R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
1.. Averaged switch modeling with the simple approximation i 1 (t) i (t) i (t) v g (t) v 1 (t) v (t) C R v(t) Switch network Averaged terminal waveforms, CCM: v (t) Ts = d(t) v 1 (t) Ts The simple approximation: i (t) Ts (t) Ts i 1 (t) Ts = d(t) i (t) Ts Fundamentals of Power Electronics 33 Chapter 1: Current Programmed Control
CPM averaged switch equations v (t) Ts = d(t) v 1 (t) Ts i (t) Ts (t) Ts i 1 (t) Ts = d(t) i (t) Ts Eliminate duty cycle: i 1 (t) Ts = d(t) (t) Ts = v (t) Ts v 1 (t) Ts (t) Ts i 1 (t) Ts v 1 (t) Ts = (t) Ts v (t) Ts = p(t) Ts So: Output port is a current source Input port is a dependent power sink Fundamentals of Power Electronics 34 Chapter 1: Current Programmed Control
CPM averaged switch model i 1 i i p v g v 1 v C R v Averaged switch network Fundamentals of Power Electronics 35 Chapter 1: Current Programmed Control
Results for other converters Boost i v g p C R v Averaged switch network Averaged switch network Buck-boost p v g C R v i Fundamentals of Power Electronics 36 Chapter 1: Current Programmed Control
Perturbation and linearization to construct small-signal model et v 1 (t) Ts = V 1 v 1 (t) i 1 (t) Ts = I 1 i 1 (t) v (t) Ts = V v (t) i (t) Ts = I i (t) (t) Ts = I c (t) Resulting input port equation: V 1 v 1 (t) I 1 i 1 (t) = I c (t) V v (t) Small-signal result: i 1 (t)= (t) V V 1 v (t) I c V 1 v 1 (t) I 1 V 1 Output port equation: î = î c Fundamentals of Power Electronics 37 Chapter 1: Current Programmed Control
Resulting small-signal model Buck example i 1 i v v V g i C R V I 1 1 c v c i v V c v 1 V 1 I 1 Switch network small-signal ac model i 1 (t)= (t) V V 1 v (t) I c V 1 v 1 (t) I 1 V 1 Fundamentals of Power Electronics 38 Chapter 1: Current Programmed Control
Origin of input port negative incremental resistance i 1 Power source characteristic v 1 i 1 = p Quiescent operating point I 1 1 r 1 = I 1 V 1 V 1 v 1 Fundamentals of Power Electronics 39 Chapter 1: Current Programmed Control
Expressing the equivalent circuit in terms of the converter input and output voltages i g i v g D 1 s R D R D v R C R v i 1 (s)=d 1s R (s) D R v(s)d R v g(s) Fundamentals of Power Electronics 40 Chapter 1: Current Programmed Control
Predicted transfer functions of the CPM buck converter i g i v g D 1 s R D R D v R C R v G vc (s)= v(s) (s) vg =0 = R 1 sc G vg (s)= v(s) v g (s) ic =0 =0 Fundamentals of Power Electronics 41 Chapter 1: Current Programmed Control
imitations of the simple model i g i v g D 1 s R R D D R v C R v Model of the current-programmed buck converter This model predicts that G vg = 0 The simple model is very useful, but it doesn t predict some things that we often need to know: ine-to-output transfer function of buck converter High-frequency (inductor) dynamics Effects of large ripple and artificial ramp ECEN 5807, Spring 015 1
1.3 A More Accurate Model l l l l l l 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: ine-to-output transfer function of the buck converter Dynamics at frequencies approaching f s More accurate model accounts for nonideal operation of current mode controller built-in feedback loop Converter duty-cycle-controlled model, plus block diagram that accurately models equations of current mode controller Simulation of current mode control Comparison of performance: duty-cycle control vs. current-mode control Fundamentals of Power Electronics 4 Chapter 1: Current Programmed Control
Current-programmed controller model Compute the average inductor current using the definition: hi (t)i Ts = 1 T s Z tts / t T s / i ( )d, i (t) m a i pk m i i 1 i (dt s ) Ts i 4 m i 1 3 i (t) Ts (d 0.5)T s (d 0.5)T s 0 dt s T s t T s ECEN 5807, Spring 015
