Simulating Power Supplies with SPICE

Simulating Power Supplies with SPICE

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AgendaWhy simulating power supplies?Average modeling techniquesThe PWM switch concept, CCMThe PWM switch concept, DCMThe voltage-mode model at workCurrent-mode modelingThe current-mode model at workPower factor correctionSwitching modelsEMI filteringConclusion

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Simulation helps feeling how the product behaves before breadboardExperiment 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 equipmentsAvoid 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.

Why Simulate Switch Mode Power Supplies?

SPICE does NOTNOT replace the breadboard!

SPICE

Calm down!

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An average model is made of equations that are continuous in timeThe switching component has disappeared, leading to:a simpler ac analysis of the power supplythe study of the stability margins in various conditionsthe assessment of the ESRs contributions in the loop stabilitya flashing simulation time!

Why AverageAverage Simulations?

Averagemodeling

1

Rload3

5

Resr100m

Cout220uF

2

VgAC = 0

Vout

4

DutyV2AC = 1

3

Vin Vout

Gnd Ctrl

AC model

DRS = 20mFS = 50kVOUT = 5RL = 3VIN = 11X1RI = 0.33L = 37.5u

0

0

0

0.458 0

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An switching model is like breadboarding on the PCThe switching component is back in place, leading to:the analysis of current and voltage stressesthe study of leakage and stray elements impactsthe analysis of the input current signature – EMIa longer simulation time…

Why SwitchingSwitching Simulations?

Switchingapproach

1 vout 2 il

3.50

4.50

5.50

6.50

7.50

vout

in v

olts

Plot

1 1

50.0u 150u 250u 350u 450utime in seconds

600m

1.00

1.40

1.80

2.20

il in

am

pere

sPl

ot2

2

7

4

10

1

6

2

14

123

8

19

5

15

RESR_C30.3702

C2100UIC = 11.9999

C3220u/DIVIC = 230

C41n

R_X2_SEC13M

D5DN4934

C50.68u/DIVIC = 0

RLOAD657

D6MUR130

R4100K

R522K

R81.0MEG

L_X2_PR320U

R911K

R_X2_PR0.162

R100.1

X3MTP8N50

VOUTVCC Vsw

Zero_CD

CS * Pout

I_LOAD

Vdrv

FB

CTRL

Ct

CS ZCD

GND

DRV

Vcc

X1Config_1

VCTRL

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Average Modeling, the SSAStateState--SpaceSpace AveragingAveraging (SSA)Introduced by Slobodan Ćuk in the 80’Long and painful processFails to predict sub-harmonic oscillations

uL

xLdt

dx 1211+−=

2.

1112 xCoutRload

xCoutdt

dx−=

211 xLdt

dx−=

2.

1112 xCoutRload

xCoutdt

dx−=

VoutL

C R

x1

x2u1

VoutL

C R

x1

x2

offon

Apply smoothing process Linearize

Pfffff!ON OFF

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Average Modeling, the PWM SwitchThe PWM SwitchPWM SwitchIntroduced by Vatché Vorpérian in the mid-80’Easy to derive and fully invariantNo auto-toggling mode modelsCan predict sub-harmonic oscillations in CCMDCM model in current-mode was never published!

ca

p

d

d'

Ia(t) Ic(t)

Vap(t) Vcp(t)

a c

PWM switch p

d

d'

Vin

L

C R Vout

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The PWM Switch Concept

Vin

L

C R Voutac

PW

M s

witc

hp

d

d'

Linear network

Linear network

off

on

diode + transistor = guilty for non-linearity!

