Small-Signal Analysis
of the Constant Peak Current-Operated
Flyback Converter in Frequency
Foldback
Christophe Basso
1 • Chris Basso
Reducing the Switching Frequency
New generation controllers reduce frequency in light load
The peak current is frozen and frequency is controlled
The relationship between vˆout and iˆp is known at constant Fsw
What about the relationship linking vˆout to Fˆsw at Ipeak constant?
2 • Chris Basso
Where do We Start From?
We have a DCM peak-current mode control large-signal model
The PWM switch controls the peak current at a fixed Fsw
Why not fixing the peak current and controlling Fsw?
a
dI c
c
∆I L
2
Vc
Ri
a
c
(1 − D )
Vc
Ri
ton FSW I c
2 LFSW
p
p
Fsw constant
I peak constant
Vout
Fˆsw
iˆp
The peak current is frozen and the frequency is controlled
3 • Chris Basso
A Transient Load Step Response First
We can check the cycle-by-cycle response to a load step
C11
680p
R18
80
30
Rprim
0.5
i nt
15
L3
2.2u
R5
10m
12
Vout
Vout
19
D2A
mbr20200ctp
vdd
ILprim
3
Lpri m
600u
V6
5
vdd
X4
XFMR-AUX
RATIO_POW = -0.25
RATIO_AUX = -0.18
R13
20k
VDS
VCO
maxdc
FB
vdd
26
D
Q
Clk
Q
Cout1
470u
IC = 18
C7
100u
IC = 18
X8
PSW1
paramete rs
Load step
Vout=19
P1=15
P2=20
Vout
14
X3
PSW1
Qb
X9
F_FLOP
Ramp Gnd
Vclk
IOpto
X10
OR2
V5
Vsense
11
10
ramp
Rled
1k
FB
VFB
R
36
vdd
C5
1nF
V4
unknown
20
6
S
Clk
EN
32
Vcc
C6
10u
IC = 17
Vdrv
maxDC
FB
R7
250m
R1=Vout^2/P1
R2=Vout^2/P2
X1
krakenclock
Krake n Clock
Resr1
60m
31
aux D4
1N4937
4
Vin
330
R19
10
R10
1k
34
1
5
R2
0.8
X2
Optocoupler
R1
1k
35
maxdc
C4
220pF
X11
COMPARHYS
9
-
+
D3
1N967
33
V7
1
Cycle by cycle model
4 • Chris Basso
Fsw = 51 kHz
A Transient Load Step Response First
And compare it to that of a modified PWM switch model
B1
Voltage
5.09V
-82.9V
R5
10m
L1
2.2u
20.7V
vout
Vout
19
20.7V
c
Vin
330
-622m+V(Vc)*1.58
X3
XFMR
RATIO = -250m
p
X2
PWMDCMCM
PWM switch CM
duty-cycle
Fsw
0V
5
a
115m V
vc
10
3
20.7V
12
Resr1
60m
R7
250m
21
20.7V
330V
20.7V
4
V4
unknown
20
15
L3
600u
Cout1
470u
C7
100u
parameters
X8
PSW1
Load step
vout
Vout=19
P1=15
P2=20
V2
5
C5
1nF
R1=Vout^2/P1
R2=Vout^2/P2
13
R4
20k
R6
1
vc
Rled
1k
5.00V
2
3.61V
3.61V
19.4V
1
X1
Optocoupler
R1
1k
*
.param {Vc}=1 Current frozen to 1 Rsense
*
Bdc dcx 0 V =
{Vc}*V(Fsw)*10k/({Se}+(abs(v(a,c))*{Ri})/{L}+1u)
Xdc dcx dc limit params: clampH=0.99
clampL=7m
BVcp 6 p V=(V(dc)/(V(dc)+V(d2)))*V(a,p)
BIap a p I=(V(dc)/(V(dc)+V(d2)))*I(VM)
Bd2 d2X 0 V=(2*I(VM)*{L}v(a,c)*V(dc)^2*1/(V(Fsw)*10k)) / (
+v(a,c)*V(dc)*1/(V(Fsw)*10k)+1u )
Xd2 d2X dc d2 limit2
Rdum1 dc 0 1Meg
Rdum2 vc 0 1Meg
1 V = 10 kHz
RS 7 c 1u
VM 6 7
*
.ENDS
18.3V
17
Modified netlist of the PWM switch
D3
1N967
Fsw = 50.9 kHz
Averaged model
5 • Chris Basso
You Need to Adjust the VCO Modulator
The control voltage to the switching frequency is the VCO gain
