Chapter 11 Current Programmed Control


 Joleen Shields
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1 Chapter 11 Current Programmed Control Buck converter v g i s Q 1 D 1 L i L C v R The peak transistor current replaces the duty cycle as the converter control input. Measure switch current R f i s Clock 0 T s Control signal i c i s R f i c R f Analog comparator S Q R Latch m 1 Switch current i s Control input Currentprogrammed controller Transistor status: 0 dt s T s on off t Compensator v Clock turns transistor on Comparator turns transistor off v ref Conventional output voltage controller Fundamentals of Power Electronics 1 Chapter 11: Current Programmed Control
2 Current programmed control vs. duty cycle control Advantages of current programmed control: Simpler dynamics inductor pole is moved to high frequency Simple robust output voltage control, with large phase margin, can be obtained without use of compensator lead networks It is always necessary to sense the transistor current, to protect against overcurrent failures. We may as well use the information during normal operation, to obtain better control Transistor failures due to excessive current can be prevented simply by limiting i c Transformer saturation problems in bridge or pushpull converters can be mitigated A disadvantage: susceptibility to noise Fundamentals of Power Electronics 2 Chapter 11: Current Programmed Control
3 Chapter 11: Outline 11.1 Oscillation for D > A simple firstorder model Simple model via algebraic approach Averaged switch modeling 11.3 A more accurate model Current programmed controller model: block diagram CPM buck converter example 11.4 Discontinuous conduction mode 11.5 Summary Fundamentals of Power Electronics 3 Chapter 11: Current Programmed Control
4 11.1 Oscillation for D > 0.5 The current programmed controller is inherently unstable for D > 0.5, regardless of the converter topology Controller can be stabilized by addition of an artificial ramp Objectives of this section: Stability analysis Describe artificial ramp scheme Fundamentals of Power Electronics 4 Chapter 11: Current Programmed Control
5 Inductor current waveform, CCM Inductor current slopes m 1 and m 2 i L i c i L (0) m1 i L (T s ) m 2 buck converter m 1 = v g v m L 2 = L v boost converter m 1 = v g m L 2 = v g v L buckboost converter 0 dt s T s t m 1 = v g L m 2 = v L Fundamentals of Power Electronics 5 Chapter 11: Current Programmed Control
6 Steadystate inductor current waveform, CPM First interval: i L i L (dt s )=i c =i L (0) m 1 dt s Solve for d: d = i c i L (0) m 1 T s Second interval: i L (T s )=i L (dt s )m 2 d't s =i L (0) m 1 dt s m 2 d't s In steady state: 0=M 1 DT s M 2 D'T s M 2 M 1 = D D' i c i L (0) m1 i L (T s ) m 2 0 dt s T s t Fundamentals of Power Electronics 6 Chapter 11: Current Programmed Control
7 Perturbed inductor current waveform i L i c I L0 i L (0) m 1 i L (0) m 2 m 1 I L0 dt s i L (T s ) m 2 Steadystate waveform Perturbed waveform 0 D d T s DT s T s t Fundamentals of Power Electronics 7 Chapter 11: Current Programmed Control
8 Change in inductor current perturbation over one switching period magnified view i c m 1 i L (0) i L (T s ) m 2 Steadystate waveform m 1 dt s m 2 Perturbed waveform i L (0) = m 1 dt s i L (T s )=m 2 dt s i L (T s )=i L (0) D D' i L (T s )=i L (0) m 2 m 1 Fundamentals of Power Electronics 8 Chapter 11: Current Programmed Control
