R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
Sampled-data response: i L /i c Sampled-data transfer function : î L (s) î c (s) = (1 ) 1 e st 1 e st st = M 2 M a M 1 M a Use Matlab to plot frequency response Buck converter example: M a = M 2 /3 α = 0.5 f s = 200 khz Note peaking and second-order response near f s /2 ECEN 5807, Spring 2015 9
The second-order Padé approximation î L (s) î c (s) e st s = (1 ) 1 e st 1 2 1 2 1! e st st! 2 s! s /2 s! s /2 s! s /2 s! s /2!! 2! Result: î L (s) î c (s) 1 1 1 s 2 1! s /2 s! s /2! 2 ECEN 5807, Spring 2015 10
Approximating the sampled-data response: i L /i c Second-order Padé approximation Original Buck converter example: M a = M 2 /10 α = 0.818 D = 0.5 f s = 200 khz ECEN 5807, Spring 2015 11
Addition of pole to F m d F m 1 s ω p Addition of simple pole within F m block Leads to second-order response in i L /i c f p = 4 1 2D 2D M! a M 2 f s v g F g F v v i L i c ECEN 5807, Spring 2015 10
Determination of ω p ECEN 5807, Spring 2015 13
Inclusion of f p in G vc (s) f s /2 Switching frequency = 200 khz G vd G vc, with f p G vc, no f p ECEN 5807, Spring 2015 14
Simulink model for current mode control Start with averaged model of converter power stage, then append block containing current mode controller model: Current mode controller block requires information from power stage model: Average inductor current Slope m 1 Slope m 2 These three quantities are placed in a vector current data ECEN 5807, Spring 2015 15
Current mode controller block! The CPM controller solves for the duty cycle: d hi L i = i c m a dt s m 1 m 2 2 dd 0 T s F m Solve for d: v g F g "" F v v 1 d = 1! 0 2 m a 1 m 1 m 2 B@ vut 2 (i c i L )(m 1 m 2 ) 1 m a m 1 m 2 2 Ts CA 2 " i c i L ECEN 5807, Spring 2015 16
Optional correction: high-frequency pole v g d F m 1 s ω p F g F v v f p = 4 1 2D 2D M! a M 2 f s Addition of the high frequency pole in this manner is valid for small signals only. This model does not correctly predict the high-frequency components of the large-signal transient response. i L See buck CPM models on course web page i c ECEN 5807, Spring 2015 17
Small-signal G vc predicted by Simulink model Compare with plots from last lecture The buck model includes losses ECEN 5807, Spring 2015 18
B.3 Simulation of Current Mode Controllers Develop a model of the currentprogrammed controller, which can be combined with existing CCM- DCM averaged switch models V g 12 V L 1 2 3 i L R L 4 5 1 4 2 CCM-DCM1 3 d 35 µh X switch f s = 200 khz L = 35 µη 0.05 Ω C 100 µf i LOAD R 10 Ω v Controller model outputs a duty cycle, in response to control input i c (or v c ) and the sensed converter voltages and currents v c CPM control current 1 2 d R f i L v(1)v(3) v(3) E i E 1 E 2 X cpm R f = 1 Ω f s = 200 khz L = 35 µη V a = 0.6 V Fundamentals of Power Electronics 72 Chapter 12: Current Programmed Control
Averaged controller waveforms i L (t) i c i L (t) i c m a i pk m a i pk m 1 m 2 m1 m 2 dt s d 2 T s =(1 d)t s t dt s d 2 T s 0 t v L (t) v 1 (t) Ts v L (t) v 1 (t) Ts v 2 (t) Ts 0 t v 2 (t) Ts 0 t T s T s CCM DCM Fundamentals of Power Electronics 73 Chapter 12: Current Programmed Control
