Ch3slides.pdf

Chapter 3. Steady-State Equivalent Circuit Modeling, Losses, and Efficiency

3.1. The dc transformer model 3.2. Inclusion of inductor copper loss 3.3. Construction of equivalent circuit model 3.4. How to obtain the input port of the model 3.5. Example: inclusion of semiconductor conduction losses in the boost converter model 3.6. Summary of key points

Fundamentals of Power Electronics

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Chapter 3: Steady-state equivalent circuit modeling, ...

3.1. The dc transformer model Ig

Basic equations of an ideal dc-dc converter: Pin = Pout

Power

(η = 100%)

input

Vg I g = V I

V = M(D) Vg

I

+

Switching

+

Vg

dc-dc

V

–

converter

–

Power output

D

(ideal conversion ratio)

I g = M(D) I Control input

These equations are valid in steady-state. During transients, energy storage within filter elements may cause Pin ≠ Pout Fundamentals of Power Electronics

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Chapter 3: Steady-state equivalent circuit modeling, ...

Equivalent circuits corresponding to ideal dc-dc converter equations Pin = Pout

Vg I g = V I

V = M(D) Vg

Dependent sources

I g = M(D) I

DC transformer Ig

Ig

1 : M(D)

+

I

I + Power

Power

Power input

+ Vg

+ M(D)I

M(D)Vg

–

+ –

V

input

Power output

–

Vg

V

–

–

output

D Control input

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Chapter 3: Steady-state equivalent circuit modeling, ...

The DC transformer model Ig

1 : M(D)

+

I + Power

Power input

Vg

V

–

– D

output

Models basic properties of ideal dc-dc converter: • conversion of dc voltages and currents, ideally with 100% efficiency • conversion ratio M controllable via duty cycle

Control input

• Solid line denotes ideal transformer model, capable of passing dc voltages and currents • Time-invariant model (no switching) which can be solved to find dc components of converter waveforms Fundamentals of Power Electronics

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Chapter 3: Steady-state equivalent circuit modeling, ...

Example: use of the DC transformer model 1. Original system

3. Push source through transformer

R1

M 2 (D) R1

+

+

+

Switching V1

+ –

dc-dc

Vg

V

R

M (D) V1 + –

converter –

–

V

R

–

D

4. Solve circuit

2. Insert dc transformer model R1

V1

+ –

1 : M(D) +

+

Vg

V

–

–

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V = M(D) V1

R R + M 2(D) R 1

R

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Chapter 3: Steady-state equivalent circuit modeling, ...

3.2. Inclusion of inductor copper loss Dc transformer model can be extended, to include converter nonidealities. Example: inductor copper loss (resistance of winding): L

RL

Insert this inductor model into boost converter circuit: L

RL

2 +

i Vg

1

+ –

C

R

v –

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Chapter 3: Steady-state equivalent circuit modeling, ...

Analysis of nonideal boost converter L

RL

2 +

i Vg

1

+ –

C

v

R

–

switch in position 1 i

L

RL

+ vL – Vg

switch in position 2

+ –

i

+

iC C

R

+ vL – Vg

v

+ –

RL +

iC C

R

v –

–

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L

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Chapter 3: Steady-state equivalent circuit modeling, ...

Circuit equations, switch in position 1

Inductor current and capacitor voltage: vL(t) = Vg – i(t) RL

L

i

RL

+ vL – Vg

iC(t) = –v(t) / R

+ –

+

iC C

R

v –

Small ripple approximation: vL(t) = Vg – I RL iC(t) = –V / R

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Chapter 3: Steady-state equivalent circuit modeling, ...

Circuit equations, switch in position 2 i

L

RL

+ vL – Vg

+

iC

+ –

C

R

v –

vL(t) = Vg – i(t) RL – v(t) ≈ Vg – I RL – V iC(t) = i(t) – v(t) / R ≈ I – V / R

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Chapter 3: Steady-state equivalent circuit modeling, ...

Inductor voltage and capacitor current waveforms

Average inductor voltage:

vL(t)

Vg – IRL

T

s vL(t) = 1 v (t)dt Ts 0 L = D(Vg – I RL) + D'(Vg – I RL – V)

Inductor volt-second balance:

DTs

D'Ts t Vg – IRL – V

iC(t)

I – V/R

0 = Vg – I RL – D'V

Average capacitor current:

–V/R

t

iC(t) = D ( – V / R) + D' (I – V / R)

Capacitor charge balance: 0 = D'I – V / R

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Chapter 3: Steady-state equivalent circuit modeling, ...

Solution for output voltage 5

We now have two equations and two unknowns:

3.5

0 = D'I – V / R

V = 1 1 Vg D' (1 + RL / D' 2R)

RL /R = 0.01

4

0 = Vg – I RL – D'V

RL /R = 0.02

3

V/ Vg

Eliminate I and solve for V:

RL /R = 0

4.5

2.5 2

RL /R = 0.05

1.5

RL /R = 0.1

1 0.5 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

D

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Chapter 3: Steady-state equivalent circuit modeling, ...

