small-signal modeling of the boost converter operated in...

1 • Chris Basso – CM Boost Study

Small-Signal Modeling of the Boost Converter Operated in CM

Christophe Basso

IEEE Senior Member

2 • Chris Basso – – CM Boost Study

Course Agenda

The PWW Switch Concept

Small-Analysis in Continuous Conduction Mode

Small-Signal Response in Discontinuous Mode

EMI Filter Output Impedance

Cascaded Converters Operation


Course Agenda







Non-Linearity in a Switching Converter

A switching converter is ruled by linear equations…

…combining so-called state variables

SWL

C R1u

1u

L

C R

L C R

1y

during

during 1 swD T

swDT

on

off

1x

1x

2x

2x


State Variables

State variables describe the mathematical state of a system n state variables for n independent storage elements knowing variables state at t0 helps compute outputs for t > t0

System described by state variables

1 2 3, , ..., nx x x x

1u t

2u t

3u t

1y t

2y t

3y t

Input vector u Output vector y

x1 is the inductor current and x2 is the capacitor voltage Differentiation gives the state variable rate of change

1

Ldi tx

dt Predict future

system state 1 Lx i t

2

Cdv tx

dt 2 Cx v t


Describe the System During the On-Time

Observe the system during the on-time duration or dTSW:

1 1 2

LC

di tu L v t Lx x

dt

2 1 2

1CC

dv ti t C Cx x x

dt R

1x

Ci t 2x

R

RC 2x

1 2 1

2 1 2

1 1

1 1

x x uL L

x x xC RC

1u

1 1 1

2 2 2

0 1 1 0

1 1 0 0

x x uL L

x x uC RC

State coefficients Source coefficients

L


Describe the System During the Off-Time

Repeat the exercise during the off-time duration or 1-dTSW

1x

Ci t

2xRCL Lv t

1Lv t Lx 2Lv t x

2 1x Lx

Ri t

1 1 2R Ci t x i t x Cx

2 Rx i t R

2 1 2x R x Cx

1 2

2 1 2

10

1 1

x xL

x x xC RC

1 1 1

2 2 2

0 1 0 0

1 1 0 0

x x uL

x x uC RC

State coefficients Source coefficients


Make it Fit the State Equation Format

Arrange expressions to make them fit the format:

x x t u t A B

1 1 1

2 2 2

0 1 1 0

1 1 0 0

x x uL L

x x uC RC

1 1 1

2 2 2

0 1 0 0

1 1 0 0

x x uL

x x uC RC

1 1

0 1 1 0

1 1 0 0

L L

C RC

A B

2 2

0 1 0 0

1 1 0 0

L

C RC

A B

on-time network

off-time network

How do we link matrixes A1 and A2, B1 and B2? We smooth the discontinuity by weighting them by D and 1-D

1 2 1 21 1x D D x t D D u t A A B B

State equation


The State-Space Averaging Method (SSA)

We now have a continuous large-signal equation We need to linearize it via perturbations

0ˆD D d 0 ˆx x x 0 ˆu u u

1 2 1 21 1x D D x t d D u t A A B B

01 2 1 10

2 1 2

ˆ1ˆ ˆ ˆ

1 1ˆ ˆ ˆ

D dx x u u

L L L

x x xC RC

C

L

2x

0D d

1x

1u

10ˆu d

Canonical small-signal model


The State-Space Averaging Method (SSA)

SSA was applied to switching converters by Dr Ćuk in 1976 It is a long, painful process, manipulating numerous terms What if you add an EMI filter for instance?

4 state variables and you have to re-derive all equations!

SW2L

2C R1u

1y1L

1C on

off

1L

1C2L

2C R

2L 2C R

1L

1C

1u

1u


The PWM Switch Model in Voltage Mode

We know that non-linearity is brought by the switching cell

L

Why don't we linearize the cell alone?

1u C R

d

a c

PWM switch VM p

a

p

c

a

p

c

a: activec: commonp: passive

Switching cell Small-signal model(CCM voltage-mode)

. .

V. Vorpérian, "Simplified Analysis of PWM Converters using Model of PWM Switch, parts I and II"IEEE Transactions on Aerospace and Electronic Systems, Vol. 26, NO. 3, 1990


Replace the Switches by the Model

Like in the bipolar circuit, replace the switching cell…

L

1u C R

a

p

c

…and solve a set of linear equations!

. . L1u C R

..


d

a c

PWM switch VM p

dac

PW

M s

witc

h V

Mp

dac

PW

M s

witc

h V

Mp

dac

PW

M s

witc

h V

Mp

An Invariant Model

The switching cell made of two switches is everywhere!

buck

buck-boost

boost

Ćuk

d

a c

PWM switch VM p


Smoothing the Discontinuity

A CRT TV displays frames at a certain rate, 50 per second

1 23

4 5 6 7

8

910

The optic nerve time constant is larger than an interval A succession of discrete events is seen as continuous

11

12

See "phi phenomena", www.wikipedia.org

Low-frequency filteringIntegration

1t


Averaging Waveforms

The keyword in the PWM switch is averagingaveraging

t

v t

A

swDT

swT

v t

0

1 11 '

sw sw

sw

sw

T T

peak peak peakTsw sw DT

v t v t dt V dt V D V DT T

swT

peakV

The resulting function is continuous in time


From Steps to a Continuous Function

Some functions require mathematical abstraction: duty ratio

At the modulation frequency scale, points look contiguous Link them through a continuous-time ripple-free function d(t)

swT

1t 2t 3t 4t 5t

v t

1D

Discrete values of D2D 3D

4D

5D

nn

sw

tD

T

mod swf F

Average and continuous evolution of d(t)