Identify the guilty network: the transistor and the diodeAverage their voltage and current waveforms: large-signal modelLinearize the equations around a dc point: small-signal model

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1

4

2

Q1

5

Rb_upper1Meg

Rb_lower100k

Vg

Vout

8

3

7

h11 Beta.Ib

Rc10k

Re150

Ce1nF

ReqRb_upper//Rb_lower

b c

e

ib ic

ie

VinRc10k

Re150

Ce1nF

Vin

Vout

Ve

Remember the bipolarsEbers-Moll model…

Replace Q1 by its small-signal model

The transistor is a highly non-linear device:Replace the transistor with its small-signal modelSolve a system of linear equations

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a c

PWM switch p

d

d'

Vin

L

C R Vout

Vin

L

C R Voutac

PW

M s

witc

hp

d

d'

Vin

LC R Vout

ac

PWM

sw

itch

p

d

d'

ca

p

d

d'

Ia(t) Ic(t)

Vap(t) Vcp(t)BuckBuck

BoostBoostBuckBuck--boostboost

The PWM switch model works in all two switch converters:Rotate the model to match the switch and diode connectionsSolve a system of linear equations

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ca

p

d

d'

Ia(t) Ic(t)

Vap(t) Vcp(t)Vin

L

C RVout

0

1( ) ( ) ( )sw

sw sw

T

a a a c cT Tsw

I t I I t dt d I t dIT

= = = =∫0

1( ) ( ) ( )sw

sw sw

T

cp cp cp ap apT Tsw

V t V V t dt d V t dVT

= = = =∫

The keyword with average modeling: waveforms averaging

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ca

p

Ia(t) Ic(t)

V=d.V(a,p)

p

I=d.Ic

Large-signal (non-linear) model

The obtained set of equations is that of a transformerA CCM two-switch DC-DC can be modeled like a 1:D transformer!

V1 V2

I1 I21:N

2 1V NV=

1 2I NI=

ca

p

Ia(t) Ic(t)

1 dVap Vcp

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Ld

100uH

Vbias0.4AC = 1

C1100u

R110

Vout

Vin10

a

cp10.0 10.0

16.7

0.400

Always verify the dc operating point!

SPICE only deals with linear equationsIt first computes a bias point then it linearizes the network

1 10 100 1k 10k 100k

-30.0

-10.0

10.0

30.0

50.0

123

4

d = 40%, Vout = 16.7 V

d = 10%, Vout = 11.1 V

dB

( )( )

outV sd s

No equations, result appears in a second!Make sure the bias point is correct…

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The PWM Switch ConceptWe have a set of non-linear equations: can’t derive transfer functions!We need a small-signal model: linearize the equations by handtwo options: perturbation or partial derivatives…

Pertubation Partial derivatives

0

0

0

ˆ

ˆ

ˆ

a c

a a a

c c c

I dI

I I i

I I i

d d d

=

= +

= +

= +

0

0

ˆˆ

cp ap

cp cp cp

V dV

V V v

d d d

=

= +

= +

( )( )0 0 0ˆˆ ˆ

a a c cI i d d I i+ = + +

0 0 0

0 0ˆˆ ˆ

a c

a c c

I d I

i d i dI

=

= +ac and dcequations

ˆˆ ˆ

a c

a aa c

c

I dII Ii i dI d

=∂ ∂

= +∂ ∂

ˆˆ ˆ

cp ap

cp cpcp ap

ap

V dV

V Vv v d

V d

=

∂ ∂= +∂ ∂

0 0ˆˆ ˆ

a c ci d i dI= +

0 0 0

0 0ˆˆ ˆ

cp ap

cp ap ap

V d V

v d v dV

=

= +

same

0 0ˆˆ ˆcp ap apv d v dV= +

ac equationsNo dc point

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The PWM Switch ConceptPut the small-signal sources in the large-signal modelYou obtain the small-signal model of the CCM PWM switch

a c

p

0 0 0

0 0ˆˆ ˆ

cp ap

cp ap ap

V d V

v d v dV

=

= +

( )0ˆ

apV d d

0 ˆap apV v+

0ˆ

c cI i+

1V0 0cd I

( )0ˆ ˆ

c cd I i+

You can now analytically find the dc bias and the ac response!