Sampling at t = dt s The current-mode controller samples and determines the duty cycle at time t = dt s. Hence we need to compute the average inductor current at this sampling instant: i (t) hi i Ts = i (dt s )i Ts = 1 T s Z (d0.5)ts (d 0.5)T s i ( )d. m a i pk m i i 1 i (dt s ) Ts i 4 m i 1 3 i (t) Ts (d 0.5)T s (d 0.5)T s 0 dt s T s T s t Sketched for the case d < 0.5 ECEN 5807, Spring 015 3
Computation i (t) m a i pk i 4 m i 1 3 (d 0.5)T s (d 0.5)T s 0 dt s T s m i i 1 i (dt s ) Ts i (dt s )i Ts = (0.5 d)i 3 di 1 d 0 i (0.5 d)i 4, i (dt s )i Ts = di 1 d 0 i (0.5 d)(i 4 i 3 ). Z T s t i (t) Ts Simplification: Time between interval midpoints i 3 and i 4 is T s Time between interval midpoints i 1 and i is T s / Slope between midpoint values is the same, shown by dotted line: i 4 i 3 = (i i 1 ). With this result, the expression for the average current becomes: hi i Ts = di 1 d 0 i (0.5 d)(i i 1 ) = d 0 i 1 di. ECEN 5807, Spring 015 4
Computation, slide i (t) m a i pk m i i 1 i (dt s ) Ts i 4 m i 1 3 i (t) Ts The midpoint currents are: i 1 = i pk m 1 dt s, i = i pk m d0 T s (d 0.5)T s (d 0.5)T s 0 dt s T s T s t i pk = m a dt s. Substitute into result from previous slide, to find the relationship between the average current and : hi i Ts = m a dt s m 1 m dd 0 T s. ECEN 5807, Spring 015 5
Small-signal model Now perturb and linearize hi i Ts = I î (t) h i Ts = ==I c î c (t) d(t) = D ˆd(t) m 1 = M 1 ˆm 1 (t) m = M ˆm (t) Buck converter ˆm 1 = ˆv g ˆv ˆm = ˆv Boost converter ˆm 1 = ˆv g ˆm = ˆv ˆv g Buck-boost converter ˆm 1 = ˆv g ˆm = ˆv inearization leads to: î (t) = î c (t) î (t) = î c (t) M a M 1 M M a M 1 M (1 D) T s ˆd(t) T s ˆd(t) 0 DD 0 T s Equilibrium relationship: DD 0 T s ˆm 1 (t) (ˆm 1 (t) ˆm (t)) DD 0 T s ˆm (t) DM 1 = D 0 M, ECEN 5807, Spring 015 6
Current mode controller small-signal block diagram î (t) = î c "(t) Solve for duty cycle: ˆd(t) = Which is of the form: M a M 1 M T s ˆd(t) 1 M a M 1 M T s " î c (t) î (t) ˆd(t) = F m hîc (t) î (t) F gˆv g (t) F vˆv(t) i DD 0 T s DD 0 T s ˆm 1 (t) ˆm 1 (t) DD 0 T s DD 0 T s ˆm (t) # ˆm (t) F m d 1 F m = M a M 1 M. T s F and F, for the basic buck, boost, and b v g F g "" " F v i v ECEN 5807, Spring 015 7
Summary of result: small-signal block diagram ˆd(t) = F m hîc (t) î (t) F gˆv g (t) F vˆv(t) i d F m 1 F m = M a M 1 M. T s F and F, for the basic buck, boost, and b v g F g "" F v v Table 17. Current programmed controller gains for basic converters " i Converter F g F v Buck DD 0 T s 0 Boost 0 DD 0 T s Buck-boost DD 0 T s DD 0 T s ECEN 5807, Spring 015 8
Block diagram of the current programmed controller d(t)=f m (t)i (t)f g v g (t)f v v(t) v g F m F g F v v d i 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 Fundamentals of Power Electronics 47 Chapter 1: Current Programmed Control
CPM buck converter model 1 : D V g d(t) i (t) v g (t) Id(t) C v(t) R d F m T v v g F g F v v T i i Fundamentals of Power Electronics 48 Chapter 1: Current Programmed Control
CPM boost converter model Vd(t) D' : 1 i (t) v g (t) Id(t) C v(t) R d F m T v v g F g F v v T i i Fundamentals of Power Electronics 49 Chapter 1: Current Programmed Control
CPM buck-boost converter model V g V d(t) 1 : D D' : 1 i (t) v g (t) Id(t) Id(t) C v(t) R d F m T v v g F g F v v T i i Fundamentals of Power Electronics 50 Chapter 1: Current Programmed Control