The Common Passive Configuration

The PWM switch is a single-pole double-throw model

a c

p

d

'd

C R

L

inV

ci t ai t

apv t cpv toutV

Install it in a buck converter and draw the waveforms

a c

p

d

'd ai t ci t

apv t cpv t

CCM



Average the current waveforms across the PWM switch

ai t

ci t

0

0

t

t

swDT

sw

c Ti t

sw

c Ti t

0

1 sw

sw sw

dT

a a a c cT Tsw

i t I i t dt D i t DIT

a cI DI

Averagedvariables

CCM



Average the voltage waveforms across the PWM switch

cpv t

apv t

0

0

t

t

swDT

sw

ap Tv t

sw

cp Tv t

0

1 sw

sw sw

dT

cp cp cp ap apT Tsw

v t V v t dt D v t DVT

cp apV DV

Averagedvariables

CCM


A Two-Port Representation

We have a link between input and output variables

Two-portcell

a

p

c

p

cDI cI

apDVapV

It can further be illustrated with current and voltage sources

a

p

c

p

cI

apDVapV

aI

cpVcDI

d

d

CCM


A Transformer Representation

The PWM switch large-signal model is a dc "transformer"!

It can be immediately plugged into any 2-switch converter

a c

p

1 D

cIaI

ac

II

Da cI DI

cp

ap

VV

D cp apV DV

dc equations!

. .

1

D..

LLr

inV C Ra

cp

CCM


Simulate Immediately with this Model

SPICE can get you the dc bias point

4

L1100u

C1470u

R110

7

Vg10

Rdum1u

a

Vout5

VICR2100m

c c

V(a,p)*V(d)

p p

a

V(d)*I(VIC)V30.3AC = 1

d

9.80V14.0V

10.0V

300mV

9.80V

-40.0

-20.0

0

20.0

40.0

10 100 1k 10k 100k

-360

-180

0

180

360

…but also the ac response as it linearizes the circuitdB °

Hz

H f

arg H f

CCM


We Want Transfer Functions

Derive the dc transfer function: open caps., short inductors

1

D

.Lr

inV Ra

c

p

aI

cIcI

.

outV

cpV

out outV I R

outI

out a cV I I R

a cI DI

1out cV I D R

in L c cp outV r I V V

in L c out outV r I DV V

1out in

c

L

V D VI

r

1

11

out

Lin

V

rVD

D R

2

1 1

'1

'

out

Lin

V

rV D

RD

CCM


Plotting Transfer Functions

Plot the lossy boost transfer function in a snapshot

0 0.2 0.4 0.6 0.8 10

1

2

3

4

5

f d 0.1( )

f d 0.2( )

f d 0.3( )

f d 0.4( )

f d 0.5( )

f d 0.6( )

f d 0.7( )

f d 0.8( )

f d 0.9( )

f d 1( )

d

0.1ΩLr

0.2ΩLr

1ΩLr

out

in

V

V

Duty ratio

10ΩR

Above a certain conversion ratio, latch-up occursCCM


Course Agenda







A Boost Converter in Current Mode

Identify the diode and switch position in a boost CM

vc

a c

PWM switch CM p

duty-cycle

Replace switches by the small-signal PWM switch model

Current mode

a

pc


A Small Signal Model

The model includes current sources and conductances

1o

i

kR

'

2sw

f o

DD Tg Dg

L

1'

2sw a

o

n

T Sg D D

L S

i

i

Dk

R c

i f

ap

Ig D g

V

cr o

ap

Ig g D

V

1

ig

a c

p

ˆc iv k ˆ

cp rv g ˆap fv g ˆ

c ov k1

og sC

cI

V. Vorpérian,"Analysis of Current-Controlled PWM Converters using the Model of the PWM Switch", PCIM Conference ,1990


Start with a Large Signal Response

Use the original large-signal (nonlinear) PWMCM model

5 V/1 AFsw = 1 MHz

1

2

Cout200uF

Resr60m

Vin2.7

4 5

L15u

7

duty_ratio

Vout

R41.5m

R91k

Rload1v

cac

PW

M s

witch

CM

p

duty

-cyc

le

X4PWMCCMCML = 5uFs = 1MegRi = -50mSe = 0

V1471mAC = 1

Vin

5.00V

5.00V

2.70V

2.69V

2.69V

462mV

471mV


Frequency Response of the CM Boost

This frequency response becomes the reference

1.00

13.0

25.0

37.0

49.0

10 100 1k 10k 100k 1Meg

frequency in hertz

-324

-252

-180

-108

-36.0

out

c

V f

V f

out

c

V f

V f

(dB)

(°)

Subharmonicpoles0 14.5 dBH


Plug the Linearized Small-Signal Model

Use the linearized model to check coefficients and response

4

C2

Cs

VIC

parameters

Fsw =1Meg

Tsw =1/Fsw

L=5u

Cs=1/(L*(Fsw *3.14)^2)

Ri=-50m

Se=0

Vin=2.7

Vout=5

Sn=(Vac/L)*Ri

Sf=(Vap/L)*Ri

Vc=471m

Vac=-Vin

Vap=-Vout

Vcp=(-Vout+Vin)

Ic=(Vc/Ri)-D*Tsw *Se-Vcp*(1-D)*Tsw /(2*L)

D=Vcp/Vap

D'=1-D

ki=D/Ri

gi=D*(gf-Ic/Vap)

gr=(Ic/Vap)-go*D

ko=1/Ri

go=(Tsw /L)*(D'*Se/Sn+0.5-D)

gf=D*go-(D*D')*Tsw /(2*L)

B3

Current

V(a,p)*gf

B4

Current

V(v c)*ko

c

B5

Current

V(v c)*ki

B6

Current

V(c,p)*gr

R4

1/gi

R5

1/go

15

L2

L

16

C3

200uF

R3

1v c

R6

60m

Vout

R7

1u

B2

Voltagev (c,p)/v (a,p)

1

R8

1.5m

a

p

a

p

c

Vin

Vin

V2

471m

AC = 1

D

2.69V 5.00V

2.69V

4.30uV

2.69V

0V

462mV

2.70V

471mV

These coefficients arecomputed by the macro


Small-Signal Response and Original Model

Validate the small-signal approach by comparing responses

1.00

13.0

25.0

37.0

49.0

10 100 1k 10k 100k 1Meg

frequency in hertz

-324

-252

-180

-108

-36.0

out

c

V f

V f

out

c

V f

V f

(dB)

(°)

0 14.5 dBH

Good to go!