3.3. Construction of equivalent circuit model Results of previous section (derived via inductor volt-sec balance and capacitor charge balance): vL = 0 = Vg – I RL – D'V iC = 0 = D'I – V / R

View these as loop and node equations of the equivalent circuit. Reconstruct an equivalent circuit satisfying these equations

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Chapter 3: Steady-state equivalent circuit modeling, ...

Inductor voltage equation vL = 0 = Vg – I RL – D'V

• Derived via Kirchhoff’s voltage law, to find the inductor voltage during each subinterval

Vg

• Average inductor voltage then set to zero • This is a loop equation: the dc components of voltage around a loop containing the inductor sum to zero

+ –

L

RL

+ 〈vL 〉 – =0

+ IRL –

I

+ –

D'V

• IRL term: voltage across resistor of value RL having current I • D’V term: for now, leave as dependent source

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Chapter 3: Steady-state equivalent circuit modeling, ...

Capacitor current equation Node

iC = 0 = D'I – V / R

V/R

• Derived via Kirchoff’s current law, to find the capacitor current during each subinterval

D'I

• Average capacitor current then set to zero

〈iC 〉 =0

+

C

V

R

–

• This is a node equation: the dc components of current flowing into a node connected to the capacitor sum to zero

• V/R term: current through load resistor of value R having voltage V • D’I term: for now, leave as dependent source

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Chapter 3: Steady-state equivalent circuit modeling, ...

Complete equivalent circuit Dependent sources and transformers

The two circuits, drawn together:

I1

RL

+ +

Vg

+ –

D'V

I

+ –

nV2 V

D'I

R

+ –

nI1 V2 –

–

n:1

The dependent sources are equivalent to a D′ : 1 transformer: RL

+ –

+

– V

R

• sources have same coefficient • reciprocal voltage/current dependence

–

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+ V2

D' : 1

I Vg

I1

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Chapter 3: Steady-state equivalent circuit modeling, ...

Solution of equivalent circuit Converter equivalent circuit RL

D' : 1 +

I Vg

+ –

V

R

–

Refer all elements to transformer secondary:

Solution for output voltage using voltage divider formula:

RL /D' 2

D'I Vg /D'

+ –

+

V= V

R

Vg D'

R R+

RL D' 2

=

Vg D'

1 1+

RL D' 2 R

– Fundamentals of Power Electronics

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Chapter 3: Steady-state equivalent circuit modeling, ...

Solution for input (inductor) current RL

D' : 1 +

I Vg

+ –

V

R

–

Vg Vg 1 I= = 2 2 D' R + RL D' 1 + RL D' 2 R

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Chapter 3: Steady-state equivalent circuit modeling, ...

Solution for converter efficiency RL

Pin = (Vg) (I)

Pout = (V) (D'I)

+

I Vg

D' : 1

+ –

V

R

–

η=

Pout (V) (D'I) V = = D' Pin Vg (Vg) (I)

1

η= 1+

RL D' 2 R

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Chapter 3: Steady-state equivalent circuit modeling, ...

Efficiency, for various values of RL 100%

η=

1

90%

R 1 + 2L D' R

80%

0.002 0.01 0.02

70%

0.05

60%

η

50%

RL /R = 0.1

40% 30% 20% 10% 0% 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

D

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Chapter 3: Steady-state equivalent circuit modeling, ...

3.4. How to obtain the input port of the model Buck converter example —use procedure of previous section to derive equivalent circuit ig

1

L

iL

RL +

+ vL – Vg

+ –

2 C

vC

R

–

Average inductor voltage and capacitor current: vL = 0 = DVg – I LRL – VC

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iC = 0 = I L – VC/R

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Chapter 3: Steady-state equivalent circuit modeling, ...

Construct equivalent circuit as usual

vL = 0 = DVg – I LRL – VC

iC = 0 = I L – VC/R

RL +

+ 〈vL〉 – =0 DVg

+ –

〈iC〉 =0

IL

VC

VC /R R

–

What happened to the transformer? • Need another equation

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Chapter 3: Steady-state equivalent circuit modeling, ...

Modeling the converter input port Input current waveform ig(t): ig(t)

iL (t) ≈ IL area = DTs IL 0 0

Ts

DTs

t

Dc component (average value) of ig(t) is Ig = 1 Ts

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Ts

ig(t) dt = DI L 0

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Chapter 3: Steady-state equivalent circuit modeling, ...

Input port equivalent circuit Ig = 1 Ts

Ts

ig(t) dt = DI L 0

Vg

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+ –

Ig

DIL

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Chapter 3: Steady-state equivalent circuit modeling, ...