9 Change in inductor current perturbation over many switching periods i L (T s )=i L (0) D D' i L (2T s )=i L (T s ) D D' = i L (0) D D' 2 i L (nt s )=i L ((n 1)T s ) D D' = i L (0) D D' n i L (nt s ) 0 when D D' when D D' <1 >1 For stability: D< 0.5 Fundamentals of Power Electronics 9 Chapter 11: Current Programmed Control
10 Example: unstable operation for D = 0.6 α = D D' = = 1.5 i L i c I L0 i L (0) 1.5i L (0) 2.25i L (0) 3.375i L (0) 0 T s 2T s 3T s 4T s t Fundamentals of Power Electronics 10 Chapter 11: Current Programmed Control
11 Example: stable operation for D = 1/3 i L α = D D' = 1/3 2/3 = 0.5 i c I L0 i L (0) 1 2 i L(0) 1 4 i L(0) 1 8 i L(0) 1 16 i L(0) 0 T s 2T s 3T s 4T s t Fundamentals of Power Electronics 11 Chapter 11: Current Programmed Control
12 Stabilization via addition of an artificial ramp to the measured switch current waveform Buck converter i s L i L i a Q 1 v g D 1 C v R m a Measure switch current i s i s R f R f m a i a R f Artificial ramp Clock 0 T s S Q 0 T s 2T s Now, transistor switches off when i a (dt s )i L (dt s )=i c t i c R f Analog comparator R Latch or, i L (dt s )=i c i a (dt s ) Control input Currentprogrammed controller Fundamentals of Power Electronics 12 Chapter 11: Current Programmed Control
13 Steady state waveforms with artificial ramp i L (dt s )=i c i a (dt s ) (i c i a ) i c i L m a m 1 I L0 m 2 0 dt s T s t Fundamentals of Power Electronics 13 Chapter 11: Current Programmed Control
14 Stability analysis: perturbed waveform (i c i a ) i c I L0 i L (0) m 1 i L (0) i L(T s ) m a m 2 m 1 m 2 I L0 dt s Steadystate waveform Perturbed waveform 0 D d T s DT s T s t Fundamentals of Power Electronics 14 Chapter 11: Current Programmed Control
15 Stability analysis: change in perturbation over complete switching periods First subinterval: i L (0) = dt s m 1 m a Second subinterval: i L (T s )=dt s m a m 2 Net change over one switching period: i L (T s )=i L (0) m 2 m a m 1 m a After n switching periods: i L (nt s )=i L ((n 1)T s ) m 2m a m 1 m a Characteristic value: α = m 2m a m 1 m a i L (nt s ) = i L (0) m 2 m a m 1 m a 0 when α <1 when α >1 n = il (0) α n Fundamentals of Power Electronics 15 Chapter 11: Current Programmed Control
16 The characteristic value α For stability, require α < 1 Buck and buckboost converters: m 2 = v/l So if v is wellregulated, then m 2 is also wellregulated A common choice: m a = 0.5 m 2 This leads to α = 1 at D = 1, and α < 1 for 0 D < 1. The minimum α that leads to stability for all D. Another common choice: m a = m 2 This leads to α = 0 for 0 D < 1. Deadbeat control, finite settling time α = 1m a m 2 D' D m a m 2 Fundamentals of Power Electronics 16 Chapter 11: Current Programmed Control
17 11.2 A Simple FirstOrder Model Switching converter v g R v d Current programmed controller Converter voltages and currents i c Compensator v v ref Fundamentals of Power Electronics 17 Chapter 11: Current Programmed Control
18 The firstorder approximation i L Ts = i c Neglects switching ripple and artificial ramp Yields physical insight and simple firstorder model Accurate when converter operates well into CCM (so that switching ripple is small) and when the magnitude of the artificial ramp is not too large Resulting smallsignal relation: i L (s) i c (s) Fundamentals of Power Electronics 18 Chapter 11: Current Programmed Control
19 Simple model via algebraic approach: CCM buckboost example Q 1 D 1 v g i L L C R v i L i c v g L v L 0 dt s T s t Fundamentals of Power Electronics 19 Chapter 11: Current Programmed Control