Fsw
y = ax + b
y=
65 kHz
∆Fsw 64.9k
=
= 15.8 kHz V
∆Vc
4.1
100 = ax + b = 0.4 ×15.8k + b
1 10000
100 Hz
Vc
0.4 V
b = 0.10 − 0.4 × 15.8 = −622 mV
4.5 V
VFSW
×10k
VCO gain GVCO is 15.8 kHz V
6 • Chris Basso
Scaling
factor
−0.622 + 15.8 × VFB
VFB
Compare the Transient Responses
Responses are identical, the averaged version looks correct
v
vout ( t )
21.00
Cycle by cycle
vout ( t )
20.90
average
20.80
20.70
20.60
10.6m
12.7m
14.8m
16.9m
t (s )
19.0m
7 • Chris Basso
Keys of Small-Signal Analysis
Rather than going full speed into small-signal analysis:
break down the system into smaller parts
run simplifications whenever you can
go step by step and verify the answer always fits the original
a
Ic
d1
d1 + d 2
Need to be computed
Ia
c
Vc
Ri
Iµ
d 2Tsw
Vcp  d1 + d 2 
1−
L 
2 
p
Start from here, DCM model
"Switch-mode Power Supplies: SPICE Simulations and Practical Designs", C. Basso, McGraw-Hill 2008, page 161
8 • Chris Basso
Simplifying and Compacting the Model
Compact the DCM model to suppress variables computing
I a = Ic
d1
d1 + d 2
Iµ =
d 2Vcp  d1 + d 2 
1−
Fsw L p 
2 
d1 =
L pVc Fsw
d2 =
Vac Ri
2 I c Fsw L p
d1Vac
− d1
substitute
"Let the craziness begin"
Ia =
Iµ = −
I a = Ic
d1
d1 + d 2
Fsw L pVc 2
2 Ri 2Vac
(
Vcp (Vc − I c Ri ) Fsw L pVc 2 − 2 I c Ri 2Vac
)
2
Fsw L p RiVc Vac
9 • Chris Basso
The New Model Looks Simpler
The equations no longer include a computed variable
Fsw
Vc
a
c
1
R8
1Meg
B2
Current
Ia
I
B4 µ
Current
-V(c,p)*({Vc}-I(Vc)*{Ri})*(V(Fsw)*{k}*{Lp}*{Vc}*{Vc}-2*I(Vc)*{Ri}*{Ri}*V(a,c))
+/((V(Fsw)*{k}*{Lp}*{Ri}*{Vc}*{Vc}*V(a,c))+1u)
B3
Current
{Vc}/{Ri}
p
(({Vc}/{Ri})^2)*{Lp}*V(Fsw)*{k}/((2*V(a,c))+1u)
PWM switch CM
vc
c
p
duty-cycle
a
Check against the
complete model
X2
PWMDCMCM
10 • Chris Basso
Ac Responses are Similar
The curves perfectly superimpose, 1st step is ok
dB
5.00
-5.00
H(f)
-15.0
-25.0
-35.0
°
-10.0
-30.0
-50.0
-70.0
∠H ( f )
-90.0
10
100
1k
f ( Hz )
10k
100k
11 • Chris Basso
Second Step, Linearize the Sources
To apply Laplace equations, we need linear elements
Linearization can be done in different ways:
perturb all equations with a small quantity (the "hat" notation)
re-arrange the terms and collect dc and ac contributors
can be tedious to re-arrange, you neglect cross products
V1 = R1 I1 + DV3
(
) (
)
V1 + vˆ1 = R1 I1 + iˆ1 + dˆ + D (V3 + vˆ3 )
V1 = R1 I1 + DV3
dc equation (bias point)
ˆ + dv
ˆ ˆ + Dvˆ ac equation
vˆ1 = R1iˆ1 + dV
3
3
3
≈0
12 • Chris Basso
Second Step, Linearize the Sources
A second option is to calculate partial derivative coefficients
the process can be automated by Mathcad®
you only have ac terms, no sort needed
V1 ( I1 , D, V3 ) = R1 I1 + DV3
∂V ( I1 , D,V )1
 ∂V ( I , D,V ) 
∂V ( I , D,V )
dV1 =  1 1
dI1 + 1 1
dD +