Rearrange Sources to Improve Readability

It is important to lay the network out properly

4

C2

Cs

VIC

B3

Current

V(0,out)*gf

B4

Current

V(v c)*ko

B5

Current

V(v c)*ki

B6

Current

V(c,out)*gr

R4

1/gi

R5

1/go

15

L2

L

16

C3

200uF

R3

1v c

R6

60m

Vout

B2

Voltagev (c,out)/v (0,out)

1

R8

1.5m c

Vin

Vin

V2

471m

AC = 1

D

out

2.69V

2.69V

2.69V

0V

462mV

2.70V

471mV

5.00V

Make sure the frequency response remains unchanged

Storage element

Storage element

Storage element


What is the Converter Order?

Count storage elements: L and two C The denominator of the transfer function is of 3rd order

1 2

1

0 2

0 0

1 11

1 1

z z

p

s s

s sH s H

s s ss Q

Dc gain

Subharmonicpoles

Zeros


Start with the Static Analysis

For s = 0, short inductors and open capacitors

1

VIC

parametersFsw=1MegTsw=1/FswL=5uCs=1/(L*(Fsw*3.14)^2)Ri=-50mSe=0Vin=2.7Vout=5

Sn=(Vac/L)*RiSf=(Vap/L)*Ri

Vc=471mVac=-VinVap=-VoutVcp=(-Vout+Vin)

Ic=(Vc/Ri)-D*Tsw*Se-Vcp*(1-D)*Tsw/(2*L)

D=Vcp/VapD'=1-Dki=D/Rigi=D*(gf-Ic/Vap)gr=(Ic/Vap)-go*Dko=1/Rigo=(Tsw/L)*(D'*Se/Sn+0.5-D)gf=D*go-(D*D')*Tsw/(2*L)

B3

Current

V(0,out)*gf

B4

Current

V(vc)*ko

B5

Current

V(vc)*ki

B6

Current

V(c,out)*gr

R4

1/gi

R5

1/go

R3

1

vc

Vout

B2

Voltagev(c,out)/v(0,out)

R8

1.5m

c

V2

471m

AC = 1

D

out

Neglect rL

0c

V

0c

V


Simplify and Rearrange

Rearrange sources to make the circuit look simpler

B3CurrentV(out)*gf

B4CurrentV(vc)*ko

B5CurrentV(vc)*ki

R41/gi

R51/go

R31

Vout

out

B1CurrentV(out)*gr

1

outout f c o out o i out r c i

VV g V k V g g V g V k

R

00

0

1

1i

f i r

k kH

g g g gR

H0 20 logko ki

gf go gi gr1

RL

14.563 dB

Mathcad® calculations


Put Storage Elements Back in Place

C2Cs

B3Current

V(out)*gf

B4Current

V(vc)*koB5Current

V(vc)*ki

B6Current

V(c,out)*gr

R41/gi

R51/go

L2L

16

C3200uF

R31

vc R660m

Vout

cV2471mAC = 1

out

Rearrange sources to make the circuit look simpler

Label each storage element with a time constant Time to call the FACTs! Set excitation to 0, Vc = 0

FACTS: Fast Analytical Circuit TechniqueS

1

2

3


Polynomial Form of a 3rd Function

A transfer function is made of a numerator and a denominator

N sH s

D s

zerospoles

The denominator combines the circuit time constants

1 1 2 2 1 12 31 2 3 1 2 1 3 2 3 1 2 31D s s s s

Time constantwith Cs

Time constantwith L

Time constantwith Cout

To calculate time constants, suppress the excitation, all is dc


Calculate the First Time Constant

B3Current

V(out)*gf

B6Current

V(out)*gr

R41/gi

R51/go

R31

1

R660m

out

V20AC = 1

Vc

tau1

L shorted

tau3C open

R11f

R2100G

c

B5CurrentV(vc)*ki

B4CurrentV(vc)*ko

?R

= 0

= 0

What resistance does Cs « see » ?

The excitation is 0 so several sources go away You can further simplify the circuit and test it in SPICE


Calculate the First Time Constant

Check the result with a simple dc operating point calculation

B3

Current

V(out)*gf

B6

CurrentV(out)*gr

R4

1/gi

R5

1/goR3

11

R6

60m

out

V2

0

AC = 1

Vctau1

L shorted

tau3C open

R1

1f

R2

100G

c

I1

1

V(vc)*ki

B5

Current

V(vc)*ko

B4

Current

R7

1/gi

R8

1/go

R9

1I2

1

out2

V(out2)*(gr-gf)

B2

Current

495mV

297f V

0V

495mVeqR

1

1 1|| ||eq L

i o

R Rg g

TVTI

outT out r f

eq

VI V g g

R

1

1outT

T Tr f

eq

VVR

I I g gR

1

1

Reqgr gf

0.495

1 sRC


Calculate the Second Time Constant

B3Current

V(out)*gf

B4Current

V(vc)*koB5Current

V(vc)*ki

B6Current

V(c,out)*gr

R41/gi

R51/go

R31

vc

1

R660m

Vout

cV2471mAC = 1

out

tau3C open

tau1Cs open

What resistance does L « see » ?

