voltages of the switch network. It can be applied to non-PWM switch networks. For the CCM PWM case, µ = d.
A generalization of the duty cycle d(t) s
6
Q1
D2
Switch network
D1
–
v2(t) Cr
v1(t) ≈ v1(t)
i 2(t) ≈ I 2 v1(t) ≈ V1
µ=
µ=
Ts
Ts
V2 I 1 = V1 I 2
v1r(t)
v2(t)
=
i 2r(t)
i 1(t)
Ts
Ts
〈 i2(t)〉T
Chapter 20: Quasi-Resonant Converters
Ts
Ts
Half-wave ZCS quasi-resonant switch cell
–
v1r(t)
i 2(t) ≈ i 2(t)
+ –
+
i1(t)
In steady state:
〈 v1(t)〉T
Lr
i2r(t) +
Chapter 20: Quasi-Resonant Converters
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Fundamentals of Power Electronics
〈 v2(t) 〉Ts and 〈 i1(t) 〉Ts
Modeling objective: find the average values of the terminal waveforms
i 2(t) ≈ I 2 v1(t) ≈ V1
In steady state, we can further approximate these quantities by their dc values:
v1(t) ≈ v1(t)
i 2(t) ≈ i 2(t)
It is assumed that the converter filter elements are large, such that their switching ripples are small. Hence, we can make the small ripple approximation as usual, for these elements:
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s
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v2(t) Cr
i2r(t) +
Half-wave ZCS quasi-resonant switch cell
–
v1r(t)
Q1
+ –
Fundamentals of Power Electronics
V1
i1(t)
–
v2(t)
+ I2
8
Diode D2 is initially conducting the filter inductor current I2. Transistor Q1 turns on, and the tank inductor current i1 starts to increase. So all semiconductor devices conduct during this subinterval, and the circuit reduces to: Lr
s
〈 i2(t)〉T
I2
i1(t)
β
Q1 D1
α
Q1 D1 D2
2
δ
–
X
ω0Ts
Vc1
3
I2 Cr ξ
D2
4
θ = ω0t
R0 = where
Lr Cr
V1 V t = ω 0t 1 Lr R0
i 1(α) = α
α=
I 2R0 V1 Chapter 20: Quasi-Resonant Converters
V1 = I2 R0
This subinterval ends when diode D2 becomes reverse-biased. This occurs at time ω0t = α, when i1(t) = I2.
i 1(t) =
Solution:
Circuit equations: di 1(t) V1 with i1(0) = 0 = Lr dt
Chapter 20: Quasi-Resonant Converters
Conducting devices:
v2(t)
1
V1 Lr
Waveforms:
Subinterval:
6XELQWHUYDO
7
Each switching period contains four subintervals
+ –
i1(t) +
D1
Fundamentals of Power Electronics
〈 v1(t)〉T
Lr
The half-wave ZCS quasi-resonant switch cell, driven by the terminal quantities 〈 v1(t)〉 Ts and 〈 i2(t)〉 Ts.
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+ –
i1(t)
–
v2(t)
+
V1 R0
I2
V1 sin β = 0 R0
Fundamentals of Power Electronics
10
Must use care to select the correct branch of the arcsine function. Note (from the i1(t) waveform) that β > π.
I R sin β = – 2 0 V1
i 1(α + β) = I 2 +
I2
i1(t)
2 β ω0Ts
4
I 2 R0 V1
3 δ
ξ
V1 R0
θ = ω0t
Chapter 20: Quasi-Resonant Converters
I2 <
– π < sin – 1 x ≤ π 2 2
β = π + sin – 1
Hence
1 α
V1 Lr
Chapter 20: Quasi-Resonant Converters
The dc components of these waveforms are the dc solution of the circuit, while the sinusoidal components have magnitudes that depend on the initial conditions and on the characteristic impedance R0.
Subinterval:
v2(ω0t) = V1 1 – cos ω0t – α
V1 sin ω0t – α R0
FRQWLQXHG
V1 sin ω0t – α R0
v2(ω0t) = V1 1 – cos ω0t – α
i 1(ω0t) = I 2 +
The solution is
6XELQWHUYDO
9
v2(α) = 0 i 1(α) = I 2
i 1(ω0t) = I 2 +
Cr
ic(t)
This subinterval ends at the first zero crossing of i1(t). Define β = angular length of subinterval 2. Then
I 1pk = I 2 +
Peak inductor current:
Fundamentals of Power Electronics
Lr
di 1(ω0t) = V1 – v2(ω0t) dt dv (ω t) C r 2 0 = i 1(ω0t) – I 2 dt
The circuit equations are
V1
Lr
Diode D2 is off. Transistor Q1 conducts, and the tank inductor and tank capacitor ring sinusoidally. The circuit reduces to:
6XELQWHUYDO
V1 R0
Cr
I2
Fundamentals of Power Electronics
12
The solution is v2(ω0t) = Vc1 – I 2 R0 ω0t – α – β
Cr
dv2(ω0t) = – I2 dt v2(α + β) = Vc1
The circuit equations are
–
v2(t)
+
1–
2
IR 1– 2 0 V1
2
Chapter 20: Quasi-Resonant Converters
V V δ = c1 = 1 1 – I 2R0 I 2R0
v2(α + β + δ) = Vc1 – I 2R0δ = 0
Subinterval 3 ends when the tank capacitor voltage reaches zero, and diode D2 becomes forward-biased. Define δ = angular length of subinterval 3. Then
Chapter 20: Quasi-Resonant Converters
I 2 R0 V1
6XELQWHUYDO
11
All semiconductor devices are off. The circuit reduces to:
Fundamentals of Power Electronics
v2(α + β) = Vc1 = V1 1 +
Capacitor voltage at the end of subinterval 2 is
The resonant switch operates with zero current switching only for load currents less than the above value. The characteristic impedance must be sufficiently small, so that the ringing component of the current is greater than the dc load current.
is violated, then the inductor current never reaches zero. In consequence, the transistor cannot switch off at zero current.