20 Smallsignal equations of CCM buckboost, duty cycle control L di L dt C dv dt = Dv g D'v V g V d =D'i L v R I Ld i g =Di L I L d Derived in Chapter 7 Fundamentals of Power Electronics 20 Chapter 11: Current Programmed Control
21 Transformed equations Take Laplace transform, letting initial conditions be zero: sli L (s)=dv g (s)d'v(s) V g V d(s) scv(s)=d'i L (s) v(s) R I Ld(s) i g (s)=di L (s)i L d(s) Fundamentals of Power Electronics 21 Chapter 11: Current Programmed Control
22 The simple approximation Now let i L (s) i c (s) Eliminate the duty cycle (now an intermediate variable), to express the equations using the new control input i L. The inductor equation becomes: sli c (s) Dv g (s)d'v(s) V g V d(s) Solve for the duty cycle variations: d(s)= sli c(s)dv g (s)d'v(s) V g V Fundamentals of Power Electronics 22 Chapter 11: Current Programmed Control
23 The simple approximation, continued Substitute this expression to eliminate the duty cycle from the remaining equations: scv(s)=d'i c (s) v(s) R I L sli c (s)dv g (s)d'v(s) V g V i g (s)=di c (s)i L sli c (s)dv g (s)d'v(s) V g V Collect terms, simplify using steadystate relations: scv(s)= sld D'R D' i c(s) D R 1 R v(s) i g (s)= sld D'R D i c(s) D R v(s) D2 D'R v g(s) D2 D'R v g(s) Fundamentals of Power Electronics 23 Chapter 11: Current Programmed Control
24 Construct equivalent circuit: input port i g (s)= sld D'R D i c(s) D R v(s) D2 D'R v g(s) i g v g D'R D 2 D 2 D'R v g D 1 sl D'R i c D R v Fundamentals of Power Electronics 24 Chapter 11: Current Programmed Control
25 Construct equivalent circuit: output port scv(s)= sld D'R D' i c(s) D R 1 R v(s) D2 D'R v g(s) Node D 2 D'R v g D' 1 sld D' 2 R i c R D D R v scv C v R R Fundamentals of Power Electronics 25 Chapter 11: Current Programmed Control
26 CPM Canonical Model, Simple Approximation i g v g r 1 f 1 (s) i c g 1 v g 2 v g f 2 (s) i c r 2 C R v Fundamentals of Power Electronics 26 Chapter 11: Current Programmed Control
27 Table of results for basic converters Table 11.1 Current programmed mode smallsignal equivalent circuit parameters, simple model Converter g 1 f 1 r 1 g 2 f 2 r 2 Buck D R D 1 sl R R D Boost D'R D' 1 sl D' 2 R R Buckboost D R D 1 sl D'R D'R D 2 D2 D'R D' 1 sdl D' 2 R R D Fundamentals of Power Electronics 27 Chapter 11: Current Programmed Control
28 Transfer functions predicted by simple model i g v g r 1 f 1 (s) i c g 1 v g 2 v g f 2 (s) i c r 2 C R v Controltooutput transfer function G vc (s)= v(s) i c (s) vg =0 = f 2 r 2 R 1 sc Result for buckboost example G vc (s)=r D' 1D 1s DL D' 2 R 1s RC 1D Fundamentals of Power Electronics 28 Chapter 11: Current Programmed Control
29 Transfer functions predicted by simple model i g v g r 1 f 1 (s) i c g 1 v g 2 v g f 2 (s) i c r 2 C R v Linetooutput transfer function G vg (s)= v(s) v g (s) ic =0 =g 2 r 2 R 1 sc Result for buckboost example G vg (s)= D 2 1 1D 2 1s RC 1D Fundamentals of Power Electronics 29 Chapter 11: Current Programmed Control
30 Transfer functions predicted by simple model i g v g r 1 f 1 (s) i c g 1 v g 2 v g f 2 (s) i c r 2 C R v Output impedance Z out (s)=r 2 R 1 sc Result for buckboost example Z out (s)= R 1D 1 1s RC 1D Fundamentals of Power Electronics 30 Chapter 11: Current Programmed Control
31 Averaged switch modeling with the simple approximation i 1 i 2 L i L v g v 1 v 2 C R v Switch network Averaged terminal waveforms, CCM: v 2 Ts = d v 1 Ts The simple approximation: i 2 Ts i c Ts i 1 Ts = d i 2 Ts Fundamentals of Power Electronics 31 Chapter 11: Current Programmed Control
