Elements of Power Electronics PART I: Bases Fabrice Frébel (fabrice.frebel@ulg.ac.be) September 21 st, 2017
Goal and expectations The goal of the course is to provide a toolbox that allows you to: understand power electronics concepts and topologies, to model a switching converter, to build it (including its magnetic components) and, to control it (with digital control). Power Electronics is a huge area and the correct approach is to focus on understanding concepts.
Organization and material The course is divided in three parts: PART I: Bases PART II: Topologies and applications PART III: Digital control In PART I and PART II, chapters are numbered according to the reference book [1]. In PART III, chapters are numbered according to the reference book [2].
PART I: Bases Chapter 1: Introduction Chapter 2: Principles of Steady-State Converter Analysis Chapter 3: Steady-State Equivalent Circuit Modeling, Losses, and Efficiency Chapter 13: Basic Magnetics Theory Chapter 4: Switch Realization Chapter 5: The Discontinuous Conduction Mode
Chapter 1: Introduction ELEC0055: Elements of Power Electronics - Fall 2017
Second edition Robert W. Erickson Dragan Maksimovic University of Colorado, Boulder 1 Chapter 1: Introduction
1.1 Introduction to Power Processing Power input Switching converter Power output Control input Dc-dc conversion: Change and control voltage magnitude Ac-dc rectification: Possibly control dc voltage, ac current Dc-ac inversion: Produce sinusoid of controllable magnitude and frequency Ac-ac cycloconversion: Change and control voltage magnitude and frequency 3 Chapter 1: Introduction
Control is invariably required Power input Switching converter Power output feedforward Control input Controller feedback reference 4 Chapter 1: Introduction
High efficiency is essential 1 η = P out P in P loss = P in P out = P out 1 η 1 η 0.8 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 0.6 0.4 0.2 0 0.5 1 1.5 P loss / P out 5 Chapter 1: Introduction
A high-efficiency converter P in Converter P out A goal of current converter technology is to construct converters of small size and weight, which process substantial power at high efficiency 6 Chapter 1: Introduction
Devices available to the circuit designer DT s T Linearmode Switched-mode s Resistors Capacitors Magnetics Semiconductor devices 7 Chapter 1: Introduction
Devices available to the circuit designer DT s T Linearmode Switched-mode s Resistors Capacitors Magnetics Semiconductor devices Signal processing: avoid magnetics 8 Chapter 1: Introduction
Devices available to the circuit designer DT s T Linearmode Switched-mode s Resistors Capacitors Magnetics Semiconductor devices Power processing: avoid lossy elements 9 Chapter 1: Introduction
Power loss in an ideal switch Switch closed: v(t) = 0 Switch open: i(t) = 0 In either event: p(t) = v(t) i(t) = 0 Ideal switch consumes zero power v(t) i(t) 10 Chapter 1: Introduction
A simple dc-dc converter example 100V I 10A Dc-dc R converter 5Ω V 50V Input source: 100V Output load: 50V, 10A, 500W How can this converter be realized? 11 Chapter 1: Introduction
Dissipative realization Resistive voltage divider 100V P in = 1000W 50V P loss = 500W R 5Ω I 10A V 50V P out = 500W 12 Chapter 1: Introduction
Dissipative realization Series pass regulator: transistor operates in active region 50V I 10A linear amplifier R V and base driver 100V 5Ω 50V P loss 500W P in 1000W P out = 500W V ref 13 Chapter 1: Introduction
Use of a SPDT switch 1 I 10 A 100 V 2 v s (t) R v(t) 50 V v s (t) DT s (1 D) T s t switch position: 1 2 1 0 V s = D 14 Chapter 1: Introduction
The switch changes the dc voltage level v s (t) V s = D D = switch duty cycle 0 D 1 DT s (1 D) T s t switch position: 1 2 1 0 T s = switching period f s = switching frequency = 1 / T s DC component of v s (t) = average value: V s = 1 T s 0 T s v s (t) dt = D 15 Chapter 1: Introduction
Addition of low pass filter Addition of (ideally lossless) L-C low-pass filter, for removal of switching harmonics: 100 V P in 500 W 1 2 v R s (t) C Choose filter cutoff frequency f 0 much smaller than switching frequency f s This circuit is known as the buck converter L P loss small i(t) v(t) P out = 500 W 16 Chapter 1: Introduction
Addition of control system for regulation of output voltage Power input Switching converter Load i v g v H(s) Sensor gain Transistor gate driver δ δ(t) Pulse-width modulator v c G c (s) Compensator Error signal v e Hv dt s T s t Reference input v ref 17 Chapter 1: Introduction
The boost converter 2 L 1 C R V 5 4 V 3 2 0 0 0.2 0.4 0.6 0.8 1 18 D Chapter 1: Introduction
A single-phase inverter 1 2 v s (t) v(t) load 2 1 v s (t) t H-bridge Modulate switch duty cycles to obtain sinusoidal low-frequency component 19 Chapter 1: Introduction
1.2 Several applications of power electronics Power levels encountered in high-efficiency converters less than 1 W in battery-operated portable equipment tens, hundreds, or thousands of watts in power supplies for computers or office equipment kw to MW in variable-speed motor drives 1000 MW in rectifiers and inverters for utility dc transmission lines 20 Chapter 1: Introduction
Chapter 2: Principles of Steady-State Converter Analysis ELEC0055: Elements of Power Electronics - Fall 2017
Chapter 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 ripple in converters containing twopole low-pass filters 2.6. Summary of key points 1 Chapter 2: Principles of steady-state converter analysis
2.1 Introduction Buck converter SPDT switch changes dc component 1 2 v s (t) R v(t) Switch output voltage waveform v s (t) DT s D'T s Duty cycle D: 0 D 1 complement D : D = 1 - D 0 DT s T s t Switch position: 1 2 1 0 2 Chapter 2: Principles of steady-state converter analysis
Dc component of switch output voltage v s (t) area = DT s v s = D 0 DT s T s 0 t Fourier analysis: Dc component = average value v s = T 1 T s v s (t) dt s 0 v s = 1 T s (DT s )=D 3 Chapter 2: Principles of steady-state converter analysis
Insertion of low-pass filter to remove switching harmonics and pass only dc component 1 L 2 v s (t) C R v(t) V v v s = D 0 0 1 D 4 Chapter 2: Principles of steady-state converter analysis
Three basic dc-dc converters Buck (a) 1 2 i L(t) L C R v M(D) 1 0.8 0.6 0.4 M(D) =D 0.2 0 0 0.2 0.4 0.6 0.8 1 D Boost (b) i L(t) L 1 2 C R v M(D) 5 4 3 2 M(D) = 1 1D 1 0 0 0.2 0.4 0.6 0.8 1 D Buck-boost (c) 1 2 i V L(t) g L C R v M(D) 0 1 2 3 D 0 0.2 0.4 0.6 0.8 1 4 M(D) = D 1D 5 5 Chapter 2: Principles of steady-state converter analysis
Objectives of this chapter 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 6 Chapter 2: Principles of steady-state converter analysis
2.2. Inductor volt-second balance, capacitor charge balance, and the small ripple approximation Actual output voltage waveform, buck converter Buck converter containing practical low-pass filter 1 2 i L (t) L v L (t) C i C (t) R v(t) Actual output voltage waveform v(t) Actual waveform v(t) = V v ripple (t) v(t)=v v ripple (t) V dc component V 0 t 7 Chapter 2: Principles of steady-state converter analysis