I2 <
If the requirement
%RXQGDU\RI]HURFXUUHQWVZLWFKLQJ
13
Chapter 20: Quasi-Resonant Converters
fs f0
Fundamentals of Power Electronics
14
1–
I 2 R0 V1
2
Chapter 20: Quasi-Resonant Converters
2π ≥ I 2 R0 + π + sin – 1 I 2 R0 + V1 1 – V1 V1 F I 2 R0
Substitute previous solutions for subinterval lengths:
ω0Ts ≥ α + β + δ
So the minimum switching period is
F=
where the normalized switching frequency F is defined as
The angular length of the switching period is 2π f0 2π ω0Ts = α + β + δ + ξ = = F fs
The length of the fourth subinterval cannot be negative, and the switching period must be at least long enough for the tank current and voltage to return to zero by the end of the switching period.
0D[LPXPVZLWFKLQJIUHTXHQF\
Fundamentals of Power Electronics
The length of subinterval 4 can be used as a control variable. Increasing the length of this interval reduces the average output voltage.
Transistor Q1 is off, and the input current i1(t) is equal to zero.
The tank capacitor voltage v2(t) is equal to zero.
Diode D2 conducts the filter inductor current I2
Subinterval 4, of angular length ξ, is identical to the diode conduction interval of the conventional PWM switch network.
6XELQWHUYDO
= 1 Ts Ts t
t + Ts
i 1(t)dt =
q1 + q2 Ts
0
α ω0
α i 1(t)dt = 12 ω 0 I2
I2
i1(t)
+ – –
v1r(t)
+
i1(t)
Q1
D2
Switch network
D1
–
v2(t) Cr
i2r(t) +
α ω0
q1
〈 i1(t)〉T
s
α ω0
i 1(t)dt
α ω0
i C(t)dt +
α ω0
α+β ω0
I 2 dt
β I 2 dt = I 2 ω 0
Fundamentals of Power Electronics
α ω0
α+β ω0
16
I2
i1(t)
Second term is integral of constant I2:
q2 =
α+β ω0
Substitute:
i 1(t) = i C(t) + I 2
Node equation for subinterval 2:
q2 =
α+β ω0
V1
+ –
α ω0
q1
t
i1(t)
–
v2(t)
+
Cr
ic(t)
〈 i1(t)〉Ts t
Chapter 20: Quasi-Resonant Converters
Circuit during subinterval 2
Lr
α+β ω0
q2
I2
s
〈 i2(t)〉T
Chapter 20: Quasi-Resonant Converters
α+β ω0
q2
Half-wave ZCS quasi-resonant switch cell
Lr
&KDUJHDUJXPHQWVFRPSXWDWLRQRIq2
Fundamentals of Power Electronics
q1 =
s
15
〈 v1(t)〉T
q1 and q2 are the areas under the current waveform during subintervals 1 and 2. q1 is given by the triangle area formula:
i 1(t)
Averaged switch modeling: we need to determine the average values of i1(t) and v2(t). The average switch input current is given by
7KHDYHUDJHWHUPLQDOZDYHIRUPV
α ω0
i C(t)dt + α ω0
α+β ω0
I 2 dt
i 1(t) µ= I2
Ts
=
Fundamentals of Power Electronics
I 2 R0 V1
18
δ
I2 Cr ξ
Ts
=
αI 2 CVc1 βI 2 + + Ts ω 0T s 2ω0Ts
Chapter 20: Quasi-Resonant Converters
1 – J 2s
Chapter 20: Quasi-Resonant Converters
i 1(t)
Substitute into expression for average switch input current:
µ = F 1 12 J s + π + sin – 1(J s) + 1 1 + Js 2π
Js =
β
–
β q 2 = CVc1 + I 2 ω 0
α + CVc1 + β 2ω0Ts I 2Ts ω0Ts Eliminate α, β, Vc1 using previous results:
where
α
Vc1
Substitute results for the two integrals:
v2(t)
6ZLWFKFRQYHUVLRQUDWLRµ
17
i C(t)dt = C Vc1 – 0 = CVc1
α+β α i C(t)dt = C v2 ω – v2 ω 0 0
Fundamentals of Power Electronics
α ω0
α+β ω0
α ω0
α+β ω0
First term: integral of the capacitor current over subinterval 2. This can be related to the change in capacitor voltage :
q2 =
α+β ω0
FRQWLQXHG
&KDUJHDUJXPHQWV
I 2 R0 V1
0
0.2
0.4
0.6
0.8
1
0
0.2
F = 0.1
0.4
0.3
µ
0.4
0.6
0.6
0.5
ZCS boundary
0.2
Fundamentals of Power Electronics
Js
19
0
2
4
6
8
10
0
0.4
Js
0.6
20
0.7
0.8
0.8
0.9
1
0.8
Chapter 20: Quasi-Resonant Converters
0.2
1 – J 2s
JsF 4π