∂I1
∂D
∂V3

 D ,V3
I1 ,V3
vˆ1 =
∂V1 ( I1 , D, V )
∂I1
iˆ1 +
∂V1 ( I1 , D, V )
D ,V3
∂D
I1 ,V3
∂V ( I , D, V )
dˆ + 1 1
∂V3
dV3
D , I1
vˆ3
D , I1
ˆ
vˆ1 = R1iˆ1 + Dvˆ3 + dV
3
ac equation, no cross products
13 • Chris Basso
Second Step, Linearize the Sources
Now, identify the variables in each source
Ia =
Fsw L pVc 2
Fsw
2 Ri 2Vac
Vac
Iµ = −
(
Vcp (Vc − I c Ri ) Fsw L pVc 2 − 2 I c Ri 2Vac
)
Fsw L p RiVc 2Vac
Vcp
Vac
Fsw
Ic
6 variables imply six partial derivatives, 6 coefficients
You must identify these static variables first
Look at the PWM switch configuration
14 • Chris Basso
Identify the Variables in the Schematic
the average voltage across L3 is 0: point c is grounded.
X3
XFMR
RATIO = -250m
B1
Voltage
dc
X2
PW MDCMCM
Vout
p
1
-622m+V(Vc)*1.58
PWM switch CM
duty-cycle
a
vc
6
Vac = Vin
1:-N
c
Vin
330
3
Vcp =
4
Vout
N
5
vL ( t )
Tsw
=0
L3
600u
15 • Chris Basso
Sources Derivation
We can now individually derive all these sources
Ia =
 ∂I 
 ∂I 
iˆa =  a  Fˆsw +  a  vˆac
∂
F
 sw Vac
 ∂Vac  Fsw
Fsw L pVc 2
2
2 Ri Vac
k1 =
Iµ = −
L pVc 2
2 Ri 2Vac
(
k2 = −
Vcp (Vc − I c Ri ) Fsw L pVc 2 − 2 I c Ri 2Vac
Fsw L pVc 2
2 Ri 2Vac 2
)
2
Fsw L p RiVc Vac
 ∂I µ
iˆµ = 
 ∂Vcp