Equations Need to write large-signal equations of controller, in a form that leads to convergence of simulator and that works for both CCM and DCM Basic equations: d 2 =1d (CCM) m 1 = v 1 (t) Ts L m 2 = v 2 (t) Ts L i pk = i c m a dt s d 2 = i pk m 2 T s (DCM) Average inductor current: d 2 = MIN 1 d, i L (t) Ts = di pk m 1dT s 2 Artificial ramp amplitude: V a = m a T s R f d 2 i pk m 2d 2 T s 2 Substitute and solve (partially) for d: i pk m 2 T s (CCM and DCM) d = 2i c (d d 2 )2 i L (t) Ts m 2 d 2 2 T s 2m a (d d 2 )T s m 1 dt s Fundamentals of Power Electronics 74 Chapter 12: Current Programmed Control
CPM controller subcircuit model Output: duty cycle d.subckt CPM control current 1 2 d params: L=100e-6 fs=1e5 Va=0.5 Rf=1 * generate d2 for CCM or DCM Inputs: Ed2 d2 0 table d CPM control current 1 2 v c (t) Ts R f i L (t) Ts v 1 (t) Ts v 2 (t) Ts {MIN(L*fs*(v(control)-va*v(d))/Rf/(v(2)),1-v(d))} (0,0)(1,1) * generate inductor current slopes, see Eqs.(B.24) and (B.26) Em1 m1 0 value={rf*v(1)/l/fs} Em2 m2 0 value={rf*v(2)/l/fs} * compute duty cycle d, see Eq.(B.32) Eduty d 0 table {2*(v(control)*(v(d)v(d2))-v(current)-v(m2)*v(d2)*v(d2)/2) /(v(m1)*v(d)2*va*(v(d)v(d2)))} (0.01,0.01) (0.99,0.99).ends Fundamentals of Power Electronics 75 Chapter 12: Current Programmed Control
CPM buck example CPM buck converter.param Va=0.6.param fs=200khz.param L=35uH.ac DEC 101 10 100KHz iout 0 4 ac 0.lib switch.lib Vg 1 0 12V ac 0 Xswitch 1 2 2 0 5 CCM-DCM1 PARAMS: L={L} fs={fs} L1 2 3 {L} RL1 3 4 0.05 C1 4 0 100uF Rload 4 0 10 Xcpm ctr ni nm1 nm2 5 CPM PARAMS: L={L} fs={fs} va={va} Rf=1 Ei ni 0 value={i(l1)} Em1 nm1 0 value={v(1)-v(3)} Em2 nm2 0 value={v(3)} Vic ctr 0 dc 1.4V ac 1.probe.end V g 12 V v c L 1 2 3 i L R L 4 5 1 4 2 CCM-DCM1 3 CPM 35 µh control current 1 2 d d X switch f s = 200 khz L = 35 µη R f i L v(1)v(3) v(3) E i E 1 E 2 X cpm 0.05 Ω C 100 µf R f = 1 Ω f s = 200 khz L = 35 µη V a = 0.6 V i LOAD R 10 Ω v Fundamentals of Power Electronics 76 Chapter 12: Current Programmed Control
Control-to-output frequency response Duty cycle control vs current programmed control G 40 db 20 db 0 db 20 db G vc G vd G In both cases: V = 8.1 V D = 0.676 For CPM: V c = 1.4 V 40 db G vd 0 V a = 0.6 V 60 db G vc 90 180 10 Hz 100 Hz 1 khz 10 khz 100 khz f Fundamentals of Power Electronics 77 Chapter 12: Current Programmed Control
Line-to-output frequency response Duty cycle control vs current programmed control G vg 20 db 0 db 20 db 40 db 60 db 80 db Duty cycle control d(t) = constant Current programmed mode v c (t) = constant In both cases: V = 8.1 V D = 0.676 For CPM: V c = 1.4 V V a = 0.6 V 100 db 10 Hz 100 Hz 1 khz 10 khz 100 khz f Fundamentals of Power Electronics 78 Chapter 12: Current Programmed Control
Output impedance Duty cycle control vs current programmed control Z out 20 dbω 0 dbω 20 dbω Current programmed mode v c (t) = constant Duty cycle control d(t) = constant In both cases: V = 8.1 V D = 0.676 For CPM: V c = 1.4 V V a = 0.6 V 40 dbω 10 Hz 100 Hz 1 khz 10 khz 100 khz f Fundamentals of Power Electronics 79 Chapter 12: Current Programmed Control
12.4 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 Small-signal transfer functions contain a single low frequency pole DCM CPM boost and buck-boost 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 80 Chapter 12: Current Programmed Control
DCM CPM buck-boost example i L (t) i 1 (t) Switch network i 2 (t) v 1 (t) v 2 (t) i c m a i pk v g (t) C R v(t) m 1 m 2 = v 2 T s L L i L (t) v L (t) = v 1 T s L v 1 (t) Ts 0 t 0 v 2 (t) Ts Fundamentals of Power Electronics 81 Chapter 12: Current Programmed Control