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The PWM Switch in DCM

ca

p

d1

d2

Ia(t) Ic(t)

Vap(t) Vcp(t)Vin

L

C RVout

d3

IaIpeak

Vap

Ipeak

Ic

VcpVap

Vcp

d1Tsw d2Tsw d3Tsw

t

t

t

t

IaIpeak

Vap

Ipeak

Ic

VcpVap

Vcp

d1Tsw d2Tsw d3Tsw

t

t

t

t

3rd event linked to DCM

The original model could not be auto-togglingA new DCM-CCM model has been derived

( )

( )

1

1 2

1 2

1

2

2

peaka

peakc

c a

I dI

I d dI

d dI I

d

=

+=

+=

Extract andreplace

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ca

p

Ia(t) Ic(t)

1 NVap Vcp

N=d1/(d1+d2)

12 1

1 1

2 ² 2c ac sw sw c

ac sw ac

I L V d T LF Id dV d T d V−

= = −Model input

Clamp d2:d2 CCM = 1- d1

d2 DCM = 1- d1- d3

d2 < 1 - d1model is in DCM!

By clamping the d2 equation, the circuit toggles between the modes

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3

V410

4

L175u

17d

a c

PWM switch VM p

X4PWMCCMVM

GAI

N

XPWMGAINK = 0.5

10.0

5.00

0.500

In voltage-mode, the duty-cycle is built with a ramp generatorThe transition occurs when the error voltage crosses the ramp

( ) ( ) ( )onerr peak peak

sw

t tV t V V d t

T= =

Vpeak

0

Verr

0

Vramp(t)

d(t)

ton

Tsw

( ) ( )err

peak

V td t

V=

( )( )

1PWM

err peak

d td Kd V t V⎛ ⎞

= =⎜ ⎟⎜ ⎟⎝ ⎠

Vpeak = 2 V

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The Voltage-Mode Model at WorkLet us compensate a buck converter operated in CCM and DCM

1. Run an open-loop Bode plot at full load, lowest input2. Identify the excess/deficiency of gain at the selected cross over3. Place a double zero at f0, a pole at the ESR zero and a pole at Fsw/24. Check final loop gain and run a transient load test

Vout2

16

Cout220u

Resr150m

2 13

L1180u

6

Rupper38k

Rlower10k

9

5

X2AMPSIMP

V22.5

vout

d

a c

PWM switch VM p

4

7

X3PWMVML = 180uFs = 100kG

AIN

15

XPWMGAINK = 0.4

R420m

LoL1kH

10

CoL1kF

V1AC = 1

Vin

vout

Vout

12

R2R2

C1C1

C2C2

8

R3R3

C3C3

Rload3

Vin20

12.0V

12.0V

12.0V

2.50V

2.50V

1.51V

20.0V

12.1V

604mV

1.51V

1.51V 12.0V

parameters

Rupper=38kfc=7kGfc=-15

G=10^(-Gfc/20)pi=3.14159

fz1=650fz2=650fp1=7kfp2=50k

C3=1/(2*pi*fz1*Rupper)R3 =1/(2*pi*fp2*C3)

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*fz2^2c=fp2^2*fp1^2+fc^2*fp2^2+fc^2*fp1^2+fc^4

R2=sqrt(c/a)*G*fc*R3/fp1

Automatedcompensation

H(s)

T(s)

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The Voltage-Mode Model at Work

-24.0

-12.0

0

12.0

24.0

-180

-90.0

0

90.0

180

2

1

10 100 1k 10k 100k

-180

-90.0

0

90.0

180

-40.0

-20.0

0

20.0

40.0

3

4

Pm = 80°

fc = 7 kHz

|H(7 kHz)| =-15 dB

Arg H(7 kHz) =-121°

Arg T(s)

|H(s)|

9.91m 11.0m 12.1m 13.3m 14.4m

11.6

11.8

12.0

12.2

12.4

12

The Bode plot reveals a gain loss of -15 dB at 7 kHzThe compensator provides a +15 dB gain increase plus phase boost

The final loop gain shows a comfortable phase marginThe transient response at both input levels shows a stable signal

|T(s)|

arg|H(s)|

Vout(t)

Iout = 200 mA to 4 Ain 10 µs

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Current-Mode OperationIn voltage-mode, the error signal directly controls the duty cycleIn current mode, the error voltage sets the inductor peak currentTo derive a model, observe the current signals and average them!