k3
16 • Chris Basso

 ∂I µ 
 ∂I µ 
 ∂I µ 
vˆcp + 
iˆc + 
Fˆsw + 
vˆac





 ∂I c  Fsw ,Vcp ,Vac
 ∂Fsw  Ic ,Vcp ,Vac
 ∂Vac Vcp , I c , Fsw
 I c , Fsw ,Vac
k4
k5
k6
Sources Derivation
Yes, Mathcad® or an equivalent software is of great help…
k3 = −
k4 =
k5 =
k6 =
(Vc − I c Ri ) ( Fsw L pVc 2 − 2 Ic Ri 2Vac )
Fsw L p RiVc 2Vac
(
Vcp Fsw L pVc 2 − 2 I c Ri 2Vac
2
) + 2R V
i cp
(Vc − I c Ri )
Fsw L pVc 2
Fsw L pVc Vac
(
Vcp (Vc − I c Ri ) Fsw L pVc 2 − 2 I c Ri 2Vac
Fsw2 L p RiVc 2Vac
(
Vcp (Vc − I c Ri ) Fsw L pVc 2 − 2 I c Ri 2Vac
2
Fsw L p RiVc Vac
2
) −V
cp
(Vc − I c Ri )
Fsw RiVac
) + 2I R V
c i cp
(Vc − Ic Ri )
Fsw L pVc 2Vac
17 • Chris Basso
Evaluate all These Coefficients
Select a converter at a certain operating point
Vout = 21.1 V
Rload = 18 Ω
Ri = 0.8 Ω
Vc = 1V
Vac = 330 V
rC = 0.06 Ω
Fsw =
Ic =
Cout = 470µF
N1 = 0.25
Vcp =
Vout
= 84.4 V
N1
L p = 600µH
k F = 10000
18 • Chris Basso
Ia =
2 I out Ri 2Vout
L pVc 2
= 52.8 kHz
(
Fsw L pVc 2 Vac + Vcp
2
2 Ri VacVcp
Fsw L pVc 2
2 Ri 2Vac
I µ = 0.882 A
) = 0.368 A
= 0.075 A
Test the Coefficient Values with the Sources
Capture a new schematic with the linearized sources
parameters
X3
XFMR
RATIO = -250m
Vout
a
16.3V
-65.0V
c
Res r1
60m
B1
Voltage
(-622m +V(Vc)*1.58)
Vc=1
Lp=600u
Ri=0.8
Lp=600u
k=10k
N=250m
Fsw=52.766k
Iout=P1/Vout
Vac=330
Vcp=Vout/N
Ia=Fsw*Lp*Vc^2/(2*Ri^2*Vac)
k1=(Lp*Vc^2)/(2*Ri^2*Vac)
k2=(Fsw*Lp*Vc^2)/(2*Ri^2*Vac^2)
AA=Fsw*Lp*Vc^2*Vac+Fsw*Lp*Vc^2*Vcp
BB=2*Ri^2*Vac*Vcp
Ic=AA/BB
A=-Vcp*(Vc-Ic*Ri)*(Fsw*Lp*Vc^2-2*Ic*Ri^2*Vac)
B=Fsw*Lp*Ri*Vc^2*Vac
Imju=A/B
AAA=-(Vc-Ic*Ri)*(Fsw*Lp*Vc^2-2*Ic*Ri^2*Vac)
BBB=Fsw*Lp*Ri*Vc^2*Vac
k3=AAA/BBB
k4A=Vcp*(Fsw*Lp*Vc^2-2*Ic*Ri^2*Vac)
k4B=Fsw*Lp*Vc^2*Vac
k4C=2*Ri*Vcp*(Vc-Ic*Ri)/(Fsw*Lp*Vc^2)
k4=(k4A/k4B)+k4C
k5A=Vcp*(Vc-Ic*Ri)*(Fsw*Lp*Vc^2-2*Ic*Ri^2*Vac)
k5B=Fsw^2*Lp*Ri*Vc^2*Vac
k5C=Vcp*(Vc-Ic*Ri)/(Fsw*Ri*Vac)
k5=(k5A/k5B)-k5C
k6A=Vcp*(Vc-Ic*Ri)*(Fsw*Lp*Vc^2-2*Ic*Ri^2*Vac)
k6B=Fsw*Lp*Ri*Vc^2*Vac^2
k6C=2*Ic*Ri*Vcp*(Vc-Ic*Ri)/(Fsw*Lp*Vc^2*Vac)
k6=(k6A/k6B)+k6C
vout
Fs w p
330V
0V
15
L3
600u
Vin
330
R22
{R1}
16.3V
Cout1
470u
parameters
vout
Vout=21.1
P1=24.73
R1=Vout^2/P1
V2
5
13
R4
20k
LoL
1kH
vc
Fsw
Rled
1k
5.00V
C5
1nF
2
5.00V
7.28V
5.00V
16.3V
1
R8
1Meg
CoL
1kF
0V
X1
Optocoupler
7
R1
1k
16.3V
V1
AC = 1
17
D3
1N967
Linearized sources
a
Vc
c
0V
8
BIa_k2
Current
{k2}*V(a,c)
BIa_k1
Current
{k1}*V(Fs w)*{k}
BIa_dc
Current
{Ia}
DCM version
BIpeak
Current
{Vc}/{Ri}
Bm ju
Current
{Im ju}
Bm juk3
Current
{k3}*V(c,p)
Bm juk4
Current
{k4}*I(Vc)
Mathcad® coefficients
Bmjuk5
Bmjuk5x
Current
Current
{k5}*V(Fsw)*10k {k6}*V(a,c)
p
19 • Chris Basso
Responses with Previous Models are Similar
The curves perfectly superimpose, 2nd step is ok
dB °
5.00 -10.0
H(f)
-5.00 -30.0
-15.0 -50.0
-25.0 -70.0
∠H ( f )
-35.0 -90.0
10
20 • Chris Basso
100
1k
f ( Hz )
10k
100k
Combine and Arrange the Sources
Now, re-arrange the sources in a more convenient way
parameters
Ac response is ok!
Fsw
Vc
V1
3.61
AC = 1
k2 is
positive
V(a,c) = -V(c) if V(a)=0
Reversed because V(a)=0
B1
Voltage
V(Vc)*1.58*{k}
VCO modulator
X3
XFMR
RATIO = -N
Vout
p
8
Resr1
60m
R22
{R1}
15
i2
BIa_k1
Current
{k1}*V(Fsw )
Bmjuk4
Current
{k4}*I(Vc)
R3
{1/k3}
BIa_k2
Current
{k2}*V(c)
Bmjuk5
Current
{k5}*V(Fsw )
DCM version
