R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

Transcription

1 . W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

2 2.4 Cuk converter example L 1 C 1 L 2 Cuk converter, with ideal switch i 1 i v C 2 v 2 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 Fundamentals of Power Electronics 28 Chapter 2: Principles of steady-state converter analysis

3 Analysis strategy This converter has two inductor currents and two capacitor voltages, that can be expressed as L 1 C 1 L 2 i 1 i v C 2 v 2 i 1 (t)=i 1 i 1-ripple (t) i 2 (t)=i 2 i 2-ripple (t) v 1 (t)=v 1 v 1-ripple (t) v 2 (t)=v 2 v 2-ripple (t) To solve the converter in steady state, we want to find the dc components I 1, I 2, V 1, and V 2, when the ripples are small. Strategy: Apply volt-second balance to each inductor voltage Apply charge balance to each capacitor current Simplify using the small ripple approximation Solve the resulting four equations for the four unknowns I 1, I 2, V 1, and V 2. Fundamentals of Power Electronics 29 Chapter 2: Principles of steady-state converter analysis

4 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 i 2 i 1 v L1 i v C1 L2 L 2 v 1 C 1 C 2 i C2 v 2 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 Fundamentals of Power Electronics 30 Chapter 2: Principles of steady-state converter analysis

5 Waveforms during subinterval 1 MOSFET conduction interval Inductor voltages and capacitor currents: v L1 = L 1 i 2 i 1 v L1 i v C1 L2 L 2 i C2 v 1 C 1 C 2 v 2 v L2 =v 1 v 2 i C1 = i 2 i C2 = i 2 v 2 Small ripple approximation for subinterval 1: v L1 = v L2 =V 1 V 2 i C1 = I 2 i C2 = I 2 V 2 Fundamentals of Power Electronics 31 Chapter 2: Principles of steady-state converter analysis

6 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 i C2 = i 2 v 2 Small ripple approximation for subinterval 2: v L1 = V 1 v L2 =V 2 i C1 = I 1 i C2 = I 2 V 2 Fundamentals of Power Electronics 32 Chapter 2: Principles of steady-state converter analysis

7 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) Volt-second balance on L 1 : DT s D'T s t v L1 = D D'( V 1 )=0 V 1 Fundamentals of Power Electronics 33 Chapter 2: Principles of steady-state converter analysis

8 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 Fundamentals of Power Electronics 34 Chapter 2: Principles of steady-state converter analysis

9 Equate average values to zero Capacitor current i C2 (t) waveform i C2 (t) DT s I 2 V 2 / (= 0) D'T s t i C2 = I 2 V 2 =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 /. When ripple is neglected, i C2 is constant and equal to zero. Fundamentals of Power Electronics 35 Chapter 2: Principles of steady-state converter analysis

10 Solve for steady-state inductor currents and capacitor voltages The four equations obtained from volt-sec and charge balance: v L1 = D D' V 1 =0 v L2 = D V 1 V 2 D' V 2 =0 i C1 = DI 2 D'I 1 =0 i C2 = I 2 V 2 =0 Solve for the dc capacitor voltages and inductor currents, and express in terms of the known, D, and : V 1 = D' V 2 = D D' I 1 = D D' I 2 = I 2 = V 2 = D D' D D' 2 Fundamentals of Power Electronics 36 Chapter 2: Principles of steady-state converter analysis

11 Cuk converter conversion ratio M = V/ 0 D M(D) M(D)= V 2 = D 1D -5 Fundamentals of Power Electronics 37 Chapter 2: Principles of steady-state converter analysis

12 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 Fundamentals of Power Electronics 38 Chapter 2: Principles of steady-state converter analysis

13 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 Fundamentals of Power Electronics 39 Chapter 2: Principles of steady-state converter analysis

14 ipple 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'C 1 Q: How large is the output voltage ripple? Fundamentals of Power Electronics 40 Chapter 2: Principles of steady-state converter analysis

15 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 (t) 2 C v 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 Fundamentals of Power Electronics 41 Chapter 2: Principles of steady-state converter analysis

16 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 Fundamentals of Power Electronics 42 Chapter 2: Principles of steady-state converter analysis

17 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) Fundamentals of Power Electronics 43 Chapter 2: Principles of steady-state converter analysis

18 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. Fundamentals of Power Electronics 44 Chapter 2: Principles of steady-state converter analysis

19 Inductor current ripple in two-pole filters Example: problem 2.9 L 1 Q 1 i T i 1 i 2 L 2 C C 1 v C1 2 D 1 v v L (t) Total flux linkage v t T s /2 i L (t) I DT s i D'T s i can use similar arguments, with = L (2 i) = inductor flux linkages = inductor volt-seconds t Fundamentals of Power Electronics 45 Chapter 2: Principles of steady-state converter analysis

20 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 Fundamentals of Power Electronics 1 Chapter 3: Steady-state equivalent circuit modeling,...