32 CPM averaged switch equations v 2 Ts = d v 1 Ts i 2 Ts i c Ts i 1 Ts = d i 2 Ts Eliminate duty cycle: i 1 Ts = d i c Ts = v 2 Ts v 1 Ts i c Ts i 1 Ts v 1 Ts = i c Ts v 2 Ts = p Ts So: Output port is a current source Input port is a dependent current sink Fundamentals of Power Electronics 32 Chapter 11: Current Programmed Control
33 CPM averaged switch model i 1 Ts i 2 Ts L i L Ts p Ts v g Ts v 1 Ts i c Ts v 2 Ts C R v Ts Averaged switch network Fundamentals of Power Electronics 33 Chapter 11: Current Programmed Control
34 Results for other converters Boost i L Ts L v g Ts i c Ts p Ts C R v Ts Averaged switch network Averaged switch network Buckboost p Ts i c Ts v g Ts C R v Ts L i L Ts Fundamentals of Power Electronics 34 Chapter 11: Current Programmed Control
35 Perturbation and linearization to construct smallsignal model Let v 1 Ts = V 1 v 1 i 1 Ts = I 1 i 1 v 2 Ts = V 2 v 2 i 2 Ts = I 2 i 2 i c Ts = I c i c Resulting input port equation: V 1 v 1 I 1 i 1 = I c i c V 2 v 2 Smallsignal result: i 1 =i c V 2 V 1 v 2 I c V 1 v 1 I 1 V 1 Output port equation: î 2 = î c Fundamentals of Power Electronics 35 Chapter 11: Current Programmed Control
36 Resulting smallsignal model Buck example i 1 i 2 L v v V g i 2 C R V I 1 1 c v c 2 i v V c 2 v 1 V 1 I 1 Switch network smallsignal ac model i 1 =i c V 2 V 1 v 2 I c V 1 v 1 I 1 V 1 Fundamentals of Power Electronics 36 Chapter 11: Current Programmed Control
37 Origin of input port negative incremental resistance i 1 Ts Power source characteristic v 1 Ts i 1 Ts = p Ts Quiescent operating point I 1 1 r 1 = I 1 V 1 V 1 v 1 Ts Fundamentals of Power Electronics 37 Chapter 11: Current Programmed Control
38 Expressing the equivalent circuit in terms of the converter input and output voltages i g L i L v g i c D 1 sl R D2 R D v R i c C R v i 1 (s)=d 1s L R i c(s) D R v(s)d2 R v g(s) Fundamentals of Power Electronics 38 Chapter 11: Current Programmed Control
39 Predicted transfer functions of the CPM buck converter i g L i L v g C R i D2 D c D 1 sl v i c v R R R G vc (s)= v(s) i c (s) vg =0 G vg (s)= v(s) v g (s) ic =0 = R 1 sc =0 Fundamentals of Power Electronics 39 Chapter 11: Current Programmed Control
40 11.3 A More Accurate Model The simple models of the previous section yield insight into the lowfrequency behavior of CPM converters Unfortunately, they do not always predict everything that we need to know: Linetooutput transfer function of the buck converter Dynamics at frequencies approaching f s More accurate model accounts for nonideal operation of current mode controller builtin feedback loop Converter dutycyclecontrolled model, plus block diagram that accurately models equations of current mode controller Fundamentals of Power Electronics 40 Chapter 11: Current Programmed Control
41 11.3.1Current programmed controller model m 1 dt s 2 m a dt s m 2 d't s 2 i c m a (i c i a ) m 1 i L dts i L d'ts m 2 i L i L Ts = d i L dts d' i L d'ts 0 dt s T s t i L Ts = i c Ts m a dt s d m 1dT s 2 = i c Ts m a dt s m 1 d 2 T s 2 m 2 d' m 2d'T s 2 d' 2 T s 2 Fundamentals of Power Electronics 41 Chapter 11: Current Programmed Control
42 Perturb Let i L Ts = I L i L i c Ts = I c i c d=dd m 1 =M 1 m 1 m 2 =M 2 m 2 It is assumed that the artificial ramp slope does not vary. Note that it is necessary to perturb the slopes, since these depend on the applied inductor voltage. For basic converters, buck converter m 1 = v g v L boost converter m 2 = v L m 1 = v g m L 2 = v v g L buckboost converter m 1 = v g L m 2 = v L Fundamentals of Power Electronics 42 Chapter 11: Current Programmed Control