The small ripple approximation v(t)=v v ripple (t) v(t) V Actual waveform v(t) = V v ripple (t) dc component V 0 In a well-designed converter, the output voltage ripple is small. Hence, the waveforms can be easily determined by ignoring the ripple: t v ripple < V v(t) V 8 Chapter 2: Principles of steady-state converter analysis
Buck converter analysis: inductor current waveform original converter 1 2 i L (t) L v L (t) C i C (t) R v(t) switch in position 1 switch in position 2 i L (t) L L v L (t) i C (t) v L (t) i C (t) C R v(t) i L (t) C R v(t) 9 Chapter 2: Principles of steady-state converter analysis
Inductor voltage and current Subinterval 1: switch in position 1 Inductor voltage v L = v(t) Small ripple approximation: i L (t) L v L (t) C i C (t) R v(t) v L V Knowing the inductor voltage, we can now find the inductor current via v L (t)=l di L(t) dt Solve for the slope: di L (t) dt = v L(t) L V L The inductor current changes with an essentially constant slope 10 Chapter 2: Principles of steady-state converter analysis
Inductor voltage and current Subinterval 2: switch in position 2 Inductor voltage v L (t)=v(t) Small ripple approximation: L v L (t) i L (t) C i C (t) R v(t) v L (t) V Knowing the inductor voltage, we can again find the inductor current via v L (t)=l di L(t) dt Solve for the slope: di L (t) dt V L The inductor current changes with an essentially constant slope 11 Chapter 2: Principles of steady-state converter analysis
Inductor voltage and current waveforms v L (t) V DT s D'T s V Switch position: 1 2 1 i L (t) I i L (0) V L i L (DT s ) V L t i L v L (t)=l di L(t) dt 0 DT s T s t 12 Chapter 2: Principles of steady-state converter analysis
Determination of inductor current ripple magnitude i L (t) i L (DT s ) I i L (0) V L V L i L 0 DT s T s t (change in i L )=(slope)(length of subinterval) 2 i L = V DT L s i L = V 2L DT s L = V 2 i L DT s 13 Chapter 2: Principles of steady-state converter analysis
Inductor current waveform during turn-on transient i L (t) i L (T s ) i L (0) = 0 0 DT s T s v(t) L v(t) L i L (nt s ) 2T s nt s (n 1)T s i L ((n 1)T s ) t When the converter operates in equilibrium: i L ((n 1)T s )=i L (nt s ) 14 Chapter 2: Principles of steady-state converter analysis
The principle of inductor volt-second balance: Derivation Inductor defining relation: v L (t)=l di L(t) dt Integrate over one complete switching period: i L (T s )i L (0) = 1 L 0 T s v L (t) dt In periodic steady state, the net change in inductor current is zero: T s 0= v L (t) dt 0 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= 1 T s v T L (t) dt = v L s 0 The average inductor voltage is zero in steady state. 15 Chapter 2: Principles of steady-state converter analysis
Inductor volt-second balance: Buck converter example Inductor voltage waveform, previously derived: v L (t) V DT s Total area λ t V Integral of voltage waveform is area of rectangles: T s λ = v L (t) dt =( V)(DT s )(V)(D'T s ) 0 Average voltage is v L = T λ = D( V)D'( V) s Equate to zero and solve for V: 0=D (D D')V = D V V = D 16 Chapter 2: Principles of steady-state converter analysis
The principle of capacitor charge balance: Derivation Capacitor defining relation: i C (t)=c dv C(t) dt Integrate over one complete switching period: v C (T s )v C (0) = 1 C 0 T s i C (t) dt In periodic steady state, the net change in capacitor voltage is zero: 0= T 1 T s i C (t) dt = i C s 0 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. 17 Chapter 2: Principles of steady-state converter analysis
2.3 Boost converter example Boost converter with ideal switch i L (t) L v L (t) 1 2 i C (t) C R v L D 1 Realization using power MOSFET and diode i L (t) v L (t) DT s T s Q 1 i C (t) C R v 18 Chapter 2: Principles of steady-state converter analysis
Boost converter analysis original converter i L (t) L v L (t) 1 2 i C (t) C R v switch in position 1 switch in position 2 L L i L (t) v L (t) i C (t) i L (t) v L (t) i C (t) C R v C R v 19 Chapter 2: Principles of steady-state converter analysis
Subinterval 1: switch in position 1 Inductor voltage and capacitor current v L = i C =v / R i L (t) L v L (t) i C (t) Small ripple approximation: C R v v L = i C =V / R 20 Chapter 2: Principles of steady-state converter analysis
Subinterval 2: switch in position 2 Inductor voltage and capacitor current v L = v i C = i L v / R i L (t) L v L (t) i C (t) Small ripple approximation: C R v v L = V i C = I V / R 21 Chapter 2: Principles of steady-state converter analysis
Inductor voltage and capacitor current waveforms v L (t) DT s D'T s V t i C (t) I V/R DT s V/R D'T s t 22 Chapter 2: Principles of steady-state converter analysis
Inductor volt-second balance Net volt-seconds applied to inductor over one switching period: 0 T s v L (t) dt =( ) DT s ( V) D'T s v L (t) DT s D'T s V t Equate to zero and collect terms: (D D') VD'=0 Solve for V: V = D' The voltage conversion ratio is therefore M(D)= V = 1 D' = 1 1D 23 Chapter 2: Principles of steady-state converter analysis
Conversion ratio M(D) of the boost converter 5 4 M(D)= 1 D' = 1 1D M(D) 3 2 1 0 0 0.2 0.4 0.6 0.8 1 D 24 Chapter 2: Principles of steady-state converter analysis
Determination of inductor current dc component i C (t) I V/R Capacitor charge balance: 0 T s i C (t) dt =( V R ) DT s (I V R ) D'T s DT s V/R D'T s t Collect terms and equate to zero: V (D D') ID'=0 R Solve for I: I = V D' R Eliminate V to express in terms of : I = D' 2 R I /R 8 6 4 2 0 0 0.2 0.4 0.6 0.8 1 D 25 Chapter 2: Principles of steady-state converter analysis
Determination of inductor current ripple Inductor current slope during subinterval 1: di L (t) dt di L (t) dt = v L(t) L = v L(t) L = L Inductor current slope during subinterval 2: = V L i L (t) I L V L 0 DT s T s i L t Change in inductor current during subinterval 1 is (slope) (length of subinterval): 2 i L = L DT s Solve for peak ripple: i L = 2L DT s Choose L such that desired ripple magnitude is obtained 26 Chapter 2: Principles of steady-state converter analysis
Determination of capacitor voltage ripple Capacitor voltage slope during subinterval 1: dv C (t) dt Capacitor voltage slope during subinterval 2: dv C (t) dt = i C(t) C = i C(t) C = V RC = I C V RC v(t) V V RC I C V RC 0 DT s T s v t Change in capacitor voltage during subinterval 1 is (slope) (length of subinterval): 2 v = V RC DT s Solve for peak ripple: v = V 2RC DT s Choose C such that desired voltage ripple magnitude is obtained In practice, capacitor equivalent series resistance (esr) leads to increased voltage ripple 27 Chapter 2: Principles of steady-state converter analysis