Chapter 20: Quasi-Resonant Converters
µ≤1–
Js ≤ 1
Mode boundary:
µ = FP12 J s
Switch characteristics:
&KDUDFWHULVWLFVRIWKHKDOIZDYH=&6UHVRQDQWVZLWFK
Fundamentals of Power Electronics
1 – J 2s
2
P1 J s
µ = F 1 12 J s + π + sin – 1(J s) + 1 1 + Js 2π
P12 J s = 1 1 J + π + sin – 1(J s) + 1 1 + Js 2π 2 s
µ = FP12 J s
This is of the form
with J s =
Switch conversion ratio:
$QDO\VLVUHVXOWVZLWFKFRQYHUVLRQUDWLRµ
ry max F bounda
1
FIR0 4π
21
0
0.2
0.4
0
Vg
+ –
Ig
L
Fundamentals of Power Electronics
22
P12 J s = 1 12 J s + π + sin – 1(J s) + 1 1 + Js 2π
µ = FP12 J s
Half-wave ZCS equations:
I 1–µ
I g R0 I R Js = 2 0 = V1 V
M= V = 1 Vg 1 – µ
For the boost converter,
Ig =
Js
0.6
0.8
1
0.2
F = 0.1
0.4
µ
0.8
1
Q1
D1
i1(t)
–
D2
+
+
–
v1(t)
Cr
v2(t)
C
R
–
V
+
Chapter 20: Quasi-Resonant Converters
1 – J 2s
Lr
i2(t)
0.6
0.6
0.7
0.8
0.9
Chapter 20: Quasi-Resonant Converters
0.2
0.3
0.4
0.5
ZCS boundary
%RRVWFRQYHUWHUH[DPSOH
Fundamentals of Power Electronics
0 ≤ V ≤ Vg –
Output voltage varies over the range
Vg I≤ R0
ZCS occurs when
IR Js = 0 Vg
For the buck converter,
M = V =µ Vg
Conversion ratio of the buck converter is (from inductor volt-second balance):
%XFNFRQYHUWHUFRQWDLQLQJKDOIZDYH=&6TXDVLUHVRQDQWVZLWFK
ry max F bounda
D2
Cr
i2r(t)
D2
Switch network
Q1
D1
Cr
i2r(t)
I2
i1(t)
1
1
V1 Lr
V1 Lr
2
2
3
3
4
4
(full wave)
1 – J 2s V1 1 –
i 1(t) I2 Fundamentals of Power Electronics
µ=
Ts
=
24
α + CVc1 + β 2ω0Ts I 2Ts ω0Ts Chapter 20: Quasi-Resonant Converters
(half wave)
1 – J 2s
(full wave)
2π – sin – 1 J s V1 1 +
(half wave)
In either case, µ is given by
Vc1 =
β=
Analysis in the full-wave case is nearly the same as in the half-wave case. The second subinterval ends at the second zero crossing of the tank inductor current waveform. The following quantities differ:
π + sin – 1 J s
θ = ω0t
θ = ω0t
Chapter 20: Quasi-Resonant Converters
I2
i1(t)
– Subinterval:
v2(t)
+
i2(t)
– Subinterval:
v2(t)
i2(t) +
$QDO\VLVIXOOZDYH=&6
23
Full-wave ZCS quasi-resonant switch cell
–
–
+ v1r(t)
Lr
v1(t)
i1(t) +
Q1
Switch network
D1
Half-wave ZCS quasi-resonant switch cell
–
–
+ v1r(t)
Lr
v1(t)
i1(t) +
Fundamentals of Power Electronics
Full wave
Half wave
7KHIXOOZDYH=&6TXDVLUHVRQDQWVZLWFKFHOO
f µ≈F= s f0
25
Js
0
0.2
0.4
0.6
0.8
1
0
0.2
F = 0.2
µ
0.6
0.6
D2
Switch network
SW Cr
i2r(t)
ZCS quasi-resonant switch cell
–
–
+ v1r(t)
Lr
v1(t)
i1(t) +
–
v2(t)
i2(t) +
Fundamentals of Power Electronics
26
1
Chapter 20: Quasi-Resonant Converters
Can be connected in other ways that preserve high-frequency components of tank waveforms
Connection of resonant elements:
• Current-bidirectional two-quadrant switch for full-wave cell
• Voltage-bidirectional two-quadrant switch for half-wave cell
SPST switch SW:
Basic ZCS switch cell:
0.8
0.8
Chapter 20: Quasi-Resonant Converters
0.4
0.4
ZCS boundary
1 – J 2s
5HVRQDQWVZLWFKWRSRORJLHV
Fundamentals of Power Electronics
so
P1 J s ≈ 1
Full-wave case: P1 can be approximated as
µ = FP1 J s
P1 J s = 1 12 J s + 2π – sin – 1(J s) + 1 1 – Js 2π
)XOOZDYHFHOOVZLWFKFRQYHUVLRQUDWLRµ
ry max F bounda
Vg
Vg
i1(t)
+ –
+ –
i1(t)
SW
D2
SW
D2
27
ZCS quasi-resonant switch
Lr
Cr
ZCS quasi-resonant switch
Lr
L
L
C
C
R
R
Fundamentals of Power Electronics
If the converter contains a ZCS quasi-resonant switch, then the result of these operations is
28
The resonant switch network remains.
