Lecture 9 More on modeling DCMecee.colorado.edu/~ecen5807/course_material/slides2-5.pdf · 2015. 1. 5. · CCM/DCM averaged switch model For DCM: 1–µ : µ The CCM averaged switch

Fundamentals of Power Electronics 1Chapter 11: AC and DC equivalent circuit modelingof the discontinuous conduction mode

Lecture 9More on modeling DCM

11.2.1 Frequency response example: boost converter

Appendix B.2

Switch conversion ratio µ

Combined CCM-DCM simulation model

SEPIC example

11.3. High-Frequency Dynamics of Converters in DCM


Small-signal DCM switch model parameters

–

+

+–v1 r1 j1d g1v2

i1

g2v1 j2d r2

i2

v2

Tabl e 11.2. Small -signal DCM switch model parameters

Switch type g1 j1 r1 g2 j2 r2

Buck,Fig. 11.16(a)

1Re

2(1 – M)V1

DRe

Re 2 – MMRe

2(1 – M)V1

DMRe

M 2Re

Boost,Fig. 11.16(b)

1(M – 1)2 Re

2MV1

D(M – 1)Re

(M – 1)2

MRe

2M – 1(M – 1)2 Re

2V1

D(M – 1)Re

(M – 1)2Re

General two switch,Fig. 11.7(a)

0 2V1

DRe

Re 2

MRe 2V1

DMRe

M 2Re


Small-signal ac model, DCM buck-boost example

+

–

+–

v1 r1 j1d g1v2

i1

g2v1 j2d r2

i2

v2

–

+

L

C R

Switch network small-signal ac model

+

–vg v

iL


A more convenient way to model the buck and boostsmall-signal DCM switch networks

+

v2(t)

–

i1(t) i2(t)

+

v1(t)

–

+

v2(t)

–

i1(t) i2(t)

+

v1(t)

–

+

–

+

–

v1 r1 j1d g1v2

i1

g2v1 j2d r2

i2

v2

In any event, a small-signal two-port model is used, of the form


Small-signal ac models of the DCM buck and boostconverters (more convenient forms)

+

–

+– v1 r1 j1d g1v2

i1

g2v1 j2d r2

i2

v2

+

–

L

C R

DCM buck switch network small-signal ac model

+

–

vg v

iL

+

–

+– v1 r1 j1d g1v2

i1

g2v1 j2d r2

i2

v2

+

–

L

C R

DCM boost switch network small-signal ac model

+

–

vg v

iL


DCM small-signal transfer functions

l When expressed in terms of R, L, C, and M (not D), the small-signal transfer functions are the same in DCM as in CCM

l Hence, DCM boost and buck-boost converters exhibit two polesand one RHP zero in control-to-output transfer functions

l But , value of L is small in DCM. Hence

RHP zero appears at high frequency, usually greater thanswitching frequency

Pole due to inductor dynamics appears at high frequency, nearto or greater than switching frequency

So DCM buck, boost, and buck-boost converters exhibitessentially a single-pole response

l A simple approximation: let L → 0


The simple approximation L → 0

Buck, boost, and buck-boost converter models all reduce to

+

–

+– r1 j1d g1v2 g2v1 j2d r2 C R

DCM switch network small-signal ac model

vg v

Transfer functions

Gvd(s) =v

dvg = 0

=Gd0

1 + sωp

Gd0 = j2 R || r2

ωp = 1R || r2 C

Gvg(s) =v

vg d = 0

=Gg0

1 + sωp

Gg0 = g2 R || r2 = M

withcontrol-to-output

line-to-output


Transfer function salient features

Tabl e 11.3. Salient features of DCM converter small -signal transfer functions

Converter Gd0 Gg0 ωp

Buck 2VD

1 – M2 – M M 2 – M

(1 – M)RC

Boost 2VD

M – 12M – 1 M 2M – 1

(M– 1)RC

Buck-boost VD M 2

RC


11.2.1 DCM boost example

R = 12 Ω

L = 5 µH

C = 470 µF

fs = 100 kHz

The output voltage is regulated to be V = 36 V. It is desired to determine Gvd(s) at the

operating point where the load current is I = 3 A and the dc input voltage is Vg = 24 V.