'( )( ) ( ) ( )

2

(1 )2

f swc i c sw e

c sw e swc cp

i i

S d t TI t R V t d t T S

V T S TI d V dR R L

= − −

= − − −

1 2

3

ca

p

Ia(t) Ic(t)

I=Vc/Ri

p

I=d.Ic I=Iu Cs

'( )( ) ( ) ( )

2

(1 )2

f swc i c sw e

c sw e swc cp

i i

S d t TI t R V t d t T S

V T S TI d V dR R L

= − −

= − − −

1 2

3

ca

p

Ia(t) Ic(t)

I=Vc/Ri

p

I=d.Ic I=Iu Cs

1

2

3

CCM

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Current-Mode OperationDo the same for DCM signalsMatch the previous structure to build a CCM/DCM model

ca

p

Ia(t) Ic(t)

I=Vc/Ri

p

I=(d1/(d1+d2)).Ic I=Iu

3rd event linked to DCM

1

12

1 1 22 1

2

c sw epeak

i

c sw ec sw f

i

cpsw esw

i

V d T SIR

V d T SI d T SR

Vd T S d dI d TR Lμ

α

−=

−= −

+⎛ ⎞= + −⎜ ⎟⎝ ⎠

α

DCM

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The Current-Mode Model at WorkTo study a converter, we can write down the equationsOr use a SPICE simulation to get the Bode plot in a secondTake the example of a current-mode flyback converter

( ) 2

1

2 22

1 3

10 0 2 2 22

1 1 1120log

1 1

z z z

p n n p

f f ff f f

H f Gf f ff f f Q

⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞⎢ ⎥

+ + +⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥= ⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎛ ⎞+ ⎜ ⎟ ⎜ ⎟− + ⎜ ⎟⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦

( )1 2 3 1

1 1 1 1 12

1arg tan tan tan tan tan

1z z z p n p

n

f f f f fH ff f f f f Q f

f

− − − − −

⎛ ⎞⎜ ⎟

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟= − + − −⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎛ ⎞⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎜ ⎟− ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

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Stabilizing a CCM Flyback ConverterCapture a SPICE schematic with an averaged model

1

C56600u

R1014.4m

2Vin90AC = 0

3 4

X2xXFMRRATIO = -0.25 vout

vout

6DC

8

13

L1Lp

D1Ambr20200ctp

B1Voltage

V(errP)/3 > 1 ?1 : V(errP)/3

Rload4

vcac

PW

M s

witc

h C

Mp

duty

-cyc

le

X9PWMCML = LpFs = 65kRi = 0.25Se = 0

19.0V

19.0V

90.0V

-78.8V 19.7V

467mV

0V

786mV

Look for the bias points values: Vout = 19 V, okVsetpoint < 1 V, enough margin on current sense

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18

Verr

Cpole2Cpole

109

11

vout

X8OptocouplerFp = PoleCTR = CTR

Czero1Czero

RpullupRpullup

5

RledRled

Rlower2Rlower

Rupper2Rupper

X10TL431_G

Vdd5

errP

X4POLEFP = poleK = 1

K

S+A

R71

err

14

LoL1kH

VstimAC = 1

15

CoL1kF

2.36V

0V

2.36V

2.36V

2.36V 18.7V

2.49V

17.7V

5.00V

parameters

Vout=19Ibridge=250uRlower=2.5/IbridgeRupper=(Vout-2.5)/IbridgeLp=350uSe=20kfc=1kpm=60Gfc=-22pfc=-71G=10^(-Gfc/20)boost=pm-(pfc)-90pi=3.14159K=tan((boost/2+45)*pi/180)Fzero=fc/kFpole=k*fcRpullup=20kRLED=CTR*Rpullup/GCzero=1/(2*pi*Fzero*Rupper)Cpole=1/(2*pi*Fpole*Rpullup)CTR=1.5Pole=6k

fromBode

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-32.0

-16.0

0

16.0

32.0

4

10 100 1k 10k 100k

-180

-90.0

0

90.0

180

6

Gain at 1 kHz-22 dB

Phase at 1 kHz-71 °

|H(s)|

argH(s)