?R

= 0

= 0

The excitation Vc is 0 so simplify the circuit Rearrange the network and test it in SPICE



B3

Current

V(out)*gf

B4

Current

V(vc)*koB5

CurrentV(vc)*ki

B6

Current

V(c,out)*gr

R4

1/gi

R5

1/go

R3

1

vc

Voutx

cV2

0

AC = 1

out

I1

1

B1

Current

V(out2)*gf

B8

Current

V(c1)*gr

R1

1/gi

R2

1/go

R6

1

Vout2

I2

1

out2

c1

R7

1/gr

0V

-52.9V

-48.8V

-52.9V -48.8V

TVTI T

T

VR

I



Express IT and Vout then rearrange to unveil VT/IT

0

1T out

T out f

V VI V g

g

out T T r eqV I V g R

1 eq f oT

T o eq r f o

R g gVR

I g R g g g

21 eq f o

o eq r f o

L

R g g

g R g g g

1 Req2 gf go go Req2 gr gf go

52.897 Same result as dc point simulation

2

1 1|| ||eq L

i r

R Rg g


Calculate the Third Time Constant

What resistance does Cout « see » ?

B3

Current

V(out)*gf

B4

CurrentV(vc)*ko

B5

Current

V(vc)*ki

B6

CurrentV(c,out)*gr

R41/gi

R5

1/go

R3

1

vc

1

R6

60m

Vout

cV2

471mAC = 1

out

tau1Cs open

tau2L shorted

?R= 0

= 0

The circuit is very close to that of 1

Rearrange the network and test it in SPICE


Calculate the Third Time Constant

R71/gi

R81/go

R91

out2

V(out2)*(gr-gf)

B2Current

6

R260m

I31

495mV

-60.0mV

Same resistance as for 1 plus rC in series

3

1

1

1 C out

r f

eq

r Cg g

R

1

1 1|| ||eq L

i o

R Rg g


First Coefficients a1 and a2

FACTs tell us that a1 sums up all time constants

1 1 2 3a

V. Vorpérian, “Fast Analytical Techniques for Electrical and Electronic Circuits”, Cambridge Press, 2002

For a2, we multiply combined-time constants

Dimension is time

1 1 22 1 2 1 3 2 3a Dimension is time2

What is this new time constants definition, ?

12

sC (HF)

L C (dc)

?R

13

sC (HF)

L C(dc)

?R

23

L (HF)

C sC (dc)

?R

12


Calculate the Terms for a2

B3

Current

V(out)*gf

B4

CurrentV(v c)*ko

B5

CurrentV(v c)*ki

B6

CurrentV(c,out)*gr

R4

1/gi

R5

1/goR3

11

R6

60m

out

V2

0

AC = 1

Vc

tau1

Cs HF

R?

tau3

C openR1

1u

I1

1

c

12

sC (HF)

L C (dc)

?R

What resistance does L while Cs is a short and Cout is open?

Go for an intermediate step


A Simplified Drawing for a Simple Answer

What R? does L see while Cs is a short and Cout is open?

B1

Current

V(out2)*gf

B8

Current

V(c2)*gr

R2

1/gi

R7

1/go

R8

1

tau1Cs HF

R?

R10

1u

I2

1

R11

1/gr

out2

c2

R9

1/gi

R13

1

I3

1 R15

1/gr

out3

B7

Current

V(out3)*gr

12

1

1

1

1 1|| ||

r

r i

L

g

Rg g


The Mid Term is Easy to Get

13

sC (HF)

L C(dc)

?R

R21/gi

R81

R111/gr

out

1

R760m

2

L and Cs ensure a complete short over RL

?R

13 C outr C

Cs is HF

L is dc


For the Last Term, L is open

23

L (HF)

C sC (dc)

?R

B3Current

V(out)*gf

B4

CurrentV(vc)*ko

B5CurrentV(vc)*ki

B6CurrentV(c,out)*gr

R4

1/gi

R51/go

R3

11

R660m

out

V2

0AC = 1

Vc

tau1Cs dc R?

tau2L is HF

I11

c

What resistance does Cout see when Cs and L are open?

Too complex! Go for another combination.

2 32 3 3 2


Reorder Time Constants for Simpler Sketch

B1

Current

V(out2)*gf

B8

Current

V(c2)*gr

R1

1/gi

R2

1/goR7

1R8

60m

R?I2

1

out2

R9

1/gr

VT

B2

Current

V(out)*gf

B3

CurrentV(T)*gr

R3

1/go

Req

R?I1

1

VT

out

The schematic now looks simpler…

32

outC (HF)

L sC (dc)

?R


Extract the New Time Constant Definition

1 1|| || ||eq C L

i r

R r Rg g

2 T T ri I V g

1T out fI i V g

1 1T out

o

V Vi

g

1T out

T out f

o

V VI V g

g

out eq T T rV R I V g

2

B1Current

V(out)*gf

B8Current

V(c2)*gr

R21/go

Req

R?I21

out

0

1 eq f o

eq r f o

R g gR

g R g g g

32

L

R

substitute

TI

TI

2i1i

TV


…And a3 is ?

For a3, we multiply by a third time-constant

1 1,23 1 2 3a Dimension is time3

What is this new time constant definition?