43 Linearize I L i L = I c i c M a T s D d M 1 m 1 D d 2 T s 2 The firstorder ac terms are M 2 m 2 D' d 2 T s 2 i L =i c M a T s DM 1 T s D'M 2 T s d D2 T s 2 m 1 D'2 T s 2 m 2 Simplify using dc relationships: i L =i c M a T s d D2 T s 2 Solve for duty cycle variations: m 1 D'2 T s 2 m 2 d= 1 M a T s i c i L D2 T s 2 m 1 D'2 T s 2 m 2 Fundamentals of Power Electronics 43 Chapter 11: Current Programmed Control
44 Equation of the current programmed controller The expression for the duty cycle is of the general form d=f m i c i L F g v g F v v F m = 1/M a T s Table Current programmed controller gains for basic converters Converter F g F v Buck D 2 T s 2L 12D T s 2L Boost 2D 1 T s 2L D' 2 T s 2L Buckboost D 2 T s 2L D' 2 T s 2L Fundamentals of Power Electronics 44 Chapter 11: Current Programmed Control
45 Block diagram of the current programmed controller d=f m i c i L F g v g F v v v g F m F g F v v d i L Describes the duty cycle chosen by the CPM controller Append this block diagram to the duty cycle controlled converter model, to obtain a complete CPM system model i c Fundamentals of Power Electronics 45 Chapter 11: Current Programmed Control
46 CPM buck converter model 1 : D V g d L i L v g Id C v R d F m T v v g F g F v v T i i L i c Fundamentals of Power Electronics 46 Chapter 11: Current Programmed Control
47 CPM boost converter model Vd L D' : 1 i L v g Id C v R d F m T v v g F g F v v T i i L i c Fundamentals of Power Electronics 47 Chapter 11: Current Programmed Control
48 CPM buckboost converter model V L g V d 1 : D D' : 1 i L v g Id Id C v R d F m T v v g F g F v v T i i L i c Fundamentals of Power Electronics 48 Chapter 11: Current Programmed Control
49 Z o (s){ Example: analysis of CPM buck converter 1 : D V D d L i L v g Id Z i (s) C v R d Z o = R 1 sc Z i = sl R 1 sc v g F m F g F v v T v i L (s)= D Z i (s) v g (s) V D 2 d(s) i L T i v(s)=i L (s)z o (s) i c Fundamentals of Power Electronics 49 Chapter 11: Current Programmed Control
50 Block diagram of entire CPM buck converter v g D i L Z i (s) V D 2 F m d F g F v v i c i L T v Z o (s) T i Voltage loop gain: T v (s)=f m V D 2 Current loop gain: T i (s)= 1 Z o (s)f v T i (s)= F m 1F m V D 2 D Zi (s) Z o(s)f v T v (s) 1T v (s) V D 2 D Zi (s) D Zi (s) Z o(s)f v Transfer function from î c to î L : i L (s) i c (s) = T i(s) 1T i (s) Fundamentals of Power Electronics 50 Chapter 11: Current Programmed Control
51 Discussion: Transfer function from î c to î L When the loop gain Ti(s) is large in magnitude, then î L is approximately equal to î c. The results then coincide with the simple approximation. i L (s) i c (s) = T i(s) 1T i (s) The controltooutput transfer function can be written as G vc (s)= v(s) i c (s) =Z o(s) T i (s) 1T i (s) which is the simple approximation result, multiplied by the factor T i /(1 T i ). Fundamentals of Power Electronics 51 Chapter 11: Current Programmed Control
52 Loop gain T i (s) Result for the buck converter: T i (s)= K 1α 1α 1sRC 1s L R 2DM a M 2 1α 1α s2 LC 2DM a M 2 1α 1α Characteristic value K = 2L/RT s CCM for K > D α = 1m a m 2 D' D m a m 2 Controller stable for α < 1 Fundamentals of Power Electronics 52 Chapter 11: Current Programmed Control
53 Loop gain 1 s ω z Q T i (s)=t i0 1 s Qω 0 s ω 0 2 0dB T i T i 1T i T i0 f 0 f z f f c T i0 =K 1α 1α ω z = RC 1 ω 0 = 1 LC 2DM a 1α M 2 1α Q=R C M 2 1α L 2DM a 1α Fundamentals of Power Electronics 53 Chapter 11: Current Programmed Control