2.4 Cuk converter example Cuk converter, with ideal switch L 1 C 1 L 2 i 1 i v 1 2 1 2 C 2 v 2 R Cuk converter: practical realization using MOSFET and diode L 1 C 1 L 2 i 1 i v 1 2 Q 1 D 1 C 2 v 2 R 28 Chapter 2: Principles of steady-state converter analysis
Cuk converter circuit with switch in positions 1 and 2 Switch in position 1: MOSFET conducts Capacitor C 1 releases energy to output L 1 L 2 i 2 i 1 v L1 i C1 v L2 i C2 v 1 C 1 C 2 v 2 R Switch in position 2: diode conducts Capacitor C 1 is charged from input i1 L 1 L 2 i 2 i C1 v L1 v L2 i C2 C 1 v 1 C 2 v 2 R 29 Chapter 2: Principles of steady-state converter analysis
Waveforms during subinterval 1 MOSFET conduction interval Inductor voltages and capacitor currents: v L1 = v L2 =v 1 v 2 i C1 = i 2 i C2 = i 2 v 2 R Small ripple approximation for subinterval 1: L 1 L 2 i 2 i 1 v L1 i v C1 L2 i C2 v 1 C 1 C 2 v 2 R v L1 = v L2 =V 1 V 2 i C1 = I 2 i C2 = I 2 V 2 R 30 Chapter 2: Principles of steady-state converter analysis
Waveforms during subinterval 2 Diode conduction interval Inductor voltages and capacitor currents: v L1 = v 1 v L2 =v 2 i C1 = i 1 i1 L 1 L 2 i 2 i C1 v L1 v L2 i C2 C 1 v 1 C 2 v 2 R i C2 = i 2 v 2 R Small ripple approximation for subinterval 2: v L1 = V 1 v L2 =V 2 i C1 = I 1 i C2 = I 2 V 2 R 31 Chapter 2: Principles of steady-state converter analysis
Equate average values to zero 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: Inductor voltage v L1 (t) v L1 (t) DT s D'T s t Volt-second balance on L 1 : v L1 = D D'( V 1 )=0 V 1 32 Chapter 2: Principles of steady-state converter analysis
Equate average values to zero Inductor L 2 voltage v L2 (t) V 2 DT s D'T s V 1 V t 2 Average the waveforms: Capacitor C 1 current i C1 (t) I 1 v L2 = D(V 1 V 2 )D'( V 2 )=0 i C1 = DI 2 D'I 1 =0 DT s I 2 D'T s t 33 Chapter 2: Principles of steady-state converter analysis
Equate average values to zero Capacitor current i C2 (t) waveform i C2 (t) DT s I 2 V 2 / R (= 0) D'T s t i C2 = I 2 V 2 R =0 Note: during both subintervals, the capacitor current i C2 is equal to the difference between the inductor current i 2 and the load current V 2 /R. When ripple is neglected, i C2 is constant and equal to zero. 34 Chapter 2: Principles of steady-state converter analysis
Cuk converter conversion ratio M = V/ 0 D 0 0.2 0.4 0.6 0.8 1-1 M(D) -2-3 -4 M(D)= V 2 = D 1D -5 35 Chapter 2: Principles of steady-state converter analysis
Inductor current waveforms Interval 1 slopes, using small ripple approximation: di 1 (t) dt di 2 (t) dt = v L1(t) L 1 = L 1 = v L2(t) L 2 = V 1 V 2 L 2 i 1 (t) I 1 i 1 L 1 V 1 L 1 DT s T s t Interval 2 slopes: DT s T s t di 1 (t) dt di 2 (t) dt = v L1(t) L 1 = V 1 L 1 = v L2(t) L 2 = V 2 L 2 I 2 i 2 (t) V 1 V 2 L 2 V 2 L 2 i2 36 Chapter 2: Principles of steady-state converter analysis
Capacitor C 1 waveform Subinterval 1: dv 1 (t) dt Subinterval 2: dv 1 (t) dt = i C1(t) C 1 = I 2 C 1 = i C1(t) C 1 = I 1 C 1 v 1 (t) v 1 V 1 I 2 C 1 DT s I 1 C 1 T s t 37 Chapter 2: Principles of steady-state converter analysis
Ripple magnitudes Analysis results i 1 = DT s 2L 1 i 2 = V 1 V 2 2L 2 v 1 = I 2DT s 2C 1 DT s Use dc converter solution to simplify: i 1 = DT s 2L 1 i 2 = DT s 2L 2 v 1 = D 2 T s 2D'RC 1 Q: How large is the output voltage ripple? 38 Chapter 2: Principles of steady-state converter analysis
2.5 Estimating ripple in converters containing two-pole low-pass filters Buck converter example: Determine output voltage ripple L 1 i L (t) i C (t) i R (t) 2 C v R C (t) Inductor current waveform. What is the capacitor current? i L (t) I i L (0) V L i L (DT s ) V L 0 DT s T s i L t 39 Chapter 2: Principles of steady-state converter analysis
Capacitor current and voltage, buck example i C (t) Must not neglect inductor current ripple! Total charge q T s /2 i L t DT s D'T s If the capacitor voltage ripple is small, then essentially all of the ac component of inductor current flows through the capacitor. v C (t) V v v t 40 Chapter 2: Principles of steady-state converter analysis
Estimating capacitor voltage ripple v i C (t) v C (t) Total charge q DT s T s /2 D'T s i L t Current i C (t) is positive for half of the switching period. This positive current causes the capacitor voltage v C (t) to increase between its minimum and maximum extrema. During this time, the total charge q is deposited on the capacitor plates, where V v v t q = C (2 v) (change in charge)= C (change in voltage) 41 Chapter 2: Principles of steady-state converter analysis
Estimating capacitor voltage ripple v i C (t) Total charge q T s /2 i L t The total charge q is the area of the triangle, as shown: q = 1 2 i L T s 2 DT s D'T s Eliminate q and solve for v: v C (t) v = i L T s 8 C V v v t Note: in practice, capacitor equivalent series resistance (esr) further increases v. 42 Chapter 2: Principles of steady-state converter analysis
Inductor current ripple in two-pole filters Example: problem 2.9 L 1 i Q T 1 i 1 i 2 L 2 C C 1 v C1 D 1 2 R v v L (t) Total flux linkage λ v t T s /2 i L (t) I DT s i D'T s i can use similar arguments, with λ = L (2 i) λ = inductor flux linkages = inductor volt-seconds t 43 Chapter 2: Principles of steady-state converter analysis
2.6 Summary of Key Points 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. 44 Chapter 2: Principles of steady-state converter analysis
Summary of Chapter 2 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. 45 Chapter 2: Principles of steady-state converter analysis
Chapter 3: Steady-State Equivalent Circuit Modeling, Losses, and Efficiency ELEC0055: Elements of Power Electronics - Fall 2017
Chapter 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 1 Chapter 3: Steady-state equivalent circuit modeling,...
3.1. The dc transformer model Basic equations of an ideal dc-dc converter: P in = P out (η = 100%) I g = V I Power input I g Switching dc-dc converter I V Power output V = M(D) I g = M(D) I (ideal conversion ratio) D Control input These equations are valid in steady-state. During transients, energy storage within filter elements may cause P in P out 2 Chapter 3: Steady-state equivalent circuit modeling,...
Equivalent circuits corresponding to ideal dc-dc converter equations P in = P out I g = V I V = M(D) I g = M(D) I Dependent sources DC transformer Power V input g I g M(D)I M(D) I V Power output Power input I g 1 : M(D) I V Power output D Control input 3 Chapter 3: Steady-state equivalent circuit modeling,...
The DC transformer model Power input I g 1 : M(D) D Control input I V Power output Models basic properties of ideal dc-dc converter: conversion of dc voltages and currents, ideally with 100% efficiency conversion ratio M controllable via duty cycle Solid line denotes ideal transformer model, capable of passing dc voltages and currents Time-invariant model (no switching) which can be solved to find dc components of converter waveforms 4 Chapter 3: Steady-state equivalent circuit modeling,...
