Simulating Power Supplies with SPICE
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Simulating Power Supplies with
SPICE
Agenda
Why simulating power supplies?
Average modeling techniques
The PWM switch concept, CCM
The PWM switch concept, DCM
The voltage-mode model at work
Current-mode modeling
The current-mode model at work
Power factor correction
Switching models
EMI filtering
Conclusion
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2
Agenda
Why simulating power supplies?
Average modeling techniques
The PWM switch concept, CCM
The PWM switch concept, DCM
The voltage-mode model at work
Current-mode modeling
The current-mode model at work
Power factor correction
Switching models
EMI filtering
Conclusion
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3
Why Simulate Switch Mode Power Supplies?
Simulation helps feeling how the product behaves before breadboard
Experiment What If? at any level. Power libraries do not blow!
Easily shows impact of parameter variations: ESR, Load etc.
Draw Bode plots without using costly equipments
Avoid trials and errors: compensate the loop on the PC first!
Use SPICE to assess current amplitudes, voltage stresses etc.
Go to the lab. and check if the assumptions were valid.
SPICE does NOT replace the breadboard!
SPICE
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4
Calm
down!
Why Average Simulations?
An average model is made of equations that are continuous in time
The switching component has disappeared, leading to:
a simpler ac analysis of the power supply
the study of the stability margins in various conditions
the assessment of the ESRs contributions in the loop stability
a flashing simulation time!
Average
modeling
AC model
Vin
Vout
0
2
D Gnd
Vg
AC = 0
0.458
4
Duty
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5
Ctrl
0
3
V2
AC = 1
Vout
RS = 20m
FS = 50k
VOUT = 5
RL = 3
VIN = 11
X1
RI = 0.33
L = 37.5u
Resr
100m 0
1
0
5
Cout
220uF
Rload
3
Why Switching Simulations?
An switching model is like breadboarding on the PC
The switching component is back in place, leading to:
the analysis of current and voltage stresses
the study of leakage and stray elements impacts
the analysis of the input current signature – EMI
a longer simulation time…
Switching
approach
1 vout 2 il
C2
100U
IC = 11.9999
R_X2_PR
0.162
R4
100K
7.50
6.50
6
C5
{0.68u/DIV}
IC = 0
14
X1
Config_1
FB
DRV
Ct
GND
CS
1
5.50
VOUT
D5
DN4934
4.50
3
Vcc
CTRL
L_X2_PR
320U Vsw
2
D6
MUR130
X3
MTP8N50
15
ZCD
R5
22K
VCTRL
Vdrv
8
5
R10
0.1
Zero_CD
10
2.20
7
CS
1
C4
1n
3.50
C3
{220u/DIV}
IC = 230
R8
1.0MEG
R9
11K
*
Pout
1.80
RESR_C3
0.3702
I_LOAD
RLOAD
657
Plot2
il in amperes
12
19
R_X2_SEC
13M
VCC
Plot1
vout in volts
4
1.40
1.00
2
600m
50.0u
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150u
250u
time in seconds
350u
450u
Agenda
Why simulating power supplies?
Average modeling techniques
The PWM switch concept, CCM
The PWM switch concept, DCM
The voltage-mode model at work
Current-mode modeling
The current-mode model at work
Power factor correction
Switching models
EMI filtering
Conclusion
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7
Average Modeling, the SSA
State-Space Averaging (SSA)
Introduced by Slobodan Ćuk in the 80’
Long and painful process
Fails to predict sub-harmonic oscillations
x1
L
Vout
x1
L
Vout
on
off
dx1
1
1
= − x2 + u
dt
L
L
u1
x2
C
R
dx 2
1
1
=
x1 −
x2
dt
Cout
Rload .Cout
ON
x2
C
dx1
1
= − x2
dt
L
R dx 2
1
1
=
x1 −
x2
dt
Cout
Rload .Cout
OFF
Pfffff!
Apply smoothing process
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Linearize
/
Average Modeling, the PWM Switch
The PWM Switch
Introduced by Vatché Vorpérian in the mid-80’
Easy to derive and fully invariant
No auto-toggling mode models
Can predict sub-harmonic oscillations in CCM
DCM model in current-mode was never published!
a
c
L
d
a
Ia(t)
Ic(t)
d
c
d'
Vap(t)
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p
d'
PWM switch
Vin
Vcp(t)
p
C
R
Vout
The PWM Switch Concept
Identify the guilty network: the transistor and the diode
Average their voltage and current waveforms: large-signal model
Linearize the equations around a dc point: small-signal model
L
Linear network
a
Vin
PWM switch
d
d'
p
c
on
off
Linear network
diode + transistor = guilty for non-linearity!