54 Crossover frequency f c High frequency asymptote of T i is T i (s) T i0 s ωz s ω 0 2 = T i0 ω 0 2 sω z Equate magnitude to one at crossover frequency: 2 f T i (j2π f c ) T 0 i0 =1 2π f c f z Solve for crossover frequency: f c = M 2 M a f s 2πD Fundamentals of Power Electronics 54 Chapter 11: Current Programmed Control
55 Construction of transfer functions i L (s) i c (s) = T i(s) 1T i (s) 1 1 s ω c G vc (s)=z o (s) T i (s) 1T i (s) R 1 1sRC 1 s ω c the result of the simple firstorder model, with an added pole at the loop crossover frequency Fundamentals of Power Electronics 55 Chapter 11: Current Programmed Control
56 Exact analysis G vc (s)=r T i0 1 1T i0 T 1s i0 1 1T ωz 1 1 s i0 1T i0 Qω 0 1T ω0 i0 2 for T i0 >> 1, G vc (s) R 1 1 s ω z s2 T i0 ω 0 2 Low Q approximation: G vc (s) R 1 1 s ω z 1 sω z T i0 ω 0 2 which agrees with the previous slide Fundamentals of Power Electronics 56 Chapter 11: Current Programmed Control
57 Linetooutput transfer function Solution of complete block diagram leads to v(s)=i c (s)z o (s) T i (s) 1T i (s) v g(s) G g0 (s) 1T i (s) with G g0 (s)=z o (s) D Zi (s) 1 V D 2 F mf g 1 V D 2 F mf v Z o (s) D Zi (s) Fundamentals of Power Electronics 57 Chapter 11: Current Programmed Control
58 Linetooutput transfer function G vg (s)= v(s) v g (s) ic (s)=0 = G g0(s) 1T i (s) G vg (s)= 1s 1 (1 α) M a 1T i0 (1 α) 2D2 1 M 2 2 T i0 1T i0 1 ωz 1 1T i0 Qω 0 1 1T i0 s ω0 2 Poles are identical to controltooutput transfer function, and can be factored the same way: G vg (s) G vg (0) 1 s ω z 1 s ω c Fundamentals of Power Electronics 58 Chapter 11: Current Programmed Control
59 Linetooutput transfer function: dc gain G vg (0) 2D2 K M a M for large T i0 For any T i0, the dc gain goes to zero when M a /M 2 = 0.5 Effective feedforward of input voltage variations in CPM controller then effectively cancels out the v g variations in the direct forward path of the converter. Fundamentals of Power Electronics 59 Chapter 11: Current Programmed Control
60 11.4 Discontinuous conduction mode Again, use averaged switch modeling approach Result: simply replace Transistor by power sink Diode by power source Inductor dynamics appear at high frequency, near to or greater than the switching frequency Smallsignal transfer functions contain a single low frequency pole DCM CPM boost and buckboost are stable without artificial ramp DCM CPM buck without artificial ramp is stable for D < 2/3. A small artificial ramp m a 0.086m 2 leads to stability for all D. Fundamentals of Power Electronics 60 Chapter 11: Current Programmed Control
61 DCM CPM buckboost example i L i 1 Switch network i 2 v 1 v 2 i c m a i pk v g C R v m 1 m 2 = v 2 T s L L i L v L = v 1 T s L v 1 Ts 0 t 0 v 2 Ts Fundamentals of Power Electronics 61 Chapter 11: Current Programmed Control
62 Analysis i pk = m 1 d 1 T s i L m 1 = v 1 Ts L i c m a i pk i c = i pk m a d 1 T s = m 1 m a d 1 T s v L m 1 = v 1 T s L v 1 Ts m 2 = v 2 T s L 0 t d 1 = i c 0 m 1 m a T s v 2 Ts Fundamentals of Power Electronics 62 Chapter 11: Current Programmed Control
63 Averaged switch input port equation i 1 Ts = 1 T s i 1 Ts = 1 2 i pkd 1 t t T s i 1 (τ)dτ = q 1 T s i 1 Area q 1 i pk i 1 Ts i 1 Ts = 1 2 m 1d 1 2 T s i 1 Ts = v 1 Ts i 1 Ts v 1 Ts = 1 2 Li 2 c f s 1 m a m Li 2 c f s 1 m a m = p T s t d 1 T s d 2 T s d 3 T s T s Fundamentals of Power Electronics 63 Chapter 11: Current Programmed Control i 2 i 2 Ts i pk Area q 2