Example: use of the DC transformer model 1. Original system 3. Push source through transformer V 1 R 1 Switching dc-dc converter V R M(D)V 1 M 2 (D)R 1 V R 2. Insert dc transformer model D 4. Solve circuit R 1 1 : M(D) V = M(D) V 1 R R M 2 (D) R 1 V 1 V R 5 Chapter 3: Steady-state equivalent circuit modeling,...
3.2. Inclusion of inductor copper loss Dc transformer model can be extended, to include converter nonidealities. Example: inductor copper loss (resistance of winding): L R L Insert this inductor model into boost converter circuit: L i R L 1 C R 2 v 6 Chapter 3: Steady-state equivalent circuit modeling,...
Analysis of nonideal boost converter L i R L 1 C R 2 v switch in position 1 switch in position 2 i L R L v L i C i L R L v L i C C R v C R v 7 Chapter 3: Steady-state equivalent circuit modeling,...
Circuit equations, switch in position 1 Inductor current and capacitor voltage: i L R L v L i C v L (t)= i(t) R L i C (t)=v(t)/r C R v Small ripple approximation: v L (t)= IR L i C (t)=v / R 8 Chapter 3: Steady-state equivalent circuit modeling,...
Circuit equations, switch in position 2 i L R L v L i C C R v v L (t)= i(t) R L v(t) IR L V i C (t)=i(t)v(t)/r I V / R 9 Chapter 3: Steady-state equivalent circuit modeling,...
Inductor voltage and capacitor current waveforms Average inductor voltage: v L (t) IR L T s v L (t) = T 1 v L (t)dt s 0 = D( IR L )D'( IR L V) DT s D'T s IR L V t Inductor volt-second balance: 0= IR L D'V i C (t) I V/R Average capacitor current: i C (t) = D (V / R)D'(I V / R) Capacitor charge balance: 0=D'I V / R V/R t 10 Chapter 3: Steady-state equivalent circuit modeling,...
Solution for output voltage We now have two equations and two unknowns: 5 4.5 4 R L /R = 0 R L /R = 0.01 0= IR L D'V 3.5 0=D'I V / R 3 R L /R = 0.02 Eliminate I and solve for V: V = V 1 1 g D' (1 R L / D' 2 R) V/ 2.5 2 1.5 1 R L /R = 0.05 R L /R = 0.1 0.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 D 11 Chapter 3: Steady-state equivalent circuit modeling,...
3.3. Construction of equivalent circuit model Results of previous section (derived via inductor volt-sec balance and capacitor charge balance): v L i C =0= IR L D'V =0=D'I V / R View these as loop and node equations of the equivalent circuit. Reconstruct an equivalent circuit satisfying these equations 12 Chapter 3: Steady-state equivalent circuit modeling,...
Inductor voltage equation Derived via Kirchhoff s voltage law, to find the inductor voltage during each subinterval Average inductor voltage then set to zero This is a loop equation: the dc components of voltage around a loop containing the inductor sum to zero v L =0= IR L D'V L v L = 0 I R L IR L D'V IR L term: voltage across resistor of value R L having current I D V term: for now, leave as dependent source 13 Chapter 3: Steady-state equivalent circuit modeling,...
Capacitor current equation i C =0=D'I V / R Node Derived via Kirchoff s current law, to find the capacitor current during each subinterval Average capacitor current then set to zero This is a node equation: the dc components of current flowing into a node connected to the capacitor sum to zero D'I i C = 0 C V/R term: current through load resistor of value R having voltage V D I term: for now, leave as dependent source V V/R R 14 Chapter 3: Steady-state equivalent circuit modeling,...
Complete equivalent circuit The two circuits, drawn together: R L V g I D'V D'I V R The dependent sources are equivalent to a D : 1 transformer: R L D' : 1 I V R Dependent sources and transformers I 1 nv 2 ni 1 V 2 I 1 n : 1 V 2 sources have same coefficient reciprocal voltage/current dependence 15 Chapter 3: Steady-state equivalent circuit modeling,...
Solution of equivalent circuit Converter equivalent circuit I R L D' : 1 V R Refer all elements to transformer secondary: R L /D' 2 D'I V g /D' V R Solution for output voltage using voltage divider formula: V = D' R R R = L D' D' 2 1 1 R L D' 2 R 16 Chapter 3: Steady-state equivalent circuit modeling,...
Solution for input (inductor) current I R L D' : 1 V R I = D' 2 R R L = D' 2 1 1 R L D' 2 R 17 Chapter 3: Steady-state equivalent circuit modeling,...
Equation correction I = D 2 R R L = D 2 R 1 1 R L D 2 R (1)
Solution for converter efficiency P in =( )(I) I R L D' : 1 P out =(V)(D'I) V R η = P out = (V)(D'I) = P in ( )(I) V V D' g η = 1 1 R L D' 2 R 18 Chapter 3: Steady-state equivalent circuit modeling,...
Efficiency, for various values of R L 100% η = 1 1 R L D' 2 R 90% 80% 70% 0.02 0.01 0.002 60% 0.05 η 50% R L /R = 0.1 40% 30% 20% 10% 0% 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 19 Chapter 3: Steady-state equivalent circuit modeling,... D
3.4. How to obtain the input port of the model Buck converter example use procedure of previous section to derive equivalent circuit i g 1 L R L i L v L 2 C R v C Average inductor voltage and capacitor current: v L =0=D I L R L V C i C =0=I L V C /R 20 Chapter 3: Steady-state equivalent circuit modeling,...
Construct equivalent circuit as usual v L =0=D I L R L V C i C =0=I L V C /R R L D v L = 0 I L i C = 0 V C V C /R R What happened to the transformer? Need another equation 21 Chapter 3: Steady-state equivalent circuit modeling,...
Modeling the converter input port Input current waveform i g (t): i g (t) i L (t) I L 0 area = DT s I L DT s 0 T s t Dc component (average value) of i g (t) is I g = 1 T s 0 T s i g (t) dt = DI L 22 Chapter 3: Steady-state equivalent circuit modeling,...
Input port equivalent circuit I g = 1 T s 0 T s i g (t) dt = DI L I DI g L 23 Chapter 3: Steady-state equivalent circuit modeling,...
Complete equivalent circuit, buck converter Input and output port equivalent circuits, drawn together: I g I L R L DI L D V C R Replace dependent sources with equivalent dc transformer: I g 1 : D R L I L V C R 24 Chapter 3: Steady-state equivalent circuit modeling,...
3.5. Example: inclusion of semiconductor conduction losses in the boost converter model Boost converter example i L i C C R v DT s T s Models of on-state semiconductor devices: MOSFET: on-resistance R on Diode: constant forward voltage V D plus on-resistance R D Insert these models into subinterval circuits 25 Chapter 3: Steady-state equivalent circuit modeling,...
Boost converter example: circuits during subintervals 1 and 2 i L i C C R v DT s T s switch in position 1 switch in position 2 i L R L v L i C i L R L R D v L i C V D R on C R v C R v 26 Chapter 3: Steady-state equivalent circuit modeling,...
Average inductor voltage and capacitor current v L (t) IR L IR on i C (t) DT s D'T s IR L V D IR D V I V/R t V/R t v L i C = D( IR L IR on )D'( IR L V D IR D V)=0 = D(V/R)D'(I V/R)=0 27 Chapter 3: Steady-state equivalent circuit modeling,...