small-signal model
ac only
4
Vc
Cout1
470u
Bmjuk6
Current
{k6}*V(c)
i1
parameters
c
L3
600u
Vout=21.1
P1=24.73
R1=Vout^2/P1
Linearized PWM switch
Vc=1
Lp=600u
Ri=0.8
Lp=600u
k=10k
N=250m
Fsw=52.766k
Iout=P1/Vout
Vac=330
Vcp=Vout/N
Ia=Fsw*Lp*Vc^2/(2*Ri^2*Vac)
k1=(Lp*Vc^2)/(2*Ri^2*Vac)
k2=(Fsw*Lp*Vc^2)/(2*Ri^2*Vac^2)
AA=Fsw*Lp*Vc^2*Vac+Fsw*Lp*Vc^2*Vcp
BB=2*Ri^2*Vac*Vcp
Ic=AA/BB
A=-Vcp*(Vc-Ic*Ri)*(Fsw*Lp*Vc^2-2*Ic*Ri^2*Vac)
B=Fsw*Lp*Ri*Vc^2*Vac
Imju=A/B
AAA=-(Vc-Ic*Ri)*(Fsw*Lp*Vc^2-2*Ic*Ri^2*Vac)
BBB=Fsw*Lp*Ri*Vc^2*Vac
k3=AAA/BBB
Rk3=1/k3
k4A=Vcp*(Fsw*Lp*Vc^2-2*Ic*Ri^2*Vac)
k4B=Fsw*Lp*Vc^2*Vac
k4C=2*Ri*Vcp*(Vc-Ic*Ri)/(Fsw*Lp*Vc^2)
k4=(k4A/k4B)+k4C
k5A=Vcp*(Vc-Ic*Ri)*(Fsw*Lp*Vc^2-2*Ic*Ri^2*Vac)
k5B=Fsw^2*Lp*Ri*Vc^2*Vac
k5C=Vcp*(Vc-Ic*Ri)/(Fsw*Ri*Vac)
k5=(k5A/k5B)-k5C
k6A=Vcp*(Vc-Ic*Ri)*(Fsw*Lp*Vc^2-2*Ic*Ri^2*Vac)
k6B=Fsw*Lp*Ri*Vc^2*Vac^2
k6C=2*Ic*Ri*Vcp*(Vc-Ic*Ri)/(Fsw*Lp*Vc^2*Vac)
k6=(k6A/k6B)+k6C
21 • Chris Basso
Go for Mesh and Node Analysis
Express the current and voltage in the primary side
Z eq ( s )
X3
XFMR
RATIO = -N
ip
p
Vout
8
−
Resr1
8.5m
Vout
N
R22
{R1}
15
Cout1
1360u
i2
Bmjuk4
Current
{k4}*I(Vc)
R1
{1/k3}
BIa_k1
Current
{k1}*V(Fsw)*{k}
BIa_k2
Current
{k2}*V(c)
4
i1
Vc
c
L3
600u
22 • Chris Basso
Bmjuk5
Current
{k5}*V(Fsw)
Bmjuk6
Current
{k6}*V(c)
Mesh and Node Analysis
KCL: the sum of currents arriving at a node equals the sum
of currents leaving the node:
k4i1 + k5V ( Fsw ) + i1 = i2 + sk6 L p i1
−
i1 =
i2 = i1 + i1k4 + k5V ( Fsw ) − sL p i1k6
Vout
− R1i2
N1
sL p
i1 ( s ) = −
Vout ( s ) + N1 R1i1 ( s ) + N1 R1i1 ( s ) k4 + N1 R1k5V ( Fsw ) − sL p R1i1 ( s ) k6 N1
L p N1s
solve for i1:
i1 ( s ) = −
Vout ( s ) + N1 R1k5V ( Fsw )
N1 R1 (1 + k4 ) + sL p N1 (1 − k6 R1 )
23 • Chris Basso
Mesh and Node Analysis
Apply similar technique to get the primary current:
i p ( s ) = k1V ( Fsw ) + k2i1 ( s ) sL p − i1 ( s )
ip ( s ) =
Vout ( s ) − L pVout ( s ) k2 s + N1 R1V ( Fsw )( k1 + k5 + k1k4 ) + sL p N1V ( Fsw )( k1 − R1k1k6 − R1k2 k5 )
iout = −
Vout ( s ) =
(
N1 R1 + R1k 4 + sL p − sL p R1k6
ip
)
Vout = iout Z eq
N1
Vout ( s ) − L pVout ( s ) k2 s + N1 R1V ( Fsw )( k1 + k5 + k1k4 ) + sL p N1V ( Fsw )( k1 − R1k1k6 − R1k2 k5 )
(
N12 R1 + R1k4 + sL p − sL p R1k6
)
( N1R1 ( k1 + k5 + k1k4 ) + sLp N1 ( k1 − R1k1k6 − R1k2 k5 ) ) Z s
eq ( )
V ( Fsw ) N12 ( sR1 L p k6 − sL p − R1k 4 − R1 ) − Z eq ( s ) + sL p k2 Z eq ( s )
Vout ( s )
24 • Chris Basso
=
Z eq ( s )
Fast Analytical Techniques
Fast analytical techniques unveil Zeq in a second!
rC
V1
I1
Rload
1st-order
system
Z eq ( s ) = Z 0
1 + s ωz
1+ s ωp
C
1. In dc, open the capacitor: Z 0 = Rload
2. What prevents the excitation I1 from reaching the output V1?
A short-circuit between rC and C:
rC +
sr C + 1
1
= C
=0
sC
sC
ωz =
1
rC C
25 • Chris Basso
Fast Analytical Techniques
Get the time constant by putting the excitation to zero:
open the current source and look at the cap. driving R
?
rC
Rload
τ = ( rC + Rload ) C
The equivalent impedance is therefore:
Z eq ( s ) = Rload
1 + srC C
1 + sC ( rC + Rload )
You cannot beat equation-solving by inspection!
26 • Chris Basso
Almost There…
Develop the expression with Zeq(s), cry and re-arrange:
H ( s ) = H0
N (s)
D (s)
H 0 = −GVCO
N1 Rload R1 ( k1 + k5 + k1k4 )
Rload +
N12 R1