64 Discussion: switch network input port Averaged transistor waveforms obey a power sink characteristic During first subinterval, energy is transferred from input voltage source, through transistor, to inductor, equal to W = 1 2 Li 2 pk This energy transfer process accounts for power flow equal to p Ts = Wf s = 1 2 Li 2 pk f s which is equal to the power sink expression of the previous slide. Fundamentals of Power Electronics 64 Chapter 11: Current Programmed Control
65 Averaged switch output port equation i 2 Ts = 1 T s t t T s i 2 (τ)dτ = q 2 T s i 1 Area q 1 i pk q 2 = 1 2 i pkd 2 T s d 2 =d 1 v 1 Ts i 1 Ts v 2 Ts i 2 i 2 Ts = p T s v 2 Ts i pk Area q 2 i 2 Ts v 2 Ts = 1 2 Li 2 c f s 1 m a m 1 2 = p T s T s i 2 Ts d 1 T s d 2 T s d 3 T s t Fundamentals of Power Electronics 65 Chapter 11: Current Programmed Control
66 Discussion: switch network output port Averaged diode waveforms obey a power sink characteristic During second subinterval, all stored energy in inductor is transferred, through diode, to load Hence, in averaged model, diode becomes a power source, having value equal to the power consumed by the transistor power sink element Fundamentals of Power Electronics 66 Chapter 11: Current Programmed Control
67 Averaged equivalent circuit i 1 Ts i 2 Ts v 1 Ts p Ts v 2 Ts v g Ts C R v Ts L Fundamentals of Power Electronics 67 Chapter 11: Current Programmed Control
68 Steady state model: DCM CPM buckboost V g P R V Solution V 2 R = P P = 1 2 LI 2 c f s 1 M a M 1 2 V= PR = I c RL f s 2 1 M a M 1 for a resistive load 2 Fundamentals of Power Electronics 68 Chapter 11: Current Programmed Control
69 Models of buck and boost Buck L v g Ts p Ts C R v Ts Boost L v g Ts p Ts C R v Ts Fundamentals of Power Electronics 69 Chapter 11: Current Programmed Control
70 Summary of steadystate DCM CPM characteristics Table Steadystate DCM CPM characteristics of basic converters Converter M I crit Stability range when m a = 0 Buck P load P P load 1 2 I c Mm a T s 0 M< 2 3 Boost P load P load P I c M M 1 2M m at s 0 D 1 Buckboost Depends on load characteristic: P load = P I c M M 1 m a T s 2 M1 0 D 1 I > I crit for CCM I < I crit for DCM Fundamentals of Power Electronics 70 Chapter 11: Current Programmed Control
71 Buck converter: output characteristic with m a = 0 I I c I = P V g V V P = P V 1 V g V CCM unstable for M > 1 2 CPM buck characteristic with m a = 0 resistive load line I = V/R with a resistive load, there can be two operating points the operating point having V > 0.67V g can be shown to be unstable 1 2 I c B CCM DCM 4P V g A DCM unstable for M > V g 2 3 V g V g V Fundamentals of Power Electronics 71 Chapter 11: Current Programmed Control
72 Linearized smallsignal models: Buck i 1 i 2 L i L v g v 1 r 1 f 1 i c g 1 v 2 g 2 v 1 f 2 i c r 2 v 2 C R v Fundamentals of Power Electronics 72 Chapter 11: Current Programmed Control
73 Linearized smallsignal models: Boost i L L i 1 i 2 v g v 1 r 1 f 1 i c g 1 v 2 g 2 v 1 f 2 i c r 2 v 2 C R v Fundamentals of Power Electronics 73 Chapter 11: Current Programmed Control
74 Linearized smallsignal models: Buckboost i 1 i 2 v 1 r 1 f 1 i c g 1 v 2 g 2 v 1 f 2 i c r 2 v 2 v g C R v L i L Fundamentals of Power Electronics 74 Chapter 11: Current Programmed Control
75 DCM CPM smallsignal parameters: input port Table Current programmed DCM smallsignal equivalent circuit parameters: input port Converter g 1 f 1 r 1 Buck 1 R M 2 1M 1 m a m 1 1 m a m 1 2 I 1 m I c R 1M 1 a m 1 M 2 1 m a m 1 Boost 1 R M M 1 2 I I c R M 2 2M M1 2 m a m 1 1 m a m 1 Buckboost 0 2 I 1 I c 1 R M 2 m a m 1 1 m a m 1 Fundamentals of Power Electronics 75 Chapter 11: Current Programmed Control