Analysis i pk = m 1 d 1 T s i L (t) m 1 = v 1(t) 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 (t) m 1 = v 1 T s L v 1 (t) Ts m 2 = v 2 T s L 0 t d 1 (t)= i c (t) 0 m 1 m a T s v 2 (t) Ts Fundamentals of Power Electronics 82 Chapter 12: Current Programmed Control
Averaged switch input port equation i 1 (t) Ts = 1 T s i 1 (t) Ts = 1 2 i pk(t)d 1 (t) t t T s i 1 (τ)dτ = q 1 T s i 1 (t) Area q 1 i pk i 1 (t) Ts i 1 (t) Ts = 1 2 m 1d 1 2 (t)t s i 1 (t) Ts = v 1 (t) Ts 1 2 Li 2 c f s 1 m a m 1 2 i 2 (t) i 2 (t) Ts i pk Area q 2 i 1 (t) Ts v 1 (t) Ts = 1 2 Li 2 c f s 1 m a m 1 2 = p(t) T s t d 1 T s d 2 T s d 3 T s T s Fundamentals of Power Electronics 83 Chapter 12: Current Programmed Control
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(t) 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 84 Chapter 12: Current Programmed Control
Averaged switch output port equation i 2 (t) Ts = 1 T s t t T s i 2 (τ)dτ = q 2 T s i 1 (t) Area q 1 i pk q 2 = 1 2 i pkd 2 T s d 2 (t)=d 1 (t) v 1 (t) Ts i 1 (t) Ts v 2 (t) Ts i 2 (t) i 2 (t) Ts = p(t) T s v 2 (t) Ts i pk Area q 2 i 2 (t) Ts v 2 (t) Ts = 1 2 Li 2 c(t) f s 1 m a m 1 2 = p(t) T s T s i 2 (t) Ts d 1 T s d 2 T s d 3 T s t Fundamentals of Power Electronics 85 Chapter 12: Current Programmed Control
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 86 Chapter 12: Current Programmed Control
Averaged equivalent circuit i 1 (t) Ts i 2 (t) Ts v 1 (t) Ts p(t) Ts v 2 (t) Ts v g (t) Ts C R v(t) Ts L Fundamentals of Power Electronics 87 Chapter 12: Current Programmed Control
Steady state model: DCM CPM buck-boost V g P R V Solution V 2 R = P P = 1 2 LI 2 c(t) 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 88 Chapter 12: Current Programmed Control
Models of buck and boost Buck L v g (t) Ts p(t) Ts C R v(t) Ts Boost L v g (t) Ts p(t) Ts C R v(t) Ts Fundamentals of Power Electronics 89 Chapter 12: Current Programmed Control
Summary of steady-state DCM CPM characteristics Table 12.6. Steady-state 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 2 M m a T s 0 D 1 Buck-boost Depends on load characteristic: P load = P I c M M 1 m a T s 2 M 1 0 D 1 I > I crit for CCM I < I crit for DCM Fundamentals of Power Electronics 90 Chapter 12: Current Programmed Control
Buck converter: output characteristic with m a = 0 I I c I = P V g V P V = 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 > 2 3 1 2 V g 2 3 V g V g V Fundamentals of Power Electronics 91 Chapter 12: Current Programmed Control
Linearized small-signal 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 92 Chapter 12: Current Programmed Control
Linearized small-signal 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 93 Chapter 12: Current Programmed Control
Linearized small-signal models: Buck-boost 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 94 Chapter 12: Current Programmed Control
DCM CPM small-signal parameters: input port Table 12.7. Current programmed DCM small-signal equivalent circuit parameters: input port Converter g 1 f 1 r 1 Buck 1 R M 2 1 M 1 m a m 1 1 m a m 1 2 I 1 I c R 1 M 1 M 2 m a m 1 1 m a m 1 Boost 1 R M M 1 2 I I c R M 2 2 M 1 M 2 m a m 1 1 m a m 1 Buck-boost 0 2 I 1 I c 1 R M 2 m a m 1 1 m a m 1 Fundamentals of Power Electronics 95 Chapter 12: Current Programmed Control