Sub harmonicpoles

Inject rampcompensation

ramp

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Stabilizing a CCM Flyback ConverterThe easiest way to damp the poles:Calculate the equivalent quality coefficient at Fsw/2Calculate the external ramp to make Q less than 1

( )1 1 8

3.14 0.5 0.461'2

e

n

QSD DS

π= = =

× −⎛ ⎞+ −⎜ ⎟

⎝ ⎠

( )1 1 90 0.25 10.5 0.5 0.5 0.46 36

' ' 320 1 0.46 3.14n in i

ep

S V RS D D kV sD L D uπ π

×⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + = − + = − + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟× −⎝ ⎠ ⎝ ⎠ ⎝ ⎠

NCP1230internal

Rramp

Ri

2.3 Vpp

18 kΩ

CS

DRV 2.3 15315rampS kV s

u= =

36 51%70

er

n

S kMS k

= = =

0.51 70 18 4.1153

r n rampcurrent

ramp

M S R k kR kS k

× ×= = = Ω

Rcurrent

On-time slope in i

p

V RL

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Stabilizing a CCM Flyback ConverterBoost the gain by +22 dB, boost the phase at fc

11

-80.0

-40.0

0

40.0

80.0

10

10 100 1k 10k 100k

-180

-90.0

0

90.0

180

11

|T(s)|

argT(s)

Cross over1 kHz

Margin at 1 kHz60°

GM20 dB

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Stabilizing a CCM Flyback ConverterTest the response at both input levels, 90 and 265 VrmsSweep ESR values and check margins again

1.80m 5.40m 9.00m 12.6m 16.2mtime in seconds

18.79

18.87

18.95

19.03

19.11

1211

Vout(t)

Lowline

Hiline

112 mV

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Power Factor Correction

CBulk100uIC = 50

6

D51N4007

2

D61N4007

D71N4007

D81N4007

Vmains

VbulkVbulk

B2Current50/V(Vbulk)

IoutItotal

Ibulk

Vbulk

Vmains

Iin

Vbulk

Vmains

Iin

The bulk capacitor connects to a low-impedance sourceAt the bulk capacitor refueling, a narrow peak current flowsThis peak conveys a large harmonic content

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Cbulk

D1 D2

D3 D4

Mains

PWM

VbulkPFCPre-converter

store release

Cbulk

D1 D2

D3 D4

Mains

PWM

VbulkPFCPre-converter

store release

A pre-converter is installed as a front-end sectionThe pre-converter draws a sinusoidal currentThe energy is stored and released in/by the bulk capacitor

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+

-

2

1

4

+-

177

3

5

16

118

Rlimit

Ddmg

12

Dout

Cout

VoutS

R

Q

Q 6

Verr

Reset detector

Rsense

current sensecomparator

9

10

13

G1A

B

K*A*B14

X6

C1

Vref

Rupper

Rlower

RdivL

RdivU18

D3

15

D4

D5D6

CinVin

Vdem

L

Peak currentsetpoint

Error amplifier

One of the most popular techinique uses Borderline modeThe MC33262 operates in peak current mode control

The NCP1606 also operates in constant-on time

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the averageaverage inductor current is halfhalf the inductor peak current value


IL(t)

IL,peak

IL,avg

ton

IL = 0

( ) ( )( )sw

L inTI t t I t=


IL(t)

IL,peak

IL,avg

ton

IL = 0

( ) ( )( )sw

L inTI t t I t=

The core is always reset from cycle to the other

On timeis constant

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4

9C5150uIC = Vrms*1.414

R1050m

16 23

L1L

parameterVrms=100Pout=150Vout=400Ri=0.22L=850u

5Vton

26Vfsw

22

13

17

G1100u

V42.5

25

C20.68u

R51G

Verr

14

V51.7

D2N = 0.01

15

D1N = 0.01

V36.4

10

R11.6Meg

R212k C3

10n

A

B

K*A*B

24

2

K = 0.6

R6100m

V11B1Current

I(V11) > 10u ? 10u : I(V11)