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C
R
Vout
Agenda
Why simulating power supplies?
Average modeling techniques
The PWM switch concept, CCM
The PWM switch concept, DCM
The voltage-mode model at work
Current-mode modeling
The current-mode model at work
Power factor correction
Switching models
EMI filtering
Conclusion
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11
The PWM Switch Concept
The transistor is a highly non-linear device:
Replace the transistor with its small-signal model
Solve a system of linear equations
5
ib
Rc
10k
Rb_upper
1Meg
Vin
7
Beta.Ib
e
Vg
Vin
4
Q1
Vout
8
h11
Vout
1
ic
c
b
Req
Rb_upper//Rb_lower
3
Rc
10k
ie
2
Rb_lower
100k
Re
150
Remember the bipolars
Ebers-Moll model…
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Ce
1nF
Re
150
Ce
1nF
Ve
Replace Q1 by its small-signal model
The PWM Switch Concept
The PWM switch model works in all two switch converters:
Rotate the model to match the switch and diode connections
Solve a system of linear equations
p
c
PWM switch
d
p
c
Vin
d'
d'
d
a
PWM switch
L
C
Buckboost
R
a
Vin
Vout
a
Ic(t)
d
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p
c
L
d
c
d'
PWM switch
Vin
13
Vout
Boost
d'
Vap(t)
R
L
a
Ia(t)
C
Vcp(t)
Buck
p
C
R
Vout
The PWM Switch Concept
The keyword with average modeling: waveforms averaging
Ia(t)
a
Ic(t)
d
L
c
d'
Vout
Vin
I a (t )
Tsw
1
= Ia =
Tsw
Tsw
∫I
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Vap(t)
0
a
(t )dt =d I c (t )
Tsw
= dI c
p
Vcp(t)
C
Vcp (t )
R
Tsw
1
= Vcp =
Tsw
Tsw
∫V
cp
0
(t )dt =d Vap (t )
Tsw
= dVap
The PWM Switch Concept
The obtained set of equations is that of a transformer
¾ A CCM two-switch DC-DC can be modeled like a 1:D transformer!
a
c
Ia(t)
I1
Ic(t)
V1
V=d.V(a,p)
I=d.Ic
p
1:N
p
a
Vap
Ic(t)
Ia(t)
d
1
c
Vcp
p
Large-signal (non-linear) model
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I2
V2
V2 = NV1
I1 = NI 2
The PWM Switch Concept
SPICE only deals with linear equations
It first computes a bias point then it linearizes the network
Vout ( s )
d (s)
50.0
10.0
L
10.0
p
Vout
d
100uH
c
a
Vin
10
d = 40%, Vout = 16.7 V
30.0
0.400
16.7
Vbias
0.4
AC = 1
C1
100u
R1
10
d = 10%, Vout = 11.1 V
10.0
-10.0
dB
4
1
2
3
-30.0
1
10
Always verify the dc operating point!
No equations, result appears in a second!
Make sure the bias point is correct…
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100
1k
10k
100k
The PWM Switch Concept
We have a set of non-linear equations: can’t derive transfer functions!
We need a small-signal model: linearize the equations by hand
two options: perturbation or partial derivatives…
Pertubation
Partial derivatives
I a = dI c
Vcp = dVap
I a = dI c
I a = I a 0 + iˆa
I c = I c 0 + iˆc
Vcp = Vcp 0 + vˆcp
∂Vcp
∂Vcp ˆ
ˆia = ∂I a iˆc + ∂I a dˆ ˆ
vcp =
vˆap +
d
∂I c
∂d
∂Vap
∂d
d = d 0 + dˆ
(
d = d 0 + dˆ
same
)(
I a 0 + iˆa = d 0 + dˆ I c 0 + iˆc
I a 0 = d0 I c 0
)
Vcp 0 = d 0Vap 0
ˆ
ˆ
ˆ
ˆ
iˆa = d 0iˆc + dI
c 0 vcp = d 0 vap + dVap 0
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ˆ
iˆa = d 0iˆc + dI
c0
ac and dc
equations
Vcp = dVap
ˆ
vˆcp = d 0 vˆap + dV
ap 0
ac equations
No dc point
The PWM Switch Concept
Put the small-signal sources in the large-signal model
You obtain the small-signal model of the CCM PWM switch
(V
ap
d 0 ) dˆ
I c 0 + iˆc
a
Vap 0 + vˆap
d0 I c0
(
dˆ I c 0 + iˆc
)
V1
c
Vcp 0 = d 0Vap 0
ˆ
vˆcp = d 0 vˆap + dV
ap 0
p
You can now analytically find the dc bias and the ac response!