PRe(D)+– R

+

V

–

Vg

Steady-state DCM boost model


Evaluate simple model parameters

P = I V – Vg = 3 A 36 V – 24 V = 36 W

Re =V g

2

P=

(24 V)2

36 W= 16 Ω

D = 2LReTs

=2(5 µH)

(16 Ω)(10 µs)= 0.25

Gd0 = 2VD

M – 12M – 1

=2(36 V)(0.25)

(36 V)(24 V)

– 1

2(36 V)(24 V)

– 1

= 72 V ⇒ 37 dBV

fp =ωp

2π = 2M – 12π(M– 1)RC

=

2(36 V)(24 V)

– 1

2π (36 V)(24 V)

– 1 (12 Ω)(470 µF)= 112 Hz


Control-to-output transfer function, boost example

–20 dB/decade

fp112 Hz

Gd0 ⇒ 37 dBV

f

0˚0˚

–90˚

–180˚

–270˚

|| Gvd ||

|| Gvd || ∠ Gvd

0 dBV

–20 dBV

–40 dBV

20 dBV

40 dBV

60 dBV

∠ Gvd

10 Hz 100 Hz 1 kHz 10 kHz 100 kHz


B.2 Combined CCM-DCM model

+

v2(t)

–

+

v1(t)

–

i1(t) i2(t)

i2(t) Ts

+

–

v2(t) Tsv1(t) Ts

i1(t) Ts

Re(d)

+

–

p(t)Ts

i2(t) Ts

+

–

v2(t) Tsv1(t) Ts

i1(t) Ts

+

–

DCMaveraged switch model

CCMaveraged switch model

d´ : dCan we write one averaged circuitmodel that can be used insimulation for both CCM and DCM?


Switch conversion ratio µ

Define the switch conversion ratio µ:In CCM: µ = d (so µ can be thought of as an effective duty cycle)

In DCM: Define µ consistently so that the terminal characteristics ofthe switch network are correctly modeled.

Port 1 characteristic:

v1(t) Ts=

d′(t)d(t)

v2(t) Ts

Letting d = µ, we obtain:

v1(t) Ts=

1 – µµ v2(t) Ts


Derivation of µ for DCM

The port 1 characteristic for DCM is

i1(t) Ts=

v1(t) Ts

Re(d1)

Equate to CCM characteristic and solve for µ:

v1(t) Ts=

1 – µµ v2(t) Ts

= Re i1(t) Ts

µ = 1

1 +Re i1(t) Ts

v2(t) Ts

(can verify that the same result isobtained by matching the port 2characteristics)


Using the switch conversion ratio µ

µ = 1

1 +Re i1(t) Ts

v2(t) Ts

i2(t) Ts

+

–

v2(t) Tsv1(t) Ts

i1(t) Ts

+

–

CCM/DCMaveraged switch model

1–µ : µFor DCM:

The CCM averaged switch model is used for DCM, using the switchconversion ratio µ. Effectively, the turns ratio becomes a function of theapplied voltages and currents.

Equations derived for CCM can be applied to DCM as well, providedthat d is replaced by µ.


An easy way to compute µ

µ = d

µ = 1

1 +Re i1(t) Ts

v2(t) Ts

CCM:

DCM: d ≤ µ ≤ 1

• When DCM is entered at light load, the current i1 is small

• Denominator of DCM equation tends to 1 as i1 approaches zero

• At CCM-DCM boundary, both equations give the same result

• Can compute µ using both formulas, and simply select largest


Simulation model

*********************************************************** MODEL: CCM-DCM1* Application: two-switch PWM converters, CCM or DCM* Limitations: ideal switches, no transformer*********************************************************** Parameters:* L=equivalent inductance for DCM* fs=switching frequency*********************************************************** Nodes:* 1: transistor positive (drain of an n-channel MOS)* 2: transistor negative (source of an n-channel MOS)* 3: diode cathode* 4: diode anode* 5: duty cycle control input**********************************************************.subckt CCM-DCM1 1 2 3 4 5+ params: L=100u fs=1E5Et 1 2 value=(1-v(u))*v(3,4)/v(u)Gd 4 3 value=(1-v(u))*i(Et)/v(u)Ga 0 a value=MAX(i(Et),0)Va a bRa b 0 1kEu u 0 table MAX(v(5), v(5)*v(5)/(v(5)*v(5)+2*L*fs*i(Et)/v(3,4))) (0 0) (1 1).ends**********************************************************


SEPIC exampleControl-to-output frequency response

+–

D1L1

C2

+

v

–Q1

C1

L2

R

Vg

RL1

RL2

100 µH

500 µH47 µF

200 µF

0.02 Ω

0.1 Ω

120 V

D = 0.4

load

fs = 100 kHz

Load resistance R:

40 Ω (CCM) 50 Ω (DCM)


