SIM - Simulating Power Supplies with SPICE

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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34
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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37
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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38
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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41
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
www.onsemi.com
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
www.onsemi.com
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
www.onsemi.com
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
www.onsemi.com
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
www.onsemi.com
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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60
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
www.onsemi.com
62