 k − R k k − R1k 2 k5  
N ( s ) = 1 + sL p  1 1 1 6
  (1 + srC C )


 R1 ( k1 + k5 + k1k4 )  

(1 + k4 )
=
Vout
GVCO
2 Fsw
D ( s ) = 1 + as + bs 2


 N12 − Rload k2 − N12 R1k6
Rload 2
a = Cout  Rload + rC −
+
L

p
2


Rload + N12 R1 (1 + k4 ) 

 Rload + N1 R1 (1 + k4 )
Rload

k2 rC − R1 Rload k6 − R1k6 rC
 Rload + rC −
N12
b = L p N12 Cout 

Rload + N12 R1 (1 + k4 )












27 • Chris Basso
A Few More Minutes, Keep the Faith…
Put the denominator under a second-order form and identify
 s 
s
D ( s ) = 1 + as + bs = 1 +
+ 
ω0 Q  ω0 
2
2
Rload
f0 =
Q=
1
2π L p Cout
Rload + rC −
N12
Rload
N12
+ R1 (1 + k4 )
k 2 rC − R1k6 ( Rload + rC )
= 1.75 kHz
1
2



 N 2 − Rload k2 − N12 R1k6
Rload
ω0  Cout  Rload + rC −
 + Lp  1
2
2

 R

Rload + N1 R1 (1 + k4 ) 

 load + N1 R1 (1 + k 4 )

28 • Chris Basso




= 0.021
A Few More Minutes, Keep the Faith…
Extract the zeros:
f z1 =
1
 k − R (k k − k k ) 
2π L p  1 1 1 6 2 5 
 R1 ( k1 + k5 + k1k4 ) 
= −487 kHz
f z2 =
1
= 5.6 kHz
2π rC Cout
RHPZ
Extract the low-frequency poles:
f p1 =
1
1
1
=
≈
Rload  π Rload Cout



Rload 2
2π Cout  Rload + rC −
 2π Cout  Rload + rC − 2 



Rload + N12 R1 (1 + k4 ) 