PSPICE model

d

+–

L1

C2

+

v

–

C1

L2

R

Vg

RL1

RL2

100 µH

500 µH47 µF

200 µF

0.02 Ω

0.1 Ω

120V load

+–

vc

1

2

4

35

CCM-DCM1

1 2 3 4

5

0

Xswitch

L = 83.3 µHfs = 100 kHz

Re =2 L1||L2

d2Ts

SEPIC frequency response.param RL=40.op.step lin param RL 40 50 10.probe* converter netlistVg 1 0 dc 120VL1 1 2x 500uHRl1 2x 2 0.1C1 2 3 47uFL2 3x 3 100uHRl2 0 3x 0.02C2 4 0 200uFRload 4 0 RL* duty cycle inputvc 5 0 dc 0.4 ac 1* subcircuitXswitch 2 0 4 3 5 CCM-DCM1+PARAMS: L=83.33uH fs=100K.lib switch.lib* analysis setup:.ac DEC 201 5 50KHz.end


Results: SEPIC example

f

0˚

–90˚

–180˚

–270˚

|| Gvd ||

|| Gvd || ∠ Gvd

0 dBV

–20 dBV

–40 dBV

20 dBV

40 dBV

60 dBV

∠ Gvd

5 Hz 50 Hz 5 kHz 50 kHz500 Hz

80 dBV

R = 40Ω

R = 40Ω

R = 50Ω

R = 50Ω


11.3 High frequency dynamicsof converters operating in DCM

Basic averaged equations of switch network terminal waveforms:

v1(t) Ts= 1 – d1(t) vg(t) Ts

– d2(t) v(t)Ts

v2(t) Ts= d1(t) vg(t) Ts

– v(t)Ts

+ d2(t)⋅0 + d3(t) – v(t)Ts

= d1(t) vg(t) Ts– 1 – d2(t) v(t)

Ts

i1(t) Ts=

d 12(t)Ts

2Lvg(t) Ts

i2(t) Ts=

d1(t)d2(t)Ts

2Lvg(t) Ts

Now eliminate d2 without making equilibrium assumption (i.e., withoutassuming volt-second balance on L:

vL(t)Ts

= d1(t) vg(t) Ts+ d2(t) v(t)

Ts= 0


Find average inductor currentto eliminate d2

t

iL(t)

d1Ts

Ts

0

ipk

vg

L

vL

d2Ts d3Ts

⟨ iL(t) ⟩t

iL(t)Ts

= 12

i pk d(t) + d2(t) =d(t) d(t) + d2(t) Ts

2Lvg(t) Ts

d2(t) =2L iL(t)

Ts

d(t)Ts vg(t) Ts

– d(t) =Re(d) iL(t)

Ts

vg(t) Ts

– 1 d(t)Solve for d2:

(Buck-boost example)


Eliminate d2 from switch network equations

v1(t) Ts= 1 – d(t) vg(t) Ts

+ d(t) v(t)Ts

–Re(d) iL(t)

Tsv(t)

Tsd(t)

vg(t) Ts

which is a function of the form:

v1(t) Ts= γ1 vg(t) Ts

, v(t)Ts

, iL(t)Ts

, d(t)

Similar manipulation for ⟨ i2(t) ⟩t:

i2(t) Ts= iL(t)

Ts–

vg(t) Ts

Re= γ2 vg(t) Ts

, iL(t)Ts

, d(t)

Note that we can’t eliminate vg from these equations!


Perturb and linearize

v1(t) = vg(t)k g + v(t)k v + i Lr1 + d (t) f1

i 2(t) = vg(t)gg + i Lh 2 + d (t) j2

+

–

+–

L

C R

v1

i1 kgvg

h2iL

i2

vg iL v

+ – + – + – + –

kvv r1iL f1d

+ –

j2d

ggvg

v2+ –

Resulting model (buck-boost example):


Results for basic converters

Table 11.4 High-frequency pole and RHP zero of the DCM converter control-to-output transfer function Gvd(s)

Converter High-frequency pole ω2 RHP zero ωz

Buck none

Boost

Buck–boost

2M fsD(1 – M)

2(M – 1) fsD

2 fsD

2 M fsD

2 fsD


Boost example revisitedControl-to-output transfer function

Quantity Averaged switch model HF model

Low freq pole 112 Hz 112 Hz

High freq pole 56 kHz 64 kHz

RHP zero 170 kHz 127 kHz

Note that neither model accounts for aliasing and other high-frequencyphenomena

fs = 100 kHz

Documents

Lecture 9 More on modeling DCMecee.colorado.edu/~ecen5807/course_material/slides2-5.pdf · 2015. 1. 5. · CCM/DCM averaged switch model For DCM: 1–µ : µ The CCM averaged switch