• Filter inductors: open circuits
i1(t)
Lr
D2
–
Cr v2(t)
+
Chapter 20: Quasi-Resonant Converters
SW
• Independent voltage source Vg: short circuit
Replace converter elements by their high-frequency equivalents:
• Filter capacitors: short circuits
–
V
+
–
V
+
Chapter 20: Quasi-Resonant Converters
–
v2(t)
i2(t) +
–
v2(t)
+
i2(t)
$WHVWWRGHWHUPLQHWKHWRSRORJ\ RIDUHVRQDQWVZLWFKQHWZRUN
Fundamentals of Power Electronics
The ac highfrequency components of the tank waveforms are unchanged.
This simply changes the dc component of tank capacitor voltage.
Connection of tank capacitor to two other points at ac ground.
Cr
&RQQHFWLRQRIWDQNFDSDFLWRU
i1(t)
Lr
SW
D2
+
29
Fundamentals of Power Electronics
Vg
+ –
i1(t)
–
v1(t)
+
vCr(t)
Q1
D1
Cr
30
+
A full-wave version based on the PWM buck converter: –
When the previously-described operations are followed, then the converter reduces to
Lr
–
v2(t)
L
C
I
D2
R
–
V
+
Chapter 20: Quasi-Resonant Converters
D2
iLr(t) i2(t) +
SW
Cr
Lr
Chapter 20: Quasi-Resonant Converters
7KH]HURYROWDJHVZLWFKLQJ TXDVLUHVRQDQWVZLWFKFHOO
Fundamentals of Power Electronics
• Peak transistor current is (1 + Js) Vg/R0, or more than twice the PWM value.
• Zero current switching of SW: device Q1 and D1 output capacitances lead to switching loss. In full-wave case, stored charge of diode D1 leads to switching loss.
–
Cr v2(t)
• Zero voltage switching of D2 eliminates switching loss arising from D2 stored charge.
Discussion
• Tank capacitor is in parallel with diode D2; hence D2 switches at zero voltage
• Tank inductor is in series with switch; hence SW switches at zero current
ZCS quasi-resonant switch:
=HURFXUUHQWDQG]HURYROWDJHVZLWFKLQJ
full-wave
µ = 1 – FP1 1 Js
31
X
α
1
D2
β
2
δ
D1 Q1 D2
ω0Ts
3
+ –
–
v1(t)
Fundamentals of Power Electronics
Vg
i1(t) +
Q1
D1
Cs
32
D2
Lr
A half-wave version based on the PWM buck converter:
Cd
When the previously-described operations are followed, then the converter reduces to
–
θ = ω0t
L
C
SW
Cs
I
R
–
V
+
D2
Lr Cd
Chapter 20: Quasi-Resonant Converters
v2(t)
i2(t) +
Q1
ξ
I2
4
Chapter 20: Quasi-Resonant Converters
Conducting devices:
iLr(t)
Subinterval:
V1
vCr(t)
Tank waveforms
7KH=96PXOWLUHVRQDQWVZLWFK
Fundamentals of Power Electronics
A problem with the quasi-resonant ZVS switch cell: peak transistor voltage becomes very large when zero voltage switching is required for a large range of load currents.
peak transistor voltage Vcr,pk = (1 + J s) V1
Js ≥ 1
ZVS boundary
half-wave
µ = 1 – FP12 1 Js
Switch conversion ratio
=96TXDVLUHVRQDQWVZLWFKFHOO
33
ZVS
ZCS
SW
SW
Cr
Cr D2
D2
Chapter 20: Quasi-Resonant Converters
Lr
Lr
+ –
Cf
Cr
Lr
D1
Q1
D2
L
C
R
I
–
V
+
Fundamentals of Power Electronics
34
Chapter 20: Quasi-Resonant Converters
• Zero-current switching in all semiconductor devices
• Peak transistor voltage is increased
• Peak transistor current is equal to peak transistor current of PWM cell
• The basic ZCS QSW switch cell is restricted to 0 ≤ µ ≤ 0.5
Vg
Lf
$TXDVLVTXDUHZDYH=&6EXFNZLWKLQSXWILOWHU
Fundamentals of Power Electronics
When the previouslydescribed operations are followed, then the converter reduces to
4XDVLVTXDUHZDYHUHVRQDQWVZLWFKHV
+ –
Q1 D2
Cr
i2(t)
–
v2(t)
+ L
Lr
C
R
I
–
V
+
Conducting devices:
v2(t)
i2(t)
D1 Q1
V1
V1 Lr
X
ω0Ts D2
0
–
V2 Lr
+ –
v1r(t) –
–
+ v1(t)
Lr
Fundamentals of Power Electronics
vg(t)
i1(t) +
Gate driver
Q1
D1
36
D2
Frequency modulator
Cr
i2r(t)
L
C
i(t)
R
µ=F
–
v(t)
+
Chapter 20: Quasi-Resonant Converters
vc(t)
–
v2(t)
+
i2(t)
Buck example with full-wave ZCS quasi-resonant cell:
Use averaged switch modeling technique: apply averaged PWM model, with d replaced by µ
Full-wave ZCS quasi-resonant switch cell
ω0t
Chapter 20: Quasi-Resonant Converters
X
$FPRGHOLQJRITXDVLUHVRQDQWFRQYHUWHUV
35
• Zero-voltage switching in all semiconductor devices
• Peak transistor current is increased
• Peak transistor voltage is equal to peak transistor voltage of PWM cell
• The basic ZVS QSW switch cell is restricted to 0.5 ≤ µ ≤ 1
–
v1(t)
+
Fundamentals of Power Electronics
Vg
i1(t)
D1
$TXDVLVTXDUHZDYH=96EXFN
= µ i 2r(t)
= µ v1r(t)
+ –
µ=F
Lr +
–
v1r
i1
I2 f0
37
fs
1:F
Gm(s)
V fs 1 f0
vc
–
Cr v2
+
L
C
R
Fs V v (t) + fs(t) 1 f0 1r f0
I2 f0
fs
I2 f0
fs
Gm(s)
V fs 1 f0
–
Fundamentals of Power Electronics
38
vc
v2
+
—same as PWM buck, with d replaced by F
+ –
1:F
+ –
vg
i1
–
v
+
Chapter 20: Quasi-Resonant Converters
i 2r
v2(t) =
i 1(t) = fs(t)
/RZIUHTXHQF\PRGHO
fs
Ts
Ts
Linearize:
C
R
–
v
+
Chapter 20: Quasi-Resonant Converters
L
Tank dynamics occur only at frequency near or greater than switching frequency —discard tank elements
Fundamentals of Power Electronics
vg
Ts
Ts
Resulting ac model:
i 1(t)
v2(t)
Averaged switch equations:
6PDOOVLJQDODFPRGHO
+ –
+ –
fs(t) P1 js(t) f0 2
–
–
Q1
Gate driver
D1
39
D2
Frequency modulator
Cr
i2r(t)
∂µ R0 I 2 ∂ js V 21 Ki = –
∂µ R0 ∂ js V1 µ Kc = 0 Fs
Kv = –
Fundamentals of Power Electronics
40
i 1(t) = µ(t) I 2 + i 2r(t) µ 0 v2(t) = µ 0 v1r(t) + µ(t) V1
C
i(t)
R
v1r(t)
i 2r(t)
–
v(t)
+
Ts
Ts
∂µ Fs 1 1 + 1 – J 2s = – ∂ js 2π f0 2 J 2s
Chapter 20: Quasi-Resonant Converters
Linearized terminal equations of switch network:
with
µ(t) = K vv1r(t) + K i i 2r(t) + K c fs(t)
Perturbation and linearization of µ(v1r, i2r, fs):
L
js(t) = R0
Chapter 20: Quasi-Resonant Converters
vc(t)
–
v2(t)
+
i2(t)
6PDOOVLJQDOPRGHOLQJ
v1r(t)
+
v1(t)
Lr
Fundamentals of Power Electronics
vg(t)
i1(t) +
µ(t) =
Half-wave ZCS quasi-resonant switch cell
Now, µ depends on js:
([DPSOH+DOIZDYH=&6TXDVLUHVRQDQWEXFN
+ –
+
–
v1r
i1
v1r
Kc
41
++
µ
fs
Ki
µ V1
vc
–
Cr v2
+
L
C
R
–
v
+
Chapter 20: Quasi-Resonant Converters
Gm(s)
i 2r
i 2r
vg
µI
Kv
Kc
++
+
Fundamentals of Power Electronics
vg
42
µ
1 : µ0
fs
Ki
µ Vg
+ –
+ –
i1
Gm(s)
i
i
vc
C
R
–
v
+
Chapter 20: Quasi-Resonant Converters
L
/RZIUHTXHQF\PRGHOVHWWDQNHOHPHQWVWR]HUR
Kv
µ I2
1 : µ0
+
Fundamentals of Power Electronics
vg
Lr
(TXLYDOHQWFLUFXLWPRGHO
+ –
2
2
1 s 1 1 + ω + ωs Q 0 0 1 s 1 1 + ω + ωs Q 0 0
43
µ 0 + K vV g K iVg 1+ R K cV g K iVg 1+ R K iVg 1+ R L rC r K iVg 1+ R R0 + K iVg R R0 R Lr Cr Chapter 20: Quasi-Resonant Converters
R0 =
Q=
ω0 =
Gc0 =
Gg0 =
Fundamentals of Power Electronics
44
Chapter 20: Quasi-Resonant Converters
3. In the zero-current-switching quasi-resonant switch, diode D2 operates with zero-voltage switching, while transistor Q1 and diode D1 operate with zero-current switching.
2. Analysis of a resonant switch cell involves determination of the switch conversion ratio µ. The resonant switch waveforms are determined, and are then averaged. The switch conversion ratio µ is a generalization of the PWM CCM duty cycle d. The results of the averaged analysis of PWM converters operating in CCM can be directly adapted to the related resonant switch converter, simply by replacing d with µ.
1. In a resonant switch converter, the switch network of a PWM converter is replaced by a switch network containing resonant elements. The resulting hybrid converter combines the properties of the resonant switch network and the parent PWM converter.
6XPPDU\RINH\SRLQWV
Fundamentals of Power Electronics
Half-wave: effective feedback reduces Q-factor and dc gains
Full-wave: poles and zeroes are same as PWM
Gvc(s) = Gc0
Gvg(s) = Gg0
+DOIZDYH=&6EXFN
3UHGLFWHGVPDOOVLJQDOWUDQVIHUIXQFWLRQV
45
Chapter 20: Quasi-Resonant Converters
Fundamentals of Power Electronics
46
Chapter 20: Quasi-Resonant Converters
11. In the case of half-wave quasi-resonant switches, as well as other types of resonant switches, the conversion ratio µ is a strong function of the switch terminal quantities v1 and i2. This leads to effective feedback, which modifies the poles, zeroes, and gains of the transfer functions.
10. In the case of full-wave quasi-resonant switches, µ depends only on the switching frequency, and therefore the transfer function poles and zeroes are identical to those of the parent PWM converter.
9. The small-signal ac models of converters containing resonant switches are similar to the small-signal models of their parent PWM converters. The averaged switch modeling approach can be employed to show that the quantity d(t) is simply replaced by µ(t).
8. In the quasi-square-wave zero-voltage-switching resonant switches, all semiconductor devices operate with zero-voltage switching, and with peak voltages equal to those of the parent PWM converter. The switch conversion ratio is restricted to the range 0.5 ≤ µ ≤ 1.
6XPPDU\RINH\SRLQWV
Fundamentals of Power Electronics
7. In the zero-voltage-switching multiresonant switch, all semiconductor devices operate with zero-voltage switching. In consequence, very low switching loss is observed.
6. Half-wave versions of the quasi-resonant switch exhibit conversion ratios that are strongly dependent on the load current. These converters typically operate with wide variations of switching frequency.
5. Full-wave versions of the quasi-resonant switches exhibit very simple control characteristics: the conversion ratio µ is essentially independent of load current. However, these converters exhibit reduced efficiency at light load, due to the large circulating currents. In addition, significant switching loss is incurred due to the recovered charge of diode D1.
4. In the zero-voltage-switching quasi-resonant switch, the transistor Q1 and diode D1 operate with zero-voltage switching, while diode D2 operates with zero-current switching.
6XPPDU\RINH\SRLQWV
Fundamentals of Power Electronics
Appendix 2
Appendix 2. Magnetics Design Tables
A 2 . 1 Pot core data A
H
Magnetics Design Tables
Geometrical data for several standard ferrite core shapes are listed here. The geometrical constant K g is a measure of core size, useful for designing inductors and transformers which attain a given copper loss [1]. The K g method for inductor design is described in Chapter 13. Kg is defined as
Kg =
A 2c WA MLT
(A2.1)
where Ac is the core cross-sectional area, WA is the window area, and MLT is the winding mean-length-per-turn. The geometrical constant K gfe is a similar measure of core size, which is useful for designing ac inductors and transformers when the total copper plus core loss is constrained. The Kgfe method for magnetics design is described in Chapter 14. K gfe
Core type
Geometrical constant
Geometrical constant
(AH) (mm)
Kg 5 cm
K gfe x cm -6
1.61·10
-6
Crosssectional area Ac 2 (cm )
Bobbin winding area WA 2 (cm )
– 1 / β) WA A 2(1 c u(β) MLT l 2e / β
0.070
0.22·10
1.46
1.0
0.5
0.101
0.034
1.90
1.26
1.0
1107
0.667·10-3
554·10-6
0.167
0.055
2.30
1.55
1408
2.107·10
-3
1.1·10-3
0.251
0.097
2.90
2.00
100
3.2
1811
9.45·10-3
2.6·10-3
0.433
0.187
3.71
2.60
60
7.3
2213
27.1·10
-3
-3
0.635
0.297
4.42
3.15
38
13
2616
69.1·10-3
8.2·10-3
0.948
0.406
5.28
3.75
30
20
3019
0.180
14.2·10-3
1.38
0.587
6.20
4.50
23
34
3622
0.411
21.7·10-3
2.02
0.748
7.42
5.30
19
57
4229
1.15
41.1·10-3
2.66
1.40
8.60
6.81
13.5
104
4.9·10
(A2.2)
u(β) =
– 2 β+2
(g)
256·10-6
defined as
β + 2
R th (˚C/W)
0.738·10
(A2.3) For modern ferrite materials, β typically lies in the range 2.6 to 2.8. The quantity u(β) is
β β+2
Core weight
0.183·10-3
Pfe = K fe B βmax
–
Thermal resistance
704
where le is the core mean magnetic path length, and β is the core loss exponent:
β 2
Magnetic path length lm (cm)
905
is defined as
K gfe =
-3
Mean length per turn MLT (cm)
β+2 β
(A2.4)
u(β) is equal to 0.305 for β = 2.7. This quantity varies by roughly 5% over the range 2.6 ≤ β ≤ 2.8. Values of Kgfe are tabulated for β = 2.7; variation of Kgfe over the range 2.6 ≤ β ≤ 2.8 is typically quite small.
2
1.8
Appendix 2. Magnetics Design Tables
Appendix 2. Magnetics Design Tables
A 2 . 4 ETD core data A 2 . 2 EE core data A A
(A) (mm)
Kg cm5
K gfe cmx
Crosssectional area Ac (cm2)
Bobbin winding area WA (cm2)
Mean length per turn MLT (cm)
Magnetic path length lm (cm)
Thermal resistance
Core weight
Mean length per turn MLT (cm)
Magnetic path length lm (cm)
Core weight
(g)
ETD29
0.0978
8.5·10-3
0.76
0.903
5.33
7.20
0.458·10-3
0.14
0.085
2.28
2.7
2.34
ETD34
0.193
13.1·10-3
0.97
1.23
6.00
7.86
19
40
-3
0.19
0.190
3.40
3.45
3.29
ETD39
0.397
19.8·10-3
1.25
1.74
6.86
9.21
15
60
Geometrical constant
(A) (mm)
Kg 5 cm
K gfe x cm
-3
Geometrical constant
Bobbin winding area WA 2 (cm )
Geometrical constant
0.731·10-3
Geometrical constant
Crosssectional area Ac 2 (cm )
Core type
EE12
Core type
0.842·10
R th (˚C/W)
(g) 30
EE16
2.02·10
EE19
4.07·10-3
1.3·10-3
0.23
0.284
3.69
3.94
4.83
ETD44
0.846
30.4·10-3
1.74
2.13
7.62
10.3
12
94
EE22
8.26·10-3
1.8·10-3
0.41
0.196
3.99
3.96
8.81
ETD49
1.42
41.0·10-3
2.11
2.71
8.51
11.4
11
124
EE30
85.7·10-3
6.7·10-3
1.09
0.476
6.60
5.77
32.4
EE40
0.209
11.8·10-3
1.27
1.10
8.50
7.70
50.3
EE50
0.909
28.4·10
-3
2.26
1.78
10.0
9.58
116
EE60
1.38
36.4·10-3
2.47
2.89
12.8
11.0
135
EE70/68/19
5.06
127·10-3
3.24
6.75
14.0
9.0
280
A 2 . 5 PQ core data A1
A 2 . 3 EC core data
2D
A
Core type (A) (mm)
Geometrical constant Kg 5 cm
Geometrical constant K gfe x cm
Crosssectional area Ac 2 (cm )
Bobbin winding area WA 2 (cm )
Mean length per turn MLT (cm)
Magnetic path length lm (cm)
Thermal resistance R th (˚C/W)
Core weight
(g)
Core type
Geometrical constant
Geometrical constant
(A1/2D) (mm)
Kg 5 cm
K gfe x cm
PQ 20/16
22.4·10-3
PQ 20/20 PQ 26/20
83.9·10
PQ 26/25
0.125
Crosssectional area Ac 2 (cm )
Bobbin winding area WA 2 (cm )
Mean length per turn MLT (cm)
Magnetic path length lm (cm)
Core weight
3.7·10-3
0.62
0.256
4.4
3.74
13
33.6·10-3
4.8·10-3
0.62
0.384
4.4
4.54
15
-3
7.2·10-3
1.19
0.333
5.62
4.63
31
9.4·10-3
1.18
0.503
5.62
5.55
36
-3
42
(g)
EC35
0.131
9.9·10-3
0.843
0.975
5.30
7.74
18.5
35.5
PQ 32/20
0.203
11.7·10
1.70
0.471
6.71
5.55
EC41
0.374
19.5·10-3
1.21
1.35
5.30
8.93
16.5
57.0
PQ 32/30
0.384
18.6·10-3
1.61
0.995
6.71
7.46
55
EC52
0.914
31.7·10-3
1.80
2.12
7.50
10.5
11.0
111
PQ 35/35
0.820
30.4·10-3
1.96
1.61
7.52
8.79
73
EC70
2.84
56.2·10-3
2.79
4.71
12.9
14.4
7.5
256
PQ 40/40
1.20
39.1·10-3
2.01
2.50
8.39
10.2
95
3
4
Appendix 3: Averaged switch modeling of a CCM SEPIC
Appendix 2. Magnetics Design Tables
5
1
Duty d(t) cycle
D1 Q1 –
v1(t)
+ i1(t)
+
v2(t)
Switch network
iL2(t)
L2
–
i2(t)
–
R C2 vC2(t)
+ + vC1(t) –
Fundamentals of Power Electronics
vg(t)
+ –
iL1(t)
Diameter, cm 1.168 1.040 0.927 0.825 0.735 0.654 0.583 0.519 0.462 0.411 0.366 0.326 0.291 0.267 0.238 0.213 0.190 0.171 0.153 0.137 0.122 0.109 0.0948 0.0874 0.0785 0.0701 0.0632 0.0566 0.0505 0.0452 0.0409 0.0366 0.0330 0.0294 0.0267 0.0241 0.0236 0.0191 0.0170 0.0152 0.0140 0.0124 0.0109 0.0096 0.00863 0.00762 0.00685
C1
Resistance, 10-6 Ω/cm 1.608 2.027 2.557 3.224 4.065 5.128 6.463 8.153 10.28 13.0 16.3 20.6 26.0 32.9 41.37 52.09 69.64 82.80 104.3 131.8 165.8 209.5 263.9 332.3 418.9 531.4 666.0 842.1 1062.0 1345.0 1687.6 2142.7 2664.3 3402.2 4294.6 5314.9 6748.6 8572.8 10849 13608 16801 21266 27775 35400 43405 54429 70308
L1
0000 000 00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43
Bare area, 10-3 cm2 1072.3 850.3 674.2 534.8 424.1 336.3 266.7 211.5 167.7 133.0 105.5 83.67 66.32 52.41 41.60 33.08 26.26 20.02 16.51 13.07 10.39 8.228 6.531 5.188 4.116 3.243 2.508 2.047 1.623 1.280 1.021 0.8046 0.6470 0.5067 0.4013 0.3242 0.2554 0.2011 0.1589 0.1266 0.1026 0.08107 0.06207 0.04869 0.03972 0.03166 0.02452
SEPIC example: write circuit with switch network explicitly identified
AWG#
Appendix 3: Averaged switch modeling of a CCM SEPIC
A 2 . 6 American wire gauge data
2
0
0
Ts
iL1 + iL2
0
v1(t)
dTs
dTs
0
i 1(t)
T2
vC1 + vC2
Fundamentals of Power Electronics
0
i1(t)
0
v1(t)
Ts
Ts
t
t
3
0
i2(t)
0
v2(t)
0
0
Sketch terminal waveforms of switch network Port 1
0
i 2(t)
dTs
dTs iL1 + iL2
0
T2
Ts
Ts
t
t
Appendix 3: Averaged switch modeling of a CCM SEPIC
Ts
vC1 + vC2
Port 2
v2(t)
Appendix 3: Averaged switch modeling of a CCM SEPIC
SEPIC CCM waveforms
Fundamentals of Power Electronics
the control input.
the average values of the remaining switch network terminal waveforms, and
• The object is simply to write the averaged equations of the switch network; i.e., to express the average values of half of the switch network terminal waveforms as functions of
be nonpulsating.
coincide with inductor currents or capacitor voltages of the converter, or
• It is not necessary that some of the switch network terminal quantities
the switch network contains no reactive elements.
its terminal voltages and currents are independent, and
• The switch network can be defined arbitrarily, as long as
A few points regarding averaged switch modeling
Ts
Ts
Ts
Ts
= d'(t)
= d(t)
= d(t)
= d'(t)
i L1(t)
vC1(t)
i L1(t)
vC1(t)
Ts
Ts
Ts
+ vC2(t)
+ i L2(t)
+ vC2(t)
+ i L2(t)
Ts
Ts
Ts
Ts
Ts
4
Appendix 3: Averaged switch modeling of a CCM SEPIC
Ts
=
Fundamentals of Power Electronics
i 2(t)
Ts
Ts
d(t)
v2(t)
d(t)
Ts
=
=
Ts
Ts
d'(t) i (t) d(t) 1
+ vC2(t)
+ i L2(t)
Ts
Ts
Ts
i 1(t)
d'(t) v1(t) T = v (t) s d(t) 2
Hence
vC1(t)
i L1(t)
We can write
5
s
Ts
+ –
Ts
s
〈i2(t)〉T
d'(t) i (t) d(t) 1
Averaged switch network
d'(t) v (t) d(t) 2
〈i1(t)〉T
+
〈v2(t)〉Ts
–
Appendix 3: Averaged switch modeling of a CCM SEPIC
Modeling the switch network via averaged dependent sources
–
s
〈v1(t)〉T
+
Result
Derivation of switch network equations (Algebra steps)
Fundamentals of Power Electronics
Need next to eliminate the capacitor voltages and inductor currents from these expressions, to write the equations of the switch network.
i 2(t)
v2(t)
i 1(t)
v1(t)
Use small ripple approximation
Expressions for average values of switch network terminal waveforms
I2
6
Appendix 3: Averaged switch modeling of a CCM SEPIC
+ –
Fundamentals of Power Electronics
Vg
IL1
L1
I1
V2 +
V1 –
7
–
+
I2
R
Appendix 3: Averaged switch modeling of a CCM SEPIC
Can now let inductors become short circuits, capacitors become open circuits, and solve for dc conditions.
–
IL2
D' : D
C2 VC2
+ L2
+ VC1 –
C1
Replace switch network with dc transformer model
Steady-state CCM SEPIC model
Fundamentals of Power Electronics
+
V2
V1 –
–
D' : D
+
I1
Steady-state switch model: Dc transformer model
i 2(t)
v2(t)
i 1(t)
Ts
Ts
Ts
s
= I 2 + i 2(t)
= V2 + v2(t)
= I 1 + i 1(t)
d(t) = D + d(t) v1(t) T = V1 + v1(t)
8
Appendix 3: Averaged switch modeling of a CCM SEPIC
V + V2 V1 + v1 = D' V2 + v2 – d 1 D D V1 = D' V2 + v2 – d D DD'
Fundamentals of Power Electronics
9
I +I I 2 + i 2 = D' I 1 + i 1 – d 1 2 D D I2 = D' I 1 + i 1 – d D DD'
Eliminate nonlinear terms and solve for i2 terms:
D + d I 2 + i 2 = D' – d I 1 + i 1
Current equation becomes
Appendix 3: Averaged switch modeling of a CCM SEPIC
Linearization, continued
Fundamentals of Power Electronics
Eliminate nonlinear terms and solve for v1 terms:
D + d V1 + v1 = D' – d V2 + v2
Voltage equation becomes
Perturb and linearize the switch network averaged waveforms, as usual:
Small-signal model
V1 d DD'
10
D' : D
I2 d DD'
+ –
L2
V1 d DD'
11
D' : D
I L2 + i L2
VC1 + vC1
+ –
Fundamentals of Power Electronics
Vg + vg
I L1 + i L1
C1
I2 d DD'
R
Appendix 3: Averaged switch modeling of a CCM SEPIC
Can now solve this model to determine ac transfer functions
–
C2 VC2 + vC2
+
Appendix 3: Averaged switch modeling of a CCM SEPIC
Replace switch network with small-signal ac model: L1
+
V2 + v2
–
I2 + i2
Small-signal ac model of the CCM SEPIC
Fundamentals of Power Electronics
–
V1 + v1
+
I1 + i1
Reconstruct equivalent circuit in the usual manner:
Switch network: Small-signal ac model
+ –