Chapter 11 AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode

Transcription

1 Chapter 11 AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode Introduction DCM Averaged Switch Model Small-Signal AC Modeling of the DCM Switch Network High-Frequency Dynamics of Converters in DCM Summary of Key Points Fundamentals of Power Electronics 1

2 We are missing ac and dc equivalent circuit models for the discontinuous conduction mode DC AC 1 : M(D) e(s) d(s) 1 : M(D) L e CCM V g V (s) j(s) d(s) C v(s) DCM V g? V (s)? v(s) Fundamentals of Power Electronics 2

3 Change in characteristics at the CCM/DCM boundary l l l l Steady-state output voltage becomes strongly load-dependent Simpler dynamics: one pole and the HP 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 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 Fundamentals of Power Electronics 3

4 11.1 Derivation of DCM averaged switch model: buck-boost example Define switch terminal quantities v 1, i 1, v 2, i 2, as shown Let us find the averaged quantities v 1, i 1, v 2, i 2, for operation in DCM, and determine the relations between them Switch network i 1 i 2 v 1 v L L i L v 2 C v Fundamentals of Power Electronics 4

5 DCM waveforms i L i 1 L i pk v L 0 Area q 1 i pk i 1 Ts v L t v 1 v 0 v 1 Ts 0 v i 2 i pk Area q 2 Switch network i 1 i 2 v 1 v L L i L v 2 C v v 2 T s i 2 Ts v v 2 Ts v 0 d 1 T s d 2 T s d 3 T s t Fundamentals of Power Electronics 5

6 Basic DCM equations: i pk, v L, and d 2 (Approximate method) Peak inductor current: i pk = L d 1T s Average inductor voltage: v L Ts = d 1 Ts d 2 v Ts d 3 0 i 1 v 1 Area q 1 v 1 Ts i pk i 1 Ts v In DCM, the diode switches off when the inductor current reaches zero. Hence, i(0) = i(t s ) = 0, and the average inductor voltage is zero. This is true even during transients. v L Ts = d 1 Ts d 2 v Ts = 0 Solve for d 2 : d 2 =d 1 Ts v Ts Fundamentals of Power Electronics 6 i 2 v 2 0 i 2 Ts v i pk Area q 2 v 2 Ts 0 t d 1 T s d 2 T s d 3 T s T s v

7 Average switch network terminal voltages Average the v 1 waveform: i 1 Area q 1 i pk v 1 Ts = d 1 0d 2 Ts v Ts d 3 Ts i 1 Ts Eliminate d 2 and d 3 : v 1 Ts v 1 Ts = Ts v 1 v Similar analysis for v 2 waveform leads to i 2 0 v 2 Ts = d 1 Ts v Ts d 2 0d 3 v Ts i pk Area q 2 = v Ts i 2 Ts v 2 v v 2 Ts v T s 0 d 1 T s d 2 T s d 3 T s t Fundamentals of Power Electronics 7

8 Average switch network terminal currents Average the i 1 waveform: i 1 Ts = 1 T s t t T s i 1 dt = q 1 T s i 1 Area q 1 i pk i 1 Ts The integral q 1 is the area under the i 1 waveform during first subinterval. Use triangle area formula: v 1 v 1 Ts v q 1 = t t T s i 1 dt = 1 2 d 1T s i pk i 2 0 Eliminate i pk : i pk Area q 2 i 1 Ts = d 2 1 T s 2L v 1 Ts Note i 1 Ts is not equal to d i L Ts! Similar analysis for i 2 waveform leads to i 2 Ts = d T v 1 s Ts 2L v 2 Ts Fundamentals of Power Electronics 8 v 2 i 2 Ts v v 2 Ts t d 1 T s d 2 T s d 3 T s T s 0 v