1.3. Solution of the CPM transfer functions In the models of the previous slides, the output voltage v can be expressed via superposition as a function of d and v g, through the duty-cycle controlled transfer functions G vd (s) and G vg (s): v(s)=g vd (s)d(s)g vg (s)v g (s) In a similar manner, the inductor current i can be expressed via superposition as a function of d and v g, by defining the transfer functions G id (s) and G ig (s): i (s)=g id (s)d(s)g ig (s)v g (s) with G id (s)= i (s) d(s) vg (s)=0 G ig (s)= i (s) v g (s) d(s)=0 Fundamentals of Power Electronics 51 Chapter 1: Current Programmed Control
System block diagram CPM controller model F v F m d Converter transfer functions G vd (s) G id (s) v F g v g G vg (s) G ig (s) i Fundamentals of Power Electronics 5 Chapter 1: Current Programmed Control
Solution of block diagram Solve block diagram for d: d = F m 1F m G id G ig F g v g F v v Substitute into expression for v: v = F m G vd 1F m G id G ig F g v g F v v G vg v g Solve for v: v = F m G vd 1F m G id F v G vd G vg F m F g G vd F m G vg G id G ig G vd 1F m G id F v G vd v g Fundamentals of Power Electronics 53 Chapter 1: Current Programmed Control
Results Transfer functions with current-mode control Control-to-output transfer function: G vc (s)= v(s) (s) v g (s)=0 = F m G vd 1F m G id F v G vd ine-to-output transfer function: G vg-cpm (s)= v(s) v g (s) (s)=0 = G vg F m F g G vd F m G vg G id G ig G vd 1F m G id F v G vd Fundamentals of Power Electronics 54 Chapter 1: Current Programmed Control
1.3.3 Discussion The more accurate controller model accounts for the differences between i and that arise by two mechanisms: Inductor current ripple Artificial ramp d F m F g and F v blocks model smallsignal effects of inductor current ripple: how the difference between i and varies with applied voltages. For operation deep in CCM, ripple is small F g and F v blocks can then be ignored. v g F g F v v i Fundamentals of Power Electronics 55 Chapter 1: Current Programmed Control
Effect of artificial ramp Artificial ramp also causes inductor current to differ from d F m block models effect of artificial ramp. F m varies inversely with M a, and becomes infinite with zero M a With zero M a, the input to the F m block tends to zero. The controller block diagram then predicts that d F m =0= i F g v g F v v v g F m F g F v v i With negligible inductor current ripple, F g and F v blocks go to zero. The model then predicts 0= i which coincides with the ideal model of Section 1. Fundamentals of Power Electronics 56 Chapter 1: Current Programmed Control
Ideal limiting case General expression for controlto-output transfer function: G vc (s)= v(s) (s) v g (s)=0 = F m G vd 1F m G id F v G vd In the limit when F m 0, F g 0, F v 0, then the control-to-output transfer function reduces to the following: lim F m F g 0 F v 0 G vc (s)= G vd G id and the line-to-output transfer function reduces to: lim F m F g 0 F v 0 G vg-cpm (s)= G vgg id G ig G vd G id These expressions coincide with the transfer functions predicted by the ideal model of Section 1.. Fundamentals of Power Electronics 57 Chapter 1: Current Programmed Control
arge artificial ramp Duty-cycle control In the extreme case of a very large artificial ramp, the controller degenerates to duty-cycle control. For large M a (small F m ) and for F g 0 and F v 0, the control-to-output transfer function reduces to lim small F m F v 0 F g 0 G vc (s)=f m G vd (s) (current-mode controller becomes pulsewidth modulator having gain F m, and dutycycle controller gain G vd (s) is obtained) For large M a (small F m ) and for F g 0 and F v 0, the line-to-output transfer function reduces to lim F m F g 0 F v 0 G vg-cpm (s)=g vg (the duty-cycle controller gain G vg (s) is obtained) Fundamentals of Power Electronics 58 Chapter 1: Current Programmed Control