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Agenda
Why simulating power supplies?
Average modeling techniques
The PWM switch concept, CCM
The PWM switch concept, DCM
The voltage-mode model at work
Current-mode modeling
The current-mode model at work
Power factor correction
Switching models
EMI filtering
Conclusion
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The PWM Switch in DCM
The original model could not be auto-toggling
A new DCM-CCM model has been derived
Ia
Ipeak
Ia(t)
a
Ic(t)
d1
d2
t Vin
Vap
Ic
Vcp
Vap(t)
t
Vap
Ic =
Vcp
t
d3Tsw
3rd event linked to DCM
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d3
Vout
Ia =
d2Tsw
c
Vcp(t)
p
C
t
Ipeak
d1Tsw
L
I peak d1
Extract and
replace
2
I peak ( d1 + d 2 )
2
d1 + d 2 )
(
Ic = Ia
d1
R
The PWM Switch in DCM
By clamping the d2 equation, the circuit toggles between the modes
a
Vap
Ic(t)
Ia(t)
N
1
p
c
Vcp
N=d1/(d1+d2)
2 I c L − Vac d1 ²Tsw 2 LFsw I c
d2 =
=
− d1
Vac d1Tsw
d1 Vac
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Clamp d2:
d2 CCM = 1- d1
d2 DCM = 1- d1- d3
d2 < 1 - d1
model is in DCM!
Model input
Agenda
Why simulating power supplies?
Average modeling techniques
The PWM switch concept, CCM
The PWM switch concept, DCM
The voltage-mode model at work
Current-mode modeling
The current-mode model at work
Power factor correction
Switching models
EMI filtering
Conclusion
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22
The PWM Switch in DCM
In voltage-mode, the duty-cycle is built with a ramp generator
The transition occurs when the error voltage crosses the ramp
X4
PWMCCMVM
Vpeak
Verr
d(t)
0
a
0.500
10.0
3
17
V4
10
GAIN
Vramp(t)
5.00
c
d
PWM switch VM
p
XPWM
GAIN
K = 0.5
ton
0
Tsw
Verr ( t ) = V peak
Verr ( t )
d (t ) =
V peak
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ton ( t )
= V peak d ( t )
Tsw
Vpeak = 2 V
1
d ⎛ d (t ) ⎞
=
= K PWM
⎜⎜
⎟⎟
d ⎝ Verr ( t ) ⎠ V peak
L1
75u
4
The Voltage-Mode Model at Work
1.
2.
3.
4.
Let us compensate a buck converter operated in CCM and DCM
Run an open-loop Bode plot at full load, lowest input
Identify the excess/deficiency of gain at the selected cross over
Place a double zero at f0, a pole at the ESR zero and a pole at Fsw/2
Check final loop gain and run a transient load test
parameters
20.0V
a
2
R4
20m
c
4
7
Vin
20
GAIN
G=10^(-Gfc/20)
pi=3.14159
604mV
XPWM
GAIN
K = 0.4
vout
Vout2
13
12.1V
Rupper=38k
fc=7k
Gfc=-15
L1
180u
12.0V
12.0V
d
PWM switch VM
X3
PWMVM
L = 180u
Fs = 100k
Resr
150m
p
12.0V
16
Cout
220u
1.51V
Automated
compensation
15
fz1=650
fz2=650
fp1=7k
fp2=50k
C3=1/(2*pi*fz1*Rupper)
R3 =1/(2*pi*fp2*C3)
C2
{C2}
R3
{R3}
C1
{C1}
R2
{R2}
T(s)
vout
Rupper
38k
12
1.51V
2.50V
LoL
1kH
C1=1/(2*pi*fz2*R2)
C2=1/(2*pi*(fp1)*R2)
a=fc^4+fc^2*fz1^2+fc^2*fz2^2+fz1^2*fz2^2
c=fp2^2*fp1^2+fc^2*fp2^2+fc^2*fp1^2+fc^4
R2=sqrt(c/a)*G*fc*R3/fp1
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Vin
10
V1
AC = 1
C3
{C3}
6
1.51V
5
CoL
1kF
2.50V
9
Vout
X2
AMPSIMP
V2
2.5
12.0V
8
Rlower
10k
Rload
3
H(s)
The Voltage-Mode Model at Work
The Bode plot reveals a gain loss of -15 dB at 7 kHz
The compensator provides a +15 dB gain increase plus phase boost
180 24.0
|H(s)|
|H(7 kHz)| =-15 dB
Vout(t)
12.4
90.0 12.0
0
0
-90.0 -12.0
-180 -24.0
arg|H(s)|
12.2
2
Arg H(7 kHz) =-121°
0
12.0
1
2
Pm = 80°
40.0 180
20.0 90.0
1
Arg T(s)
11.8
4
0
fc = 7 kHz
-20.0 -90.0
|T(s)|
-40.0 -180
10
100
1k
10k
3
100k
Iout = 200 mA to 4 A
in 10 µs
11.6
9.91m
11.0m
12.1m
13.3m
The final loop gain shows a comfortable phase margin
The transient response at both input levels shows a stable signal
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14.4m
Agenda
Why simulating power supplies?