231,

sC (HF)

C (dc)

?R

L (HF)

B3Current

V(out)*gf

B4CurrentV(vc)*ko

B5CurrentV(vc)*ki

B6CurrentV(c,out)*gr

R41/gi

R51/go

R31

1

R660m

out

V20AC = 1

Vc

tau1Cs HF

L is in HF

R?R11u

I11

c

R121G

7.40V

7.40V

-60.0mV

0V


Rearrange and Simplify

B3

Current

V(out)*gf

B4

CurrentV(vc)*ko

B5

CurrentV(v c)*ki

B6

CurrentV(c,out)*gr

R4

1/gi

R5

1/goR3

11

R6

60m

out

V2

0

AC = 1

Vc

tau1Cs HF

L is in HF

R?R1

1uI1

1

c

R12

1G

parameters

Fsw =1Meg

Tsw =1/Fsw

L=5u

Cs=1/(L*(Fsw *3.14)^2)

Ri=-50m

Se=0

Vin=2.7

Vout=5

Sn=(Vac/L)*Ri

Sf=(Vap/L)*Ri

Vc=471m

Vac=-Vin

Vap=-Vout

Vcp=(-Vout+Vin)

Ic=(Vc/Ri)-D*Tsw *Se-Vcp*(1-D)*Tsw /(2*L)

D=Vcp/Vap

D'=1-D

ki=D/Ri

gi=D*(gf-Ic/Vap)

gr=(Ic/Vap)-go*D

ko=1/Ri

go=(Tsw /L)*(D'*Se/Sn+0.5-D)

gf=D*go-(D*D')*Tsw /(2*L)

6

R2

1/gi

R8

12

R9

60m

R?I2

1

7.40V

7.40V

-60.0mV

0V

7.40V

-60.0mV 1,23

1|| L C out

i

r r Cg


Final Denominator Expression

1

2

1 12

1 1

1 11s C out

eq f or f r f

eq eqo eq r f o

La C r C

R g gg g g gR R

g R g g g

2

2

1 1 10 2

1

1 1 1

1 1 1 1 11

1 1|| ||

s s C out C out

eq f or f r f r f

eq eq eqr eq r f o

r i

L La C C r C r C

R g gg g g g g gR R Rg g R g g g

Rg g

1 1 2 3a

1 1 2 1 1 32 1 2 1 3 2 3 1 2 1 3 3 2a

1 123 1 2 3a 3

1

1

1 1||

1 1

1

1 1|| ||

s L C outi

r feq

r

r i

La C r r C

gg gR g

Rg g


C2

Cs

B3

CurrentV(out)*gf

B4

CurrentV(vc)*ko

B5

CurrentV(vc)*ki

B6

CurrentV(c,out)*gr

R4

1/gi

R5

1/go

L2

L

1

C3

200uF

R3

1

vc R6

60m

Vout

c

V2

471m

AC = 1

out

Find the Zeros

A zero means the excitation does not generate a response

pIcI

No response means that:

0pI 1pI I 0C outr sC Short circuit

1

1z

C out

sr C

0pI 0outV

cI

aI1I


Find the Zeros

C2

Cs

B3

CurrentV(out)*gf

B4

CurrentV(vc)*ko

B5

CurrentV(vc)*ki

B6

CurrentV(c,out)*gr

R4

1/gi

R5

1/go

L2

L

1

C3

200uF

R3

1

vc R6

60m

Vout

c

V2

471m

AC = 1

out

A Capitalize on the 0-V response to remove sources

pIcI

cI

aI

0

rcv g

a pI I r c i ccv g V k I

c c o o sc cI V k v g v sC

r c i c o o sc c cv g V k V k v g v sC

c

c o o sc c

vV k v g v sC

sL

1c o

c

o s

V kv

g sCsL


Find The Zeros

After substituting v(c) into the equation rearrange it

2

2 21 1

c i o i r o s i c o

o s o s

V k Lg k s Lg k s C Lk s V k

sLg s C L sLg s C L

2i o i r o s i ok Lg k s Lg k s C Lk s k

21 0o i r o s ii o

i o i o

g k g k C Lkk k sL s

k k k k

21 0o i r o s i

i o i o

g k g k C LksL s

k k k k

For low Q values, the 2nd order polynomial can expressed as

2 21 2 1

1

1 1 1a

a s a s a s sa


Find the Zeros – Final Lap!

The first zero is a right half-plane zero

1

1 1o i r

i o z

og k g k

ks

k

sL

s

Negative value!