Construction of equivalent circuits IR L IDR on D'V D ID'R D D'V =0 R L DRon D'V D D'R D IR L IDR on ID'R D I D'V D'I V/R =0 V/R D'I V R 28 Chapter 3: Steady-state equivalent circuit modeling,...
Complete equivalent circuit R L DRon D'V D D'R D I D'V D'I V R R L DRon D'V D D'R D D' : 1 I V R 29 Chapter 3: Steady-state equivalent circuit modeling,...
Solution for output voltage R L DRon D'V D D'R D D' : 1 I V R V = 1 D' D'V D D' 2 R D' 2 R R L DR on D'R D V = 1 D' 1 D'V D 1 1 R L DR on D'R D D' 2 R 30 Chapter 3: Steady-state equivalent circuit modeling,...
Solution for converter efficiency P in =( )(I) R L DRon D'V D D'R D D' : 1 P out =(V)(D'I) I V R η = D' V = 1 D'V D 1 R L DR on D'R D D' 2 R Conditions for high efficiency: /D' > V D D' 2 R > R L DR on D'R D 31 Chapter 3: Steady-state equivalent circuit modeling,...
Accuracy of the averaged equivalent circuit in prediction of losses Model uses average currents and voltages To correctly predict power loss in a resistor, use rms values Result is the same, provided ripple is small i(t) MOSFET current waveforms, for various ripple magnitudes: I 0 (c) (b) (a) DT s 2 I 1.1 I 0 T s t Inductor current ripple MOS FET rms current Average power loss in R on (a) i = 0 (b) i = 0.1 I (c) i = I I D D I 2 R on (1.00167) I D (1.0033) D I 2 R on (1.155) I D (1.3333) D I 2 R on 32 Chapter 3: Steady-state equivalent circuit modeling,...
Summary of chapter 3 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. 33 Chapter 3: Steady-state equivalent circuit modeling,...
Chapter 13: Basic Magnetics Theory ELEC0055: Elements of Power Electronics - Fall 2017
Example: a simple inductor Faraday s law: For each turn of wire, we can write v turn (t)= dφ(t) dt Total winding voltage is v(t)=nv turn (t)=n dφ(t) dt i(t) v(t) n turns Φ core Core area A c Core permeability µ Express in terms of the average flux density B(t) = F(t)/A c v(t)=na c db(t) dt 15 Chapter 13: Basic Magnetics Theory
Inductor example: Ampere s law 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 l m. i(t) n turns H Magnetic path length l m For uniform field strength H(t), the core MMF around the path is H l m. 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). From Ampere s law, we have H(t) l m = n i(t) 16 Chapter 13: Basic Magnetics Theory
Inductor example: core material model B B sat B sat for H B sat /µ B = µh for H < B sat /µ B sat for H B sat /µ µ H Find winding current at onset of saturation: substitute i = I sat and H = B sat / into equation previously derived via Ampere s law. Result is B sat I sat = B satl m µn 17 Chapter 13: Basic Magnetics Theory
Electrical terminal characteristics We have: v(t)=na c db(t) dt H(t) l m = n i(t) B = B sat for H B sat /µ µh for H < B sat /µ B sat for H B sat /µ Eliminate B and H, and solve for relation between v and i. For i < I sat, dh(t) v(t)=µna c dt v(t)= µn2 A c di(t) l m dt which is of the form v(t)=l di(t) with dt L = µn2 A c l m an inductor For i > I sat the flux density is constant and equal to B sat. Faraday s law then predicts db v(t)=na sat c =0 saturation leads to short circuit dt 18 Chapter 13: Basic Magnetics Theory
13.1.2 Magnetic circuits Uniform flux and magnetic field inside a rectangular element: Flux Φ { Length l MMF F Area A c MMF between ends of Core permeability µ element is F = Hl H R = µa l c Since H = B / and = / A c, we can express F as F = ΦR with R = µa l c A corresponding model: F R = reluctance of element Φ R 19 Chapter 13: Basic Magnetics Theory
Magnetic circuits: magnetic structures composed of multiple windings and heterogeneous elements Represent each element with reluctance Windings are sources of MMF MMF voltage, flux current Solve magnetic circuit using Kirchoff s laws, etc. 20 Chapter 13: Basic Magnetics Theory
Magnetic analog of Kirchoff s current law Divergence of B = 0 Flux lines are continuous and cannot end Total flux entering a node must be zero Physical structure Φ 1 Φ 2 Node Φ 3 Magnetic circuit Node Φ 1 = Φ 2 Φ 3 Φ 1 Φ 3 Φ 2 21 Chapter 13: Basic Magnetics Theory
Magnetic analog of Kirchoff s voltage law Follows from Ampere s law: closed path H dl = total current passing through interior of path Left-hand side: sum of MMF s across the reluctances around the closed path 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). Total MMF s around the closed path add up to zero. 22 Chapter 13: Basic Magnetics Theory
Example: inductor with air gap i(t) v(t) n turns Core permeability µ Φ Cross-sectional area A c Air gap l g Magnetic path length l m 23 Chapter 13: Basic Magnetics Theory
Magnetic circuit model i(t) v(t) n turns Core permeability µ Φ Cross-sectional area A c Air gap l g ni(t) F c R c Φ(t) R g F g Magnetic path length l m F c F g = ni R c = l c µa c ni = Φ R c R g l g R g = µ 0 A c 24 Chapter 13: Basic Magnetics Theory
Solution of model i(t) v(t) n turns Core permeability µ Φ Cross-sectional area A c Air gap l g ni(t) F c R c Φ(t) R g F g Magnetic path length l m Faraday s law: v(t)=n dφ(t) dt Substitute for : v(t)= n 2 di(t) R c R g dt Hence inductance is n 2 L = R c R g ni = Φ R c R g l c R c = µa c l g R g = µ 0 A c 25 Chapter 13: Basic Magnetics Theory
Effect of air gap ni = Φ R c R g L = n 2 B sat A c R 1 c R g R c sat = B sat A 1 c R c R g I sat = B sat A c n R c R g Effect of air gap: decrease inductance increase saturation current inductance is less dependent on core permeability Φ = BA c ni sat1 B sat A c ni sat2 ni H c 26 Chapter 13: Basic Magnetics Theory
13.2 Transformer modeling Two windings, no air gap: R = l m µa c F c = n 1 i 1 n 2 i 2 i 1 (t) n v 1 1 (t) turns Φ n 2 turns i 2 (t) v 2 (t) ΦR = n 1 i 1 n 2 i 2 Core Magnetic circuit model: Φ R c F c n 1 i 1 n 2 i 2 27 Chapter 13: Basic Magnetics Theory
13.2.1 The ideal transformer In the ideal transformer, the core reluctance R approaches zero. MMF F c = R also approaches zero. We then obtain n 1 i 1 Φ R c F c n 2 i 2 0=n 1 i 1 n 2 i 2 Also, by Faraday s law, v 1 = n dφ 1 dt v 2 = n dφ 2 dt Eliminate : dφ = v 1 = v 2 dt n 1 n 2 Ideal transformer equations: v 1 = v 2 and n n 1 n 1 i 1 n 2 i 2 =0 2 v 1 i 1 n 1 : n 2 i 2 Ideal v 2 28 Chapter 13: Basic Magnetics Theory
13.2.2 The magnetizing inductance For nonzero core reluctance, we obtain ΦR = n 1 i 1 n 2 i 2 with v 1 = n 1 dφ dt Eliminate : v 1 = n 2 1 R dt d i 1 n 2 i n 2 1 n 1 i 1 Φ R c F c n 2 i 2 This equation is of the form v 1 = L M di M dt with L M = n 2 1 R i M = i 1 n 2 i n 2 1 v 1 i 1 n i 2 1 n 1 : n 2 i 2 L M = n 2 1 R 29 n 2 i 1 n 2 n 1 i 2 Ideal v 2 Chapter 13: Basic Magnetics Theory
Magnetizing inductance: discussion Models magnetization of core material A real, physical inductor, that exhibits saturation and hysteresis If the secondary winding is disconnected: we are left with the primary winding on the core primary winding then behaves as an inductor the resulting inductor is the magnetizing inductance, referred to the primary winding Magnetizing current causes the ratio of winding currents to differ from the turns ratio 30 Chapter 13: Basic Magnetics Theory