Average modeling techniques
The PWM switch concept, CCM
The PWM switch concept, DCM
The voltage-mode model at work
Current-mode modeling
The current-mode model at work
Power factor correction
Switching models
EMI filtering
Conclusion
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26
Current-Mode Operation
In voltage-mode, the error signal directly controls the duty cycle
In current mode, the error voltage sets the inductor peak current
To derive a model, observe the current signals and average them!
3
1
2
I c (t ) Ri = Vc (t ) − d (t )Tsw Se −
S f d '(t )Tsw
2
1
Ic =
2
3
Vc
T S
T
− d sw e − Vcp (1 − d ) sw
Ri
Ri
2L
a
c
Ia(t)
Ic(t)
I=Vc/Ri
I=d.Ic
p
CCM
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p
I=Iu
Cs
Current-Mode Operation
Do the same for DCM signals
Match the previous structure to build a CCM/DCM model
I peak =
Vc − d1Tsw Se
Ri
Vc − d1Tsw Se
Ic =
− α d 2Tsw S f
Ri
Vcp
d1Tsw Se
+ d 2Tsw
Iμ =
Ri
L
α
a
c
Ia(t)
Ic(t)
I=Vc/Ri
I=(d1/(d1+d2)).Ic
DCM
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3rd event linked to DCM
p
⎛ d1 + d 2 ⎞
⎜1 −
⎟
2 ⎠
⎝
p
I=Iu
Agenda
Why simulating power supplies?
Average modeling techniques
The PWM switch concept, CCM
The PWM switch concept, DCM
The voltage-mode model at work
Current-mode modeling
The current-mode model at work
Power factor correction
Switching models
EMI filtering
Conclusion
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29
The Current-Mode Model at Work
To study a converter, we can write down the equations
Or use a SPICE simulation to get the Bode plot in a second
Take the example of a current-mode flyback converter
⎡
2
2
2
⎛ f ⎞
⎛ f ⎞
⎛ f ⎞
⎢
1+ ⎜
⎟⎟ 1 + ⎜
⎟
⎟ 1 + ⎜⎜
⎢
f
f
f
z
z
z
1
3
⎝
⎠
⎝
⎠
⎝ 2⎠
H ( f ) = 20 log10 ⎢G0
⎢
2
⎛
⎞
f
⎢
1
+
⎜
⎢
⎜ f p ⎟⎟
⎝ 1⎠
⎢⎣
1
2
⎛ ⎛ f ⎞2 ⎞ ⎛ f
⎜1 −
⎟ +⎜
⎜ ⎜⎝ f n ⎟⎠ ⎟ ⎜⎝ f nQ p
⎝
⎠
⎞
⎟⎟
⎠
2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
⎛
⎞
⎜
⎟
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
f
f
f
f
f
1
⎟
−1
−1
−1 ⎜
arg H ( f ) = tan −1 ⎜ ⎟ − tan −1 ⎜
⎟ − tan ⎜
⎟ + tan ⎜
⎟ − tan ⎜
2 ⎟
⎜ fz ⎟
⎜ fz ⎟
⎜ fp ⎟
⎜ fz ⎟
f nQp
⎛
⎞
f
⎝ 1⎠
⎝ 2⎠
⎝ 1⎠
⎝ 3⎠
⎜
1 − ⎜ ⎟ ⎟⎟
⎜
⎝ fn ⎠ ⎠
⎝
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Stabilizing a CCM Flyback Converter
6
a
90.0V
X2x
XFMR
RATIO = -0.25
-78.8V
p
2
PWM switch CM
DC
duty-cycle
467mV
vc
Capture a SPICE schematic with an averaged model
4
vout
vout
19.7V
19.0V
c
Vin
90
AC = 0
3
D1A
mbr20200ctp
R10
14.4m
0V
X9
PWMCM
L = Lp
Fs = 65k
Ri = 0.25
Se = 0
13
L1
{Lp}
Rload
4
19.0V
786mV
8
V(errP)/3 > 1 ?