f p1 =
1
π Cout ( Rload + rC )
f p2 =
rC ≪ Rload
= 37.73Hz
1

− Rload k2 − N12 R1k6
2π L p 
2
 R
 load + N1 R1 (1 + k4 )
N12




= 152 kHz
29 • Chris Basso
Final Lap, Compare the ac Plots
Compare the original equation and its re-arranged form:
10 dB
0
(
20⋅ log(
20⋅ log H 1( i⋅ 2π ⋅ f k )
)
H final( i⋅ 2π ⋅ f k )
H(f)
100
(
)
0
− 40
− 60
10
∠H ( f )
100
1×10
(
3
f ( Hz )
1×10
4
It confirms the derivation is correct!
1×10
5
1×10
180
π
)
arg Hfinal( i⋅ 2π ⋅ f k ) ⋅
− 100
fk
30 • Chris Basso
)
arg H1( i⋅ 2π ⋅ f k ) ⋅
− 20
6
180
π
The Final Test: SPICE vs Mathcad
If all is well, the curves must perfectly superimpose
dB °
3.00 144
H(f)
-11.0 72.0
-25.0
0
-39.0 -72.0
∠H ( f )
-53.0 -144
10
100
1k
10k
f ( Hz )
100k
1Meg
If not, there is a hidden mistake: chase it (good luck)
31 • Chris Basso
The Final Transfer Function
The current-mode flyback converter is operated in DCM
The peak current is fixed and Fsw is controlled
What is the simplified transfer function?
vˆc
VCO
Fsw
Flyback
vˆout
I peak = freeze
NCP1250
H (s) =
32 • Chris Basso
vˆout
1 + s ωz
V
= H0
= out GVCO
vˆc
1 + s ω p 2 Fsw
1 + srC Cout
R
1 + s load Cout
2
A Faster Way?
If you are in a hurry and looking for a faster way:
Formulate the output current expression
Differentiate the expression to its variables
Draw an equivalent schematic and solve the equations
I out Vout
vˆout
rC
iˆout
Rload
Cout
A DCM/CCM
current-mode converter
Simplified linearized model
33 • Chris Basso
The Founding Equation is Well Known
A flyback converter running in DCM obeys:
Pout = I outVout
1
= L p I peak 2 Fsw
2
I peak
V
= c
Ri
2
I outVout
1 V 
= L p  c  Fsw
2  Ri 
2
I out
V 
L p  c  Fsw
 Ri 
=
2Vout
Two
variables
 ∂I ( F , V ) 
 ∂I ( F , V ) 
iˆout =  out sw out  vˆout +  out sw out  Fˆsw
∂Vout
∂Fsw

 Fsw

Vout
34 • Chris Basso
First Expression Comes Easily
Rework the individual coefficients to a simpler form
Fsw L p
 ∂I out ( Fsw , Vout ) 
 vˆout = −


∂Vout
2Vout 2

 Fsw
2
 Vc 
  vˆout
 Ri 
g m ⋅ vˆout =
1
vˆout
R
 Ω −1 


1
2
I peak 2 Fsw L p
Fsw L p  Vc 
2
−
 =−
2 
R
2Vout  i 
Vout 2
−
Pout
Vout
=−
2
vˆout
Rload
Vout I out
1
=−
VoutVout
Rload
35 • Chris Basso
Second Expression is Also Simple
Identify the power in the equation, simplify…
Lp
 ∂I out ( Fsw ,Vout ) 

 Fˆsw =
∂Fsw
2Vout

Vout
2
 Vc  ˆ
  Fsw
 Ri 
2
L p  Vc 
Pout ˆ
V
Vout ˆ
Vout
Fsw = out
Fsw =
Fˆsw
  Fˆsw =
2Vout  Ri 
FswVout
Rload FswVout
Rload Fsw
vˆout
rC
Vout
Fˆsw
Rload Fsw
Controlled
variable
36 • Chris Basso
Rload
Rload
Cout
Can't beat such a
simple schematic!
The Transfer Function is Immediate
The transfer function is now straightforward
vˆout ( s ) = Fˆsw ( s )
Vout
Rload || Z eq ( s )
Rload Fsw
≈
Rload
2
1 + srC C
R

1 + sC  load 
 2 
Put it under the normalized form
vˆout ( s ) = vˆc ( s ) k F GVCO
Vout
2 Fsw
1 + srC C
R

1 + sC  load 
 2 
vˆout
1 + s ωz
V
= H0
= out k F GVCO
vˆc
1 + s ω p 2 Fsw
1 + srC Cout
R
1 + s load Cout
2
No RHPZ and no high-frequency pole prediction
Good for low-frequency analysis only, but good enough for
us!
H (s) =
37 • Chris Basso
Simulation Shows the Differences
SPICE only manipulates linear equations
Large-signal
equation
Small-signal
equation
out
Vout
Resr1
60m
B1
Current
R22
{R1}
2
Cout1
470u
{Lp}*({Vc}/{Ri})^2*V(Fsw)/(2*V(out)+100n)
V1
(-622m+V(Vc)*1.58)*{kf}
3.61
AC = 1
Vc=1
Lp=600u
Ri=0.8
Lp=600u
kf=10k
Vout=21.1
P1=24.73
R1=Vout^2/P1
Fsw=50.8k
Fsw
Vc
parameters
VoutSIMP
4
Resr2
60m
B3
Current
R4
{R1}
R3
{R1}
3
Cout2
470u
({Vout}/({R1}*{Fsw}))*V(Fsw)
B2
Voltage
VCO modulator
V