Transformer saturation Saturation occurs when core flux density B(t) exceeds saturation flux density B sat. When core saturates, the magnetizing current becomes large, the impedance of the magnetizing inductance becomes small, and the windings are effectively shorted out. Large winding currents i 1 (t) and i 2 (t) do not necessarily lead to saturation. If 0=n 1 i 1 n 2 i 2 then the magnetizing current is zero, and there is no net magnetization of the core. Saturation is caused by excessive applied volt-seconds 31 Chapter 13: Basic Magnetics Theory
Saturation vs. applied volt-seconds Magnetizing current depends on the integral of the applied winding voltage: i M (t)= 1 L M Flux density is proportional: B(t)= 1 n 1 A c v 1 (t)dt v 1 (t)dt v 1 i 1 n i 2 1 n 1 : n 2 i 2 L M = n 2 1 R n 2 Ideal Flux density becomes large, and core saturates, when the applied volt-seconds 1 are too large, where λ 1 = t 2 t 1 i 1 n 2 n 1 i 2 v 1 (t)dt limits of integration chosen to coincide with positive portion of applied voltage waveform v 2 32 Chapter 13: Basic Magnetics Theory
13.2.3 Leakage inductances v 1 (t) i 1 (t) Φ l1 Φ M i 2 (t) Φ l2 v 2 (t) Φ l1 Φ l2 i 1 (t) v 1 (t) Φ M i 2 (t) v 2 (t) 33 Chapter 13: Basic Magnetics Theory
Transformer model, including leakage inductance Terminal equations can be written in the form v1 (t) v 2 (t) = L 11 L 12 L 12 L 22 d dt i 1 (t) i 2 (t) v 1 L l1 i 1 n 1 : n 2 i 2 L M = n 1 n 2 L 12 i M L l2 v 2 mutual inductance: Ideal L 12 = n 1n 2 R = n 2 n 1 L M primary and secondary self-inductances: effective turns ratio n e = L 22 L 11 L 11 = L l1 n 1 n 2 L 12 L 22 = L l2 n 2 n 1 L 12 coupling coefficient k = L 12 L 11 L 22 34 Chapter 13: Basic Magnetics Theory
13.3 Loss mechanisms in magnetic devices Low-frequency losses: Dc copper loss Core loss: hysteresis loss High-frequency losses: the skin effect Core loss: classical eddy current losses Eddy current losses in ferrite cores High frequency copper loss: the proximity effect Proximity effect: high frequency limit MMF diagrams, losses in a layer, and losses in basic multilayer windings Effect of PWM waveform harmonics 35 Chapter 13: Basic Magnetics Theory
13.3.1 Core loss Energy per cycle W flowing into n- turn winding of an inductor, excited by periodic waveforms of frequency f: W = one cycle v(t)i(t)dt i(t) v(t) n turns Φ core Core area A c Core permeability µ Relate winding voltage and current to core B and H via Faraday s law and Ampere s law: db(t) v(t)=na c dt H(t)l m = ni(t) Substitute into integral: db(t) W = na c dt one cycle = A c l m HdB 36 one cycle H(t)l m n dt Chapter 13: Basic Magnetics Theory
Core loss: Hysteresis loss W = A c l m one cycle HdB B Area one cycle HdB The term A c l m is the volume of the core, while the integral is the area of the BH loop. H (energy lost per cycle) = (core volume) (area of BH loop) P H = f A c l m HdB one cycle Hysteresis loss is directly proportional to applied frequency 37 Chapter 13: Basic Magnetics Theory
Modeling hysteresis loss Hysteresis loss varies directly with applied frequency Dependence on maximum flux density: how does area of BH loop depend on maximum flux density (and on applied waveforms)? Empirical equation (Steinmetz equation): P H = K H fb α max (core volume) The parameters K H and are determined experimentally. Dependence of P H on B max is predicted by the theory of magnetic domains. 38 Chapter 13: Basic Magnetics Theory
Core loss: eddy current loss 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 (t). The eddy currents tend to prevent flux from penetrating the core. Flux Φ(t) Eddy current i(t) Core Eddy current loss i 2 (t)r 39 Chapter 13: Basic Magnetics Theory
Modeling eddy current loss Ac flux (t) induces voltage v(t) in core, according to Faraday s law. Induced voltage is proportional to derivative of (t). In consequence, magnitude of induced voltage is directly proportional to 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. Eddy current power loss i 2 (t)r then varies with square of excitation frequency f. Classical Steinmetz equation for eddy current loss: P E = K E f 2 2 B max (core volume) Ferrite core material impedance is capacitive. This causes eddy current power loss to increase as f 4. 40 Chapter 13: Basic Magnetics Theory
Total core loss: manufacturer s data Ferrite core material 1 1MHz 500kHz 200kHz Empirical equation, at a fixed frequency: Power loss density, Watts/cm 3 0.1 100kHz 50kHz 20kHz P fe = K fe ( B) β A c l m 0.01 0.01 0.1 0.3 B, Tesla 41 Chapter 13: Basic Magnetics Theory
Core materials Core type B sat Relative core loss Applications Laminations iron, silicon steel Powdered cores powdered iron, molypermalloy Ferrite Manganese-zinc, Nickel-zinc 1.5-2.0 T high 50-60 Hz transformers, inductors 0.6-0.8 T medium 1 khz transformers, 100 khz filter inductors 0.25-0.5 T low 20 khz - 1 MHz transformers, ac inductors 42 Chapter 13: Basic Magnetics Theory
13.3.2 Low-frequency copper loss DC resistance of wire R = ρ l b A w where A w is the wire bare cross-sectional area, and l b is the length of the wire. The resistivity is equal to 1.724 10 6 cm for soft-annealed copper at room temperature. This resistivity increases to 2.3 10 6 cm at 100 C. i(t) R The wire resistance leads to a power loss of 2 P cu = I rms R 43 Chapter 13: Basic Magnetics Theory
13.4 Eddy currents in winding conductors 13.4.1 Intro to the skin and proximity effects Wire i(t) Φ(t) Eddy currents Wire Current density δ i(t) Eddy currents 44 Chapter 13: Basic Magnetics Theory
Penetration depth For sinusoidal currents: current density is an exponentially decaying function of distance into the conductor, with characteristic length known as the penetration depth or skin depth. Penetration depth δ, cm 0.1 Wire diameter #20 AWG δ = ρ πµ f For copper at room temperature: 0.01 100 C 25 C #30 AWG #40 AWG δ = 7.5 f cm 0.001 10 khz 100 khz 1 MHz Frequency 45 Chapter 13: Basic Magnetics Theory
The proximity effect 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. A multi-layer foil winding, with h >. Each layer carries net current i(t). Current density J iconductor 1 h Area Φ Conductor 2 i i i Area i Area i 46 Chapter 13: Basic Magnetics Theory
Example: a two-winding transformer Cross-sectional view of two-winding transformer example. Primary turns are wound in three layers. For this example, let s assume that each layer is one turn of a flat foil conductor. The secondary is a similar three-layer winding. Each layer carries net current i(t). Portions of the windings that lie outside of the core window are not illustrated. Each layer has thickness h >. Core i i i { Layer 1 Layer 2 Layer 3 Primary winding i i i Layer 3 { Layer 2 Layer 1 Secondary winding 47 Chapter 13: Basic Magnetics Theory