1 : V(errP)/3
1
C5
6600u
B1
Voltage
Look for the bias points values: Vout = 19 V, ok
Vsetpoint < 1 V, enough margin on current sense
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Stabilizing a CCM Flyback Converter
Capture a SPICE schematic with an averaged model
parameters
Vdd
5
5.00V
errP
2.36V
vout
5
K
Rpullup
{Rpullup}
S+A
X4
POLE
FP = pole
K=1
Rled
{Rled}
2.36V
2.36V
18.7V
Verr
Rupper2
{Rupper}
LoL
1kH
err
18
11
R7
1
X8
Optocoupler
Fp = Pole
CTR = CTR
2.36V
14
CoL
1kF
0V
15
17.7V
9
2.49V
Vstim
AC = 1
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Cpole2
{Cpole}
X10
TL431_G
10
Czero1
{Czero}
Rlower2
{Rlower}
Vout=19
Ibridge=250u
Rlower=2.5/Ibridge
Rupper=(Vout-2.5)/Ibridge
Lp=350u
Se=20k
fc=1k
from
pm=60
Bode
Gfc=-22
pfc=-71
G=10^(-Gfc/20)
boost=pm-(pfc)-90
pi=3.14159
K=tan((boost/2+45)*pi/180)
Fzero=fc/k
Fpole=k*fc
Rpullup=20k
RLED=CTR*Rpullup/G
Czero=1/(2*pi*Fzero*Rupper)
Cpole=1/(2*pi*Fpole*Rpullup)
CTR=1.5
Pole=6k
Stabilizing a CCM Flyback Converter
Capture a SPICE schematic with an averaged model
32.0
Gain at 1 kHz
-22 dB
16.0
Sub harmonic
poles
0
-16.0
4
180
Phase at 1 kHz
-71 °
90.0
Inject ramp
compensation
0
-90.0
argH(s)
-180
10
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100
1k
10k
ramp
|H(s)|
-32.0
6
100k
Stabilizing a CCM Flyback Converter
The easiest way to damp the poles:
¾ Calculate the equivalent quality coefficient at Fsw/2
¾ Calculate the external ramp to make Q less than 1
1
Q=
⎛
π ⎜D'
⎝
Se =
⎞
Se 1
+ − D⎟
Sn 2
⎠
=
1
=8
3.14 × ( 0.5 − 0.46 )
Sn ⎛ 1
90 × 0.25
⎞ Vin Ri ⎛ 1
⎞
⎛ 1
⎞
− 0.5 + 0.46 ⎟ = 36 kV s
⎜ − 0.5 + D ⎟ =
⎜ − 0.5 + D ⎟ =
⎜
D'⎝π
⎠ Lp D ' ⎝ π
⎠ 320u × (1 − 0.46 ) ⎝ 3.14
⎠
2.3 Vpp
Rramp
18 kΩ
DRV
CS
NCP1230
internal
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Se 36k
=
= 51%
Vin Ri
On-time slope
S n 70k
Lp
2.3
S ramp =
= 153 kV s
15u
M S R
0.51× 70k × 18k
Rcurrent = r n ramp =
= 4.1 k Ω
S ramp
153k
Mr =
Rcurrent
Ri
Stabilizing a CCM Flyback Converter
Boost the gain by +22 dB, boost the phase at fc
11
Cross over
1 kHz
80.0
40.0
GM
20 dB
0
-40.0
10
|T(s)|
-80.0
180
90.0
0
Margin at 1 kHz
60°
-90.0
argT(s)
-180
10
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35
100
1k
10k
11
100k
Stabilizing a CCM Flyback Converter
Test the response at both input levels, 90 and 265 Vrms
Sweep ESR values and check margins again
19.11
Vout(t)
Hi
line
19.03
12
11
18.95
112 mV
18.87
Low
line
18.79
1.80m
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36
5.40m
9.00m
time in seconds
12.6m
16.2m
Agenda
Why simulating power supplies?
Average modeling techniques
The PWM switch concept, CCM
The PWM switch concept, DCM
The voltage-mode model at work
Current-mode modeling
The current-mode model at work
Power factor correction
Switching models
EMI filtering
Conclusion
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Power Factor Correction
The bulk capacitor connects to a low-impedance source
At the bulk capacitor refueling, a narrow peak current flows
This peak conveys a large harmonic content
Itotal
Vbulk
Iout
Vbulk
D6
1N4007
2
Vmains
D5
1N4007
Ibulk
6
D7
1N4007
D8
1N4007
Vbulk
CBulk
100u
IC = 50
B2
Current
50/V(Vbulk)
Iin
Vmains
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Power Factor Correction
A pre-converter is installed as a front-end section
The pre-converter draws a sinusoidal current
The energy is stored and released in/by the bulk capacitor
PFC
Pre-converter
D1
D2
Mains
Vbulk
store
release
Cbulk
PWM
D3
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39
D4
Power Factor Correction
One of the most popular techinique uses Borderline mode
The MC33262 operates in peak current mode control
Rlimit
8
11
Reset detector
-
1
L
4
2
D3
Dout
+
D4
5
15
Vout
12
S
RdivU
Ddmg
18
Q
Vdem
Q
16
6
Rupper
Cin
Vin
R
3
D6
Cout
Error amplifier
D5
A
7
+
14
G1
-
X6
current sense
comparator
Rlower
17
K*A*B
13
9
B
Verr
Rsense
RdivL
10
Vref
C1
Peak current
setpoint
The NCP1606 also operates in constant-on time
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40
Power Factor Correction
The core is always reset from cycle to the other
I L (t )
Tsw
( t ) = I in ( t )
IL,peak
IL,avg
IL(t)
On time
is constant
IL = 0
ton
8.05m
8.15m
8.25m
time in seconds
8.35m
8.45m
the average inductor current is half the inductor peak current value
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Power Factor Correction
A 150 W BCM PFC average example with the MC33262
Iin
X1
KBU4J
Vrect
+
8
R6
100m
L1
{L}
16
23
Vout
3
p
10
5
a
Vton
26
vc
Vfsw
Current-mode
borderline model
R2
12k
4
PWM switch BCM
R1
1.6Meg
-
6
Fsw (kHz)
IN
c
Cin
1u
ton
Δ
Vin
Vin
{VRMS}
X5
PWMBCMCM2
L=L
Ri = Ri
C5
150u
IC = {Vrms*1.414}
C3
10n
K = 0.6
R10
50m
Rload
{Vout*Vout/Pout}
9
R4
1.6Meg
Vmul
A
K*A*B
24
B4
Voltage
2
B
B1 3 0 V = V(1) * V(2) * {K}>1.3 ? 1.3 : V(1) * V(2) * {K}
G1
100u
V(err)-2.05 < 0 ? 0 : V(err)-2.05
err
parameter
Vrms=100
Pout=150
Vout=400
Ri=0.22
L=850u
Verr
22
D1
N = 0.01
15
D2
N = 0.01
R5
1G
14
V3
6.4
V5
1.7
R8
23k
25
13
B1
Current
V11
C2
0.68u
I(V11) > 10u ? 10u : I(V11)
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42
17
V4
2.5
R3
10k
Power Factor Correction
2.00
8.00
1.00
4.00
0
-1.00
vton in volts
Plot1
iin in amperes
Average models can also work in transient conditions
-2.00
ton(t) - µs
3
0
-4.00
-8.00
1
Vin = 230 Vac
Plot2
vout, vout#a in volts
410
2.00
30.0
vton in volts
Plot3
iin in amperes
398
60.0
-2.00
2
6
402
4.00
0
Vout,peak = 406 V
406
Vout,valley = 398 V
394
Constant
on-time
-4.00
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Vout(t)
ton(t) - µs
5
0
-30.0
-60.0
4
Iin(t)
THD = 2%
Vin = 100 Vac
1.139
43
Iin(t)
THD = 10%
High
line
1.149
1.159
time in seconds
1.169
1.179
Low
line
Power Factor Correction
180
80.0
90.0
40.0
0
gain in db(volts)
plot1
phase in degrees
Use the model to boost the phase at the cross over point
gain
phase
Pm = 61°
0
-90.0
-40.0
-180
-80.0
1
fc = 5 Hz
Zero added
180
80.0
90.0
40.0
0
gain in db(volts)
Plot2
phase in degrees
2
phase
0
-90.0
-40.0
-180
-80.0
1m
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44
gain
4
fc = 5 Hz
No zero added
10m
100m
1
10
frequency in hertz
100
1k
10k
5
100k
Power Factor Correction
The zero improves the overshoot but degrades the THD…
440
Plot1
vout#a, vout#a#1 in volts
No zero
440 V
410
2
3
380
413 V
350
Added zero
320
122m
365m
Plot2
iin, iin#1 in amperes
800m
608m
time in seconds
851m
1.09
THD = 1.5%
No zero
400m
0
1
4
THD = 10%
-400m
Added zero
-800m
1.128
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45
Vout(t)
1.135
Iin(t)
1.142
time in seconds
1.149
1.157
Agenda
Why simulating power supplies?
Average modeling techniques
The PWM switch concept, CCM
The PWM switch concept, DCM
The voltage-mode model at work
Current-mode modeling
The current-mode model at work
Power factor correction
Switching models
EMI filtering
Conclusion
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46
Switching Models, the Breadboard on PC
Turn your PC into a virtual breadboard
Ohmic losses
Rs
100m
1:N
7
4
Primary
inductance
Vinput
300V
3
Cout
100uF
Lleak
10uH
6
Vout
2
Resr
100m
5
Leakage
inductance
9
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1
Lp
1mH
Pulse
generator
47
D1
1N5818
Capacitor
parasitics
Rload
10
Switching Models, the Breadboard on PC
Wire your device as you would do in the lab.
Iout
Isec
X1
XFMR
RATIO = -0.08
L1
2.2uH
4
12
R16
10m
31
R3
200m
Primary inductance
Vsec
D1
MBR140P
vout
Vout
17
1
R4
100m
R17
300m
6
Iprim
Iripple1
L2
3.5mH
C1
470uF
IC = 5.5
Istartup
Rload
13.4
32
9
C2
10uF
16
Leakage inductance
X5
L5
80uH
NCP1200
Vadj
Vinput
126
8
Vdrain
1
8
2
7
4
6
NCP1200
vout
19
Drv
3
R15
470
IDrain
10
Vcc
5
5
13
X2
MTD1N60E
X4
MOC8101
C3
150p
15
C5
100nF
IC = 2.5
R5
100m
Vsense
X3
TL431
R2
14k
14
18
23
Cvcc
10uF
IC = 11.99
Rsense
2.8
R7
10k
VFB
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48
Simulations (Really) Work!!
Assess the average, rms currents in your circuit
Check if enough margins exist on your semiconductors
Leakage
effects
Vdrain
100V/div
0
Vsense
500mV/div
505.00U
515.00U
525.00U
535.00U
545.00U
simulated
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49
measured
Simulations (Really) Work!!
With accurate models, the simulation results are excellent
You can then vary the parasitic terms and see their impact
16.93
16.87
30mV/div
16.81
16.75
2ms/div
16.69
4.50m
8.50m
12.5m
16.5m
simulated
www.onsemi.com
50
20.5m
measured
Agenda
Why simulating power supplies?
Average modeling techniques
The PWM switch concept, CCM
The PWM switch concept, DCM
The voltage-mode model at work
Current-mode modeling
The current-mode model at work
Power factor correction
Switching models
EMI filtering
Conclusion
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51
EMI Filtering on a DC-DC
DC-DC are highly EMI polluting systems
A filter has to be installed to avoid noise in the source
C3
11n
Need to
limit the
ac current
< 15 mA peak
R3
285
3
9
Rupper
38k
Vswitch
ICoil
Δ
Rlimit
10m
vin
V1
36
Vin
30
X1
PWMVM
PERIOD = 10u
DUTYMAX = 0.9
REF = 2.5
IMAX = 5
DUTYMIN = 0.01
VOL = 100M
VOH = 2.5
DUTYMIN = 0.01
10
C1
3.3nF
13
CMP
R2
120k
OUT
Vdrv
5
FB
GND
IMAX
Rlower
10k
4
Vout
7
Resr
69m
Rload
3
D1
mbr845
14
R30
1k
Vsense
1
ID_FW
C30
470p
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2
X2
PSW1
RON = 0.1
Verr
C2
1.3nF
R1
50m
L1
180uH
6
Iin
52
Vrect
8
VIL
B1
Voltage
I(Rlimit)
Cout
1000uF
IC = 11
EMI Filtering on a DC-DC
Use SPICE to extract the current signature
Run Fourier analysis to look at the spectrum
Iripple < 15 mA
4.00
attenuation
Plot1
iin in amperes
2.00
0
-2.00
-4.00
1
Iavg = 1.7 A
Irms = 2.7 A
3.922m
A filter <
15m
< 6m
2.45
Input current signature
3.939m
3.956m
time in seconds
3.974m
f 0 < 0.006 × Fsw < 7.7 kHz
3.991m
10
Ipeak = 2.45A
Plot1_f
mag(fft(temp)) in amperes
1
100m
2
10m
1m
100u
10u
FFT results
105k
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53
315k
525k
frequency in hertz
735k
945k
Position the cutoff
frequency of the LC filter
EMI Filtering on a DC-DC
A LC configuration offers the best efficiency
As any LC network, it is subject to resonances
L
Rlf
Vin
C
Switch
Mode
Converter
Rload
L = 100 µH
1
= 5.2 µF
C=
2
2
4π f 0 L
7.7 kHz
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54
Check
impedance
peaking
Z outFILTER
max
=
2
Z0
R1
⎛ R1 ⎞
1+ ⎜ ⎟
⎝ Z0 ⎠
Vout
2
EMI Filtering on a DC-DC
The incremental input resistance of a DC-DC in negative
A LC filter loaded by a negative resistance can oscillate!
No damping
Zout = 57.5 dBΩ
50.0
Problem!
7
30.0
ZinSMPS
p
18 dBΩ
Need to
damp this!
p
10.0
-10.0
Power
supply input
impedance
1
-30.0
Zout
10m
100m
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55
1
10
100
frequency in hertz
1k
10k
100k
EMI Filtering on a DC-DC
If the resonance is too peaky, problems can arise
180
40.0
|T(s)|
Without filter
20.0
gain
0
vdbout2, vdbout2#1 in db(volts)
Plot1
ph_vout2#1, ph_vout2 in degrees
90.0
With filter
Not stable!
0
argT(s)
phase
-90.0
-20.0
-180
-40.0
Without filter
3
With filter
10
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56
100
1k
frequency in hertz
10k
1
4
2
100k
EMI Filtering on a DC-DC
A resistor is damping the LC filter by creating losses
A dc-block capacitor is installed to limit dissipation
L
Rlf
Vin
Switch
Mode
Converter
C
L + CR1 R2 −
Rdamp = − Z i nSMPS
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57
2 Z inSMPS CR1 −
Z inSMPS
ω0
Rload
R1
ω0
+ L + CR2 R1 −
R1
ω0
Vout
EMI Filtering on a DC-DC
The right resistor prevents the overlaps between curves
50.0
7
30.0
Ok, margin is 8 dB min
ZinSMPS
18 dBΩ
Rdamp= 5 Ω
10.0
Rdamp= 4 Ω
-10.0
1
4
5
6
3
2
Rdamp= 1 Ω
Rdamp= 2 Ω
-30.0
Rdamp= 3 Ω
Zout
10m
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58
100m
1
10
100
frequency in hertz
1k
10k
100k
EMI Filtering on a DC-DC
A final check shows a noise amplitude under control
0
1
1.700
1.690
p
y (pk-pk) = 22.8m amperes
between 3.91m and 3.92m seconds
1
1.680
1.670
1.660
3.82m
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59
3.86m
3.90m
time in seconds
3.94m
3.98m
A Book on Power Supply Design
To learn more about power supplies and simulations…
886 pages, 8 chapters
Learn DC-DC converters theory
Understand average modeling
Feedback and loop control
Design examples of DC-DC and AC-DC
Power Factor Correction
Chapters on flyback and forward converters
Supplied CDROM with working examples
I already
have ideas
for the next
edition!!
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Conclusion
SPICE can be seen as a design companion
It shields us from going through complex equations
Simulation time is short and PC helps to run tests
Use SPICE before going to the bench: NO trial and error!
Once the simulation is stable, build the prototype
Simulations and laboratory debug: the success recipe!
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For More Information
•
View the extensive portfolio of power management products from ON
Semiconductor at www.onsemi.com
•
View reference designs, design notes, and other material supporting
the design of highly efficient power supplies at
www.onsemi.com/powersupplies
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62