1

i oz

o i r o

k ks

L g k g k

1

21 L

z

D R

L

The second zero is given by the capacitor ESR

1 0C outsr C

The third zero is located at high frequency

3

1 1s i

o i r o z

C k ss

g k g k s

3

o i r oz

s i

g k g ks

C k

Neglect it

2

21 L

z

D R

L

fz2

z2

23.16 10

4 kHz

V. Vorpérian,“Analytical Methods in Power Electronics", 3-day course in Toulouse,2004


Arranging the Transfer Function

The denominator has to be rearranged a little bit…

2 31 2 31D s b s b s b s

The term 1+b1s dominates at low frequency

2 3 2321 2 3 1

1 1

1 1 1bb

b s b s b s b s s sb b

1

2321 2

1 1

11 1 1

1p

n p n

bb sb s s s

b b s s

Q

1

3

21sw a

L np

out

T S

R SLM

C

1

' 0.5p

c

Qm D

nswT

1 a

cn

Sm

S

Dc gain H0

0

2

1

12

2

L

i L sw a

n

RH

R R T SM

SLM


Final Expression

The transfer function of a CCM Boost Converter in CM is

2

2

2

3

1 111 1

1 12 1221

C out

Lout L

outc i L sw a

sw an n p n

L n

Ls sr C

D RV s R

CV s R R T S s ssMT SSLM Q

R SLM

Sa is the external ramp compensation to damp subharmonic oscillations


Check SPICE Versus Mathcad

10 100 1 103

1 104

1 105

1 106

0

20

40

20 log H1 i 2 fk 10

fk

10 100 1 103

1 104

1 105

1 106

100

0

100

arg H1 i 2 fk 180

fk

Full-featured transfer functions versus SPICE

Simplified transfer functions versus SPICE

10 100 1 103

1 104

1 105

1 106

0

20

40

20 log H1 i 2 fk 10

20 log H2 i 2 fk 10

fk

10 100 1 103

1 104

1 105

1 106

100

0

100

arg H1 i 2 fk 180

arg H2 i 2 fk 180

fk

out

c

V s

V s

out

c

V s

V s

out

c

V s

V s

out

c

V s

V s


Course Agenda







The CM Boost Converter in DCM

To check the operating mode, calculate the critical load

2

2

2

1

sw outcrit

inin

out

F LVR

VV

V

1 MHz

5 µH

2.7 V

5 V

sw

in

out

F

L

V

V

74.7 ΩcritR

CCM, RL < Rcrit

DCM, RL > Rcrit

Li t

t

t

sw

L Ti t

,L peakI

,valleyLI

,L peakI

,valley 0LI deadtime


Transfer Function of DCM CM Boost

The converter is a first-order converter at low frequency

1 2

1 2

1 121 1

2 11 1

z zout

n c sw

p p

s s

V MH s

S m T D M s s

inn i

VS R

L

1

1z

C outr C

2 2load

z

R

M L

1

1 2 1

1p

L out

M

R C M

2

21

12p sw

MFD

1 ac

n

Sm

S


The Ac Response of the DCM CM Boost

10 100 1 103

1 104

1 105

1 106

40

20

0

20

40

20 log H3 i 2 fk 10

fk

10 100 1 103

1 104

1 105

1 106

100

0

100

arg H3 i 2 fk 180

fk

A RHPZ is still present in DCM with a high-frequency pole

SPICE simulation versus Mathcad

out

c

V s

V s

out

c

V s

V s


Why a RHP Zero in DCM?

The RHP Zero is present in CCM and is still there in DCM!

Tsw

D1Tsw

Id(t)

t

D2Tsw

D3Tsw

iL,peak

When D1 increases, [D1,D2] stays constant but D3 shrinks


Why a RHP Zero in DCM?

The triangle is simply shifted to the right by 1d

The capacitor refueling time is delayed and a drop occurs

Tsw

D1Tsw

t

D2Tsw

D3Tsw

1d

iL,peak

Id(t)


The Refueling Time is Shifted

If D increases, the diode current is delayed by 1d

2.02m 2.04m 2.06m 2.09m 2.11m

1.85m 1.99m 2.13m 2.27m 2.41m

2.02m 2.04m 2.06m 2.09m 2.11m

Vout(t)

Vout(t)

1d

Id(t)D(t)

Simulation, no peak increase


Course Agenda







EMI Filter Interaction

A LC filter is included to prevent the dc line pollution

gVloadR

outVL

C

inV

Switching converter

gVloadR

outVL

C

inV

Switching converter

C. Basso, “Designing Control Loops for Linear and Switching Power Supplies”, Artech House, 2012

What load does the converter offer?

?Z s


A Negative Incremental Resistance

Assume a 100%-efficient converter:

out inP P in in out outI V I V

In closed-loop operation, Pout is constant

2

out

in in in out

in in in

Pd

dI V V P

dV dV V

outin in

in

PI V

V

For a constant Pout , if Vin increases, Iin drops

10 15 20 25 300.2

0.4

0.6

0.8

1

VinV

AinI

10 15 20 25 300.2

0.4

0.6

0.8

1

VinV

AinI

The incremental input resistance is negative


A Simple LC Filter

The low-pass filter is built with a L and C components:

1

1z

Cr C

Lr LCr

C

gV

1

2

0 0

1

1

zsT s

s s

Q

0

1 L

C

r R

r RLC

0 C

load

L C L C

LC r R CQ R

L C r r r R r R L

If R is || with -R R R C

QR R L

R

A negative resistance cancels losses: poles become imaginary


A Negative Impedance Oscillator

If losses are compensated, the damping factor is zero

2

20 0

1

1

T ss s

Q

1

2Q

If ohmic losses are gone, the damping factor is zero, Q is infinite.

Without precautions, instability can happen!

3.64m 10.9m 18.2m 25.5m 32.7m

70.0

90.0

110

130

150

Output voltage

> 0 0

< 0

3.64m 10.9m 18.2m 25.5m 32.7m

70.0

90.0

110

130

150

> 0 0

< 0

Input voltage

Lr

inV

Cr

C

L

I

out

out

PI

V Negative resistance

(constant)


Filter Output Impedance

What is the output impedance of an LC filter?

L

Lr

Cr

C

R

TV s

TI s

0out

N sZ s R

D s

Ω ΩNo dimension

It follows the form

2

1 1

||

1 ||

C out

Lout L load

C loadL load C

L load L load

Ls sr C

rZ s r R

r RLs C r R r s LC

r R r R


Negative Resistance at Low Frequency Neg. resistance exists because of feedback (Pout = constant )

16

4

Cout200uF

Resr60m

7

Vin2.7

3 18

L15u

6

R1Rupper

R310k

1

V22.5

9

Verr

2

duty_cycle

vout

5

R41.5m

14

R610k

R71k

X2AMPSIMPVHIGH = 100VLOW = 10m

R91k

SwitchMode Power Supplies: SPICE Simulations and Practical DesignsChristophe Basso - McGraw-Hill, 2014

11

D1mbra210et3

13

Rdson140m

parameters

Rupper=10kfc=8k

pm=70Gfc=-20

pfc=-101

G=10^(-Gfc/20)boost=pm-(pfc)-90

pi=3.14159K=tan((boost/2+45)*pi/180)

C2=1/(2*pi*fc*G*k*Rupper)C1=C2*(K^2-1)

R2=k/(2*pi*fc*C1)

fp=1/(2*pi*R2*C2)fz=1/(2*pi*R2*C1)