21 3.1. The dc transformer model Basic equations of an ideal dc-dc converter: P in = P out I g = V I (η = 100%) 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 Fundamentals of Power Electronics 2 Chapter 3: Steady-state equivalent circuit modeling,...

22 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 input 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 Fundamentals of Power Electronics 3 Chapter 3: Steady-state equivalent circuit modeling,...

23 The DC transformer model Power input I g 1 : M(D) I V Power output Models basic properties of ideal dc-dc converter: conversion of dc voltages and currents, ideally with 100% efficiency D Control input 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 Fundamentals of Power Electronics 4 Chapter 3: Steady-state equivalent circuit modeling,...

24 Example: use of the DC transformer model 1. Original system 3. Push source through transformer V 1 1 Switching dc-dc converter V M(D)V 1 M 2 (D) 1 V 2. Insert dc transformer model D 4. Solve circuit 1 1 : M(D) V = M(D) V 1 M 2 (D) 1 V 1 V Fundamentals of Power Electronics 5 Chapter 3: Steady-state equivalent circuit modeling,...

25 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 L Insert this inductor model into boost converter circuit: L i L 1 C 2 v Fundamentals of Power Electronics 6 Chapter 3: Steady-state equivalent circuit modeling,...

26 Analysis of nonideal boost converter L i L 1 C 2 v switch in position 1 switch in position 2 i L L v L i C i L L v L i C C v C v Fundamentals of Power Electronics 7 Chapter 3: Steady-state equivalent circuit modeling,...

27 Circuit equations, switch in position 1 Inductor current and capacitor voltage: i L L v L i C v L (t)= i(t) L i C (t)=v(t)/ C v Small ripple approximation: v L (t)= I L i C (t)=v / Fundamentals of Power Electronics 8 Chapter 3: Steady-state equivalent circuit modeling,...

28 Circuit equations, switch in position 2 i L L v L i C C v v L (t)= i(t) L v(t) I L V i C (t)=i(t)v(t)/ I V / Fundamentals of Power Electronics 9 Chapter 3: Steady-state equivalent circuit modeling,...

29 Inductor voltage and capacitor current waveforms Average inductor voltage: v L (t) I L T s v L (t) = T 1 v L (t)dt s 0 = D( I L )D'( I L V) DT s D'T s I L V t Inductor volt-second balance: 0= I L D'V i C (t) I V/ Average capacitor current: i C (t) = D (V / )D'(I V / ) Capacitor charge balance: 0=D'I V / V/ t Fundamentals of Power Electronics 10 Chapter 3: Steady-state equivalent circuit modeling,...

30 Solution for output voltage We now have two equations and two unknowns: L / = 0 L / = = I L D'V 3.5 0=D'I V / 3 L / = 0.02 Eliminate I and solve for V: V/ L / = 0.05 V = 1 D' 1 (1 L / D' 2 ) L / = D Fundamentals of Power Electronics 11 Chapter 3: Steady-state equivalent circuit modeling,...

31 3.3. Construction of equivalent circuit model esults of previous section (derived via inductor volt-sec balance and capacitor charge balance): v L i C =0= I L D'V =0=D'I V / View these as loop and node equations of the equivalent circuit. econstruct an equivalent circuit satisfying these equations Fundamentals of Power Electronics 12 Chapter 3: Steady-state equivalent circuit modeling,...

32 Inductor voltage equation v L =0= I L D'V L L 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 I I L I L term: voltage across resistor of value L having current I D V term: for now, leave as dependent source D'V Fundamentals of Power Electronics 13 Chapter 3: Steady-state equivalent circuit modeling,...

33 Capacitor current equation i C =0=D'I V / 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/ term: current through load resistor of value having voltage V D I term: for now, leave as dependent source V V/ Fundamentals of Power Electronics 14 Chapter 3: Steady-state equivalent circuit modeling,...

34 Complete equivalent circuit The two circuits, drawn together: L D'V I D'I V The dependent sources are equivalent to a D : 1 transformer: L D' : 1 I V Dependent sources and transformers I 1 nv 2 ni 1 I 1 n : 1 V 2 V 2 sources have same coefficient reciprocal voltage/current dependence Fundamentals of Power Electronics 15 Chapter 3: Steady-state equivalent circuit modeling,...

35 Solution of equivalent circuit Converter equivalent circuit I L D' : 1 V efer all elements to transformer secondary: L /D' 2 D'I V g /D' V Solution for output voltage using voltage divider formula: V = D' = L D' D' L D' 2 Fundamentals of Power Electronics 16 Chapter 3: Steady-state equivalent circuit modeling,...