Distribution of currents on surfaces of conductors: two-winding example Skin effect causes currents to concentrate on surfaces of conductors Surface current induces equal and opposite current on adjacent conductor This induced current returns on opposite side of conductor Net conductor current is equal to i(t) for each layer, since layers are connected in series Circulating currents within layers increase with the numbers of layers Core Current density J Layer 1 h Layer 1 i i i Layer 2 Primary winding 48 Layer 3 i i i Layer 3 Layer 2 Φ 2Φ 3Φ 2Φ Φ i i 2i 2i 3i 3i 2i 2i i i Layer 2 Layer 3 Layer 3 Layer 2 Layer 1 Layer 1 Secondary winding Chapter 13: Basic Magnetics Theory
Estimating proximity loss: high-frequency limit The current i(t) having rms value I is confined to thickness d on the surface of layer 1. Hence the effective ac resistance of layer 1 is: R ac = h δ R dc This induces copper loss P 1 in layer 1: P 1 = I 2 R ac Power loss P 2 in layer 2 is: P 2 = P 1 4P 1 =5P 1 Power loss P 3 in layer 3 is: Current density J h Layer 1 Φ 2Φ 3Φ 2Φ Φ i i 2i 2i 3i 3i 2i 2i i i Layer 2 Primary winding Layer 3 Layer 3 Layer 2 Secondary winding Power loss P m in layer m is: P 3 = 2 2 3 2 P 1 =13P 1 P m = I 2 m 1 2 m 2 h δ R dc Layer 1 49 Chapter 13: Basic Magnetics Theory
Total loss in M-layer winding: high-frequency limit Add up losses in each layer: Μ Σ P = I 2 h δ R dc m 1 2 m 2 m =1 = I 2 h δ R M dc 2M 2 1 3 Compare with dc copper loss: If foil thickness were H =, then at dc each layer would produce copper loss P 1. The copper loss of the M-layer winding would be P dc = I 2 MR dc So the proximity effect increases the copper loss by a factor of F R = P P dc = 1 3 h δ 2M 2 1 50 Chapter 13: Basic Magnetics Theory
Discussion: design of winding geometry to minimize proximity loss Interleaving windings can significantly reduce the proximity loss when the winding currents are in phase, such as in the transformers of buckderived converters or other converters In some converters (such as flyback or SEPIC) the winding currents are out of phase. Interleaving then does little to reduce the peak MMF and proximity loss. See Vandelac and Ziogas [10]. For sinusoidal winding currents, there is an optimal conductor thickness near = 1 that minimizes copper loss. Minimize the number of layers. Use a core geometry that maximizes the width l w of windings. Minimize the amount of copper in vicinity of high MMF portions of the windings 74 Chapter 13: Basic Magnetics Theory
Litz wire A way to increase conductor area while maintaining low proximity losses Many strands of small-gauge wire are bundled together and are externally connected in parallel Strands are twisted, or transposed, so that each strand passes equally through each position on inside and outside of bundle. This prevents circulation of currents between strands. Strand diameter should be sufficiently smaller than skin depth The Litz wire bundle itself is composed of multiple layers Advantage: when properly sized, can significantly reduce proximity loss Disadvantage: increased cost and decreased amount of copper within core window 75 Chapter 13: Basic Magnetics Theory
Practical realisations and simulations Transformer (50 Hz) Transformer (20 khz) Inductors Simulations with ONELAB
Chapter 4: Switch Realization Diode MOSFET Other power semi-conductors
Forward-biased power diode i v conductivity modulation { p n - n minority carrier injection 32 Chapter 4: Switch realization
Charge-controlled behavior of the diode The diode equation: q(t)=q 0 e λv(t) 1 i v Charge control equation: dq(t) dt = i(t) q(t) τ L With: λ= 1/(26 mv) at 300 K τ L = minority carrier lifetime (above equations don t include current that charges depletion region capacitance) p n - n 33 } Total stored minority charge q In equilibrium: dq/dt = 0, and hence i(t)= q(t) τ L = Q 0 τ L e λv(t) 1 = I 0 e λv(t) 1 Chapter 4: Switch realization
Charge-control in the diode: Discussion The familiar iv curve of the diode is an equilibrium relationship that can be violated during transient conditions During the turn-on and turn-off switching transients, the current deviates substantially from the equilibrium iv curve, because of change in the stored charge and change in the charge within the reverse-bias depletion region Under forward-biased conditions, the stored minority charge causes conductivity modulation of the resistance of the lightly-doped n region, reducing the device on-resistance 34 Chapter 4: Switch realization
Diode in OFF state: reversed-biased, blocking voltage v v(t) t p n n { E v Depletion region, reverse-biased Diode is reverse-biased No stored minority charge: q = 0 Depletion region blocks applied reverse voltage; charge is stored in capacitance of depletion region i(t) 0 (1) t 35 Chapter 4: Switch realization
Turn-on transient v(t) i(t) Charge depletion region (1) (2) Diode conducts with low on-resistance Diode is forward-biased. Supply minority charge to n region to reduce on-resistance On-state current determined by converter circuit t t The current i(t) is determined by the converter circuit. This current supplies: charge to increase voltage across depletion region charge needed to support the on-state current charge to reduce on-resistance of n region 36 Chapter 4: Switch realization
Turn-off transient i (< 0) v p n - n } Removal of stored minority charge q 37 Chapter 4: Switch realization
Diode turn-off transient continued v(t) t i(t) (4) Diode remains forward-biased. Remove stored charge in n region di dt t r (5) Diode is reverse-biased. Charge depletion region capacitance. 0 t Area Q r (1) (2) (3) (4) (5) (6) 38 Chapter 4: Switch realization
The diode switching transients induce switching loss in the transistor i A fast transistor v A v B i L (t) silicon diode i B L transistor waveforms i A (t) Q r v A (t) 0 i L 0 t see Section 4.3.2 Diode recovered stored charge Q r flows through transistor during transistor turn-on transition, inducing switching loss Q r depends on diode on-state forward current, and on the rate-of-change of diode current during diode turn-off transition diode waveforms 39 p A (t) = v A i A i L 0 area Q r i B (t) v B (t) t r 0 t area ~Q r area ~i L t r t 0 t 1 t 2 t Chapter 4: Switch realization
Switching loss calculation Energy lost in transistor: W D = v A (t) i A (t) dt switching transition With abrupt-recovery diode: W D (i L i B (t)) dt switching transition transistor waveforms diode waveforms 0 i L v A (t) 0 i A (t) Q r area Q r i B (t) v B (t) i L 0 0 t t Soft-recovery diode: (t 2 t 1 ) >> (t 1 t 0 ) Abrupt-recovery diode: (t 2 t 1 ) << (t 1 t 0 ) = i L t r Q r t r Often, this is the largest component of switching loss p A (t) = v A i A area ~Q r area ~i L t r 40 t 0 t 1 t 2 t Chapter 4: Switch realization