15

R2R2

C1C1

C2C2

vca

cL

ossy

PW

M

swit

ch C

M

p

duty

-cyc

le

SW

D

PWMCM2_LX1L = 5uFs = 1MegRi = -40mSe = 16.4k

VZin

I1AC = 1

L210GH

Rload55.00V

5.00V

2.70V

2.70V2.70V

2.50V

2.50V

1.10V

536mV

99.9mV

-455mV-301mV

2.50V

2.70V


Negative Resistance at Low Frequency The resistance is truly negative up to 200 Hz

1 2

-24.0

-12.0

0

12.0

24.0

10 100 1k 10k 100k 1Meg

-144

-72.0

0

72.0

144

inZ f

inZ f

(dBΩ)

(°)

180inZ

1.02 Ω

In this regionPout constant


Negative Resistance is One Part Only

10 100 1k 10k 100k

-55.0

-25.0

5.00

35.0

65.0

36 dBΩ

outZ s

dBΩ

undamped

damped

3.6 /100 µF

10 100 1k 10k 100k

-55.0

-25.0

5.00

35.0

65.0

36 dBΩ

outZ s

dBΩ

undamped

damped

3.6 /100 µF

Negative resistance problems occur only when Beyond this region, instability is linked to gain degradation

180inZ

Negative argumentregion


Open-Loop Gain Degradation

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

-30.0

-10.0

10.0

30.0

50.0

p in

1 /

am

pe

re

1

7

Zout

No dampingZout = 57.5 dB

ZinSMPS

18 dB

Problem!


-30.0

-10.0

10.0

30.0

50.0

p in

1 /

am

pe

re

1

7

Zout

No dampingZout = 57.5 dB

ZinSMPS

18 dB

Problem!

10 100 1k 10k 100kfrequency in hertz

-40.0

-20.0

0

20.0

40.0

vdbout2

, vd

bout2

#1

in d

b(v

olts

)

-180

-90.0

0

90.0

180

ph_vo

ut2

#1,

ph_vo

ut2

in d

egre

es

Plo

t1

2

3

41

phase

gain

With filter

With filter

Without filter

Without filter

10 100 1k 10k 100kfrequency in hertz

-40.0

-20.0

0

20.0

40.0

vdbout2

, vd

bout2

#1

in d

b(v

olts

)

-180

-90.0

0

90.0

180

ph_vo

ut2

#1,

ph_vo

ut2

in d

egre

es

Plo

t1

2

3

41

phase

gain

With filter

With filter

Without filter

Without filter

Filter output impedanceversus closed-loop buck input impedance.

Buck open-loop gain withand without input filter.

T f

T f

Here the EMI filter peaks at high frequency The open-loop is severely affected and stability is at stake


Evaluating How Loop Gain is Modified

outZ s inZ s

ˆgv

ôutv

H s

G sd

The converter transfer function is evaluated at Zout = 0

0in

out

V s

V sT s H s G s

D s

ˆ 0inv For Zout = 0


In reality, the EMI output impedance makes The converter ac input voltage is no longer zero

0outZ s

outZ s ôutv

H s

G sd

ôutv

G sd

Considering the Filter Output Impedance

0outZ

0outZ

ˆ 0inv H s

0in

out

V s

V sT s

D s

0in

out

V s

V sT s

D s


Extra Element Theorem to Help

It can be shown how an EMI affects the open-loop gain:

ˆoutv

H s

G sd

outZ s

0out

out

Z s

V sT s

D s

Gain without the filter

outZ s Filter output impedance

0 0

1

1out out

out

out out N

outZ s Z s

D

Z s

V s V s Z sT s

Z sD s D s

Z s


What are ZD and ZN?

ZD and ZN come from the Extra Element Theorem, EET

0D i D s

Z s Z s

Open-loop input impedance

0out

N i V sZ s Z s

Input impedance for 0outV s

0 35%

ˆ 0

D

d

?DZ s

ˆ 0

out

out

V

v

0 35%

ˆ 0

D

d

ˆ 0

out

out

V

v

?NZ s

ˆ 0outi ˆ 0outi

Open-loop input impedance Input impedance for 0outV s Ideal control

Null


What are ZD and ZN for a Boost Converter?

These values have already been derived

2

2' 1

'N

sLZ s D R

D R

2

2 22

1' ''

1D

L LCs s

D R DZ s D RsRC

The original loop gain (without filter) is untouched if

1

1

1

out

N

out

D

Z s

Z s

Z s

Z s

out NZ s Z s

out DZ s Z s


Checking for Stability (1)

You design the filter together with the converter Plot |Zout|, |ZN | and |ZD| in the same graph Check that no overlap exists. If so, damp filter

1 10 100 1 103

1 104

1 105

1 106

60

40

20

0

20

40

60

20 logZout i 2 fk

10

20 logZD i 2 fk

10

20 logZN i 2 fk

10

fk

Need to redesign the filter!

NZ s

DZ s

outZ s


Checking for Stability (2)

You design the filter after the converter Plot |Zout|, |Zin | in the same graph Check that no overlap exists. If so, damp filter


-30.0

-10.0

10.0

30.0

50.0

7

654321

Zout

ZinSMPS

18 dB

Ok, margin is 8 dB min

Rdamp= 1

Rdamp= 2

Rdamp= 3

Rdamp= 4

Rdamp= 5


-30.0

-10.0

10.0

30.0

50.0

7

654321

Zout

ZinSMPS

18 dB

Ok, margin is 8 dB min

Rdamp= 1

Rdamp= 2

Rdamp= 3

Rdamp= 4

Rdamp= 5

Dampingnetwork

R damps

C cuts dc


Course Agenda







Cascading Converters

When cascading converters, impedances also matter

thZ s

thV s inZ s

Boost Buck

thV s

inV s

inV s+

-th inZ Z

The system can be modeled by a gain TM

1

1in th

th

in

V s V sZ s

Z s

MT s

outZ s inZ s


Check the Stability of the Minor Loop (1)