9 Input port: Averaged equivalent circuit i 1 Ts = d 2 1 T s 2L v 1 Ts i 1 Ts i 1 Ts = v 1 Ts e (d 1 ) v 1 Ts e (d 1 ) e (d 1 )= 2L d 1 2 T s Fundamentals of Power Electronics 9

10 Output port: Averaged equivalent circuit i 2 Ts = d 2 1 T s 2L 2 v 1 Ts v 2 Ts p i v 2 i 2 Ts v 2 Ts = v 1 Ts e (d 1 ) = p Ts Fundamentals of Power Electronics 10

11 The dependent power source i i vi = p p v v Must avoid open- and short-circuit connections of power sources Power sink: negative p Fundamentals of Power Electronics 11

12 How the power source arises in lossless two-port networks In a lossless two-port network without internal energy storage: instantaneous input power is equal to instantaneous output power In all but a small number of special cases, the instantaneous power throughput is dependent on the applied external source and load 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 A power source (or power sink) element is obtained Fundamentals of Power Electronics 12

13 Properties of power sources Series and parallel connection of power sources P 1 P 1 P 2 P 3 P 2 P 3 eflection of power source through a transformer P 1 n 1 : n 2 P 1 Fundamentals of Power Electronics 13

14 The loss-free resistor (LF) i 1 Ts p Ts i 2 Ts v 1 Ts e (d 1 ) v 2 Ts A two-port lossless network Input port obeys Ohm s Law Power entering input port is transferred to output port Fundamentals of Power Electronics 14

15 Averaged modeling of CCM and DCM switch networks Switch network Averaged switch model i 1 i 2 i 1 Ts 1 : d i 2 Ts CCM v 1 v 2 v 1 Ts v 2 Ts i 1 i 2 i 1 Ts p Ts i 2 Ts DCM v 1 v 2 v 1 Ts e (d 1 ) v 2 Ts Fundamentals of Power Electronics 15

16 Averaged switch model: buck-boost example Original circuit Switch network i 1 i 2 v 1 v 2 v L L i L C v Averaged model i 1 Ts v 1 Ts e (d) p Ts i 2 Ts v 2 Ts Ts C v Ts L Fundamentals of Power Electronics 16

17 Solution of averaged model: steady state Let L short circuit I 1 P C open circuit V g e (D) V Converter input power: P = V 2 g e Converter output power: Equate and solve: P = V 2 g = V 2 e P = V 2 V V g = ± e Fundamentals of Power Electronics 17

18 Steady-state LF solution V = ± is a general result, for any system that can V g e be modeled as an LF. For the buck-boost converter, we have e (D)= 2L D 2 T s Eliminate e : V V g = D 2 T s 2L = D K which agrees with the previous steady-state solution of Chapter 5. Fundamentals of Power Electronics 18

19 Steady-state LF solution with ac terminal waveforms i 1 i 2 p C e v Converter average input power: P av = V 2 g,rms e Converter average output power: P av = V 2 rms Fundamentals of Power Electronics 19 Note that no average power flows into capacitor Equate and solve: V rms V g,rms = e

20 Averaged models of other DCM converters Determine averaged terminal waveforms of switch network In each case, averaged transistor waveforms obey Ohm s law, while averaged diode waveforms behave as dependent power source Can simply replace transistor and diode with the averaged model as follows: i 1 i 2 i 1 Ts p Ts i 2 Ts v 1 v 2 v 1 Ts e (d 1 ) v 2 Ts Fundamentals of Power Electronics 20

21 DCM buck, boost Buck e (d) L e = 2L d 2 T s Ts p Ts C v Ts Boost L Ts e (d) C v Ts p Ts Fundamentals of Power Electronics 21

22 DCM Cuk, SEPIC Cuk L 1 C 1 L 2 e = 2 L 1 L 2 d 2 T s Ts e (d) p Ts C 2 v Ts SEPIC L 1 C 1 Ts e (d) L 2 p Ts C 2 v Ts Fundamentals of Power Electronics 22