DCM CPM small-signal parameters: output port Table 12.8. Current programmed DCM small-signal equivalent circuit parameters: output port Converter g 2 f 2 r 2 Buck 1 R M 1 M m a m 1 2 M M 1 m a m 1 2 I I c R 1 M 1 m a m 1 12M m a m 1 Boost 1 R M M 1 2 I 2 R M 1 I c M Buck-boost 2M R m a m 1 2 I 2 I c R 1 m a m 1 Fundamentals of Power Electronics 96 Chapter 12: Current Programmed Control
Simplified DCM CPM model, with L = 0 Buck, boost, buck-boost 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 i c =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 97 Chapter 12: Current Programmed Control
Buck ω p Plug in parameters: ω p = 1 RC 23M 1M m a m 2 M 2M 1M 1M M 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 > 0.086 leads to stability for all M 1. Output voltage feedback can also stabilize system, without an artificial ramp Fundamentals of Power Electronics 98 Chapter 12: Current Programmed Control
12.5 Summary of key points 1. In current-programmed control, the peak switch current i s (t) follows the control input i c (t). This widely used control scheme has the advantage of a simpler control-to-output transfer function. The line-to-output transfer functions of current-programmed buck converters are also reduced. 2. The basic current-programmed controller is unstable when D > 0.5, regardless of the converter topology. The controller can be stabilized by addition of an artificial ramp having slope ma. When m a 0.5 m 2, then the controller is stable for all duty cycle. 3. The behavior of current-programmed converters can be modeled in a simple and intuitive manner by the first-order approximation i L (t) T s i c (t). 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 small-signal model. Alternatively, the smallsignal converter equations derived in Chapter 7 can be adapted to cover the current programmed mode, using the simple approximation i L (t) i c (t). Fundamentals of Power Electronics 99 Chapter 12: Current Programmed Control
Summary of key points 4. The simple model predicts that one pole is eliminated from the converter line-to-output and control-to-output transfer functions. Current programming does not alter the transfer function zeroes. The dc gains become load-dependent. 5. The more accurate model of Section 12.3 correctly accounts for the difference between the average inductor current i L (t) T s and the control input i c (t). This model predicts the nonzero line-to-output transfer function G vg (s) of the buck converter. The current-programmed controller behavior is modeled by a block diagram, which is appended to the small-signal converter models derived in Chapter 7. Analysis of the resulting multiloop feedback system then leads to the relevant transfer functions. 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 line-to-output transfer function G vg (s) of the buck converter is nulled when m a = 0.5 m 2. Fundamentals of Power Electronics 100 Chapter 12: Current Programmed Control
Summary of key points 7. Current programmed converters operating in the discontinuous conduction mode are modeled in Section 12.4. The averaged transistor waveforms can be modeled by a power sink, while the averaged diode waveforms are modeled by a power source. The power is controlled by i c (t). Perturbation and linearization of these averaged models, as usual, leads to small-signal equivalent circuits. Fundamentals of Power Electronics 101 Chapter 12: Current Programmed Control