The stability test requires the comparison of Zout and Zin

The buck converter closed-loop input impedance is

1 1 1 1

1 1in N D

T s

Z s Z s T s Z s T s

0D i D s

Z s Z s

Open-loop input impedance

0out

N i V sZ s Z s

Input impedance for 0outV s

ZN and ZD are defined as

2

2

1

1D

Ls s LC

R RZ sD sRC

2N

RZ s

D

in NZ s Z sLow frequency

1T s

T s H s G s


Use SPICE Models First Simulate the boost converter output impedance

16

C5

820u

R10

18m3

Vin

11

12

L1

33u

1

R1

11.66k

R3

1.35k

6

8

X2

AMPSIMP

VHIGH = 10

VLOW = 10m

V2

2.5

Verr

2

duty_cycle

5

Rload

8.3

voutvout

11

vca

c

PW

M s

witc

h C

Mp

duty

-cyc

le

X3

PWMCM

L = 33u

Fs = 200k

Ri = -300m

Se = 82k

9

R2

50k

C2

10n

R5

7m

C3

282p

R4

1m

I1

AC = 1A

11-15 V/24 V – 3 A

1 10 100 1k 10k 100k 1Meg

-80.0

-60.0

-40.0

-20.0

0 outZ f

(dBΩ)

Closed-loop Zout


Use SPICE Models First Simulate the buck converter input impedance

Vout

2

Cout

820u

Resr

14m1

V4

24

3 4

L1

8u

7

5

duty_cycle

R4

4m

R2

416m

vout

vc

a c

PWM switch CM p

duty-cyc le

6

PWMCM

X2

L = 8u

Fs = 200k

Ri = 176m

Se = 30k

9

R1

1k

R3

1k

10

V2

2.5

11

Verr

vout

8

R6

15k

C1

3.5n

C3

1n

R7

1m

L2

1kH

Zin

I1

AC = 1

5.00V

5.00V

24.0V

5.05V 5.05V

210mV

24.0V

2.37V

2.50V

2.50V

2.37V

2.37V

24 V/5 V – 12 A

Closed-loop Zin

(dBΩ)

10 100 1k 10k 100k 1Meg

18.2

18.6

19.0

19.4

19.8

inZ f


Compare Magnitude Curves

10 100 1k 10k 100k 1Meg

-41.0

-21.0

-1.00

19.0

39.0

(dBΩ)

inZ f

outZ f

No overlap


Compare Phase Curves

1 10 100 1k 10k 100k 1Meg

-90.0

0

90.0

180

270

inZ f

outZ f

(°)

Neg. up to 1 kHz


Check Minor Loop Gain

1 10 100 1k 10k 100k 1Meg

-360

-270

-180

-90.0

0

-80.0

-70.0

-60.0

-50.0

-40.0

dB °

out

in

Z f

Z f

out

in

Z f

Z f

There is no gain, . The system is stableout inZ Z


With Cascaded Converters

16

4

C5

820u

R10

18m

3

Vin

11

12

L1

33u

1

R1

11.66k

R3

1.35k

6

7

X2

AMPSIMP

VHIGH = 10

VLOW = 10m

V2

2.5

VerrBoost

2

dc_boost

VoBoost

11

vca

c

PW

M s

witc

h C

Mp

du

ty-c

ycle

X3

PWMCM

L = 33u

Fs = 200k

Ri = -300m

Se = 82k

9

R2

50k

C2

10n

R5

7m

C3

282p

VoBuck

13

Cout

820u

19

Resr

14m

14 15

L2

8u

18

10

dc_buck

R4

4m v out

vc

a c

PWM switch CM p

duty-cycle

PWMCM

X1

L = 8u

Fs = 200k

Ri = 176m

Se = 30k

21

R9

1k

R11

1k

22

V3

2.5

23

VerrBuck

v out

24

R6

15k

C1

3.5n

C6

1n

R7

1m

I1

unknown

Iout

24.1V

24.1V

11.0V

11.0V

2.50V

2.50V

2.02V

545mV

11.0V

2.50V

5.00V

5.00V

5.05V 5.05V

209mV

2.37V

2.50V

2.50V

2.37V

2.37VBoost

Buck


Check Transient Response

4.90

4.95

5.00

5.05

5.10

100u 300u 500u 700u 900u

23.90

23.95

24.00

24.05

24.10

Transient response is good for the buck

(V)

(V)

outv t

outv t

Buck

Boost

Iout from 5 to 12 A in 1 µs


Literature

“Fundamentals of Power Electronics”, R. Erickson, D. Maksimovic

“Designing Control Loops for Linear and Switching Power Supplies”, C. Basso

“Practical Issues of Input/Output Impedance Measurements”, Y. Panov,

M. Jovanović, IEEE Transactions on Power Electronics, 2005

“Physical Origins of Input Filter Oscillations in Current Programmed Converters”

, Y. Jang, R. Erickson, IEEE Transactions on Power Electronics, 1992

“Design Consideration for a Distributed Power System”, S. Schultz, B. Cho,

F. Lee, Power Electronics Specialists Conference, June 1990, pp. 611-617

“Input Filter Considerations in Design and Applications of Switching Regulators”,

R. D. Middlebrook, IAS 1976


Conclusion

The PWM switch model is an essential tool for modeling

It can be used to derive small-signal response of a boost in CM

Its SPICE implementation helps test various transfer functions

It can be used to assess open- and closed-loop parameters

Interactions with the EMI input filter can be analyzed

Cascaded converters stability can be checked with the model

Practical measurements on the bench must always be performed

Merci !Thank you!

Xiè-xie!

small-signal modeling of the boost converter operated in...

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