23 Steady-state solution: DCM buck, boost Let L short circuit C open circuit Buck V g e (D) P V Boost V g e (D) P V Fundamentals of Power Electronics 23

24 Steady-state solution of DCM/LF models Table CCM and DCM conversion ratios of basic converters Converter M, CCM M, DCM Buck D e / Boost 1 1D 1 14/ e 2 Buck-boost, Cuk SEPIC D 1D D 1D e e I>I crit I<I crit forccm fordcm I crit = 1D D V g e (D) Fundamentals of Power Electronics 24

25 11.2 Small-signal ac modeling of the DCM switch network Large-signal averaged model Perturb and linearize: let i 1 Ts p Ts i 2 Ts d=d d v 1 Ts = V 1 v 1 v 1 Ts e (d) d v 2 Ts i 1 Ts = I 1 i 1 v 2 Ts = V 2 v 2 i 2 Ts = I 2 i 2 i 1 Ts = d 2 1 T s 2L i 2 Ts = d 2 1 T s 2L v 1 Ts 2 v 1 Ts v 2 Ts Fundamentals of Power Electronics 25 i 1 = v 1 r 1 j 1 d g 1 v 2 i 2 = v 2 r j 2 d g 2 v 1 2

26 Linearization via Taylor series Given the nonlinear equation i 1 Ts = v 1 Ts e (d) = f 1 v 1 Ts, v 2 Ts, d Expand in three-dimensional Taylor series about the quiescent operating point: I 1 i 1 = f 1 V 1, V 2, D v 1 df 1 v 1, V 2, D dv 1 v 1 = V 1 (for simple notation, drop angle brackets) v 2 df 1 V 1, v 2, D dv 2 d df 1 V 1, V 2, d dd v 2 = V 2 d = D higherorder nonlinear terms Fundamentals of Power Electronics 26

27 Equate dc and first-order ac terms AC i 1 =v 1 1 r 1 v 2 g 1 d j 1 1 r = df 1 v 1, V 2, D = 1 1 dv 1 v 1 = V e (D) 1 DC I 1 = f 1 V 1, V 2, D = V 1 e (D) g 1 = df 1 V 1, v 2, D dv 2 v 2 = V 2 =0 j 1 = df 1 V 1, V 2, d dd = 2V 1 D e (D) d = D = V 1 e 2 (D) Fundamentals of Power Electronics 27 d e (d) dd d = D

28 Output port same approach i 2 Ts = 2 v 1 Ts = f 2 e (d) v 2 Ts v 1 Ts, v 2 Ts, d I 2 = f 2 V 1, V 2, D = V 1 2 e (D) V 2 DC terms i 2 =v 2 1 r 2 v 1 g 2 dj 2 Small-signal ac linearization Fundamentals of Power Electronics 28

29 Output resistance parameter r 2 1 r = df 2 V 1, v 2, D = 2 dv 2 1 = 1 v 2 = V M 2 e (D) 2 i 2 Ts Power source characteristic Quiescent operating point 1 Load characteristic Linearized model 1 r 2 v 2 Ts Fundamentals of Power Electronics 29

30 Small-signal DCM switch model parameters i 1 i 2 v 1 r 1 j 1 d g 1 v 2 g 2 v 1 j 2 d r 2 v 2 Table Small-signal DCM switch model parameters Switch type g 1 j 1 r 1 g 2 j 2 r 2 Buck, Fig (a) 1 2(1M)V 1 e 2M 2(1M)V 1 e M D e DM e e M 2 e Boost, Fig (b) 1 (M 1) 2 e 2MV 1 D(M 1) e 2 (M1) 2M1 M e (M 1) 2 2V 1 e D(M 1) e (M1) 2 e Gen two-switch Fig. 11.7(a) 0 2V 1 D e e 2M e 2V 1 DM e M 2 e Fundamentals of Power Electronics 30