76 DCM CPM smallsignal parameters: output port Table Current programmed DCM smallsignal equivalent circuit parameters: output port Converter g 2 f 2 r 2 Buck 1 R M 1M m a m 1 2M M 1 m a m 1 2 I I c R 1M 1 m a m 1 12M m a m 1 Boost 1 R M M 1 2 I 2 R M1 I c M Buckboost 2M R m a m 1 2 I 2 I c R 1 m a m 1 Fundamentals of Power Electronics 76 Chapter 11: Current Programmed Control
77 Simplified DCM CPM model, with L = 0 Buck, boost, buckboost all become v g r 1 f 1 i c g 1 v g 2 v g f 2 i c r 2 C R v G vc (s)= v i c vg=0= G c0 1 s ω p G vg (s)= v v g ic =0 = G g0 1 s ω p G c0 = f 2 R r 2 ω p = 1 R r 2 C G g0 =g 2 R r 2 Fundamentals of Power Electronics 77 Chapter 11: Current Programmed Control
78 Buck ω p Plug in parameters: ω p = 1 RC 23M 1M m a m 2 M 2M 1M 1MM m a m 2 For m a = 0, numerator is negative when M > 2/3. ω p then constitutes a RHP pole. Converter is unstable. Addition of small artificial ramp stabilizes system. m a > leads to stability for all M 1. Output voltage feedback can also stabilize system, without an artificial ramp Fundamentals of Power Electronics 78 Chapter 11: Current Programmed Control
79 11.5 Summary of key points 1. In currentprogrammed control, the peak switch current i s follows the control input i c. This widely used control scheme has the advantage of a simpler controltooutput transfer function. The linetooutput transfer functions of currentprogrammed buck converters are also reduced. 2. The basic currentprogrammed controller is unstable when D > 0.5, regardless of the converter topology. The controller can be stabilized by addition of an artificial ramp having slope ma. When m a 0.5 m 2, then the controller is stable for all duty cycle. 3. The behavior of currentprogrammed converters can be modeled in a simple and intuitive manner by the firstorder approximation i L T s i c. The averaged terminal waveforms of the switch network can then be modeled simply by a current source of value i c, in conjunction with a power sink or power source element. Perturbation and linearization of these elements leads to the smallsignal model. Alternatively, the smallsignal converter equations derived in Chapter 7 can be adapted to cover the current programmed mode, using the simple approximation i L i c. Fundamentals of Power Electronics 79 Chapter 11: Current Programmed Control
80 Summary of key points 4. The simple model predicts that one pole is eliminated from the converter linetooutput and controltooutput transfer functions. Current programming does not alter the transfer function zeroes. The dc gains become loaddependent. 5. The more accurate model of Section 11.3 correctly accounts for the difference between the average inductor current i L T s and the control input i c. This model predicts the nonzero linetooutput transfer function G vg (s) of the buck converter. The currentprogrammed controller behavior is modeled by a block diagram, which is appended to the smallsignal converter models derived in Chapter 7. Analysis of the resulting multiloop feedback system then leads to the relevant transfer functions. 6. The more accurate model predicts that the inductor pole occurs at the crossover frequency fc of the effective current feedback loop gain T i (s). The frequency f c typically occurs in the vicinity of the converter switching frequency f s. The more accurate model also predicts that the linetooutput transfer function G vg (s) of the buck converter is nulled when m a = 0.5 m 2. Fundamentals of Power Electronics 80 Chapter 11: Current Programmed Control
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