R310k

R41.6Meg

R823k

err

B4VoltageV(err)-2.05 < 0 ? 0 : V(err)-2.05

B1 3 0 V = V(1) * V(2) * K>1.3 ? 1.3 : V(1) * V(2) * K

vcac

PWM

sw

itch

BCM

p

ton

Fsw

(kH

z)

X5PWMBCMCM2L = LRi = Ri

3

6

VinVRMS

+

-

IN

8

X1KBU4J

Iin

ΔVin

Vrect

Cin1u

Vmul

Vout

RloadVout*Vout/Pout

A 150 W BCM PFC average example with the MC33262

Current-modeborderline model

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-8.00

-4.00

0

4.00

8.00vt

on in

vol

ts

-2.00

-1.00

0

1.00

2.00

iin in

am

pere

sP

lot1

1

3

394

398

402

406

410

vout

, vou

t#a

in v

olts

Plo

t2

62

1.139 1.149 1.159 1.169 1.179time in seconds

-60.0

-30.0

0

30.0

60.0

vton

in v

olts

-4.00

-2.00

0

2.00

4.00

iin in

am

pere

sP

lot3

4

5

THD = 2%

THD = 10% Iin(t)

Iin(t)

Vin = 230 Vac

Vin = 100 Vac

ton(t) - µs

ton(t) - µs

Vout,peak = 406 V

Vout,valley = 398 V Vout(t)

-8.00

-4.00

0

4.00

8.00vt

on in

vol

ts

-2.00

-1.00

0

1.00

2.00

iin in

am

pere

sP

lot1

1

3

394

398

402

406

410

vout

, vou

t#a

in v

olts

Plo

t2

62

1.139 1.149 1.159 1.169 1.179time in seconds

-60.0

-30.0

0

30.0

60.0

vton

in v

olts

-4.00

-2.00

0

2.00

4.00

iin in

am

pere

sP

lot3

4

5

THD = 2%

THD = 10% Iin(t)

Iin(t)

Vin = 230 Vac

Vin = 100 Vac

ton(t) - µs

ton(t) - µs

Vout,peak = 406 V

Vout,valley = 398 V Vout(t)

Highline

Lowline

Constanton-time

Average models can also work in transient conditions

www.onsemi.com44


-80.0

-40.0

0

40.0

80.0ga

in in

db(

volts

)

-180

-90.0

0

90.0

180

phas

e in

deg

rees

plot

1

1

2

1m 10m 100m 1 10 100 1k 10k 100kfrequency in hertz

-80.0

-40.0

0

40.0

80.0

gain

in d

b(vo

lts)

-180

-90.0

0

90.0

180

phas

e in

deg

rees

Plo

t2

4

5

No zero added

Zero added

phase

gain

phase

gain

fc = 5 Hz

fc = 5 Hz

Pm = 61°

-80.0

-40.0

0

40.0

80.0ga

in in

db(

volts

)

-180

-90.0

0

90.0

180

phas

e in

deg

rees

plot

1

1

2

1m 10m 100m 1 10 100 1k 10k 100kfrequency in hertz

-80.0

-40.0

0

40.0

80.0

gain

in d

b(vo

lts)

-180

-90.0

0

90.0

180

phas

e in

deg

rees

Plo

t2

4

5

No zero added

Zero added

phase

gain

phase

gain

fc = 5 Hz

fc = 5 Hz

Pm = 61°

Use the model to boost the phase at the cross over point

www.onsemi.com45


122m 365m 608m 851m 1.09time in seconds

320

350

380

410

440vo

ut#a

, vou

t#a#

1 in

vol

tsP

lot1

32

1.128 1.135 1.142 1.149 1.157time in seconds

-800m

-400m

0

400m

800m

iin, i

in#1

in a

mpe

res

Plo

t2

41

Vout(t)

Iin(t)