31 Small-signal ac model, DCM buck-boost example i 1 Switch network small-signal ac model i 2 v 1 r 1 j 1 d g 1 v 2 g 2 v 1 j 2 d r 2 v 2 C v L i L Fundamentals of Power Electronics 31

32 A more convenient way to model the buck and boost small-signal DCM switch networks i 1 i 2 i 1 i 2 v 1 v 2 v 1 v 2 In any event, a small-signal two-port model is used, of the form i 1 i 2 v 1 r 1 j 1 d g 1 v 2 g 2 v 1 j 2 d r 2 v 2 Fundamentals of Power Electronics 32

33 Small-signal ac models of the DCM buck and boost converters (more convenient forms) i 1 DCM buck switch network small-signal ac model i 2 L i L v 1 r 1 j 1 d g 1 v 2 g 2 v 1 j 2 d r 2 v 2 C v i L L i 1 DCM boost switch network small-signal ac model i 2 v 1 r 1 j 1 d g 1 v 2 g 2 v 1 j 2 d r 2 v 2 C v Fundamentals of Power Electronics 33

34 DCM small-signal transfer functions l When expressed in terms of, L, C, and M (not D), the smallsignal transfer functions are the same in DCM as in CCM l Hence, DCM boost and buck-boost converters exhibit two poles and one HP zero in control-to-output transfer functions l But, value of L is small in DCM. Hence HP 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 l A simple approximation: let L 0 Fundamentals of Power Electronics 34

35 The simple approximation L 0 Buck, boost, and buck-boost converter models all reduce to DCM switch network small-signal ac model r 1 j 1 d g 1 v 2 g 2 v 1 j 2 d r 2 C v Transfer functions control-to-output line-to-output G vd (s)= v d vg =0= G d0 1 ω s with G d0 = j 2 r 2 p ω p = 1 r 2 C G vg (s)= v = G g0 d =0 1 ω s G g0 = g 2 r 2 = M p Fundamentals of Power Electronics 35

36 Transfer function salient features Table Salient features of DCM converter small-signal transfer functions Converter G d0 G g0 ω p Buck 2V D 1M 2M M 2M (1M)C Boost 2V D M1 2M 1 M 2M 1 (M1)C Buck-boost V D M 2 C Fundamentals of Power Electronics 36

37 DCM boost example = 12 Ω L = 5 µh C = 470 µf f s = 100 khz The output voltage is regulated to be V = 36 V. It is desired to determine G vd (s) at the operating point where the load current is I = 3 A and the dc input voltage is V g = 24 V. Fundamentals of Power Electronics 37

38 Evaluate simple model parameters P = IV V g = 3A 36 V 24 V =36W e = V 2 g P D = = (24 V)2 36 W =16Ω 2L e T s = G d0 = 2V D M 1 2M 1 f p = ω p 2π = 2M 1 2(5 µh) (16 Ω)(10 µs) = 2(36 V) (0.25) Fundamentals of Power Electronics 38 2 (36 V) (24 V) 1 2π (M1)C = 2 2π = 0.25 =72V 37 dbv (36 V) (24 V) 1 (36 V) (24 V) 1 = 112 Hz (36 V) (24 V) 1 (12 Ω)(470 µf)

39 Control-to-output transfer function, boost example 60 dbv G vd G vd 40 dbv G d0 37 dbv 20 dbv 0 dbv 20 dbv 40 dbv G vd 0 G vd f p 112 Hz 20 db/decade Hz 100 Hz 1 khz 10 khz 100 khz Fundamentals of Power Electronics 39 f

40 11.4 Summary of Key Points 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 e (d). The average diode voltage and current obey a power source characteristic, with power equal to the power effectively dissipated by 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. Fundamentals of Power Electronics 40

41 Key points 4. 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 (a) and 10.16(b). These models are also y-parameter twoport models, but have different parameter values. Fundamentals of Power Electronics 41

42 Key points 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. Fundamentals of Power Electronics 42

43 Key points 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 43