Small-Signal Modeling of the
Boost Converter Operated in CM
Christophe Basso
IEEE Senior Member
1 • Chris Basso – CM Boost Study
Course Agenda
 The PWW Switch Concept
 Small-Analysis in Continuous Conduction Mode
 Small-Signal Response in Discontinuous Mode
 EMI Filter Output Impedance
 Cascaded Converters Operation
2 • Chris Basso – – CM Boost Study
Course Agenda
 The PWW Switch Concept
 Small-Analysis in Continuous Conduction Mode
 Small-Signal Response in Discontinuous Mode
 EMI Filter Output Impedance
 Cascaded Converters Operation
3 • Chris Basso – – CM Boost Study
Non-Linearity in a Switching Converter
 A switching converter is ruled by linear equations…
x1
u1
SW
u1
L
x2 C
during
DTsw
R
x1
off
L x2 C
 …combining so-called state variables
4 • Chris Basso – – CM Boost Study
R
y1
C
on
L
R
during
1  D  Tsw
State Variables
 State variables describe the mathematical state of a system
 n state variables for n independent storage elements
 knowing variables state at t0 helps compute outputs for t > t0
System described
by state variables
u1  t 
u2  t 
u3  t 
y1  t 
y2  t 
y3  t 
 x1 , x2 , x3 ..., xn 
Input vector u
Output vector y
 x1 is the inductor current and x2 is the capacitor voltage
 Differentiation gives the state variable rate of change
x1 
x1  iL  t 
diL  t 
Predict future
system state
dt
x2  vC  t 
x2 
dvC  t 
dt
5 • Chris Basso – – CM Boost Study
Describe the System During the On-Time
 Observe the system during the on-time duration or dTSW:
x1
u1
L
iC  t 
x2
R
C
R
iC  t   C
1
1
x2  u1
L
L
1
1
x2  x1 
x2
C
RC
6 • Chris Basso – – CM Boost Study
dt
 Cx2  x1 
1
x2
R
x2
u1  L
x1  
dvC  t 
 x1   0
 x   1 C
 2 
diL  t 
dt
 vC  t   Lx1  x2
1 L   x1  1 L 0   u1 

1 RC   x2   0 0  u2 
State coefficients
Source
coefficients
Describe the System During the Off-Time
 Repeat the exercise during the off-time duration or 1-dTSW
x1
iR  t 
iC  t 
C
vL  t  L
vL  t    x2 vL  t   Lx1
x2   Lx1
R
x2
iR  t   x1  iC  t   x1  Cx2
x2  i R  t  R
x2  R  x1  Cx2 
x1  0 
x2 
1
x2
L
1
1
x1 
x2
C
RC
 x1   0
 x   1 C
 2 
1 L   x1   0 0   u1 

1 RC   x2   0 0  u2 
State coefficients
Source
coefficients
7 • Chris Basso – – CM Boost Study
Make it Fit the State Equation Format
 Arrange expressions to make them fit the format:
x  Ax  t   Bu  t 
State equation
on-time network
 x1   0
 x   1 C
 2 
1 L   x1  1 L 0   u1 

1 RC   x2   0 0  u2 
 x1   0
 x   1 C
 2 
1 L   x1   0 0   u1 

1 RC   x2   0 0  u2 
 0
A1  
1 C
1 L 
1 L 0 
B1  

1 RC 
 0 0
off-time network
 0
A2  
1 C
1 L 
0 0
B2  

1 RC 
0 0
 How do we link matrixes A1 and A2, B1 and B2?
 We smooth the discontinuity by weighting them by D and 1-D
x   A1 D  A 2 1  D   x  t    B1 D  B 2 1  D   u  t 
8 • Chris Basso – – CM Boost Study
The State-Space Averaging Method (SSA)
 We now have a continuous large-signal equation
 We need to linearize it via perturbations
x  x0  xˆ
u  u0  uˆ
D  D  dˆ
0
x   A1 D  A 2 1  D   x  t    B1d  B 2 1  D   u  t 
u10 dˆ
L
x̂1
C
u1
D
1
dˆ
xˆ1  xˆ2  0 uˆ1  u10
L
L
L
1
1
xˆ2  xˆ1 
xˆ2
C
RC
D0
x̂2
d̂
Canonical smallsignal model
9 • Chris Basso – – CM Boost Study
The State-Space Averaging Method (SSA)
 SSA was applied to switching converters by Dr Ćuk in 1976
 It is a long, painful process, manipulating numerous terms
 What if you add an EMI filter for instance?
SW
L1
u1
C1
L2
on
off
C2
L1
u1
y1
R
C1
L2
C2
R
L2 C2
R
L1
u1 C1
 4 state variables and you have to re-derive all equations!
10 • Chris Basso – – CM Boost Study
The PWM Switch Model in Voltage Mode
 We know that non-linearity is brought by the switching cell
a
c
L
C
u1
R
p
a: active
c: common
p: passive
 Why don't we linearize the cell alone?
a
c
a
c
. .
d
PWM switch VM
p
Small-signal model
(CCM voltage-mode)
Switching cell
p
V. Vorpérian, "Simplified Analysis of PWM Converters using Model of PWM Switch, parts I and II"
IEEE Transactions on Aerospace and Electronic Systems, Vol. 26, NO. 3, 1990
11 • Chris Basso – – CM Boost Study
Replace the Switches by the Model
 Like in the bipolar circuit, replace the switching cell…
a
c
L
u1
C
. .
R
p
 …and solve a set of linear equations!
u1
12 • Chris Basso – – CM Boost Study
. .
L
C
R
An Invariant Model
 The switching cell made of two switches is everywhere!
a
d
p
a
buck
a
buck-boost
d
PWM switch VM
PWM switch VM
p
c
c
boost
c
d
Ćuk
d
c
p
c
PWM switch VM
a
d
PWM switch VM
p
p
a
PWM switch VM
13 • Chris Basso – – CM Boost Study
Smoothing the Discontinuity
 A CRT TV displays frames at a certain rate, 50 per second
9
8
t1
1
2
3
6 7
4 5
10
11
12
 The optic nerve time constant is larger than an interval
 A succession of discrete events is seen as continuous
Low-frequency filtering
Integration
See "phi phenomena", www.wikipedia.org
14 • Chris Basso – – CM Boost Study
Averaging Waveforms
 The keyword in the PWM switch is averaging
v t 
A
V peak
v t 
Tsw
t
Tsw
DTsw
v t 
Tsw
Tsw
1

Tsw

0
1
v  t   dt 
Tsw
Tsw

V peak  dt  V peak 1  D   V peak D '
DTsw
 The resulting function is continuous in time
15 • Chris Basso – – CM Boost Study
From Steps to a Continuous Function
 Some functions require mathematical abstraction: duty ratio
Tsw
v t 
t1
t2
t3
D2
t5
Discrete values of D
D3
Average and continuous
evolution of d(t)
D4
D1
Dn 
t4
tn
Tsw
D5
f mod  Fsw
 At the modulation frequency scale, points look contiguous
 Link them through a continuous-time ripple-free function d(t)
16 • Chris Basso – – CM Boost Study
The Common Passive Configuration
 The PWM switch is a single-pole double-throw model
d
c
a
ia  t 
ic  t 
d'
vap  t 
vcp  t 
p
 Install it in a buck converter and draw the waveforms
ia  t 
ic  t 
d
c
a
L
d'
Vin
C
R
vap  t 
vcp  t 
p
Vout
CCM
17 • Chris Basso – – CM Boost Study
The Common Passive Configuration
 Average the current waveforms across the PWM switch
ic  t 
ic  t 
0
Tsw
t
ia  t 
I a  DI c
ic  t 
Tsw
Averaged
variables
t
0
DTsw
ia  t 
Tsw
 Ia 
1
Tsw
dTsw
 i  t  dt  D i  t 
a
0
c
Tsw
 DI c
CCM
18 • Chris Basso – – CM Boost Study
The Common Passive Configuration
 Average the voltage waveforms across the PWM switch
vap  t 
vap  t 
0
Tsw
t
vcp  t 
vcp  t 
t
0
DTsw
vcp  t 
Tsw
 Vcp 
1
Tsw
Vcp  DVap
Tsw
Averaged
variables
dTsw

vcp  t  dt  D vap  t 
0
Tsw
 DVap
CCM
19 • Chris Basso – – CM Boost Study
A Two-Port Representation
 We have a link between input and output variables
DI c
d
a
Vap
p
Two-port
cell
c
Ic
DVap
p
 It can further be illustrated with current and voltage sources
Ia
d
a
Vap
c
Vcp
DI c DVap
p
Ic
p
CCM
20 • Chris Basso – – CM Boost Study
A Transformer Representation
 The PWM switch large-signal model is a dc "transformer"!
Ia
Ic
a
c
I
I a  DI c
Ic  a
. .
D
1 D
dc equations!
Vcp
Vcp  DVap
Vap 
D
p
 It can be immediately plugged into any 2-switch converter
c
.
.
L
rL
p
1
a
Vin
D
R
C
CCM
21 • Chris Basso – – CM Boost Study
Simulate Immediately with this Model
 SPICE can get you the dc bias point
R2
L1
100m 9.80V 100u
10.0V
7
d
Vg
10
VIC
9.80V
c
4
c
300mV
a a
V3
0.3
AC = 1
Rdum
1u
V(a,p)*V(d)
p p
5
14.0V
Vout
C1
470u
V(d)*I(VIC)
R1
10
 …but also the ac response as it linearizes the circuit
dB
°
40.0
360
Hf
20.0
0
-20.0
180
0
arg H  f 
-180
-40.0
10
22 • Chris Basso – – CM Boost Study
100
1k
Hz
10k
-360
100k
CCM
We Want Transfer Functions
 Derive the dc transfer function: open caps., short inductors
Vcp
c
I out
D Ic
.
rL
p
Ic
.
1
a
Vin
Vout
R
Ia
Vout

Vin
Vin  rL I c  Vcp  Vout
Vout  I out R
1
1  D  
Vin  rL I c  DVout  Vout
Vout   I a  I c  R
I a  DI c
Vout  I c  D  1 R
Ic 
Vout 1  D   Vin
Vout
1

Vin
D'
rL
rL
 D  1 R
1
r
1 L 2
RD '
CCM
23 • Chris Basso – – CM Boost Study
Plotting Transfer Functions
 Plot the lossy boost transfer function in a snapshot
5
rL  0.1Ω
R  10 Ω
f ( d 0.1)
f ( d 0.2)
4
rL  0.2 Ω
f ( d 0.3)
f ( d 0.4)
Vout
Vin
f ( d 0.5) 3
f ( d 0.6)
f ( d 0.7)
f ( d 0.8) 2
f ( d 0.9)
f ( d 1)
1
rL  1Ω
0
0
0.2
0.4
Duty ratio
0.6
0.8
1
d
 Above a certain conversion ratio, latch-up occurs
CCM
24 • Chris Basso – – CM Boost Study
Course Agenda
 The PWW Switch Concept
 Small-Analysis in Continuous Conduction Mode
 Small-Signal Response in Discontinuous Mode
 EMI Filter Output Impedance
 Cascaded Converters Operation
25 • Chris Basso – – CM Boost Study
A Boost Converter in Current Mode
 Identify the diode and switch position in a boost CM
c
a
duty-cycle
p
c
vc
PWM switch CM
p
a
Current mode
 Replace switches by the small-signal PWM switch model
26 • Chris Basso – – CM Boost Study
A Small Signal Model
 The model includes current sources and conductances
Ic
a
c
1
gi
vˆcp g r
vˆc ki
vˆap g f
vˆc ko
1
go
Cs
p
ko 
1
Ri
ki 
D
Ri
g f  Dg o 
DD ' Tsw
2L

I 
gi  D  g f  c 

Vap 

go  
gr 

Tsw  Sa 1
 D'   D
L  Sn 2

Ic
 go D
Vap
V. Vorpérian,"Analysis of Current-Controlled PWM Converters using the Model of the PWM Switch", PCIM Conference ,1990
27 • Chris Basso – – CM Boost Study
Start with a Large Signal Response
 Use the original large-signal (nonlinear) PWMCM model
R4
1.5m
L1
5u
Vin
4
5
2.69V
c
2.69V
R9
1k
vc
7
5.00V
X4
PWMCCMCM
L = 5u
Fs = 1Meg
Ri = -50m
Se = 0
PWM switch CM
462mV
duty-cycle
duty_ratio
a
Vin
2.7
p
2.70V
5.00V
471mV
V1
471m
AC = 1
Vout
2
Resr
60m
Rload
1
1
Cout
200uF
5 V/1 A
Fsw = 1 MHz
28 • Chris Basso – – CM Boost Study
Frequency Response of the CM Boost
 This frequency response becomes the reference
49.0
37.0
Subharmonic
poles
(dB) 25.0 H 0  14.5 dB
13.0
Vout  f 
1.00
Vc  f 
-36.0
(°)
-108

Vout  f 
Vc  f 
-180
-252
-324
10
100
1k
10k
frequency in hertz
100k
1Meg
29 • Chris Basso – – CM Boost Study
Plug the Linearized Small-Signal Model
 Use the linearized model to check coefficients and response
R8
1.5m
2.70V
1
parameters
Fsw =1Meg
Tsw =1/Fsw
L=5u
Cs=1/(L*(Fsw *3.14)^2)
Ri=-50m
Se=0
Vin=2.7
Vout=5
L2
{L}
15
c
2.69V
2.69V
c
VIC
C2
{Cs}
2.69V
5.00V
4
R5
{1/go}
Vc=471m
Vac=-Vin
Vap=-Vout
Vcp=(-Vout+Vin)
B4
Current
V(v c)*{ko}
Vin
{Vin}
These coefficients are
computed by the macro
Sn=(Vac/L)*Ri
Sf=(Vap/L)*Ri
B3
Current
Ic=(Vc/Ri)-D*Tsw *Se-Vcp*(1-D)*Tsw /(2*L)
p
V(a,p)*{gf }
D=Vcp/Vap
D'=1-D
ki=D/Ri
gi=D*(gf-Ic/Vap)
gr=(Ic/Vap)-go*D
ko=1/Ri
go=(Tsw /L)*(D'*Se/Sn+0.5-D)
gf=D*go-(D*D')*Tsw /(2*L)
B6
Current
V(c,p)*{gr}
4.30uV
B5
Current
p
Vout
V(v c)*{ki}
C3
200uF
R4
{1/gi}
a
a
0V
R7
1u
30 • Chris Basso – – CM Boost Study
16
vc
471mV
462mV
V2
471m
AC = 1
D
B2
Voltage
v (c,p)/v (a,p)
R6
60m
R3
1
Small-Signal Response and Original Model
 Validate the small-signal approach by comparing responses
49.0
37.0
(dB) 25.0 H 0  14.5 dB
13.0
Vout  f 
1.00
Vc  f 
Good to go!
-36.0
-108
(°)

Vout  f 
Vc  f 
-180
-252
-324
10
100
1k
10k
100k
1Meg
frequency in hertz
31 • Chris Basso – – CM Boost Study
Rearrange Sources to Improve Readability
 It is important to lay the network out properly
B3
Current
Storage element
V(0,out)*{gf }
B4
Current
2.69V
V(v c)*{ko}
R8
1.5m
1
L2
{L}
c
R5
{1/go}
VIC
Storage element
out
5.00V
Vout
15
2.69V
C2
{Cs}
2.70V
C3
200uF
4
2.69V
Vin
{Vin}
Storage element
B6
Current
V(c,out)*{gr}
B5
Current
V(v c)*{ki}
R4
{1/gi}
471mV
0V
16
vc
462mV
V2
471m
AC = 1
D
R6
B2
60m
Voltage
v (c,out)/v (0,out)
 Make sure the frequency response remains unchanged
32 • Chris Basso – – CM Boost Study
R3
1
What is the Converter Order?
 Count storage elements: L and two C
 The denominator of the transfer function is of 3rd order
Zeros
Dc gain

s 
s
1
1  

s z1  s z2
H  s   H0 
s
1
s p1



1
 s 
s
1
 
0 Q  0 
2
Subharmonic
poles
33 • Chris Basso – – CM Boost Study
Start with the Static Analysis
 For s = 0, short inductors and open capacitors
B3
Current
parameters
Fsw=1Meg
Tsw=1/Fsw
L=5u
Cs=1/(L*(Fsw*3.14)^2)
Ri=-50m
Se=0
Vin=2.7
Vout=5
V(0,out)*{gf}
B4
Current
V(vc)*{ko}
R5
{1/go}
Sn=(Vac/L)*Ri
Sf=(Vap/L)*Ri
Vc=471m
Vac=-Vin
Vap=-Vout
Vcp=(-Vout+Vin)
Vout
VIC
34 • Chris Basso – – CM Boost Study
B6
Current
V(c,out)*{gr}
V c  0
c
Ic=(Vc/Ri)-D*Tsw*Se-Vcp*(1-D)*Tsw/(2*L)
D=Vcp/Vap
D'=1-D
ki=D/Ri
gi=D*(gf-Ic/Vap)
gr=(Ic/Vap)-go*D
ko=1/Ri
go=(Tsw/L)*(D'*Se/Sn+0.5-D)
gf=D*go-(D*D')*Tsw/(2*L)
out
1
R8
1.5m
Neglect rL
D
vc
V2
471m
AC = 1
B2
Voltage
v(c,out)/v(0,out)
V c  0
B5
Current
V(vc)*{ki}
R4
{1/gi}
R3
1
Simplify and Rearrange
 Rearrange sources to make the circuit look simpler
out
Vout
B3
Current
V(out)*{gf}
B4
Current
V(vc)*{ko}
B1
Current
V(out)*{gr}
R5
{1/go}
Vout g f  Vc ko  Vout  g o  gi   Vout g r  Vc ki 
Vout
R1
k 0  ki
H0 
g f   g 0  gi   g r 
B5
Current
V(vc)*{ki}
R3
1
R4
{1/gi}
Mathcad® calculations
ko  ki


  14.563 dB
H0  20 log
g  g  g  g  1 
 f  o i r RL 


1
R1
35 • Chris Basso – – CM Boost Study
Put Storage Elements Back in Place
 Rearrange sources to make the circuit look simpler
out
Vout
B3
Current
B4
Current
V(out)*{gf}
V(vc)*{ko}
R5
{1/go}
C2
{Cs}
1
C3
200uF
B6
Current
V(c,out)*{gr}
vc
V2
471m
AC = 1
c
L2
{L}
B5
Current
V(vc)*{ki}
R4
{1/gi}
3
16
R3
1
R6
60m
2
 Label each storage element with a time constant 
 Time to call the FACTs! Set excitation to 0, Vc = 0
FACTS: Fast Analytical Circuit TechniqueS
36 • Chris Basso – – CM Boost Study
Polynomial Form of a 3rd Function
 A transfer function is made of a numerator and a denominator
H s 
zeros
poles
N s
D s
 The denominator combines the circuit time constants
D  s   1  1   2   3  s  1 21   1 31   2 32  s 2   1 21 312  s 3
Time constant
Time constant
with Cs
with Cout
Time constant
with L
 To calculate time constants, suppress the excitation, all is dc
37 • Chris Basso – – CM Boost Study
Calculate the First Time Constant
 What resistance does Cs « see » ?
out
R?
=0
B3
Current
V(out)*{gf}
B4
Current
V(vc)*{ko}
=0
R5
{1/go}
tau1
Vc
c
V2
0
AC = 1
L shorted
B6
Current
V(out)*{gr}
B5
Current
V(vc)*{ki}
tau3
C open
R2
100G
R3
1
R4
{1/gi}
1
R6
60m
R1
1f
 The excitation is 0 so several sources go away
 You can further simplify the circuit and test it in SPICE
38 • Chris Basso – – CM Boost Study
Calculate the First Time Constant
 Check the result with a simple dc operating point calculation
out
495m V
B3
Current
V(out)*{gf }
0V
B4
Current
V(vc)*{ko}
R5
{1/go}
tau3
C open
I1
1
B6
Current
V(out)*{gr}
tau1
Vc
B5
Current
V(vc)*{ki}
R4
{1/gi}
L shorted
R3
1
297f V
1
R6
60m
c
V2
0
AC = 1
R2
100G
R1
1f
IT  Vout  g r  g f  
out2
IT
Req
495mV
VT
I2
1
R
R7
{1/gi}
B2
Current
R8
{1/go}
R9
1
V(out2)*({gr}-{gf })
VT Vout


1
IT
IT
gr  g f 
Req
1
1 1
Req1  RL || ||
gi go
1
Req
Vout
Req
1
 0.495
 1  RCs
 g r  gf
39 • Chris Basso – – CM Boost Study
Calculate the Second Time Constant
 What resistance does L « see » ?
out
Vout
=0
B3
Current
V(out)*{gf}
B4
Current
V(vc)*{ko}
R5
{1/go}
tau1
Cs open
=0
B6
Current
V(c,out)*{gr}
B5
Current
V(vc)*{ki}
tau3
C open
R4
{1/gi}
vc
c
V2
471m
AC = 1
R?
 The excitation Vc is 0 so simplify the circuit
 Rearrange the network and test it in SPICE
40 • Chris Basso – – CM Boost Study
R3
1
1
R6
60m
Calculate the Second Time Constant
out
Voutx
-48.8V
B4
Current
B3
Current
V(out)*{gf}
R5
{1/go}
V(vc)*{ko}
B6
Current
V(c,out)*{gr}
vc
0V
B5
Current
V(vc)*{ki}
R3
1
R4
{1/gi}
-52.9V
c
V2
0
AC = 1
I1
1
-52.9V
B1
Current
out2
-48.8V
Vout2
V(out2)*{gf }
c1
R2
{1/go}
IT
VT
B8
Current
V(c1)*{gr}
I2
1
R7
{1/gr}
R1
{1/gi}
R6
1
R
VT
IT
41 • Chris Basso – – CM Boost Study
Calculate the Second Time Constant
 Express IT and Vout then rearrange to unveil VT/IT
IT  Vout g f 
VT  Vout
1
g0
R
Vout   IT  VT gr  Req
1  Req  g f  go 
VT

IT go  Req g r  g f  go 
Req 2  RL ||
2 
1 1
||
gi g r
L
1  Req  g f  go 
go  Req g r  g f  go 
1  Req2   g f  g o 
g o  Req2 g r  gf  g o
42 • Chris Basso – – CM Boost Study
 52.897
Same result as dc point simulation
Calculate the Third Time Constant
 What resistance does Cout « see » ?
out
Vout
=0
B3
Current
V(out)*{gf}
B4
Current
V(vc)*{ko}
=0
tau1
Cs open
R5
{1/go}
B6
Current
V(c,out)*{gr}
vc
B5
Current
V(vc)*{ki}
R?
1
R6
60m
c
V2
471m
AC = 1
R3
1
R4
{1/gi}
tau2
L shorted
 The circuit is very close to that of 1
 Rearrange the network and test it in SPICE
43 • Chris Basso – – CM Boost Study
Calculate the Third Time Constant
 Same resistance as for 1 plus rC in series
out2
495mV
I3
1
B2
Current
V(out2)*({gr}-{gf})
R7
{1/gi}
R8
{1/go}
R9
1
-60.0mV
6
R2
60m




1
3  
 rC  Cout
g g  1

f
 r

R
eq
1


44 • Chris Basso – – CM Boost Study
Req1  RL ||
1 1
||
gi g o
First Coefficients a1 and a2
 FACTs tell us that a1 sums up all time constants
a1   1   2   3
Dimension is time
 For a2, we multiply combined-time constants
a2  1 21   1 31   2 32
Dimension is time2
1
 What is this new time constants definition, 2 ?
Cs (HF)
Cs (HF)
 12
L (HF)
 31
L C (dc)
 32
L (dc) C
R?
C Cs (dc)
R?
R?
V. Vorpérian, “Fast Analytical Techniques for Electrical and Electronic Circuits”, Cambridge Press, 2002
45 • Chris Basso – – CM Boost Study
Calculate the Terms for a2
 What resistance does L while Cs is a short and Cout is open?
out
Cs (HF)
 12
B3
Current
V(out)*{gf }
L C (dc)
R?
B4
Current
V(v c)*{ko}
R5
{1/go}
tau1
Cs HF
R1
1u
tau3
C open
B6
Current
V(c,out)*{gr}
Vc
V2
0
AC = 1
c
I1
1
R?
Go for an intermediate step
46 • Chris Basso – – CM Boost Study
B5
Current
V(v c)*{ki}
R3
1
R4
{1/gi}
1
R6
60m
A Simplified Drawing for a Simple Answer
 What R? does L see while Cs is a short and Cout is open?
out2
B1
Current
V(out2)*{gf }
R7
{1/go}
out3
tau1
Cs HF
R10
1u
B8
Current
R11
{1/gr}
R2
{1/gi}
I3
1
R8
1
B7
Current
V(c2)*{gr}
R15
{1/gr}
V(out3)*{gr}
c2
I2
1
R?
L
1
 12 
1
 gr
1 1
R1 ||
||
g r gi
47 • Chris Basso – – CM Boost Study
The Mid Term is Easy to Get
 L and Cs ensure a complete short over RL
out
Cs (HF)
 31
R?
Cs is HF
L (dc) C
R?
R11
{1/gr}
R8
1
1
R7
60m
2
L is dc
 31  rC Cout
48 • Chris Basso – – CM Boost Study
R2
{1/gi}
R9
{1/gi}
R13
1
For the Last Term, L is open
 What resistance does Cout see when Cs and L are open?
out
L (HF)
 32
B3
Current
V(out)*{gf}
C Cs (dc)
B4
Current
V(vc)*{ko}
tau1
Cs dc
I1
1
R?
B6
Current
V(c,out)*{gr}
B5
Current
V(vc)*{ki}
R4
{1/gi}
Vc
R3
1
1
R6
60m
c
V2
0
AC = 1
R?
R5
{1/go}
tau2
L is HF
Too complex! Go for another combination.
 2 32   3 23
49 • Chris Basso – – CM Boost Study
Reorder Time Constants for Simpler Sketch
 The schematic now looks simpler…
out
out2
B1
C urrent
B2
Current
V(out)*{gf }
R2
{1/go}
V(out2)*{gf }
B8
Current
V(c2)*{gr}
R9
{1/gr}
R1
{1/gi}
R8
60m
R3
{1/go}
R7
1
B3
Current
V(T)*{gr}
VT
I2
1
R?
Cout (HF)
 23
L Cs (dc)
R?
50 • Chris Basso – – CM Boost Study
VT
I1
1
R?
Req
Extract the New Time Constant Definition
out
i1
B1
Current
Req  rC || RL ||
IT
i2
R2
{1/go}
V(out)*{gf}
1 1
||
gi g r
i2  IT  VT g r
2
B8
Current
IT
IT  i1  Vout g f
Req
V(c2)*{gr}
I2
1
R?
i1 
VT
Vout  Req  IT  VT g r 
substitute
IT 
VT  Vout
 Vout g f
1
go
VT  Vout
1
go
R

1  Req g f  go


g0  Req g r g f  g o
 23 

L
R
51 • Chris Basso – – CM Boost Study
…And a3 is ?
 For a3, we multiply by a third time-constant
a3   1 12 31,2
Dimension is time3
 What is this new time constant definition?
Cs (HF) L (HF)
 31,2
out
C (dc)
7.40V
B3
Current
V(out)*{gf}
R?
0V
Vc
B4
Current
V(vc)*{ko}
R5
{1/go}
tau1
Cs HF
R1
1u
7.40V
V2
0
AC = 1
52 • Chris Basso – – CM Boost Study
c
R12
1G
L is in HF
B6
Current
V(c,out)*{gr}
B5
Current
V(vc)*{ki}
I1
1
R?
R4
{1/gi}
-60.0mV
1
R6
60m
R3
1
Rearrange and Simplify
out
7.40V
parameters
Fsw =1Meg
Tsw =1/Fsw
L=5u
Cs=1/(L*(Fsw *3.14)^2)
Ri=-50m
Se=0
Vin=2.7
Vout=5
B3
Current
V(out)*{gf }
0V
Vc
B4
Current
V(v c)*{ko}
tau1
Cs HF
R5
{1/go}
R1
1u
B6
Current
V(c,out)*{gr}
B5
Current
V(v c)*{ki}
c
R12
1G
R3
1
-60.0mV
1
R6
60m
7.40V
V2
0
AC = 1
I1
1
R?
R4
{1/gi}
L is in HF
Sn=(Vac/L)*Ri
Sf=(Vap/L)*Ri
Vc=471m
Vac=-Vin
Vap=-Vout
Vcp=(-Vout+Vin)
7.40V
Ic=(Vc/Ri)-D*Tsw *Se-Vcp*(1-D)*Tsw /(2*L)
D=Vcp/Vap
D'=1-D
ki=D/Ri
gi=D*(gf-Ic/Vap)
gr=(Ic/Vap)-go*D
ko=1/Ri
go=(Tsw /L)*(D'*Se/Sn+0.5-D)
gf=D*go-(D*D')*Tsw /(2*L)
I2
1
R?
6
R2
{1/gi}
-60.0mV
R8
1
2
R9
60m
 1

|| rL  rC  Cout
 gi

 31,2  
53 • Chris Basso – – CM Boost Study
Final Denominator Expression
a1   1   2   3








1
L
1



a1 
C 

 rC  Cout
1  s
1



1  Req 2 g f  g o
 gr  g f  R 
 gr  g f  R

eq
1
eq
1




g o  Req 2 g r g f  g o




a2  1 12   1 31   2 32   1 12  1 31   3 23




1

C
a2 
1  s

 gr  g f  R 
eq1 

1 12
1 2 3
a3    
L
1
1
1 1
R1 ||
||
g r gi








1
1
L




Cr C 
 rC  Cout
1  s C out 
1


1  Req 2 g f  g o
 gr  g f  R 
 gr  g f  R

eq1 
eq1


 gr 
g 0  Req 2 g r g f  g o





1

C
a3 
1  s

g

g

r
f

Req1 

L
1
R1 ||
54 • Chris Basso – – CM Boost Study


1
 gr
1 1
||
g r gi
 1

 || rL  rC  Cout
 gi


Find the Zeros
 A zero means the excitation does not generate a response
Ip
Ic
out
Vout
Ia
B3
Current
V(out)*{gf}
B4
Current
V(vc)*{ko}
R5
{1/go}
I1
C2
{Cs}
C3
200uF
c
B6
Current
V(c,out)*{gr}
Ic
vc
V2
471m
AC = 1
B5
Current
V(vc)*{ki}
R3
1
1
R4
{1/gi}
R6
60m
L2
{L}
No response means that:
Ip  0
I p  I1
Short circuit
Ip  0
Vout  0
s z1  
rC  sCout  0
1
rC Cout
55 • Chris Basso – – CM Boost Study
Find the Zeros
 A Capitalize on the 0-V response to remove sources
Ic
Ip
out
0
Vout
Ia
B3
Current
V(out)*{gf}
B4
Current
V(vc)*{ko}
R5
{1/go}
C2
{Cs}
C3
200uF
c
B6
Current
V(c,out)*{gr}
Ic
vc
V2
471m
AC = 1
L2
{L}
I c  Vc ko  v c  g o  v c  sCs
v c  g r  Vc ki  Vc ko  v c  g o  v c  sC s
56 • Chris Basso – – CM Boost Study
R3
1
R6
60m
v c  g r
v c  g r  Vc ki  I c
Ia  I p
1
R4
{1/gi}
B5
Current
V(vc)*{ki}
v c 
sL
 Vc ko  v c  g o  v c  sCs
v c  
Vc ko
1
 g o  sCs
sL
Find The Zeros
 After substituting v(c) into the equation rearrange it

Vc ki  Lg o ki s  Lg r ko s  Cs Lki s 2
2
1  sLg o  s Cs L

Vc ko
1  sLg o  s 2 Cs L
ki  Lg o ki s  Lg r ko s  Cs Lki s 2  ko

 g o ki  g r ko  2 Cs Lki
s
ki  k o
 ki  ko 
 ki  ko  1  sL 


  0

 g k  g r ko  2 Cs Lki
1  sL  o i
0
s
ki  ko
 ki  ko 
For low Q values, the 2nd order polynomial can expressed as
 a 
1  a1 s  a2 s 2  1  a1 s   1  2 s 
a1 

57 • Chris Basso – – CM Boost Study
Find the Zeros – Final Lap!
 The first zero is a right half-plane zero
 g k  g r ko 
s
1  sL  o i
  1
s z1
 ki  ko 
s z1 
ki  ko
L  g o ki  g r ko 
 z1 
1  D 2 RL
L
Negative value!
 The second zero is given by the capacitor ESR
1  srC Cout  0
 z2 
1  D 2 RL
L
 The third zero is located at high frequency
1 s
C s ki
s
 1
g o ki  g r k o
s z3
s z3 
g o ki  g r k o
C s ki
Neglect it
z2
4
fz2 
 3.16  10 kHz
2
V. Vorpérian,“Analytical Methods in Power Electronics", 3-day course in Toulouse,2004
58 • Chris Basso – – CM Boost Study
Arranging the Transfer Function
 The denominator has to be rearranged a little bit…
D  s   1  b1 s  b2 s 2  b3 s3
 The term 1+b1s dominates at low frequency
 b

b
1  b1 s  b2 s 2  b3 s 3  1  b1 s   1  2 s  3 s 2 
b1 
 b1

1  b1s  1 

 p1
 
b
b2
s
s  3 s 2   1 

b1
b1    p1
T  S 
2
 sw 3  1  a 
R
LM  S n 
 L
Cout
Qp 
1
  mc D ' 0.5 
n 

Tsw




Dc gain H0
1
1
 s 
s


n Q p  n 
mc  1 
2
H0 
RL
Ri
1
RLTsw  1 S a 
2M 
 

LM 2  2 S n 
Sa
Sn
59 • Chris Basso – – CM Boost Study
Final Expression
 The transfer function of a CCM Boost Converter in CM is


L
1  s
 1  srC Cout 
2

1  D  RL 
Vout  s  RL
1
1


2
Cout
Vc  s 
Ri
RLTsw  1 S a 
 s 
s
1 s
2M 



1




T  S 
2
LM 2  2 Sn 
n Q p   n 
 sw 3  1  a 
RL LM  Sn 
Sa is the external ramp compensation to damp subharmonic oscillations
60 • Chris Basso – – CM Boost Study
Check SPICE Versus Mathcad
 Full-featured transfer functions versus SPICE
Vout  s 


100
Vc  s 
40

20 log H1 i 2  f k  10


arg H 1 i 2  f k  
20
180
Vout  s 
Vc  s 
0

 100
0
10
100
3
110
110
4
110
5
6
110
10
100
3
110
fk
 Simplified transfer functions versus SPICE
Vout  s 
40


100
Vc  s 


180


180
arg H1 i 2  f k  

20
20 log H 1 i 2  f k  10
20 log H 2 i 2  f k  10
arg H2 i 2  f k  

110
4
5
110
110
fk
Vout  s 
Vc  s 

0

 100
0
10
100
110
3
110
4
110
5
110
6
10
100
3
4
110
110
5
110
fk
fk
61 • Chris Basso – – CM Boost Study
Course Agenda
 The PWW Switch Concept
 Small-Analysis in Continuous Conduction Mode
 Small-Signal Response in Discontinuous Mode
 EMI Filter Output Impedance
 Cascaded Converters Operation
62 • Chris Basso – – CM Boost Study
6
110
6
The CM Boost Converter in DCM
 To check the operating mode, calculate the critical load
Fsw  1 MHz
2 Fsw LVout 2


V 
Vin 2  1  in 
 Vout 
Rcrit
Rcrit  74.7 Ω
L  5 µH
Vin  2.7 V
Vout  5 V
iL  t 
I L , peak
iL  t 
I L ,valley
CCM, RL < Rcrit
Tsw
t
I L , peak
DCM, RL > Rcrit
t
I L ,valley  0
deadtime
63 • Chris Basso – – CM Boost Study
Transfer Function of DCM CM Boost
 The converter is a first-order converter at low frequency
H s 
Sn 
 p2
1
S n mcTsw
Vin
Ri
L

s
1 

2Vout M  1   z1
D 2M  1 
s
1 
 p

1
 z1 
1

1 M
 2 Fsw 
 D

64 • Chris Basso – – CM Boost Study





1
rC Cout
 z2 
2
mc  1 
Sa
Sn

s 
 1 

  z 
2 


s 
 1 

 p 

2 
Rload
2
M L
 p1 
1 2M  1
RL Cout M  1
The Ac Response of the DCM CM Boost
 A RHPZ is still present in DCM with a high-frequency pole
40
Vout  s 


20 log H 3 i 2  f k  10
180
arg H3 i 2  f k  


0
 20
 40
10

100
Vc  s 
20

Vout  s 
Vc  s 
0
 100
100
3
4
110
110
5
1 10
110
6
10
100
110
3
110
4
110
5
SPICE simulation versus Mathcad
65 • Chris Basso – – CM Boost Study
Why a RHP Zero in DCM?
 The RHP Zero is present in CCM and is still there in DCM!
Id(t)
iL,peak
D3Tsw
t
D1Tsw
D2Tsw
Tsw
 When D1 increases, [D1,D2] stays constant but D3 shrinks
66 • Chris Basso – – CM Boost Study
6
110
fk
fk
Why a RHP Zero in DCM?
 The triangle is simply shifted to the right by d̂1
Id(t)
iL,peak
D3Tsw
d̂1
t
D1Tsw
D2Tsw
Tsw
 The capacitor refueling time is delayed and a drop occurs
67 • Chris Basso – – CM Boost Study
The Refueling Time is Shifted
 If D increases, the diode current is delayed by d̂1
D(t)
Id(t)
2.02m
2.04m
1.85m
1.99m
d̂1
2.06m
2.09m
2.11m
2.13m
2.27m
2.41m
Vout(t)
Vout(t)
2.02m
68 • Chris Basso – – CM Boost Study
2.04m
2.06m
2.09m
2.11m
Simulation, no
peak increase
Course Agenda
 The PWW Switch Concept
 Small-Analysis in Continuous Conduction Mode
 Small-Signal Response in Discontinuous Mode
 EMI Filter Output Impedance
 Cascaded Converters Operation
69 • Chris Basso – – CM Boost Study
EMI Filter Interaction
 A LC filter is included to prevent the dc line pollution
L
Vg
Vin
Vout
C
Rload
Z s?
Switching converter
 What load does the converter offer?
C. Basso, “Designing Control Loops for Linear and Switching Power Supplies”, Artech House, 2012
70 • Chris Basso – – CM Boost Study
A Negative Incremental Resistance
 Assume a 100%-efficient converter:
Pout  Pin
I inVin  I outVout
 In closed-loop operation, Pout is constant
I in Vin  
Pout
Vin
 For a constant Pout , if Vin increases, Iin drops
1
0.8
Iin  A 
dI in Vin 
0.6
dVin
P 
d  out 
V
P
  in    out2
dVin
Vin
0.4
0.2
10
15
20
25
The incremental input
resistance is negative
30
Vin  V 
71 • Chris Basso – – CM Boost Study
A Simple LC Filter
 The low-pass filter is built with a L and C components:
rL
1  s  z1
T s 
L
rC
Vg
1
s
Q
72 • Chris Basso – – CM Boost Study
R   R  C

RR
L
1
0 
C
Q
z 
0Q  0 
R
If R is || with -R
 s 
 
2
1
rC C
1
LC
rL  R
rC  R
LC0  rC  R 
C
 Rload
L  C  rL rC  rL R  rC R 
L
A negative resistance cancels
losses: poles become imaginary
A Negative Impedance Oscillator
 If losses are compensated, the damping factor is zero
T s 
1
s2
02
Q
s

1
0Q
1
2
If ohmic losses are gone, the
damping factor is zero, Q is infinite.
 Without precautions, instability can happen!
Input voltage
150
rL
L
rC
Vin
130
I
110
C
I
Output voltage
90.0
 0
 >0
70.0
Pout (constant)
Vout
 <0
Negative resistance
3.64m
10.9m
18.2m
25.5m
32.7m
73 • Chris Basso – – CM Boost Study
Filter Output Impedance
 What is the output impedance of an LC filter?
VT  s 
L
rC
IT  s 
It follows
the form
Z out  s   R0
R
rL
C
Ω
N s
D s
Ω
No dimension
Z out  s    rL || Rload 
74 • Chris Basso – – CM Boost Study

L
1 s 
 rL  Rload

L
1  s  1  srC Cout 
rL 



r  Rload 
 C  rL || Rload   rC    s 2  LC C

rL  Rload 


Negative Resistance at Low Frequency
 Neg. resistance exists because of feedback (Pout = constant )
R4 2.70V
1.5m
2.70V
VZin
5
L1
5u
2.70V
3
18
PWMCM2_L
X1
L = 5u
Fs = 1Meg
Ri = -40m
Se = 16.4k
c
L2
10GH
p
2.70V
7
a
D
Resr
60m
SW
2
R9
1k
vc
536mV
vout
4
5.00V
duty_cycle
duty-cycle
Vin
2.7
L ossy PWM
switch CM
I1
AC = 1
-455mV
-301mV
5.00V
Rload
5
16
11
Cout
200uF
13
Rdson
140m
D1
mbra210et3
param eters
C2
{C2}
Rupper=10k
fc=8k
pm=70
Gfc=-20
pfc=-101
C1
{C1}
R7
1k
99.9mV
R2
{R2}
15
14
R1
{Rupper}
2.50V
R6
10k
G=10^(-Gfc/20)
boost=pm-(pfc)-90
pi=3.14159
K=tan((boost/2+45)*pi/180)
C2=1/(2*pi*fc*G*k*Rupper)
C1=C2*(K^2-1)
R2=k/(2*pi*fc*C1)
2.50V
6
1.10V
9
Verr
2.50V
1
V2
2.5
R3
10k
X2
AMPSIMP
VHIGH = 100
VLOW = 10m
fp=1/(2*pi*R2*C2)
fz=1/(2*pi*R2*C1)
Sw itchMode Pow er Supplies: SPICE Simulations and Practical Designs
Christophe Basso - McGraw -Hill, 2014
75 • Chris Basso – – CM Boost Study
Negative Resistance at Low Frequency
 The resistance is truly negative up to 200 Hz
1
2
24.0
12.0
(dBΩ)
0
-12.0
1.02 Ω
Zin  f 
-24.0
144
72.0
(°)
Z in  f 
In this region
Pout  constant
0
-72.0
-144
Z in  180
10
76 • Chris Basso – – CM Boost Study
100
1k
10k
100k
1Meg
Negative Resistance is One Part Only
 Negative resistance problems occur only when Z in  180
 Beyond this region, instability is linked to gain degradation
dBΩ
65.0
undamped
36 dBΩ
35.0
Negative
argument
region
5.00
damped
3.6 /100 µF
-25.0
Z out  s 
-55.0
10
100
1k
10k
100k
77 • Chris Basso – – CM Boost Study
Open-Loop Gain Degradation
 Here the EMI filter peaks at high frequency
 The open-loop is severely affected and stability is at stake
No damping
Zout = 57.5 dB
Problem!
40.0
90.0
20.0
Tf 
Without filter
7
30.0
gain
Plot1
ph_vout2#1, ph_vout2 in degrees
ZinSMPS
p in 1 / ampere
180
18 dB
10.0
-10.0
1
-30.0
0
vdbout2, vdbout2#1 in db(volts)
50.0
With filter
0
-90.0
-20.0
-180
-40.0
T  f 
phase
Without filter
3
Zout
10m
100m
With filter
1
10
100
frequency in hertz
1k
10k
100k
Filter output impedance
versus closed-loop buck input impedance.
78 • Chris Basso – – CM Boost Study
10
100
1k
frequency in hertz
10k
Buck open-loop gain with
and without input filter.
1
4
2
100k
Evaluating How Loop Gain is Modified
 The converter transfer function is evaluated at Zout = 0
H  s
Z out  s  Z in  s 
vˆout
vˆg
vˆin  0
d̂
For Zout = 0
T  s  H sG  s 
Vout  s 
D  s V
in
G s
 s  0
79 • Chris Basso – – CM Boost Study
Considering the Filter Output Impedance
 In reality, the EMI output impedance makes Z out  s   0
 The converter ac input voltage is no longer zero
Z out  0
vˆin  0
H s
vˆout
H s
vˆout
Z out  s 
Z out  0
d̂
T s 
Vout  s 
D s
d̂
G s
T s 
Vin  s   0
80 • Chris Basso – – CM Boost Study
Vout  s 
D s
Vin  s   0
G s
Extra Element Theorem to Help
 It can be shown how an EMI affects the open-loop gain:
H s
Vout  s 
without the filter
vˆout T  s   D  s  Z  sGain
0
out
Z out  s 
Z out  s 
d̂
T  s 
Filter output impedance
G s
Vout  s 
D  s Z
Z out  s 
V s
ZN  s
 out
Z s
D  s  Z s 0
out  
1  out
ZD  s 
1
out
 s  0
81 • Chris Basso – – CM Boost Study
What are ZD and ZN?
 ZD and ZN come from the Extra Element Theorem, EET
Z D  s   Zi  s  D s 0
Open-loop input impedance
Z N  s   Zi  s  V
Input impedance for Vout  s   0
 
out
 s 0
ZD  s  ?
Null
ZN  s  ?
Vout
vˆout  0
D0  35%
dˆ  0
iˆout  0
Open-loop input impedance
82 • Chris Basso – – CM Boost Study
Vout
vˆout  0
Ideal control
D0  35%
dˆ  0
iˆout  0
Input impedance for Vout  s   0
What are ZD and ZN for a Boost Converter?
 These values have already been derived
sL 

Z N  s    D '2 R  1  2 
 D' R 
L
LC 

1  s 2  s2 2 

D' R
D'
Z D  s    D '2 R 

1

sRC




 The original loop gain (without filter) is untouched if
Z out  s 
ZN  s
1
Z out  s 
1
ZD  s 
Z out  s   Z N  s 
1
Z out  s   Z D  s 
83 • Chris Basso – – CM Boost Study
Checking for Stability (1)
 You design the filter together with the converter
 Plot |Zout|, |ZN | and |ZD| in the same graph
 Check that no overlap exists. If so, damp filter
60
40
 Z out  i 2  f k 

20 log
10 20



 Z D i 2  f k 

20 log
10 



 Z N i 2  f k 
20 log


ZN  s
ZD  s
0

10   20

Z out  s 
 40
Need to redesign the filter!
 60
1
10
100
3
110
fk
84 • Chris Basso – – CM Boost Study
110
4
5
110
6
110
Checking for Stability (2)
 You design the filter after the converter
 Plot |Zout|, |Zin | in the same graph
 Check that no overlap exists. If so, damp filter
50.0
7
30.0
R damps
Ok, margin is 8 dB min
ZinSMPS
18 dB
Rdamp= 5 
10.0
Rdamp= 4 
C cuts dc
-10.0
Damping
network
1
4
5
6
3
2
Rdamp= 1 
Rdamp= 2 
-30.0
Rdamp= 3 
Zout
10m
100m
1
10
100
frequency in hertz
1k
10k
85 • Chris Basso – – CM Boost Study
Course Agenda
 The PWW Switch Concept
 Small-Analysis in Continuous Conduction Mode
 Small-Signal Response in Discontinuous Mode
 EMI Filter Output Impedance
 Cascaded Converters Operation
86 • Chris Basso – – CM Boost Study
100k
Cascading Converters
 When cascading converters, impedances also matter
Vin  s 
Z th  s 
Z in  s 
Vth  s 
Z out  s  Z in  s 
Boost
Buck
 The system can be modeled by a gain TM
Vin  s   Vth  s 
1
Z th  s 
1
Z in  s 
Vth  s 
+
Vin  s 
-
Z th Z in
TM  s 
87 • Chris Basso – – CM Boost Study
Check the Stability of the Minor Loop (1)
 The stability test requires the comparison of Zout and Zin
 The buck converter closed-loop input impedance is
1
Z in  s 

T s
1
1

ZN  s 1 T  s ZD  s 1 T  s
1
Low frequency
T  s   1
Z in  s   Z N  s 
T s  H sG s
 ZN and ZD are defined as
Z D  s   Z i  s  D s 0
Open-loop input impedance
Z N  s   Zi  s  V
Input impedance for Vout  s   0
 
out
R
ZD  s  2
D
 s  0
L 2
 s LC
R
1  sRC
1 s
88 • Chris Basso – – CM Boost Study
ZN s  
R
D2
Use SPICE Models First
 Simulate the boost converter output impedance
L1
33u
R5
7m
12
Closed-loop Zout
p
c
11
2
X3
PWMCM
L = 33u
Fs = 200k
Ri = -300m
Se = 82k
vc
Vin
11
vout
vout
PWM switch CM
3
a
duty-cycle
duty_cycle
R10
18m
Z out  f 
0
Rload
8.3
16
I1
AC = 1A
C5
820u
-20.0
C3
282p
-40.0
C2
10n
R2
50k
R1
11.66k
9
R4
1m
-60.0
1
5
8
6
R3
1.35k
Verr
V2
2.5
-80.0
1
X2
AMPSIMP
VHIGH = 10
VLOW = 10m
(dBΩ)
10
100
1k
10k
100k 1Meg
11-15 V/24 V – 3 A
89 • Chris Basso – – CM Boost Study
Use SPICE Models First
 Simulate the buck converter input impedance
Zi n
duty_cycl e
Closed-loop Zin
210mV
L2
1kH
5
24.0V
a
duty-c yc le
c
6
24.0V
2.37V
1
I1
AC = 1
3
R4
4m
vout
5.00V
PWM switch CM
Vout
4
vc
7
V4
24
L1
8u5.05V
5.05V
19.8
Re sr
14m
p
PWM CM
X2
L = 8u
Fs = 2 00k
Ri = 1 76m
Se = 30k
5.00V
R2
416m
2
Cout
820u
19.4
C3
1n
vout
19.0
C1
3.5 n
R6
15k
2.37V
R1
1k
8
R7
1m
Z in  f 
2.50V
18.6
9
2.37V
11
2.50V
10
Verr
V2
2.5
R3
1k
18.2
10
24 V/5 V – 12 A
90 • Chris Basso – – CM Boost Study
(dBΩ)
100
1k
10k
100k 1Meg
Compare Magnitude Curves
39.0
19.0
Z in  f 
(dBΩ)
No overlap
-1.00
-21.0
Z out  f 
-41.0
10
100
1k
10k
100k
1Meg
91 • Chris Basso – – CM Boost Study
Compare Phase Curves
270
Neg. up to 1 kHz
180
Z in  f 
(°)
90.0
0
Z out  f 
-90.0
1
92 • Chris Basso – – CM Boost Study
10
100
1k
10k
100k
1Meg
Check Minor Loop Gain
 There is no gain, Z out  Z in . The system is stable
dB °
-40.0
0
-50.0 -90.0
-60.0 -180

-70.0 -270
Z out  f 
Z out  f 
Z in  f 
Zin  f 
-80.0 -360
1
10
100
1k
10k
100k
1Meg
93 • Chris Basso – – CM Boost Study
With Cascaded Converters
L1
33u
11.0V
R5
7m
12
dc_buck
10
c
a
p
2
X3
PWMCM
L = 33u
Fs = 200k
Ri = -300m
Se = 82k
vc
545mV
a
duty-cycle
dc_boost
24.1V
PWM switch CM
3
4
duty-cycle
5.05V
c
R4
4m
L2
8u5.05V
11
11.0V
Vin
11
Iout
209mV
VoBoost
11.0V
14
VoBuck
19
2.37V
R10
18m
24.1V
18
vc
PWM switch CM
Resr
14m
p
PWMCM
X1
L = 8u
Fs = 200k
Ri = 176m
Se = 30k
16
C5
820u
5.00V
C6
1n
v out
Boost
C2
10n
R2
50k
2.50V
R7
1m
1
2.02V
2.50V
21
2.37V
23
2.50V
2.50V
6
X2
AMPSIMP
VHIGH = 10
VLOW = 10m
V2
2.5
R9
1k
24
2.50V
7
C1
3.5n
R6
15k2.37V
R1
11.66k
9
94 • Chris Basso – – CM Boost Study
I1
unknown
13
Cout
820u
C3
282p
VerrBoost
v out5.00V
15
R3
1.35k
Buck
VerrBuck
22
V3
2.5
R11
1k
Check Transient Response
 Transient response is good for the buck
5.10
Buck
5.05
(V)
5.00
4.95
Iout from 5 to 12 A in 1 µs
4.90
vout  t 
24.10
24.05
(V)
24.00
23.95
Boost
23.90
100u
300u
500u
vout  t 
700u
900u
95 • Chris Basso – – CM Boost Study
Literature
“Fundamentals of Power Electronics”, R. Erickson, D. Maksimovic
“Designing Control Loops for Linear and Switching Power Supplies”, C. Basso
“Practical Issues of Input/Output Impedance Measurements”, Y. Panov,
M. Jovanović, IEEE Transactions on Power Electronics, 2005
“Physical Origins of Input Filter Oscillations in Current Programmed Converters”
, Y. Jang, R. Erickson, IEEE Transactions on Power Electronics, 1992
“Design Consideration for a Distributed Power System”, S. Schultz, B. Cho,
F. Lee, Power Electronics Specialists Conference, June 1990, pp. 611-617
“Input Filter Considerations in Design and Applications of Switching Regulators”,
R. D. Middlebrook, IAS 1976
96 • Chris Basso – – CM Boost Study
Conclusion
 The PWM switch model is an essential tool for modeling
 It can be used to derive small-signal response of a boost in CM
 Its SPICE implementation helps test various transfer functions
 It can be used to assess open- and closed-loop parameters
 Interactions with the EMI input filter can be analyzed
 Cascaded converters stability can be checked with the model
 Practical measurements on the bench must always be performed
Merci !
Thank you!
Xiè-xie!
97 • Chris Basso – – CM Boost Study