440 V

413 V

THD = 1.5%

THD = 10%

122m 365m 608m 851m 1.09time in seconds

320

350

380

410

440vo

ut#a

, vou

t#a#

1 in

vol

tsP

lot1

32

1.128 1.135 1.142 1.149 1.157time in seconds

-800m

-400m

0

400m

800m

iin, i

in#1

in a

mpe

res

Plo

t2

41

Vout(t)

Iin(t)

440 V

413 V

THD = 1.5%

THD = 10%

Added zero

No zero

Added zero

No zero

The zero improves the overshoot but degrades the THD…

www.onsemi.com46


www.onsemi.com47

4

5

1 2

D11N5818

3Cout100uF

Resr100m Rload

10

Vout

Lp1mH

7

Rs100m

6

Lleak10uH

9

Vinput300V

Pulsegenerator

1:NOhmic losses

Primaryinductance

Leakageinductance

Capacitorparasitics

Switching Models, the Breadboard on PCTurn your PC into a virtual breadboard

www.onsemi.com48

Switching Models, the Breadboard on PC

18

Rsense2.8

12

6

R3200m

L23.5mH

1

9

R4100m

C1470uFIC = 5.5

Vout

IprimIripple1

L580uH

Vdrain

10

IDrain

Vsense

19

R15470

4

X1XFMRRATIO = -0.08

Rload13.4

15

X4MOC8101

Vinput126

13

Drv

VFB

31

L12.2uH

17

R1610m

32

R17300m

C210uF

IsecVsec

23 Cvcc10uFIC = 11.99

5 Vcc

16

Istartup

8Vadj

vout

Iout

R5100m

D1MBR140P

C3150p

14X3TL431

R214k

R710k

vout

C5100nFIC = 2.5

1

2

3

4 5

8

6

7

NCP1200

X5NCP1200

X2MTD1N60E

Primary inductance

Leakage inductance

Wire your device as you would do in the lab.

www.onsemi.com49

Simulations (Really) Work!!

505.00U 515.00U 525.00U 535.00U 545.00U

0

Vdrain100V/div

Vsense500mV/div

Assess the average, rms currents in your circuit Check if enough margins exist on your semiconductors

Leakageeffects

simulated measured

www.onsemi.com50

Simulations (Really) Work!!

4.50m 8.50m 12.5m 16.5m 20.5m

16.69

16.75

16.81

16.87

16.93

30mV/div

2ms/div

With accurate models, the simulation results are excellent You can then vary the parasitic terms and see their impact

simulated measured

www.onsemi.com51


www.onsemi.com52

EMI Filtering on a DC-DC

7

5

10

8

2 4

Vdrv

Verr

Vsense VIL

Vrect

Vout

ID_FW

3

ICoil

9

R3285

C311n

C21.3nF

13R2120k

CMP

FBOUT

GNDIMAX

X1PWMVMPERIOD = 10uDUTYMAX = 0.9REF = 2.5IMAX = 5DUTYMIN = 0.01VOL = 100MVOH = 2.5DUTYMIN = 0.01

6X2PSW1RON = 0.1

C13.3nF

D1mbr845

Rupper38k

Rlimit10m

1

R150m

L1180uH

14Cout1000uFIC = 11

Resr69m

C30470p

R301k

B1VoltageI(Rlimit)

Rlower10k

Δ

Vswitch

Iin

36

V1

Vin30

vin

Rload3

Need tolimit the

ac current< 15 mA peak

DC-DC are highly EMI polluting systemsA filter has to be installed to avoid noise in the source

www.onsemi.com53



-4.00

-2.00

0

2.00

4.00

iin in

am

pere

sP

lot1

1

105k 315k 525k 735k 945kfrequency in hertz

10u

100u

1m

10m

100m

1

10

mag

(fft(t

emp)

) in

am

pere

sPl

ot1_

f

2

FFT results

Input current signature

Ipeak = 2.45A

Iavg = 1.7 AIrms = 2.7 A


-4.00

-2.00

0

2.00

4.00

iin in

am

pere

sP

lot1

1

105k 315k 525k 735k 945kfrequency in hertz

10u

100u

1m

10m

100m

1

10

mag

(fft(t

emp)