36 Solution for input (inductor) current I L D' : 1 V I = D' 2 L = D' L D' 2 Fundamentals of Power Electronics 17 Chapter 3: Steady-state equivalent circuit modeling,...

37 Solution for converter efficiency P in =( )(I) I L D' : 1 P out =(V)(D'I) V η = P out P in = (V)(D'I) ( )(I) = V D' η = 1 1 L D' 2 Fundamentals of Power Electronics 18 Chapter 3: Steady-state equivalent circuit modeling,...

38 Efficiency, for various values of L 100% η = 1 1 L D' 2 90% 80% 70% % 0.05 η 50% L / = % 30% 20% 10% 0% Fundamentals of Power Electronics 19 Chapter 3: Steady-state equivalent circuit modeling,... D

39 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 i L L v L 2 C L v C Average inductor voltage and capacitor current: v L =0=D I L L V C i C =0=I L V C / Fundamentals of Power Electronics 20 Chapter 3: Steady-state equivalent circuit modeling,...

40 Construct equivalent circuit as usual v L =0=D I L L V C i C =0=I L V C / L D v L = 0 I L i C = 0 V C V C / What happened to the transformer? Need another equation Fundamentals of Power Electronics 21 Chapter 3: Steady-state equivalent circuit modeling,...

41 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 Fundamentals of Power Electronics 22 Chapter 3: Steady-state equivalent circuit modeling,...

42 Input port equivalent circuit I g = 1 T s 0 T s i g (t) dt = DI L I DI g L Fundamentals of Power Electronics 23 Chapter 3: Steady-state equivalent circuit modeling,...

43 Complete equivalent circuit, buck converter Input and output port equivalent circuits, drawn together: I g I L L DI L D V C eplace dependent sources with equivalent dc transformer: I g 1 : D I L L V C Fundamentals of Power Electronics 24 Chapter 3: Steady-state equivalent circuit modeling,...

44 3.5. Example: inclusion of semiconductor conduction losses in the boost converter model Boost converter example i L i C C v DT s T s Models of on-state semiconductor devices: MOSFET: on-resistance on Diode: constant forward voltage V D plus on-resistance D Insert these models into subinterval circuits Fundamentals of Power Electronics 25 Chapter 3: Steady-state equivalent circuit modeling,...

45 Boost converter example: circuits during subintervals 1 and 2 i L i C C v DT s T s switch in position 1 switch in position 2 i L L v L i C i L L D v L i C V D on C v C v Fundamentals of Power Electronics 26 Chapter 3: Steady-state equivalent circuit modeling,...

46 Average inductor voltage and capacitor current v L (t) I L I on DT s D'T s I L V D I D V t i C (t) I V/ V/ t v L i C = D( I L I on )D'( I L V D I D V)=0 = D(V/)D'(I V/)=0 Fundamentals of Power Electronics 27 Chapter 3: Steady-state equivalent circuit modeling,...

47 Construction of equivalent circuits I L ID on D'V D ID' D D'V =0 L Don D'V D D' D I L ID on ID' D I D'V D'I V/ =0 V/ D'I V Fundamentals of Power Electronics 28 Chapter 3: Steady-state equivalent circuit modeling,...

48 Complete equivalent circuit L Don D'V D D' D I D'V D'I V L Don D'V D D' D D' : 1 I V Fundamentals of Power Electronics 29 Chapter 3: Steady-state equivalent circuit modeling,...

49 Solution for output voltage L Don D'V D D' D D' : 1 I V V = 1 D' D'V D D' 2 D' 2 L D on D' D V = 1 D' 1 D'V D 1 1 L D on D' D D' 2 Fundamentals of Power Electronics 30 Chapter 3: Steady-state equivalent circuit modeling,...

50 Solution for converter efficiency P in =( )(I) L Don D'V D D' D D' : 1 P out =(V)(D'I) I V η = D' V = 1 D'V D 1 L D on D' D D' 2 Conditions for high efficiency: /D' > V D D' 2 > L D on D' D Fundamentals of Power Electronics 31 Chapter 3: Steady-state equivalent circuit modeling,...

51 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 esult 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 MOSFET rms current Average power loss in on (a) i = 0 (b) i = 0.1 I (c) i = I I D D I 2 on ( ) I D (1.0033) D I 2 on (1.155) I D (1.3333) D I 2 on Fundamentals of Power Electronics 32 Chapter 3: Steady-state equivalent circuit modeling,...

52 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. Fundamentals of Power Electronics 33 Chapter 3: Steady-state equivalent circuit modeling,...