Ringing induced by diode stored charge see Section 4.3.3 v i (t) i L (t) L v L (t) silicon diode i B (t) v B (t) C v i (t) 0 i L (t) V 1 V 2 t Diode is forward-biased while i L (t) > 0 Negative inductor current removes diode stored charge Q r When diode becomes reverse-biased, negative inductor current flows through capacitor C. Ringing of L-C network is damped by parasitic losses. Ringing energy is lost. 44 0 v B (t) 0 V 2 area Q r t 1 t 2 t 3 Chapter 4: Switch realization t t
Energy associated with ringing Recovered charge is Q r = i L (t) dt Energy stored in inductor during interval t 2 t t 3 : t 3 W L = v L (t) i L (t) dt t 2 t 2 t 3 v i (t) 0 i L (t) V 1 V 2 t Applied inductor voltage during interval t 2 t t 3 : v L (t)=l di L(t) =V dt 2 Hence, t 3 W L = L di t L(t) 3 i dt L (t) dt = (V 2 ) i L (t) dt t 2 t 2 0 v B (t) 0 area Q r t t W L = 1 2 Li 2 L(t 3 )=V 2 Q r V 2 t 1 t 2 t 3 45 Chapter 4: Switch realization
Diode: static characteristic example Excerpt of IXYS DSEP29-12A diode data-sheet:
Diode: recovery characteristics example Excerpt of IXYS DSEP29-12A diode data-sheet:
Types of power diodes Standard recovery Reverse recovery time not specified, intended for 50/60Hz Fast recovery and ultra-fast recovery Reverse recovery time and recovered charge specified Intended for converter applications Schottky diode 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) 41 Chapter 4: Switch realization
Characteristics of several commercial power rectifier diodes Part number Rated max voltage Rated avg current V F (typical) t r (max) Fast recovery rectifiers 1N3913 400V 30A 1.1V 400ns SD453N25S20PC 2500V 400A 2.2V 2µs Ultra-fast recovery rectifiers MUR815 150V 8A 0.975V 35ns MUR1560 600V 15A 1.2V 60ns RHRU100120 1200V 100A 2.6V 60ns Schottky rectifiers MBR6030L 30V 60A 0.48V 444CNQ045 45V 440A 0.69V 30CPQ150 150V 30A 1.19V 42 Chapter 4: Switch realization
4.2.2. The Power MOSFET Gate Source Gate lengths approaching one micron n p n n - n p n Consists of many small enhancementmode parallelconnected MOSFET cells, covering the surface of the silicon wafer n Vertical current flow n-channel device is shown Drain 46 Chapter 4: Switch realization
MOSFET: Off state source p-n - junction is reverse-biased n p n n p n off-state voltage appears across n - region depletion region n - n drain 47 Chapter 4: Switch realization
MOSFET: on state source p-n - junction is slightly reversebiased n p n n p n positive gate voltage induces conducting channel channel n - drain current flows through n - region and conducting channel drain n drain current on resistance = total resistances of n - region, conducting channel, source and drain contacts, etc. 48 Chapter 4: Switch realization
MOSFET body diode n p source n Body diode n p n p-n - junction forms an effective diode, in parallel with the channel negative drain-tosource voltage can forward-bias the body diode drain n - n diode can conduct the full MOSFET rated current diode switching speed not optimized body diode is slow, Q r is large 49 Chapter 4: Switch realization
Typical MOSFET characteristics I D 10A 5A off state V DS = 200V V DS = 10V on state V DS = 2V V DS = 1V V DS = 0.5V 0A 0V 5V 10V 15V V GS Off state: V GS < V th On state: V GS >> V th MOSFET can conduct peak currents well in excess of average current rating characteristics are unchanged on-resistance has positive temperature coefficient, hence easy to parallel 50 Chapter 4: Switch realization
A simple MOSFET equivalent circuit D C gs : large, essentially constant C gd : small, highly nonlinear G C gd C gs C ds C ds : intermediate in value, highly nonlinear switching times determined by rate at which gate driver charges/discharges C gs and C gd S C ds (v ds )= C 0 1 v ds V 0 C ds(v ds) C 0 V 0 v ds = C 0 ' v ds 51 Chapter 4: Switch realization
Switching loss caused by semiconductor output capacitances Buck converter example C ds C j Energy lost during MOSFET turn-on transition (assuming linear capacitances): W C = 1 2 (C ds C j ) 2 52 Chapter 4: Switch realization
MOSFET nonlinear C ds Approximate dependence of incremental C ds on v ds : V C ds (v ds ) C 0 0 v = C ' 0 ds v ds Energy stored in C ds at v ds = V DS : V DS W Cds = v ds i C dt = v ds C ds (v ds ) dv ds 0 V DS W Cds = C ' 0 (v ds ) v ds dv ds = 2 3 C 2 ds(v DS) V DS 0 same energy loss as linear capacitor having value 4 3 C ds(v DS ) 53 Chapter 4: Switch realization
Equation correction W Cds = VDS 0 C 0 vds dv ds = 2 3 C ds(v ds )V 2 ds (2)
MOSFET: hard switching losses (waveforms) Excerpt of [4]: The area P on and P off represent the switching losses.
MOSFET: hard switching losses (explanations) (a) Turn-On: 1. The gate voltage V GS rises from 0 to V th. 2. During t CR, the drain current I DS rises according to V GS change (linked by the transconductance). 3. Once I DS reaches I L, the transistor carries the full load current. 4. V GS stays at the plateau voltage V pl due to the Miller effect and the drain voltage V DS falls linearly during t VF. 5. Once V DS reaches 0V, V GS continues to rise up to V DR. (b) Turn-Off: 1. V GS falls from V DR to V pl. 2. When V GS reaches V pl, V DS starts rising linearly during t VR. V GS stays at V pl due to the Miller effect. 3. Once V DS reaches V BUS, V GS starts falling and I DS also starts falling accordingly (linked by the transconductance) during t CF. 4. Once V GS reaches V th, I DS reaches 0V. 5. Finally, V GS goes to 0V.
MOSFET: static characteristics example Excerpt of IR IRFP4668PbF MOSFET data-sheet:
MOSFET: dynamic characteristics example Excerpt of IR IRFP4668PbF MOSFET data-sheet:
Characteristics of several commercial power MOSFETs Part number Rated max voltage Rated avg current R on Q g (typical) IRFZ48 60V 50A 0.018Ω 110nC IRF510 100V 5.6A 0.54Ω 8.3nC IRF540 100V 28A 0.077Ω 72nC APT10M25BNR 100V 75A 0.025Ω 171nC IRF740 400V 10A 0.55Ω 63nC MTM15N40E 400V 15A 0.3Ω 110nC APT5025BN 500V 23A 0.25Ω 83nC APT1001RBNR 1000V 11A 1.0Ω 150nC 54 Chapter 4: Switch realization
MOSFET: conclusions 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 55 Chapter 4: Switch realization
Other power semi-conductors (brief overview) Thyristor: high voltage, high current, switches off at zero current, GTO (gate turn off Thyristor): similar to Thyristor but can be switched off with the gate signal, IGBT (Isolated Gate Bipolar Transistor): high voltage, high current, controlled like a MOSFET, BJT transistor: not often used, replaced by MOSFET, Schottky diode: diode with higher conduction and switching performances but lower breakdown voltage, SiC diode: emerging component that could/will replace diodes, SiC transistor: emerging component that could/will replace IGBT, GaN transistor: emerging component that could/will replace MOSFET.
Other power semi-conductors Excerpt of [5]: