# APEC 2014, 3-hour professional seminar held in Fort Worth, TX ```Small-Signal Modeling at Work
with Power Converters
Christophe Basso
IEEE Senior Member
1 • Chris Basso – APEC 2014
Course Agenda
Introducing the PWM Switch Model
CCM, DCM and BCM in Voltage Mode
Pulse Width Modulator Gain
The PWM Switch Model in Current Mode
PWM Switch at Work in a Buck Converter
A Simplified Approach to Modeling a DCM Boost
Transfer Function of a BCM Boost in Current Mode
Small-Signal Model of The Active Clamp Forward
2 • Chris Basso – APEC 2014
Course Agenda
Introducing the PWM Switch Model
CCM, DCM and BCM in Voltage Mode
Pulse Width Modulator Gain
The PWM Switch Model in Current Mode
PWM Switch at Work in a Buck Converter
A Simplified Approach to Modeling a DCM Boost
Transfer Function of a BCM Boost in Current Mode
Small-Signal Model of The Active Clamp Forward
3 • Chris Basso – APEC 2014
Manipulating Linear Networks
A switching converter is made of linear elements!
rDS ( on )
Vout
rL
L
L
C
Vin
Vin
rd
on time
rL
C
Vout
off time
The non-linearity or discontinuity is coming from transitions
linear
Cannot
differentiate
DTsw
linear
linear
vDRV ( t )
4 • Chris Basso – APEC 2014
(1 − D ) Tsw
on time
off time
Singularity
State Space Averaging (SSA)
Despite linear networks, equation is discontinuous in time
Introduced in 76, SSA weights on and off expressions
weighted
during (1-D)Tsw
xɺ =  A1 D + A 2 (1 − D )  x ( t ) +  B1 D + B 2 (1 − D )  u ( t )
weighted
during DTsw
A is the state coefficient matrix
B is the source coefficient matrix
This equation is now continuous in time: singularity is gone
However, it became a non-linear equation
You need to linearize it by perturbation (or differentiation)
If you add a new element, you have to restart from scratch!
S.Ćuk, &quot;Modeling, Analysis and Design of Switching Converters&quot;, Ph. D. Thesis, Caltech November 1976
5 • Chris Basso – APEC 2014
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
Switching cell
p
Small-signal model
(CCM voltage-mode)
p
V. Vorp&eacute;rian, &quot;Simplified Analysis of PWM Converters using Model of PWM Switch, parts I and II&quot;
IEEE Transactions on Aerospace and Electronic Systems, Vol. 26, NO. 3, 1990
6 • Chris Basso – APEC 2014
The Bipolar Small-Signal Model
A bipolar transistor is a highly non-linear system
Replace it by its small-signal model to get the response
Rb1
Rc
b
Vout
Vg
Vin
Ib
Ic
rπ
β Ib
Q1
Ie
e
Rb2
Re
Ce
Ebers-Moll model
7 • Chris Basso – APEC 2014
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
8 • Chris Basso – APEC 2014
. .
L
C
R
c
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
9 • Chris Basso – APEC 2014
Course Agenda
Introducing the PWM Switch Model
CCM, DCM and BCM in Voltage Mode
Pulse Width Modulator Gain
The PWM Switch Model in Current Mode
PWM Switch at Work in a Buck Converter
A Simplified Approach to Modeling a DCM Boost
Transfer Function of a BCM Boost in Current Mode
Small-Signal Model of The Active Clamp Forward
10 • Chris Basso – APEC 2014
CCM, DCM and BCM Operations
Three types of conduction modes exist
I peak
iL ( t )
iL ( t )
iL ( t )
I peak
iL ( t )
Tsw
I peak
iL ( t )
iL ( t )
Tsw
Discontinuous
Conduction
iL ( t )
Mode
Tsw
Boundary
Conduction
Mode
Third event
iL ( t )
Continuous
Conduction
Mode
iL ( t )
Tsw
Tsw
Tsw
&gt;
I peak
&lt;
I peak
=
I peak
2
2
2
Each mode has its own small-signal characteristics
A model is needed for these three modes!
11 • Chris Basso – APEC 2014
CCM Common Passive Configuration
The PWM switch is a single-pole double-throw model
d
c
a
ia ( t )
ic ( t )
d'
vap ( t )
p
vcp ( t )
Install it in a buck and draw its terminals waveforms
ia ( t )
ic ( t )
d
c
a
L
d'
Vin
C
R
vap ( t )
vcp ( t )
p
Vout
CCM VM
12 • Chris Basso – APEC 2014
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
c
0
= DI c
Tsw
CCM VM
13 • Chris Basso – APEC 2014
The Common Passive Configuration
Average the voltage waveforms across the PWM switch
vap ( t )
vap ( t )
0
t
vcp ( t )
vcp ( t )
t
0
DTsw
vcp ( t )
Tsw
Tsw
= Vcp =
1
Tsw
DTsw
∫
0
vcp ( t ) dt = D vap ( t )
Vcp = DVap
Tsw
Tsw
Averaged
variables
= DVap
CCM VM
14 • Chris Basso – APEC 2014
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
Ic
Vcp
DI c DVap
p
p
CCM VM
15 • Chris Basso – APEC 2014
A Transformer Representation
The PWM switch large-signal model is a dc &quot;transformer&quot;!
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 plugged into any 2-switch CCM converter
c
rL
Vin
L
a
.
.
D
1
p
C
R
Dc bias point
Ac response
CCM VM
16 • Chris Basso – APEC 2014
The Discontinuous Case
In DCM, a third timing event exists when iL(t) = 0
L
iL ( t )
D1
SW
D2
Vin
L
C
R
C
R
C
R
during
D1Tsw
D3
Vin
C
on
R
iL ( t )
off
L
iL ( t )
during
D2Tsw
iL ( t ) = 0
iL ( t ) = 0
during
D3Tsw
t
0
D1Tsw
Tsw
D2Tsw
D3Tsw
DCM VM
17 • Chris Basso – APEC 2014
The Same Configuration as in CCM
Draw the waveforms in the &quot;common passive&quot; configuration
ia ( t )
I peak
a
Ia
D1
D2
c
D3
t
vap ( t )
p
Vin
t
vcp ( t )
Vin
Vout
t
D1Tsw
Tsw
D2Tsw
18 • Chris Basso – APEC 2014
D3Tsw
C
off
t
I peak
Vout
L
on
Vin
ic ( t )
Ic
R
Average the waveforms:
Ia =
I peak
Ic =
I peak
Ic =
2
2
D1
D1 +
I peak
2
D2 =
I peak
2
2 I a D1 + D2
D + D2
= Ia 1
D1
2
D1
( D1 + D2 )
DCM VM
Derive Vcp to Unveil the New Model
The addition of the third event complicates the equations
vcp ( t )
Vcp = Vap D1 + Vcp D3
Vap
Vcp = Vap D1 + Vcp (1 − D1 − D2 )
Vcp
t
D1Tsw
D2Tsw
a
D3Tsw
1
. .
Vcp = Vap
D1
D1 + D2
Control
input
Ic
Ia
D1 + D2 + D3 = 1
c
Ndcm
I a = N dcm I c
Vap =
p
Vcp
N dcm
Ic =
Ia
N dcm
Vcp = N dcmVap
N dcm =
D1
D1 + D2
f ( D1 )
DCM VM
19 • Chris Basso – APEC 2014
Finally, Get the D2 Value
In DCM the inductor average voltage per cycle is always 0
Vcp = Vout
What is the averaged inductor peak current?
I peak =
vL ( t )
vL ( t )
D1Tsw
L
D1Tsw
D1Tsw
Vac = L
= Vac
I peak
D1Tsw
The peak current uses a previous expression
Ic =
I peak
2
( D1 + D2 )
I peak =
D2 =
20 • Chris Basso – APEC 2014
2Ic
D1 + D2
2 LFsw I c
− D1
D1 Vac
DCM VM
Variable Frequency Operation
The power supply operates in a self-oscillating mode
Vbulk
Demag
detector
Np:Ns
Lp
+- 65 mV
+
.
.
.
Vout
Cout
Resr
S
Q
Q
Vdd
R
+
Rpullup
-
FB
GFB
Rsense
CTR
Critical conduction Mode (CRM)
Boundary Conduction Mode
Borderline Conduction Mode (BCM)
21 • Chris Basso – APEC 2014
Benefits of Quasi-Resonance
Wait for core demagnetization and valley voltage
A
iD ( t )
2.10
1.50
0.9
0.3
-0.3
V
230
170
Valley
110
Core is
reset
50
vDS ( t )
0
2.039m
2.052m
2.066m
2.080m
2.093m
BCM VM
22 • Chris Basso – APEC 2014
Idealized Waveforms
Draw the PWM switch waveforms in a buck configuration
ia ( t )
I peak
t
I peak
ic ( t )
ic ( t )
Tsw
t
vcp ( t )
Vap
t
BCM VM
23 • Chris Basso – APEC 2014
Derive Founding Equations
Average current in terminal C is straightforward
I peak
ic ( t )
0
Tsw
=
I peak
2
The off time depends on the voltage across the inductor
toff =
L
I peak
Vcp
I peak = 2 I c
toff =
2 LI c
Vcp
Period and duty ratio come easily as ton is imposed
ton + toff = Tsw
D=
ton
Tsw
Need to transform the error voltage into time
24 • Chris Basso – APEC 2014
BCM VM
Generating the On Time
The on time modulator works as a PWM block
Vdd
ton (Verr ) =
I
Verr
Verr C
I
2
PWM
5
1
C
4
d
C
ton (Verr ) =
dVerr
I
DRV
The modulator small-signal gain is a simple coefficient
C = 100 pF
I = 20 &micro;A
C
= 5u
I
Verr
Assuming 1 V = 1 &micro;s
ton
5
BCM VM
25 • Chris Basso – APEC 2014
Final PWM Switch Model in BCM
Just add the on-time modulator to the VM PWM switch
Verr
k=
Vac
I peak =
Vcp
toff =
C
I
Vac ton
L
L
I peak
Vcp
ton
a
c
1
. .
D=
D
I peak
ton
ton + toff
toff
ton
p
toff
duty-cycle
a
Fsw (kHz)
Ip
c
vc
PWM switch BCM
p
X1
PWMBCMVM
C. Basso, &quot;Switch Mode Power Supplies: SPICE Simulations and Practical Design&quot;, McGraw-Hill 2008
26 • Chris Basso – APEC 2014
BCM VM
Course Agenda
Introducing the PWM Switch Model
CCM, DCM and BCM in Voltage Mode
Pulse Width Modulator Gain
The PWM Switch Model in Current Mode
PWM Switch at Work in a Buck Converter
A Simplified Approach to Modeling a DCM Boost
Transfer Function of a BCM Boost in Current Mode
Small-Signal Model of The Active Clamp Forward
27 • Chris Basso – APEC 2014
In voltage mode, the duty ratio depends on Verr
v (t )
verr ( t ) = Vp
Vp
+
vPWM ( t )
-
D (t )
verr ( t )
60
100
30
10
28 • Chris Basso – APEC 2014
ton ( t )
Tsw
=
1
Vp
t
ton ( t2 )
ton ( t1 )
D (t ) =
Tsw
verr ( t )
t
%
ton ( t )
d (t )
t
d
1
D (Verr ) = GPWM =
dVerr
Vp
Including the PWM Contribution
In a simulation fixture, insert a gain block after Verr
a
D
c
PWM switch VM
GPWM =
PWM switch
model in VM
d
p
1
Vp
Modulator gain
Verr
29 • Chris Basso – APEC 2014
Considering Feedforward
A perturbation disturbs operations and must be fought
In a power supply, the input voltage is a perturbation
iˆout Zˆ out ,OL
GVin vˆin
Vref
-
vˆerr
-
Plant
vˆout
The perturbation must affect the output to trigger action
Why not reacting before the perturbation hits?
This is the principle of feedforward
30 • Chris Basso – APEC 2014
Input Contribution in a Buck Converter
The transfer function of a CCM buck converter includes Vin
V
H ( s ) = in
Vp
1+
1+
s
sz1
 s 
s
+ 
ω0 Q  ωo 
2
Let's make Vp a function of Vin: Vp (Vin ) = kFFVin
H (s) =
Vin
k FF Vin
1+
1+
s
sz1
 s 
s
+ 
ω0 Q  ωo 
2
=
1
k FF
1+
1+
s
sz1
 s 
s
+ 
ω0 Q  ωo 
2
The transfer function no longer depends on Vin
31 • Chris Basso – APEC 2014
How to Make Vp a Function of Vin?
Make the sawtooth capacitor current depend on Vin
Vin
vC ( t )
R
Verr
Verr
PWM
HL
LL
IC
t
C
DRV
HL = hi line
LL = lo line
32 • Chris Basso – APEC 2014
Ramp Amplitude and Input Voltage
We can neglect the sawtooth ramp amplitude
IC ≈
Vin
R
The peak value, Vp, is linked to the time constant τ
Vp =
IC
Vin
V
Tsw =
Tsw = in
Cramp
Cramp Rramp
τ Fsw
In the time domain, the peak value will change
vramp ( t ) = V p
V
t
t
= in
Tsw τ Fsw Tsw
At t = ton, the error voltage equals Vramp
Verr =
Vin ton
V
= in D
τ Fsw Tsw τ Fsw
D (Verr ) = Verr
τ Fsw
Vin
33 • Chris Basso – APEC 2014
An In-Line Equation to Include Feedforward
Differentiate the expression to get the small-signal gain
GPWM =
∂D (Verr )
∂Verr
=
τ Fsw
Vin
=
1
k FF Vin
k FF =
1
Fswτ
The feedforward block requires an ABM source
in
a
R = 82 kΩ
C = 470 pF
Fsw = 500 kHz
c
d
Vin
B1
Voltage
PWM switch VM
p
V(err)/(V(in)*{kFF})
parameters:
k FF =
kFF=52m
err
1
= 52m
500k &times; 470 p &times; 82k
ABM: Analog Behavioral Model
34 • Chris Basso – APEC 2014
Course Agenda
Introducing the PWM Switch Model
CCM, DCM and BCM in Voltage Mode
Pulse Width Modulator Gain
The PWM Switch Model in Current Mode
PWM Switch at Work in a Buck Converter
A Simplified Approach to Modeling a DCM Boost
Transfer Function of a BCM Boost in Current Mode
Small-Signal Model of The Active Clamp Forward
35 • Chris Basso – APEC 2014
Peak Current Mode Control
In voltage-mode, the loop controls the duty ratio
In current-mode, the inductor peak current is controlled
PWM latch
clock
vc ( t )
R
Q
vr ( t )
iL ( t ) Ri + vr ( t )
+
iL ( t ) Ri
Slope
comp.
a
Ri
D
c
D'
p
An artificial ramp is added for stabilization purposes
36 • Chris Basso – APEC 2014
We Want the Average Current Definition
The value Ic is the inductor current at half the ripple
ic ( t )
a
I peak
ic ( t )
S a Ri
∆I L
b
Tsw
S2
0
t
DTsw
(1 − D ) Tsw
Tsw
Current at point b is that of a minus half the inductor ripple
ic ( t )
Tsw
= Ic =
Vc S a
S D ' Tsw
− DTsw − 2
Ri Ri
2
CCM CM
37 • Chris Basso – APEC 2014
Define the Converter off-Slope
Use a buck configuration to see voltages at play
ia ( t )
ic ( t )
d
c
a
L
d'
Vin
C
R
vap ( t )
vcp ( t )
p
Vout
The downslope depends on the output voltage Vout:
S2 = −
Vout
L
The inductor average voltage is 0 at steady-state
Vcp = Vout
38 • Chris Basso – APEC 2014
S2 = −
Vcp
L
CCM CM
A Current Mode Generator
Update the previous equation to obtain final definition
Vc
T
S
− Vcp (1 − D ) sw − a DTsw
Ri
2 L Ri
Ic =
Peak current
setpoint
Half inductor
ripple
Group 2nd
and 3rd terms
I &micro; = Vcp (1 − D )
Tsw S a
+ DTsw
2 L Ri
Compensation
ramp
Inductor ripple and compensation ramp alter peak value
c
c
∆I L
2
Vc
Ri
Vc
Ri
Sa DTsw
Ri
p
I&micro;
p
CCM CM
39 • Chris Basso – APEC 2014
CM or VM Lead to Similar Input Currents
Average the current waveforms across the PWM switch
ic ( t )
ic ( t )
0
t
ia ( t )
ic ( t )
dTsw
Tsw
= Ia =
1
Tsw
dTsw
∫ i ( t ) dt = D i ( t )
a
0
I a = DI c
Tsw
Ia =
t
0
ia ( t )
Tsw
c
Tsw
= DI c
D=
Vcp
Vac
Ic
Vcp
Vap
CCM CM
40 • Chris Basso – APEC 2014
The PWM Switch Model in Current Mode
The final model associates three current sources
c
a
Ia
Vcp
Vin
L
Ic
Vap
Vc
Ri
Ic
I&micro;
C
R
p
This is the large-signal current-mode PWM switch model
V. Vorp&eacute;rian,&quot;Analysis of Current-Controlled PWM Converters using the Model of the PWM Switch&quot;, PCIM Conference ,1990
41 • Chris Basso – APEC 2014
Final Model Includes Subharmonic Effects
A simple capacitor is enough to mimic instability
Ia
c
a
L
Ic
Vcp
Vin
Vap
Vc
Ri
Ic
I&micro;
Cs
resonant
tank
C
R
p
As the instability is placed at half the switching frequency:
Fsw
1
=
2
2π LCs
42 • Chris Basso – APEC 2014
Cs =
1
L ( Fsw π )
2
CCM CM
The PWM Switch in DCM
The PWM CM can work in discontinuous mode
ic ( t )
I peak
ic ( t )
Vc
Ri
Current setpoint
Sa Ri
S1
Theoretical value
Sa = 0
S2
Ic
Tsw
D3Tsw
0
t
D1Tsw
D2Tsw
Tsw
The average current Ic is somewhere in the downslope S2
I peak =
Vc − D1Tsw Sa
Ri
Ic =
Vc − D1Tsw Sa
− α D2Tsw S 2
Ri
DCM CM
43 • Chris Basso – APEC 2014
Derive the Inductor Average Current
We must now obtain the value of α to get Ic
I peak − I c
α I peak
S2
Ic
Tsw
Ic
t
D2Tsw
Ic is the area under the triangle divided by the switching period
Ic =
I peak D1
2
+
I peak D2
2
= I peak
α I peak = I peak − I peak
44 • Chris Basso – APEC 2014
D1 + D2
2
D1 + D2
2
α = 1−
D1 + D2
2
DCM CM
Adopt the CCM Structure for DCM
Substitute and rearrange to get the inductor current
Ic =
Vcp
Vc D1Tsw Sa
−
− D2Tsw
Ri
Ri
L
 D1 + D2 
1 −
2 

If we stick to the original CCM architecture
Ic =
Vc
− I&micro;
Ri
I&micro; =
with
Vcp
D1Tsw Sa
+ D2Tsw
Ri
L
 D1 + D2 
1 −

2 

c
V  D + D2 
D1Tsw Sa
+ D2Tsw cp 1 − 1

Ri
L 
2 
Vc
Ri
p
DCM CM
45 • Chris Basso – APEC 2014
Discontinuous Waveforms
Let's have a look at the PWM switch voltages in DCM
ia ( t )
I peak
ia ( t )
Tsw
t
vap ( t )
Vin
t
ic ( t )
DCM
t
vcp ( t )
Vap
D3Tsw
0
t
D1Tsw
46 • Chris Basso – APEC 2014
Vout
D2Tsw
DCM CM
Derive the Duty Ratios
From the DCM voltage-mode PWM switch we have:
Vcp = Vap
D1
D1 + D2
D1 =
D2Vcp
Vap − Vcp
From the operating DCM waveforms
Ia =
I peak D1
I peak =
2
2I a
D1
I c = I peak
D1 + D2
2
I a = Ic
D1
D1 + D2
Almost there, just need to express D2
I peak =
Vac
D1Tsw
L
I peak =
and
2Ic
D1 + D2
Vac
2Ic
D1Tsw =
L
D1 + D2
D2 =
2 LFsw I c
− D1
D1 Vac
DCM CM
47 • Chris Basso – APEC 2014
The DCM Model is Complete!
We can use this model for DCM simulations
L
a
c
Ia
Ic
Vc
Ri
Ia
Vin
I&micro;
C
R
p
I a = Ic
D1
D1 + D2
I&micro; =
V  D + D2 
D1Tsw Sa
+ D2Tsw cp 1 − 1

Ri
L 
2 
DCM CM
48 • Chris Basso – APEC 2014
Current Mode Borderline
In CM, the error voltage sets the peak current
I peak =
Vc
Ri
The peak current also depends on duty ratio D
iL ( t )
I peak
DTsw
I peak =
iL ( t )
Vcp L
Vac L
Tsw
Tsw
Vac
DTsw
L
D=
Vc L
Ri VacTsw
ton =
Vc L
Ri Vac
BCM CM
49 • Chris Basso – APEC 2014
Operating Points of BCM Current Mode
The off-time duration depends on Ipeak too
I peak =
Vcp
L
(1 − D ) Tsw
toff =
Verr L
Ri Vcp
Switching frequency comes easily
Tsw = ton + toff =
Vc L  1
1 
+


Ri  Vap Vcp 
The average inductor current Ic is straigthforward
Ic =
I peak
2
=
Vc
2 Ri
The relationship between Ia and Ic is always the same
I a = DI c
BCM CM
50 • Chris Basso – APEC 2014
This Completes the BCM CM Model
The model is really simple, two current sources
L
a
c
Ia
Ic
Vc
DI c
2 Ri
Vin
C
R
p
Other ABM sources will compute D and Tsw
BCM CM
51 • Chris Basso – APEC 2014
Course Agenda
Introducing the PWM Switch Model
CCM, DCM and BCM in Voltage Mode
Pulse Width Modulator Gain
The PWM Switch Model in Current Mode
PWM Switch at Work in a Buck Converter
A Simplified Approach to Modeling a DCM Boost
Transfer Function of a BCM Boost in Current Mode
Small-Signal Model of The Active Clamp Forward
52 • Chris Basso – APEC 2014
A Buck Converter in Current Mode
Identify the diode and switch position in a buck CM
a
a
duty-cycle
c
c
vc
PWM switch CM
p
p
Current mode
Replace switches by the small-signal PWM switch model
53 • Chris Basso – APEC 2014
A Small Signal Model
The model includes current sources and conductances
Ic
a
c
1
gi
vˆc ki
vˆcp g r
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. Vorp&eacute;rian,&quot;Analysis of Current-Controlled PWM Converters using the Model of the PWM Switch&quot;, PCIM Conference ,1990
54 • Chris Basso – APEC 2014
Plug the Model and Simplify
Plug, simplify and rearrange: test in between!
Ic
a
Vin
1
gi
1
go
vˆc ko
vˆc ki
vˆcp g r
p
L
c
C
Cs
R
rC
vˆap g f
We want the control-to-output function, remove input stimulus
vˆin = 0
vˆap g f = 0
The input contribution can also disappear, no interest in Zin
55 • Chris Basso – APEC 2014
Time to Call FACTS!
End-up with a simpler and less ugly sketch
c
L
Vout ( s )
1
C
Vc ( s ) ko
1
go
Cs
2
R
rC
There are 3 storage elements: 3rd-order system
FACTS: Fast Analytical Circuit TechniqueS
56 • Chris Basso – APEC 2014
Don't Use Brute-Force Algebra!
You can use brute-force analysis…
Vout ( s )
L
Z th
 1 1 
Vth = Vc ( s ) k0 

 g o sCs 
C
R
Vth
Z (s)
 1 1 
Z th = 

 g o sCs 
rC
…or consider FACTS to write a 3rd-order system TF
H ( s) = H0
N (s)
D (s)
= H0
N (s)
1 + a1s + a2 s 2 + a3 s 3
57 • Chris Basso – APEC 2014
Consider dc and high-frequency states for L and C
impedance
ZC =
C
impedance
1
sC
Z L = sL
L
Dc state Z C = ∞
Cap. is an open circuit
HF state Z C = 0
Cap. is a short circuit
Dc state Z L = 0
Inductor is a short circuit
Inductor is an open circuit
HF state Z L = ∞
In the circuit, open capacitors and short inductors
Vout ( s )
Vc ( s ) ko
1
go

1 
H 0 = k0  R

 g0 
R
10 s!
58 • Chris Basso – APEC 2014
Identify The Zeros
Zeros prevent the excitation from reaching the output
Vin ( s )
Z (s) = ∞
Vout ( s )
rC +
response
rC
Z (s) = 0
excitation
sr C + 1
1
= C
sC
sC
srC C + 1 = 0
R
C
ω z1 =
Z (s) = 0
1
rC C
We have the zero expression, half of the work is done
H ( s ) = H0
1+
s
sz1
1 + a1s + a2 s 2 + a3 s3
59 • Chris Basso – APEC 2014
Finding the Poles
The poles are linked to the time constants of the system
These time constants solely depend on the structure
Remove the excitation signal to isolate the structure
short
circuit
voltage
source
current
source
open
circuit
The denominator order depends on the storage elements
1 storage
element
1st-order
2 storage
elements
L
C
C
2nd-order
Not always! Consider individual state variables.
60 • Chris Basso – APEC 2014
Start by Identifying the Time Constants
The excitation is zero, elements are in their dc states
3 storage elements, 3 time constants, 3 drawings
1 g0
R?
τ 1 = f ( Cs )
τ 1 = Cs 
R
τ 2 = f ( L)
τ2 =
τ 3 = f (C )
τ 3 = C  rC + 
rC
R?
1 g0
 1

R
 go 
R
rC
R?
R
1 g0
rC
L
1
+R
g0


 1

R  
 g0  
61 • Chris Basso – APEC 2014
First Coefficients a1 and a2
FACTs tell us that a1 sums up all time constants
a1 = τ 1 + τ 2 + τ 3
Dimension is time
For a2, we multiply combined-time constants
a2 = τ1τ 12 + τ 1τ 31 + τ 2τ 32
Dimension is time2
What is this new time constants definition,τ 2 ?
1
Cs (HF)
τ 12
Cs (HF)
τ 31
L C (dc)
R?
L (HF)
τ 32
L (dc) C
R?
C Cs (dc)
R?
V. Vorp&eacute;rian, “Fast Analytical Techniques for Electrical and Electronic Circuits”, Cambridge Press, 2002
62 • Chris Basso – APEC 2014
Coefficient a2 Mixes Time Constants
Drawings are key to avoid mistakes
Cs (HF)
τ 12
R?
1 g0
L C (dc)
R
τ 12 =
L
R
rC
R?
Cs (HF)
R?
τ 31
L (dc) C
R
1 g0
τ 31 = rC C
rC
R?
63 • Chris Basso – APEC 2014
(Carefully) Mixing Time Constants
The last drawing completes the a2 expression
L (HF)
τ 32
R?
C Cs (dc)
R
1 g0
τ 32 = ( rC + R ) C
rC
R?
a2 coefficient is there!
 1
L
 1

L
a2 = C s 
R  + Cs 
R  rC C +
( rC + R ) C
1
 go  R
 go 
+R
g0
64 • Chris Basso – APEC 2014
…And a3 is ?
For a3, we multiply by a third time-constant
a3 = τ 1τ 12τ 31,2
Dimension is time3
What is this new time constant definition?
Cs (HF) L (HF)
τ 31,2
R?
R
1 g0
C (dc)
τ 31,2 = ( rC + R ) C
rC
R?
The final coefficient has been identified
 1
L
a3 = Cs 
R  ( rC + R ) C
 go  R
65 • Chris Basso – APEC 2014
Time to Check Results
A Mathcad&reg; sheet can be built to verify these calculations
H ( s ) = G0
ω z1 =
1+
s
sz1
1 + a1s + a2 s + a3 s
2
3

1 
G0 = k0  R

 g0 
1
rC C

 1

 1

L
a1 = Cs 
R+
+ C  rC + 
R 


1
 go 
 g0  

+R
g0
 1
L
 1

L
a2 = Cs 
R  + Cs 
R  rC C +
( rC + R ) C
1
g
R
g
 o 
 o 
+R
g0
 1
L
a3 = Cs 
R  ( rC + R ) C
 go  R
66 • Chris Basso – APEC 2014
5 V/1 A buck
Vin = 10 V, Fsw = 100 kHz, Ri = 0.25Ω, Se = 2.5 kV s
C = 100 &micro;F, rC = 0.1Ω, L = 100&micro;H, Cs = 101nF, Vc = 1.28 V
I c = 4.94 A
ki = 2 Ω −1 k0 = 4 Ω −1
g 0 = 0.01 Ω −1
g f = −7.5 mΩ −1 g r = 0.49 Ω −1
g i = −250 mΩ −1
G0 = 12 dB
f z1 = 15.9 kHz
See What SPICE is Saying
Use the large-signal model, SPICE linearizes it for you
a
{Se}*V(D)/({Ri}*{Fsw}) + v(c,p)*(1-V(D))*({1/Fsw}/(2*{L}))
L1
VIC
{L}
4.95V
c
c
5
a
B3
Current
V(Vc)/{Ri}
C1
100uF
0V
-2.50mV
V1
10
p
Vout
4.95V
C2
B4
Current {Cs}
B2
Current
V(D)*I(VIC)
10.0V
parameters
Fsw=100kHz
L=100u
Cs=1/(L*(Fsw*3.14)^2)
Ri=250m
Se=2.5k
4.95V
1
495mV
1.28V
R3
1m
R2
B1
100m
Voltage
v(c,p)/v(a,p)
Vstim
1.28
AC = 1
p
R1
1
4
D
vc
Then compare results with those of Mathcad&reg;
67 • Chris Basso – APEC 2014
Excellent Agreement!
Superimposed curves mean transfer functions are identical
dB
&deg;
20
0
10
− 50
0
−100
− 10
−150
H(f)
− 20
10
100
68 • Chris Basso – APEC 2014
3
1&acute; 10
4
1&acute; 10
5
1&acute; 10
10
arg H ( f
)
100
5
3
1&acute; 10
4
1&acute; 10
1&acute; 10
Rearranging Expressions
The denominator is not really in a low-entropy form
D ( s ) = 1 + a1s + a2 s 2 + a3 s 3
This is a third-order polynomial form that can be factored
H(f)
D ( s ) ≈ 1 + a1s
D ( s) ≈ 1+
Low frequency
High frequency
a2
a
s + 3 s2
a1
a1
69 • Chris Basso – APEC 2014
Final Lap!
The transfer function can now unveil peaking and damping
H ( s) ≈ H0
1+
s
ωz
1
1
1+
s
s
 s 
H0 =
2
ωp 1+ ω Q +  ω 
n
 n
1
ωp =
1
ωn =
T
1
+ sw  mc (1 − D ) − 0.5
RC LC
π
Q=
Tsw
S
mc = 1 + e
Sn
ωz =
1
π  mc (1 − D ) − 0.5
1
R
1
Ri 1 + RTsw  m 1 − D − 0.5
)
c(

L 
1
rC C
20
10
0
− 10
Artificial ramp
Inductor on slope
− 20
10
H(f)
100
10
3
R. B. Ridley, “A new Continuous-Time Model for CM Control”, IEEE Transactions of Power Electronics, Vol. 6, April 1991
70 • Chris Basso – APEC 2014
10
4
10
5
Course Agenda
Introducing the PWM Switch Model
CCM, DCM and BCM in Voltage Mode
Pulse Width Modulator Gain
The PWM Switch Model in Current Mode
PWM Switch at Work in a Buck Converter
A Simplified Approach to Modeling a DCM Boost
Transfer Function of a BCM Boost in Current Mode
Small-Signal Model of The Active Clamp Forward
71 • Chris Basso – APEC 2014
A DCM Boost Converter
The goal is to regulate the LED string current
D
L
Vout
Vf
Cout
Vin
SW
rC
Ri
I out
Rsense
PWM
control
The LED current is sensed via a shunt element
72 • Chris Basso – APEC 2014
Characterize the LED String
Evaluate forward drop and dynamic resistance
n LEDs in series
A
A
A
A
IF
n
vf
rd
rLEDs
∑r
VT0
vZ
∑V
i =1
di
n
K
i =1
T0i
K
v f = I F rd + VT0
K
K
Lab. measurements require simple V and A-meters
rLEDs =
V f1 − V f 2
I F1 − I F2
=
27.5 − 26.4
= 55Ω
0.1 − 0.08
VZ ≈ V f1 − RLEDs I F1 = 27.5 − 0.1&times; 55 = 22 V
73 • Chris Basso – APEC 2014
A Simplified Approach
The LED string is driven by a current source
I out
Vout
Rac = rLEDs + Rsense
Rac
Vc
Vout = Rac I out + Vz
I out
VZ
Vout ( s ) = Rac I out ( s )
VZ sets the operating point, Rac sets the ac response
iˆo ( s )
Z (s)
vˆo
vˆc ( s )
rC
I out
Cout
74 • Chris Basso – APEC 2014
Rac
Current Source Model
It is convenient and fast to consider 1st-order models
Use the converter transfer function in DCM
Vout
=
Vin
1+ 1+
2Tsw D 2 Rdc
L
2
We purposely consider an instantaneous power response
if D changes, Pout immediately translates
this is not true for boost or buck-boost converters: RHPZ
high-frequency phenomena are also lost
1st-order
rC
R
I out
D
C
75 • Chris Basso – APEC 2014
Define the Duty Ratio
In the previous expression, D is unknown
vc ( t )
Ri
Se
Ri
Sn =
I peak
Vin
L
iD ( t )
DTsw
I peak
V S
= c − e DTsw
Ri Ri
I peak =
76 • Chris Basso – APEC 2014
DTswVin
L
D=
Vc L
SeTsw L + RiTswVin
Update the Output Current Equation
Massage the dc transfer equation to unveil Iout
Vout
=
Vin
1+ 1+
2Tsw D 2
Vout
I out
L
2
Inject the duty ratio definition, solve for Iout
I out =
2Vout LVc 2
1
2
Tsw
 2V


 out − 1 − 1 ( Se L + RiVin )2
 Vin




There are two modulated variables, Vc and Vout
vˆin = 0
77 • Chris Basso – APEC 2014
A Simple Model
You obtain two small-signal sources: iˆout
∂I (V , V )
Vin 2Vc L
g1 = out c out =
2
∂Vc
Tsw (Vout − Vin )( Se L + RiVin )
g2 =
∂I out (Vc ,Vout )
∂Vout
=−
= g1vˆc + g 2 vˆout
Vin 2Vc 2 L
2Tsw (Vin − Vout )
The circuit is quite simple
2
( Se L + RiVin )2
iˆo ( s )
vˆo
rC
vˆc ( s )
vˆo ( s )
Rac
Cout
78 • Chris Basso – APEC 2014
∂I out
∂Vc
vˆout
∂I out
∂Vout
vˆc
vˆc
vˆo
A Current Driving an Impedance
The second term is a simple resistance
I
R
V
I=
Update the final model
2Tsw (Vin − Vout ) ( Se L + RiVin )
2
V
R
R1 =
2
Vc 2Vin 2 L
Z (s)
vˆo
rC
R1
vˆc ( s )
Rac
Vout ( s ) = Vc ( s ) g1Z ( s )
Cout
79 • Chris Basso – APEC 2014
Use FACTS to Get the Impedance
Obtain the impedance (transfer function) in a snapshot
Vres ( s )
Response
Ω
I exc ( s )
rC
R1
Rac
Z (s) =
Vres ( s )
I exc ( s )
= Z0
Cout
Excitation
For dc, open the capacitor
R1
80 • Chris Basso – APEC 2014
rC
Rac
Z 0 = R1 Rac
1+
s
s z1
1+
s
s p1
No Algebra to Get the Result!
At the zero frequency, the response disappears
Cout
Short circuit
rC
rC +
sr C + 1
1
= C out
sCout
sCout
srC Cout + 1 = 0
ω z1 =
1
rC Cout
Remove the excitation and look at the resistance driving Cout
Z ( s ) = R1 Rac
rC
R1
Rac
R?
1 + srC Cout
1 + sCout ( rC + R1 Rac )
τ =  rC + ( R1 Rac )  Cout
81 • Chris Basso – APEC 2014
Final Expression
Associate the impedance expression with g1
Vout ( s )
Vc ( s )
ω z1 =
1+
= H0
1+
s
ω z1
s
H0 =
ω p1
1
rC Cout
ω p1 =
Vin 2Vc L
Tsw (Vout − Vin )( Se L + RiVin )
2
( Rac
R1 )
1
Cout ( rC + R1 Rac )
However, we want the control-to-output current expression
Vout ( s )
s
Vsense ( s )
rLEDs
Rsense
82 • Chris Basso – APEC 2014
Vsense ( s )
Vc ( s )
1+
ω z1
Rsense
=
H0
s
Rsense + rLEDs
1+
ω p1
Checking our Model Response
Ac simulation of the current source-based approach
parameters
Vout
2
R3
4m
Rd
55
R1
Ri=0.25
{R1}
vc=0.4
Se=100k
B1
6
1
Vc
Current
Fsw=1Meg
Vsense
{k1}*V(Vc)/{k2}
C1
L=3.3u
Rsense
2.2uF
Tsw=1/Fsw
11
Vin=12
Vout=32.85
k1=Vin^2*vc*L
k2=Tsw*(Vout-Vin)*(Se*L+Ri*Vin)^2
R1=2*Tsw*(Vin-Vout)^2*(Se*L+Ri*Vin)^2/(Vc^2*(Vin^2)*L)
V2
{Vc}
AC = 1
83 • Chris Basso – APEC 2014
Simplified Approach vs PWM Switch
The comprehensive model with the PWM switch
L1
3.3u
12.0V
10
12.0V
c
9
p
32.6V
Vout
Rd
55
400mV
0V
11
Vin
12
vc
a
8
PWM switch CM
405mV
dc
duty-cycle
5
X3
PWMCM
L = 3.3u
Fs = 1Meg
Ri = Ri
Se = 100k
Shunt model
R3
4m
7
23.8V
X2
IRF530M
32.6V
2
5.86V
1
23.8V
6
3
1.77V
4
X1
AMPSIMP
parameters
Ri=-0.25
vc=0.4
V2
{Vc}
AC = 1
C1
2.2uF
Vsense
Rsense
11
84 • Chris Basso – APEC 2014
V1
22
Final Results
The RHPZ effect is not modeled in the simplified approach
50.0
( dB)
30.0
10.0
Vout ( f )
-10.0
Vc ( f )
-30.0
15.0
(&deg;)
-15.0
Simple model
-45.0
-75.0
arg
-105
1
Vout ( f )
Vc ( f )
PWM switch
10
100
1k
10k
100k
1Meg
85 • Chris Basso – APEC 2014
Real Measurement vs Model
Deviation occurs, as expected, because of the RHPZ
3.3uH
Control to Output (R29 current sense)
10
0
model
prototype
0
-20
-40
Gain (dB) (Calculated)
-10
-60
Gain (dB) (Vin=12V)
Phase (o) (Calculated)
Phase (o) (Vin=12V)
-80
-20
-100
-30
RHPZ
-120
100
1,000
10,000
100,000
C. Basso, A. Laprade, &quot;Simplified Analysis of a DCM Boost Converter Driving an LED String&quot;, www.how2power.com
86 • Chris Basso – APEC 2014
Course Agenda
Introducing the PWM Switch Model
CCM, DCM and BCM in Voltage Mode
Pulse Width Modulator Gain
The PWM Switch Model in Current Mode
PWM Switch at Work in a Buck Converter
A Simplified Approach to Modeling a DCM Boost
Transfer Function of a BCM Boost in Current Mode
Small-Signal Model of The Active Clamp Forward
87 • Chris Basso – APEC 2014
Why Bordeline Operation?
More converters are using variable-frequency operation
This is known as Quasi-Square Wave Resonant mode: QR
Valley switching ensures extremely low capacitive losses
DCM operation saves losses in the secondary-side diode
Easier synchronous rectification
The Right Half-Plane Zero is pushed to high frequencies
Smooth signals
iD ( t )
Less noise
vDS ( t )
88 • Chris Basso – APEC 2014
Low CV&sup2; losses
id ( t )
vDS ( t )
What is the Principle of Operation?
The drain-source signal is made of peaks and valleys
A valley presence means:
The drain is at a minimum level, capacitors are naturally discharged
The converter is operating in the discontinuous conduction mode
V
480
360
Vin
valley
240
120
toff
ton
DT
vDS ( t )
0
2.061m
2.064m
2.066m
2.069m
2.071m
Flyback structure
BCM = Borderline or Boundary Conduction Mode
89 • Chris Basso – APEC 2014
A QR Circuit Does not Need a Clock
The system is a self-oscillating current-mode converter
L
Vout
50 mV
+
-
Vin
C
S
Q
d (t )
Vout
R
+
-
+
2.5 V
90 • Chris Basso – Huawei Technical Seminar 2012
vc ( t )
vsense ( t )
Ri
R
A Winding is Used to Detect Core Reset
When the flux returns to zero, the aux. voltage drops
Discontinuous mode is always maintained
vaux ( t )
Vbulk
V
Core is
reset
800
.
.
.
delay
400
200
iL p ( t )
0
vDS ( t )
vaux ( t ) = − N
vDS ( t )
600
V
delay
A
dϕ ( t )
10 400m
dt
0
vaux ( t )
iL p ( t )
20 800m
0
-10 -400m
ϕ =0
-20 -800m
2.251m
2.255m
2.260m
2.264m
2.269m
set
91 • Chris Basso – Huawei Technical Seminar 2012
Test the Large-Signal Model Response
Insert the PWM BCM CM model in the boost converter
L1
{L}
10.0V
X1
PWMBCMCM
Vout
Fsw (kHz)
11.0V
vc
L=22u
Ri=-0.1
PWM switch BCM p
33.4V
parameters
a
Vin
10
ton
c
10.0V
15.8V
C1
470u
500mV
V3
0.5
AC = 1
92 • Chris Basso – APEC 2014
R1
10
SPICE Gives Us the Response
SPICE linearizes the model for us around the bias point
dB
40.0
pole
G0
20.0
0
-20.0
-40.0
-1 slope
Vout ( f )
Vc ( f )
zero
&deg;
0
-40.0
RHPZ?
-80.0
-120
arg
-160
1
Vout ( f )
Vc ( f )
10
100
1k
10k
100k
1Meg
93 • Chris Basso – APEC 2014
Derive a Small-Signal Model
A Quasi Resonant model is built with the PWM switch model
Vc L  1
1 
+


Ri  Vac Vcp 
V
Ic
L
3 variables
Ia = c
Ic =
Ri VacTsw
 1
1 
Vac 
+
 Vac Vcp 
Vc


1 variable
Ic =
2 Ri
D=
c
a
Vc
2 Ri
DI c
p
Vc L
Ri VacTsw
Tsw =
These are large-signal equations that need linearization
I c = f (Vc )
iˆa =
iˆc =
∂I a (Vcp , Vac , I c )
∂Vcp
94 • Chris Basso – APEC 2014
∂I c (Vc )
∂Vc
vˆcp +
I c ,Vac
vˆc
 1
iˆc = vˆc 
 2 Ri
∂I a (Vcp , Vac , I c )
∂I c

 = vˆc kc

iˆc +
Vap ,Vac
kc =
1
2 Ri
∂I a (Vcp , Vac , I c )
∂Vac
vˆac
I c ,Vcp
Large to Small-Signal
Final steps before the small-signal model
iˆa = vˆcp
I c 0Vac 0
(V
ac 0
kcp =
(V
+ Vcp 0 )
2
+
I c 0Vac 0
ac 0
+ Vcp 0 )
Vcp 0
Vcp 0 + Vac 0
kic =
2
iˆc − vˆac
Vcp 0 I c 0
(V
+ Vcp 0 )
ac 0
Vcp 0
Vcp 0 + Vac 0
kac =
2
Vcp 0 I c 0
(V
ac 0
+ Vcp 0 )
2
A small-signal model can now be assembled
a
c
Vcp kcp
Vac kac
I c kic
conductance
Vc kc
conductance
p
conductance
95 • Chris Basso – APEC 2014
Always Check the Small-Signal Response
Always verify if coefficients are well derived!
R2
10m
parameters
33.4kV
L1
{L} 9.97V
Fsw
ton
7
Vg
10
Vin=10
Vout=15.6
M=Vout/Vin
Ic=-2.5
Vac=-10
Vap=-Vout
Vcp=Vin-Vout
kcp=Ic*Vac/(Vac+Vcp)^2
kic=Vcp/(Vcp+Vac)
kac=Vcp*Ic/(Vac+Vcp)^2
kc=1/(2*Ri)
500m V
Vc
V3
0.5
AC = 1
15.8V
a V(Vc)*{L}/({Ri}*(V(a,c)+1u))
B6
Voltage
R1
10
C1
470u
Rdum
1u
c
a
B1
Current
V(c,p)*{kcp}
Vc
9.97V
B2
Current
I(Vc)*{Kic}
B3
Current
V(a,c)*{kac}
p
96 • Chris Basso – APEC 2014
Vout
p
c 11.0uV
4
10.0V
L=22u
Ri=-0.1
9.97V
1/((V(Vc)*{L}/{Ri})*(1/V(a,c)+1/V(c,p)))
B5
Voltage
10
B4
Current
V(Vc)*{kc}
A Simple Intermediate Sanity Check
Same response as with the non-linear model: good to go
dB
40.0
20.0
0
-20.0
-40.0
Vout ( f )
Vc ( f )
&deg;
0
-40.0
-80.0
-120
arg
-160
1
Vout ( f )
Vc ( f )
10
100
1k
10k
100k
1Meg
97 • Chris Basso – APEC 2014
The Model at Work in the BCM Boost
Replace the switch/diode in the boost configuration
c
L
Vc kc
p
Vcp kcp
Vin
Vc
a
I c kic
Vac kac
C
R
In the original model, Ic leaves node c. It enters it in a boost.
98 • Chris Basso – APEC 2014
Identify Static Parameters in the Coefficients
Depending on configurations, update variables in coefficients
kcp =
Vcp 0 = Vin − Vout
Ic0 = −
I outVout
Vin
I outVout
Vin
Vin ( −Vin + Vin − Vout )2
kac = −
Vac 0 = −Vin
kic =
kcp =
1
R
kac =
I outVout
(Vin − Vout )
Vin ( −Vin + Vin − Vout )2
Vin − Vout
Vin − Vout − Vin
M 1
−
R R
kic = 1 −
1
M
kc =
1
2 Ri
99 • Chris Basso – APEC 2014
First Step is Dc Gain
Open the output capacitor, short the inductor
Vc kc
Vin
Vout ( s )
p
c
Vc ( s )
I out ( s )
Vcp kcp
I c kic
Vac kac
R
a
Write output voltage expression
I out ( s ) = Vc ( s ) kc + Vcp ( s ) kcp − I c ( s ) kic − Vac ( s ) kac
(
)
(
)
I out ( s ) = Vc ( s ) kc + V( c ) ( s ) − V( p ) ( s ) kcp − Vc ( s ) kc kic − V( a ) ( s ) − V( c ) ( s ) k ac
100 • Chris Basso – APEC 2014
Dc Gain Derivation
There is no contribution from the source, Vin(s) = 0
I out ( s ) = Vc ( s ) kc − Vout ( s ) kcp − Vc ( s ) kc kic = Vc ( s ) kc (1 − kic ) − Vout ( s ) kcp
I out ( s ) =
Vout ( s )
R
1

Vc ( s ) kc (1 − kic ) = Vout ( s )  kcp + 
R


Vout ( s ) kc (1 − kic )
=
1
Vc ( s )
kcp +
R
H0 =
R
4 MRi
We have the first term of our transfer function
s
sz1
Vout ( s )
= H0
s
Vc ( s )
1+
s p1
1+
101 • Chris Basso – APEC 2014
Deriving the Zero Position
For the zero, the stimulus does not reach the output
current in the resistance is 0, node p voltage is also 0
p
c
Vc kc
VL ( s ) L
Vc ( s )
Vout ( sz ) = 0
i p ( sz ) = 0
Vcp kcp
I c kic
Vac kac
R
a
All the inductor ac current is absorbed before reaching R
Vc ( s ) kc + V( c ) ( s ) − V( p ) ( s )  kcp = Vc ( s ) kc kic + V( a ) ( s ) − V( c ) ( s )  kac
102 • Chris Basso – APEC 2014
What is the Root to N(s) = 0?
The voltage at node c depends on the inductance
V( c ) ( s ) = −VL ( s ) = − I c ( s ) sL = −Vc ( s ) kc sL
Substitute in the previous equations and simplify
Vc ( s ) kc − Vc ( s ) kc sLkcp = Vc ( s ) kc kic + Vc ( s ) kc sLkac
Solve for s, this is the zero position
1 − kic = sL ( kac + kcp )
sz1 =
1 − kic
L ( kac + kcp )
Substitute
kic, kac, kcp
The root is positive, this is a Right Half Plane Zero
sz1 =
R
LM 2
103 • Chris Basso – APEC 2014
For the Pole, Reduce Excitation to 0
Excitation is Vc: all current sources f (Vc) are open
p
c
V( c ) ( s ) = 0
R?
L
Vcp kcp
Vac kac
R
a
Look for the resistance “seen” by the capacitor C
The source Vcpkcp can be reworked
vˆ( c ) = 0
Vcp ( s ) kcp = −V p ( s ) kcp = −
104 • Chris Basso – APEC 2014
Vout ( s )
R
Replace by a resistance R
A Really Simple Circuit
Difficult to beat in terms of problem solving
p
c
R?
L
R
R
a
The pole due to the capacitor comes immediately
R ? = R || R =
R
2
τ=
R
C
2
s p1 =
2
RC
105 • Chris Basso – APEC 2014
With a few steps, we have our transfer function
s
1−
ω z1
Vout ( s )
= H0
s
Vc ( s )
1+
H0 =
ωp
R
4 MRi
ωp =
1
2
RC
ωz =
1
R
LM 2
1
We can easily add the capacitor ESR contribution
rC +
rC
sr C + 1
1
= C
sC
sC
srC C + 1 = 0
C
Z (s) = 0
106 • Chris Basso – APEC 2014
ω z2 =
1
rC C

s 
s
 1 −
  1 +
ω
ω
Vout ( s )
z1  
z2
= H0 
s
Vc ( s )
1+
ωp
R

Pole changes: τ =  + rC  C
2

1



Time to Confront Mathcad&reg; with SPICE!
If equations are correct, curves must superimpose perfectly
&deg;
dB
40
H(f)
100
20
0
− 20
− 40
1
0
arg H ( f
10
)
− 100
100
3
10
4
10
5
10
6
10
The BCM boost response in CM is first-order with RHPZ
107 • Chris Basso – APEC 2014
Course Agenda
Introducing the PWM Switch Model
CCM, DCM and BCM in Voltage Mode
Pulse Width Modulator Gain
The PWM Switch Model in Current Mode
PWM Switch at Work in a Buck Converter
A Simplified Approach to Modeling a DCM Boost
Transfer Function of a BCM Boost in Current Mode
Small-Signal Model of The Active Clamp Forward
108 • Chris Basso – APEC 2014
Active Clamp Forward in Voltage Mode
An Active Clamp Forward (ACF) is a forward converter…
1: N
L
Cclp
Lmag
C
R
Q2
Vin
Cclp
Q1
Low-side
GND referenced
…featuring a controlled-upper-side switch for ZVS operations
109 • Chris Basso – APEC 2014
Control Strategy
A deadtime is inserted to let VDS swing towards grounds
vDS ( t )
ZVS
vQ2 ( t )
DT
vQ1 ( t )
DT
Quasi-ZVS can be implemented on the drain voltage
110 • Chris Basso – APEC 2014
Active Clamp Forward in Voltage Mode
Energy is stored in the magnetizing inductance at turn on
L
1: N
Cclp
VCclp
Q2
iL ( t )
Lmag
imag ( t )
C
R
NiL ( t )
Vin
imag ( t ) + NiL ( t )
Q1
We need to offer a path at turn off to demagnetize the core
111 • Chris Basso – APEC 2014
All Parasitic Capacitors are Charged
At turn off, the current charges the lump capacitor
L
1: N
Cclp
VCclp
Q2
Lmag
iL ( t )
imag ( t )
imag ( t )
Vin
imag ( t )
C
Q1
Clump
The drain voltage increases until it touches Vclamp
112 • Chris Basso – APEC 2014
R
Magnetizing Current Resonates
The upper-side MOSFET switches on with delay: ZVS
imag ( t )
Cclp
VCclp
L
1: N
Lmag
iL ( t )
C
R
imag ( t )
Q2
Vin
imag ( t )
Q1
Clump
Magnetizing current decreases to 0 then reverses
113 • Chris Basso – APEC 2014
The deadtime lets magnetizing current reach –Imag,peak
imag ( t )
Cclp
VCclp
Q2
L
1: N
Lmag
iL ( t )
C
R
imag ( t )
imag ( t )
Vin
imag ( t )
imag ( t )
Q1
Clump
imag ( t )
At this moment, Q2 opens and current discharges Clump
114 • Chris Basso – APEC 2014
Full ZVS at Nominal Power is Difficult
True ZVS is difficult to reach, quasi-ZVS is usually obtained
Vclamp
max
vDS ( t )
0
0
imag ( t )
− I mag , peak
Q2 opens
True ZVS
Q1 closes
2
1
1
Lmag ( I mag , peak − − NI L ) &gt; ClumpVin 2
2
2
115 • Chris Basso – APEC 2014
Building an Ac Model
A model can be built following different methods
Write large-signal equations of voltages and currents
Assemble sources to build a large-signal model
Linearize expressions and derive transfer functions
o Reveal the presence of the PWM Switch model
o Implant its already-available small-signal model
o Solve for the transfer function expression
Both approaches have pros and cons
Going along both paths helps to cross-check results
116 • Chris Basso – APEC 2014
Identify PWM Switches Pair
A pair of switches appear in primary and secondary sides
L
Cclp
C
R
Ic
Vin
c
Ip
p
a
Ic
c
a
p
Ip
117 • Chris Basso – APEC 2014
Primary-Side PWM Switch Model
Identify currents and voltages during transitions
Cclp
Vds − on
0
Vin
Cclp
Lmag
Lmag
ron 2 I mag
ron1 ( I mag + NI out )
I mag
Vds − off
0
Vclamp
I mag + NI out
During the on-time DTsw
vDS − on ( t )
DTsw
118 • Chris Basso – APEC 2014
= ( I mag + NI out ) ron1
During the off-time (1 − D ) Tsw
vDS − off ( t )
(1− D )Tsw
= Vin + VC + ron 2 I mag
The First PWM Switch Model is Here
Connect the PWM Switch the right way, check polarities
Cclp
Cclp
Lmag
Lmag
vDS ( t )
ron 2 I c
ron 2 I c
Vloss
(1 − D ) I c
Vin
Vin
V pa
V pa (1 − D )
a
V pa (1 − D )
a
d
PWM switch VM
Ic
Vap ( D − 1)
I c = I mag
c
DVap
− DI c
−Vap
c
Vclamp
(1 − D ) I c
p
p
Tsw
Vloss
119 • Chris Basso – APEC 2014
Secondary-Side PWM Switch Model
The forward converter is a buck-derived topology
N (Vin − Vloss )
Vloss
2
a
3
. .I
out
DN (Vin − Vloss )
c
Vout
5
4
D
D d
6
PW M switch VM
p
X2
PWMCCMVM
1: N
1
Vin
I p = DNI out
Vloss = ( I mag + NI out ) ron1
Ip 

Vloss =  I mag +  ron1
D

Primary-side loss
NI out
120 • Chris Basso – APEC 2014
Check the Dual-PWM Switch Response
Compare ac response with that of the large-signal version
{ron1}*(I(VLP)+(I(V5)/V(D)))
B4
Voltage
in
X2
PWMCCMVM
47.7V
48.0V
7.95V
17
Iin
V1
48
C2
0.2u
91.7V
D
out
3.30V
11
Vout
6
3.80V
3.80V
48.0V
10
VIL
{Vf}
L1
0.5u
c
16
X1
XFMR
RATIO = N
L2
50u
a
d
0V
p
PWM switch VM
C1
500u
R1
100m
1
5
{ron2}*I(VLP)
Vclp
R2
50m
VLP
B2
Voltage
48.0V
0V
2
clp
4
B5
Voltage
({ron1}*(I(VLP)+(I(V5)/V(D))))*V(D)
47.8V
V5
18
c
p
91.7V
PWM switch VM
parameters
478mV
L3
1kH
14
C3
1kF
0V
478mV
out
12
478mV
3.30V
13
15
d
V6
3.3
Vstim
AC = 1
a
N=1/6
ron1=60m
ron2=60m
Vf=0.5
E1
10k
R5
100m
D
X3
PWMCCMVM
D
Check first if dc operating points are similar
Use this fixture to also run transient simulations
121 • Chris Basso – APEC 2014
Large-Signal Equations with SPICE
Non-linear equations are linearized by SPICE
V(D)*{N}*(V(in)-{ron1}*(I(VLP)+I(VIL)*{N}))-{Vf}
L1
0.5u 3.30V
in
48.0V
0V
10
Iin
V1
48
B4
Current
I(VIL)*V(D)*{N}
L2
50u
C2
0.2u
48.0V
4
5
Vclp
clp
VLP
91.7V
VIL
3.30V
out
Vout
6
3.30V
C1
500u
B3
Voltage
0V
R1
100m
1
R2
50m
V5
48.0V
B1
Current
(1-V(D))*I(VLP)
2
B2
Voltage
parameters
(V(in)+V(clp,in)+{ron2}*I(VLP))*(1-V(D))+V(D)*{ron1}*(I(VLP)+(I(V5)/V(D)))
D
E1
10k
R5
100m
478mV
L3
1kH
14
C3
1kF
0V
478mV
N=1/6
ron1=60m
ron2=60m
Vf=0.5
out
12
478mV
13
15
3.30V
Vstim
AC = 1
V6
3.3
Proposed by Dr. Jos&eacute; Capilla, ON Semi, November 2012
122 • Chris Basso – APEC 2014
Curves Perfectly Superimpose
Plot1
vdbout#b, vdbout
30.0
Vout ( f )
D( f )
20.0
10.0
0
-10.0
Plot3
vdbclp#a, vdbclp
100
Vclamp ( f )
D( f )
80.0
Control-to-clamp-voltage transfer function
60.0
40.0
8
3
20.0
60.0
Plot4
idbin, idbin#a
I in ( f )
D( f )
50.0
Input current response
40.0
30.0
Plot2
ph_vout#a, ph_vout
20.0
Vout ( f )
D( f )
10
9
20.0
-20.0
-60.0
7
2
-100
-140
Control-to-output transfer function argument
10
100
1k
frequency in hertz
10k
100k
123 • Chris Basso – APEC 2014
Compare Transient Responses
Cycle-by-cycle results are similar to that of average model
Plot1
vout#a, vout in volts
4.50
3.50
12
2
2.50
vout ( t )
1.50
Plot3
ilmag#a, ilmag in amperes
500m
imag ( t )
4.00
2.00
0
13
11
-2.00
-4.00
vclamp ( t )
130
Plot2
va, vclp in volts
∠
6
1
Control-to-output transfer function magnitude
110
14
7
90.0
70.0
50.0
200u
124 • Chris Basso – APEC 2014
600u
1.00m
time in seconds
1.40m
1.80m
Simulate with Small-Signal Sources
Plug the small-signal models of the PWM Switch in
B8
Voltage
V(a2,p2)*V(dac)
{ron1}*(I(VLP)+I(VIL)*{N})
in
48.0V
47.7V
7.94V
13
V1
48
Iin
Imag
C2
0.2u
L2
50u
48.0V
191.7V
0
L1
0.5u
VIC2
3.80V
c2
11
c2
VIL
{Vf}
3.80V
out
3.30V
Vout
6
3.80V
X1
XFMR
RAT IO = N
X5
dcXFMR
B7
Current
I(VIC2)*V(dac)
-33.0nV
C1
500u
0V
R1
100m
1
d
VLP
48.0V
0V
2
91.7V
Vclp
9
5
B2
Voltage
{ron2}*I(VLP)
B6
Voltage
3.80V
a2 a2
47.8V
D
p2
4
B5
Voltage
R3
1n
R2
50m
p2
p2
V5
V(D)*{ron1}*(I(VLP)+{N}*I(VIL))
parameters
N=1/6
ron1=60m
ron2=60m
Vf=0.5
Iout=33
c1
c1
p1
p1
0V
Vstim
AC = 1
VIC1
VDbias
478.279m
47.8V
478mV
20
d
D
D
dac
X4
dc XFMR
176fV
19
B3
Current
I(VIC1)*V(dac)
B1
Voltage
(V(a1,p1)*V(dac))/V(D)
a1 176fV
a1
Ac
modulation
R4
1n
Dc
bias
Check if ac response is ok before proceeding!
Simplify circuitry to concentrate on control-to-output only
125 • Chris Basso – APEC 2014
Same Response Between Fixtures
Plot1
vdbout#a,vdboutin db(volts)
20.0
14.0
8.00
2.00
Vout ( f )
D( f )
-4.00
38
36
Plot2
ph_vout#a,ph_voutin degrees
20.0
-20.0
-60.0
∠
Vout ( f )
D( f )
39
37
-100
-140
10
126 • Chris Basso – APEC 2014
100
1k
10k
100k
Make the Circuit Look Friendlier
Re-arrange sources to simplify the electrical circuit
Mag. current generator
B8
Voltage
{ron1}*{N}^2*(I(VLP)/{N}+I(VIL)) (V(4,0)*V(dac))/{D}
in
15
V1
48
Iin
C2
0.2u
B6
Voltage
B7
Current
I(VIL)*V(dac)
X1
XFMR
RATIO = N
Lmag
50u
10
5
VIL
{Vf}
Lout
0.5u
14
Vout
17
3
C1
500u
4
B10
Current
I(VIL)*{D}
B11
Voltage
V(9,0)*{D}
R1
100m
1
R2
50m
VLP
B2
Voltage
{ron2}*I(VLP)
9
Isolated buck converter
2
6
B5
Voltage
{D}*{ron1}*(I(VLP)+{N}*I(VIL))+{ron1}*{Iout}*{N}*V(dac)
B4
Voltage
Vclp
13
V(7,6)*{D}
parameters
B9
Current
dac
I(VLP)*{D}
Vstim
AC = 1
7
B3
Current
I(VLP)*V(dac)
B1
Voltage
(V(6)*V(dac))/{D}
N=1/6
ron1=60m
ron2=60m
Vf=0.5
Iout=33
D=478.279m
dac = dˆ
D = D0
Run a sanity check further to any simplification
127 • Chris Basso – APEC 2014
Circuit is Looking Simpler Now
Look for ways to get simpler equations, final arrangement
VIL
{Vf}
Lout
0.5u
in
4
Vout
17
3
V1
48
C2
0.2u
Lmag
50u
B12
Voltage
10
5
R1
100m
1
R2
50m
VLP
B2
Voltage
{ron2}*I(VLP)
C1
500u
2
6
B5
(({N}*V(in)-{ron1}*{N}^2*(I(VLP)/{N}+I(VIL)))*(1+V(dac)/{D}))*{D}
Voltage
{D}*{ron1}*(I(VLP)+{N}*I(VIL))+{ron1}*{Iout}*{N}*V(dac)
B3
Voltage
Vclp
13
parameters
V(6)*((V(dac)/{D})+1)*{D}
B9
Current
I(VLP)*{D}
dac
Vstim
AC = 1
Final ac check is mandatory!
128 • Chris Basso – APEC 2014
Secondary-side
contribution –
neglected for
simplicity
N=1/6
ron1=60m
ron2=60m
Vf=0.5
Iout=33
D=478.279m
Simpler Circuit Does Not Distort Response
We are good to start analyzing the equivalent circuit
Plot1
vdbout#a, vdbout in db(volts)
30.0
20.0
10.0
0
18
15
-10.0
Plot2
ph_vout#a, ph_vout in degrees
20.0
-20.0
-60.0
19
17
-100
-140
10
100
1k
frequency in hertz
10k
100k
129 • Chris Basso – APEC 2014
For the dc transfer function, open caps and short inductors
48.0V
V1
48
6
Vin
Vclp
VLP
Imag
48.0V
1
92.0V
c
Vclamp
B9
Current
I(VLP)*{D}
B5
Voltage
V(c)*{D}
parameters
Vclp = V( c ) D = Vclamp D
N=1/6
ron1=60m
ron2=60m
Vf=0.5
Iout=33
Vin=48
D=478.279m
Vclamp = Vin + Vclp
Vclp = (Vin + Vclp ) D
Vclp = Vin
dac
0V
Vstim
AC = 1
D
0.478
= 48
≈ 44 V
1− D
1 − 0.478
Vclamp = Vin + Vclp = 92 V
This is a buck-boost dc transfer function
130 • Chris Basso – APEC 2014
We Need the Magnetizing Current
Use KVL and KCL to get the mag. current expression
Vin
48V
ac = 0
C2
0.2u
iˆmag
3
VLmag
Lmag
50u
4
B1
Voltage
{ron2}*imag
(
B2
Voltage
{D}{ron1}*Imag
The secondary side
contribution has been
purposely neglected,
Iout and &icirc;out
1
B5
Voltage
V(c)*(V(dac)+{D})
iˆmag (1 − D )
c
iˆmag D
iˆmag
B9
Current
imag*{D}
)
VLmag = V(C ) − V( C ) dˆ + D0 + D0 ron1iˆmag
2
VC
131 • Chris Basso – APEC 2014
Collect Terms and (Carefully) Simplify
Use KVL and KCL to get the mag. current expression
VLmag = V ( 2 ) − Vin
with
(
)
V ( 2 ) = V( C ) − V(C ) D0 + dˆ + D0 ron1iˆmag
The voltage at node (c) has a dc and an ac component
(
VLmag + vˆLmag = V(C ) + vˆ( C ) − V( C ) + vˆ(C )
) ( D + dˆ ) + D r
0
iˆ
0 on1 mag
− Vin − vˆin
VLmag + vˆLmag = V(C ) + vˆ( C ) − V(C ) dˆ − V(C ) D0 − vˆ( C ) dˆ − vˆ(C ) D0 + D0 ron1iˆmag −Vin − vˆin
≈0
0
Sort out an ac and a dc equation
VLmag
Tsw
=0
Vin = V(C ) (1 − D0 )
dc
VLmag = V( C ) − V(C ) D0 − Vin
ac
vˆLmag = vˆ(C ) (1 − D0 ) − Vclamp dˆ + D0 ron1iˆmag
132 • Chris Basso – APEC 2014
Vclamp =
Vin
(1 − D0 )
Express the Magnetizing Current imag
The clamp capacitor ac voltage depends on imag
 1
vˆ(C ) = iˆmag (1 − D0 ) 
 sCclp



 1
 + iˆmag ron 2 = iˆmag (1 − D0 ) 


 sCclp


 + ron 2 


substitute
iˆmag = −
vˆLmag
sLmag
=−
vˆ(C ) (1 − D0 ) − Vclamp dˆ + D0 ron1iˆmag
sLmag
Solve for imag
I mag ( s ) = D ( s ) Vclamp
sCclp
D0 − 2 D0 + 1 + sCclp ( ron 2 + D0 ron1 − D0 ron 2 ) + s 2 Lmag Cclp
2
133 • Chris Basso – APEC 2014
Identify Second-Order Coefficients
Identify terms with a second-order polynomial form
I mag ( s )
D (s)
=
Vclamp
(1 − D0 )
sCclp
2
 r (1 − D0 ) + D0 ron1  2 Lmag Cclp
1 + sCclp  on 2
+s
2
2

(1 − D0 )

(1 − D0 )
Develop and rearrange:
I mag ( s )
D (s)
M0 =
sCclp
= M0
1+
134 • Chris Basso – APEC 2014
 s 
s
+

ω0M QM  ω0M 
Vclamp
(1 − D0 )
2
=
Vin
(1 − D0 )
3
ω0M =
2
QM =
1 − D0
Cclp ron 2 (1 − D0 ) + D0 ron1
Lmag
1 − D0
Lmag Cclp
Re-Write the Expression Nicely
A tuned network offers the following transfer function
A0
Q ( dB )
H ( s ) = A0
H(f)
90&deg;
∠H ( f )
0&deg;
1
ω
s 
1+  0 + Q
 s ω0 
-90&deg;
ω0
I mag ( s )
D (s)
A0 =
I mag ( s )
sCclp
= M0
1+
 s 
s
+

ω0M QM  ω0M 
D (s)
2
Vclamp
= A0
1
ω
s
1 +  0M +
ω
s
0M

A0 (ω0M ) = 67.713 dB
ron 2 (1 − D0 ) + D0 ron1
135 • Chris Basso – APEC 2014
Time for an ac Check
SPICE can obtain the ac response in a snap-shot
48.0V
parameters
6
V1
48
C2
0.2u
92.0V
2
Lmag
50u
48.0V
5
B1
Voltage
{ron2}*I(VLP)
VLP
Imag
48.0V
1
B2
Voltage
{D}*{ron1}*I(VLP)
B5
Voltage
92.0V
c
348.0V
V(c)*(V(dac)+{D})
B9
Current
I(VLP)*{D}
136 • Chris Basso – APEC 2014
N=1/6
ron1=60m
ron2=60m
Vf=0.5
Iout=33
Vin=48
D=478.279m
dac
0V
Vstim
AC = 1

 QM

Curves Perfectly Superimpose
It is important to obtain similar plots, otherwise: error!
60
50

20⋅ log
Hclp ( i⋅ 2π ⋅ f k )

40
, 10 

A
(
20
)
arg H clp ( i⋅ 2π ⋅ f k ) ⋅
0
180
π
− 50
0
4
1&times;10
1&times;10
5
fk
Max SPICE is 63.710 dB
&quot;Can do!&quot;
137 • Chris Basso – APEC 2014
Run Another Round of Rearrangement
Now concentrate on the buck output stage only
B8
Voltage
(V(4,0)*V(dac))/{D}
{ron1}*{N}^2*(I(VLP)/{N}+I(VIL))
8
Vin
X1
XFMR
RATIO = N
5
9
B6
Voltage
B7
Current
I(VIL)*V(dac)
1
3
Vin
48V
7
7
B6
Voltage
B7
Current
I(VIL)*V(dac)
B8B8
Voltage
Voltage
V(9,0)*V(dac)
V(7,0)*V(dac)
1
1
B11
Voltage
V(9,0)*{D}
V(7,0)*{D}
dc
138 • Chris Basso – APEC 2014
(
V ( 7 ) D + dˆ
ac
6
B10
Current
I(VIL)*{D}
R1
100m
6
R2
50m
{ron1}*{N}^2*(I(VLP)/{N}+I(VIL))
5
C1
500u
B11
Voltage
V(9,0)*{D}
B10
Current
I(VIL)*{D}
 I mag

V ( 7 ) = NVin − ron1 N 
+ I out 
N


8
Vout
2
reflect
4
2
X1
XFMR
RATIO = N
VIL
{Vf}
Lout
0.5u
)
F ( s)
VIL
{Vf}
Lout
0.5u
Vout
2
3
C1
500u
R1
100m
4
R2
50m
Further Simplification is Necessary
Extract the ac current from the equation,
 iˆmag

V( 7 ) = NVin − ron1 N 2 
+ I out + iˆout 
 N



Develop V(1) keep ac terms only
(
imag ( t )
Tsw
=0

 iˆmag
V(1) =  NVin − ron1 N 2 
+ I out + iˆout
 N


)
(

)  ( D + dˆ )

ˆˆ r − DI N 2 r − DN 2 iˆ r − I N 2 dr
ˆ − N 2 di
ˆˆ r
DNVin + NVin dˆ − DNiˆmag ron1 − Ndi
mag on1
out
on1
out on1
out
on1
out on1
dc
≈0
dc
≈0
Rearranging the result, we obtain:
ron1 ≪ 1
ˆ
NVin dˆ − DNiˆmag ron1 − DI out N 2 ron1 − DN 2 iˆout ron1 − I out N 2 dr
on1
(
dˆ ≪ 1
)
2ˆ
ˆ
ˆ
dˆ NVin − I out N 2 dr
on1 − DNimag ron1 − DN iout ron1
N2 ≪1
Neglecting small terms, we finally obtain:
ˆ
ˆ
vˆ(1) = dNV
in − DNimag ron1
139 • Chris Basso – APEC 2014
Add the Second-Order Response of LC Filter
The magnetizing current definition is:
I mag ( s )
D (s)
= M0
sCclp
 s 
s
1+
+

ω0M QM  ω0M 
2
= M (s)
Once re-injected in to the previous
definition, we have:
V1 ( s ) = D ( s ) NVin − D0 Nron1 D ( s ) M ( s )
V1 ( s ) = D ( s )  NVin − D0 Nron1 M ( s ) 
The source is filtered by the 2nd-order LC filter:
1
F ( s ) = F0
2
Lout
rL
3
Cout
rC
ω0F =
140 • Chris Basso – APEC 2014
1+
4
F (s)
1+
F0 =
1
Lout Cout
QF =
s
szF
 s 
s
+

ω0F QF  ω0F 
ω zF =
2
1
rC Cout
Lout Cout ω0F ( rC + RLoad )
Lout + Cout ( rL rC + Rload ( rL + rC ) )
We Have the Final Transfer Function
The transfer function reveals the mag. current contribution
Vout ( s )
D (s)
= F ( s ) ( NVin − D0 Nron1 M ( s ) )
YES!
The mag. current substracts and explains the notch




sCclp
Vout ( s )


= F0
N  Vin − D0 ron1 M 0
2
2 
D (s)
 s 
 s  
s
s

+
+
1+
1+



ω0FQF  ω0F  
ω0M QM  ω0M  
s
szF
1+
This is the control-to-output transfer function of the ACF
141 • Chris Basso – APEC 2014
Final Sanity Check - Magnitude
Compare the analytical ac response with that of SPICE
20
10

H 1( i⋅ 2π ⋅ f k )

V
20⋅ log

, 10 

0
− 10
Vout ( f )
− 20
10
D( f )
100
3
4
1&times;10
1&times;10
1&times;10
fk
Magnitude curves superimpose perfectly
142 • Chris Basso – APEC 2014
5
1&times;10
6
Final Sanity Check - Phase
Compare the analytical ac response with that of SPICE
0
 arg H i⋅ 2π ⋅ f ⋅ 180 
 ( 1(

k))
π 

− 100
∠
Vout ( f )
D( f )
10
3
1&times;10
100
1&times;10
4
5
6
1&times;10
1&times;10
fk
Argument curves are in excellent agreement too
143 • Chris Basso – APEC 2014
The Bench is the Final Referee
Build a hardware using simple gates arrangements
D1A
m br2545ctp
20
L1
74uH
1
Vout
2
14
D1B
m br2545ctp
C5
100u
Vin
51V
C1
480uF
V15
V1
15V
V15
U1A
CD4093B
4V
50 kHz
R7
4.7k
0
5
U1B
CD4093B
V15
V15
Vgen
23
13
VEE
3
15
QA
IRF9530
Q4
2N2907
U1E
CD4093B
V15
U1C
CD4093B
V15
C8
1uF
17
21
V15
R2
100k
16
R3
14k
U1D
CD4093B
C2
330p
Q3
2N2222
Out M
7
R4
1k
U2A
LM393T
C7
0.1uF
R1
1k
26
6
C6
220n
QM
SPP04N60C3
V15
V15
VCC
25
rC = 76m
XFMR
RATIO = 0.25
V15
C9
0.1uF
V15
ac
input
Out A
11
9
8
R5
1k
C3
330p
U1F
CD4093B
R10
1k
Q1
2N2222
10
C4
Q2 0.1u
2N2907
12
D2
1N4148
R6
10k
This active clamp forward converter delivers 5 V/5 A
144 • Chris Basso – APEC 2014
R9
1
The Prototype Hardware
The circuit is assembled using available components
Make sure caps. are well characterized before soldering
With the kind help of Yann Vaquette, Application engineer
145 • Chris Basso – APEC 2014
Typical Prototype Waveforms
voutA ( t )
voutM ( t )
vDS ( t )
N-channel
146 • Chris Basso – APEC 2014
Quasi-ZVS is Ensured on the Drain
voutA ( t )
voutM ( t )
140 ns
100 V
Vin = 51 V
vDS ( t )
28 V
N-channel
147 • Chris Basso – APEC 2014
ZVS is Also Ensured on Clamp Switch
voutA ( t )
voutM ( t )
240 ns
Body diode
vDS ( t )
P-channel
148 • Chris Basso – APEC 2014
−100 V
Current in the Clamping Network
voutA ( t )
voutM ( t )
iCclamp ( t )
149 • Chris Basso – APEC 2014
SPICE Test Fixture
We have used the following SPICE simulation fixture
{ron1}*(I(VLP)+(I(V5)/V(D)))
X2
PWMCCMVM
B4
Voltage
in
51.0V
49.8V
12.5V
17
Iin
Vin
51
C2
0.23u
99.8V
L2
{Lmag}
51.0V
10
Vclp
L1
74u
c
20
X1
XFMR
RATIO = N
VIL
{Vf}
R3
60m
5.20V
11
6.16V
D
PWM switch VM
6.16V
Vout
5.85V
0V
2
4
B5
Voltage
({ron1}*(I(VLP)+(I(V5)/V(D))))*V(D)
V5
18
c
parameters
p
99.8V
X3
PWMCCMVM
R5
100m
1.98V
D
GAIN
X4
GAIN
K = 1/4
a
d
PWM switch VM
N=0.25
Lmag=580uH
ron1=0.9
ron2=0.6
Vf=0.65
19
C3
1kF
0V
L3
1kH
14
E1
10k
1.98V
out
12
1.98V
5.20V
13
15
Vstim
AC = 1
D
Check dc points versus hardware values: ok
150 • Chris Basso – APEC 2014
R1
1
1
R2
76m
51.0V
50.4V
C1
480uF
0V
p
VLP
B2
Voltage
clp
out
6
d
495mV
5
{ron2}*I(VLP)
a
16
V6
5.2
Magnitude Response
There is a slight shift but overall agreement is good
SPICE
measured
H(f)
cap. ESR is often the offender in loop measurement
151 • Chris Basso – APEC 2014
Phase Response
Good agreement between curves, expecially peaking
measured
SPICE
∠H ( f )
The notch Q depends on resistive elements rDS(on) etc.
152 • Chris Basso – APEC 2014
Comparison with Simplis Response
Excellent agreement between curves, almost no shift
Simplis
measured
H(f)
153 • Chris Basso – APEC 2014
Comparison with Simplis Response
Despite a slightly lower peaking, phase agreement is ok
Measured
∠H ( f )
154 • Chris Basso – APEC 2014
Simplis
Simulation also Confirms Damping!
Inserting a 10-Ω resistance damps the notch nicely
simulated
H(f)
measured
simulated
∠H ( f )
measured
155 • Chris Basso – APEC 2014
A Practical Case
The model helped to stabilize this 3.3-V/30-A converter
∠T ( f )
f c = 31.9 kHz
ϕm = 62.9&deg;
156 • Chris Basso – APEC 2014
T(f)
ϕm
Transient Response Test
Step-load transient response confirms stability margins
Overshoot = 42.2 mV
vOUT ( t )
3.3 V
Undershoot = 48.2 mV
VIN = 36 V, 15 A to 22.5 A - Slew rate 1 A / &micro;s
157 • Chris Basso – APEC 2014
Conclusion
The PWM switch model is an essential tool for modeling
We have seen how to derive it in different operating modes
Small-signal modeling using the PWM switch is simple and fast
When modeling converters, always proceed step by step
Always perform intermediate sanity checks (SPICE, Mathcad&reg;…)
Analytical analysis does not shield you against lab. experiments
Analysis, simulation and bench: the path to success!
Merci !
Thank you!
Xi&egrave;-xie!
158 • Chris Basso – APEC 2014
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