) in

am

pere

sPl

ot1_

f

2

FFT results

Input current signature

Ipeak = 2.45A

Iavg = 1.7 AIrms = 2.7 A

15 62.45filter

mA m< <

Iripple < 15 mAattenuation

0 0.006 7.7swf F kHz< × <

Position the cutofffrequency of the LC filter

Use SPICE to extract the current signatureRun Fourier analysis to look at the spectrum

www.onsemi.com54


Vin

L Rlf

C Rload VoutSwitchModeConverter

2 20

1 5.24

C µFf Lπ

= =

L = 100 µH

Checkimpedance

peaking

220 1

max1 0

1outFILTERZ RZR Z

⎛ ⎞= + ⎜ ⎟

⎝ ⎠

7.7 kHz

A LC configuration offers the best efficiencyAs any LC network, it is subject to resonances

www.onsemi.com55


10m 100m 1 10 100 1k 10k 100kfrequency in hertz

-30.0

-10.0

10.0

30.0

50.0

pp

1

7

Zout

No dampingZout = 57.5 dBΩ

ZinSMPS

18 dBΩ

Problem!


-30.0

-10.0

10.0

30.0

50.0

pp

1

7

Zout

No dampingZout = 57.5 dBΩ

ZinSMPS

18 dBΩ

Problem!

Need todamp this!

The incremental input resistance of a DC-DC in negativeA LC filter loaded by a negative resistance can oscillate!

Powersupply inputimpedance

www.onsemi.com56


10 100 1k 10k 100kfrequency in hertz

-40.0

-20.0

0

20.0

40.0

vdbo

ut2,

vdb

out2

#1 in

db(

volts

)

-180

-90.0

0

90.0

180

ph_v

out2

#1,

ph_v

out2

in d

egre

esP

lot1

2

3

41

phase

gainWith filter

With filter

Without filter

Without filter

10 100 1k 10k 100kfrequency in hertz

-40.0

-20.0

0

20.0

40.0

vdbo

ut2,

vdb

out2

#1 in

db(

volts

)

-180

-90.0

0

90.0

180

ph_v

out2

#1,

ph_v

out2

in d

egre

esP

lot1

2

3

41

phase

gainWith filter

With filter

Without filter

Without filter

Not stable!

If the resonance is too peaky, problems can arise

|T(s)|

argT(s)

www.onsemi.com57


Vin

L Rlf

C Rload VoutSwitchModeConverter

11 2

0

11 2 1

0 0

2damp inSMPS

inSMPSinSMPS

RL CR RR Z Z RZ CR L CR R

ω

ω ω

+ −= −

− + + −

A resistor is damping the LC filter by creating lossesA dc-block capacitor is installed to limit dissipation

www.onsemi.com58



-30.0

-10.0

10.0

30.0

50.0

7

654321

Zout

ZinSMPS

18 dBΩ

Ok, margin is 8 dB min

Rdamp= 1 Ω

Rdamp= 2 ΩRdamp= 3 Ω

Rdamp= 4 Ω

Rdamp= 5 Ω


-30.0

-10.0

10.0

30.0

50.0

7

654321

Zout

ZinSMPS

18 dBΩ

Ok, margin is 8 dB min

Rdamp= 1 Ω

Rdamp= 2 ΩRdamp= 3 Ω

Rdamp= 4 Ω

Rdamp= 5 Ω

The right resistor prevents the overlaps between curves

www.onsemi.com59



1.660

1.670

1.680

1.690

1.700

p

0 1

1

y (pk-pk) = 22.8m amperesbetween 3.91m and 3.92m seconds

A final check shows a noise amplitude under control

www.onsemi.com60

A Book on Power Supply Design

Learn DC-DC converters theoryUnderstand average modelingFeedback and loop controlDesign examples of DC-DC and AC-DCPower Factor CorrectionChapters on flyback and forward convertersSupplied CDROM with working examples

To learn more about power supplies and simulations…

I alreadyhave ideasfor the nextedition!!

886 pages, 8 chapters

www.onsemi.com61

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!

www.onsemi.com62

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

Documents

Simulating Power Supplies with SPICE