 21.1

20 log10 H 0 = 20 log10  out k F GVCO  = 20log10 
10k ⋅1.58  = 10.3dB
 2 × 50.8k

 2 Fsw

38 • Chris Basso
Simulation Shows the Differences
Ac results are identical at low frequency
10.3 dB
dB
10.0
Simple
model
H(f)
-4.00
-18.0
-32.0
Complex
model
-46.0
°
0
Simple
model
-40.0
-80.0
∠H ( f )
-120
Complex
model
-160
10
100
f ( Hz )
1k
10k
100k
1Meg
39 • Chris Basso
Test Fixture
Plant frequency response with a NCP1250
Load is reduced to force DCM where Ipeak is frozen
R11 R13
47k 47k
Vbulk
C2
10n
R14
NTC
7
R21
22
D10
24
D1
1N964
6
R24
220
1N4148
10
4
Np
C9
330pF
11
9
R22
536k
5
Q1
2N2907
C10 R5
22pF 1.6k
C3
4.7uF
21
R8
1k
16
C4
100pF
2
1 kV
R18
22k
C6
10nF
R10
66k
H (s) =
3
20
R19
1k
15
17
C5
10nF
R23
1k
IC2
TL431
R6a R6b R6c
1
1 1
R9
10k
C13
2.2nF
Type = Y1
VA ( f )
40 • Chris Basso
. .
D7
1N4937
C17
100uF
1N4148 R16
8
C15
220p
19
R100
22
1
12
Vout
.
23
13
C8
1nF
C5b
680uF
C5a
680uF
22
D6
6
4
14
Ns
D4
1N4937
NCP1250
3
5
.
18
22
2
.
Naux
D3
1N4937
C12
100uF
1
D5
MBR20200
T1
VB ( f )
VB ( f )
VA ( f )
Connections
Connect probe grounds to a quiet point: opto emitter is ok
Use an isolated current-limited dc-supply
Short primary and secondary grounds
H(f)
100 V dc
Power supply under test
Injection
transformer
41 • Chris Basso
Run the SPICE Simulation
B1
Voltage
3.84V
10
X2
PWMDCMCM
L = Lp
Ri = Ri
duty-cycle
a
vc
PWM switch CM
5
Check operating points:
vout
Vout
19.0V
p
100V
Vin
100
({b}+V(Vc)*{a})/10k
X3
XFMR
RATIO = -250m
c
174mV
dc
-76.0V
3
Resr1
8.5m
19.0V
15
L3
{Lp}
Cout1
1.36m
parameters
Vout=19
P1=6.6
R1=Vout^2/P1
Lp=600u
Ri=0.33
LoL
1kH
vc
R2
100m
797mV
797mV
12
Main contributor to errors
is the capacitor ESR
vout
797mV
13
14
CoL
1kF
0V
7
V1
AC = 1
Fmax=65k
Fmin=26k
Vfold=1.5
Vmin=0.35
a2=(Fmax-Fmin)/(Vfold-Vmin)
a=27.75k
b=Fmin-Vmin*a
19.0V
X1
AMPSIMP
V2
19
VCO slope measured to 27.8 kHz/V
42 • Chris Basso
VFB = 0.71V ( bench ) 0.79 V (SPICE )
R22
{R1}
-14.2fV
4
Fsw = 38.6 kHz ( bench ) 38.4 kHz ( SPICE )
ZL series Rubycon
Approximate ESR Extraction
A triangular output shows that ESR is the main contributor
I out
I p , peak = 1A
isec ( t )
0
rC
isec ( t ) − I out
1: N
vout ( t )
Cout
I sec, peak =
I p , peak
N
=
1
= 4A
0.25
Vout , peak ≈ I sec, peak rC
rC ≈
Vout , pp
I sec, peak
=
35m
= 8.75 mΩ
4
43 • Chris Basso
Plant Frequency Response
H(f)
∠H ( f )
Fsw = 38.6 kHz
44 • Chris Basso
Fsw 2
DS says
11.5 mΩ
Conclusion
The transfer function in DCM frequency foldback has been derived
It can be approximated in low frequency as a 1st order system
Stability in this mode is not at stake with current designs
Besides the comprehensive PWM switch approach, the linearized
output current gives good results too.
Bench measurements confirm the low-frequency results, highfrequency points start to diverge as we approach the switching
frequency.
45 • Chris Basso