Complete equivalent circuit, buck converter Input and output port equivalent circuits, drawn together: Ig

IL

RL +

Vg

+ –

+ –

DIL

VC

DVg

R

–

Replace dependent sources with equivalent dc transformer: Ig

IL

1:D

RL +

Vg

+ –

VC

R

– Fundamentals of Power Electronics

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Chapter 3: Steady-state equivalent circuit modeling, ...

3.5. Example: inclusion of semiconductor conduction losses in the boost converter model Boost converter example L

i

+ iC Vg

+ – DTs

Ts

C

+ –

R

v –

Models of on-state semiconductor devices: MOSFET: on-resistance Ron Diode: constant forward voltage VD plus on-resistance RD Insert these models into subinterval circuits

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Chapter 3: Steady-state equivalent circuit modeling, ...

Boost converter example: circuits during subintervals 1 and 2 L

i

+ iC Vg

+ – DTs

Ts

C

+ –

RL

+ vL – Vg

+ –

switch in position 2 i +

iC Ron

C

R

v –

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L

RL

RD

+ –

L

v –

switch in position 1 i

R

+ vL – Vg

+ –

VD

+

iC C

R

v –

Chapter 3: Steady-state equivalent circuit modeling, ...

Average inductor voltage and capacitor current vL(t)

Vg – IRL – IRon DTs

D'Ts t Vg – IRL – VD – IRD – V

iC(t) I – V/R t –V/R

vL = D(Vg – IRL – IRon) + D'(Vg – IRL – VD – IRD – V) = 0 iC = D(–V/R) + D'(I – V/R) = 0 Fundamentals of Power Electronics

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Chapter 3: Steady-state equivalent circuit modeling, ...

Construction of equivalent circuits Vg – IRL – IDRon – D'VD – ID'RD – D'V = 0 D'VD

DRon

D'RD

+ –

RL

+ IRL – + IDRon – Vg

+ –

+ ID'RD – + –

I

D'I – V/R = 0

D'V

V/R + D'I

V

R

–

Fundamentals of Power Electronics

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Chapter 3: Steady-state equivalent circuit modeling, ...

Complete equivalent circuit D'VD

DRon

D'RD

+ –

RL

Vg

+ –

+ D'V

I

+ –

D'I

V

R

–

D'VD

DRon

D'RD

+ –

RL

Vg

+ –

D' : 1 + V

I

R

– Fundamentals of Power Electronics

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Chapter 3: Steady-state equivalent circuit modeling, ...

Solution for output voltage D'VD

DRon

D'RD

D' : 1

+ –

RL

Vg

+ –

+ V

I

R

–

V= 1 D'

Vg – D'VD

V = 1 Vg D'

1–

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D'VD Vg

D' 2R D' 2R + RL + DRon + D'RD

1 R + DRon + D'RD 1+ L D' 2R

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Chapter 3: Steady-state equivalent circuit modeling, ...

Solution for converter efficiency D'VD

DRon

+ –

RL

Pin = (Vg) (I) Vg

+ –

I

Pout = (V) (D'I)

1+

D' : 1 + V

R

–

1–

η = D' V = Vg

D'RD

D'VD Vg

RL + DRon + D'RD D' 2R

Conditions for high efficiency:

Vg/D' > V D D' 2R > RL + DRon + D'RD Fundamentals of Power Electronics

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Chapter 3: Steady-state equivalent circuit modeling, ...

Accuracy of the averaged equivalent circuit in prediction of losses • Model uses average currents and voltages • To correctly predict power loss in a resistor, use rms values • Result is the same, provided ripple is small

MOSFET current waveforms, for various ripple magnitudes: i(t)

2I (c) (b) (a)

I

0 DTs

0 Inductor current ripple (a) ∆i = 0 (b) ∆i = 0.1 I (c) ∆i = I

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1.1 I

MOS FET rms current I

D

(1.00167) I (1.155) I

32

Ts

t

Average power loss in R on D I2 R on

D D

(1.0033) D I2 R on (1.3333) D I2 R on

Chapter 3: Steady-state equivalent circuit modeling, ...

Summary of chapter 3 1. The dc transformer model represents the primary functions of any dc-dc converter: transformation of dc voltage and current levels, ideally with 100% efficiency, and control of the conversion ratio M via the duty cycle D. This model can be easily manipulated and solved using familiar techniques of conventional circuit analysis. 2. The model can be refined to account for loss elements such as inductor winding resistance and semiconductor on-resistances and forward voltage drops. The refined model predicts the voltages, currents, and efficiency of practical nonideal converters. 3. In general, the dc equivalent circuit for a converter can be derived from the inductor volt-second balance and capacitor charge balance equations. Equivalent circuits are constructed whose loop and node equations coincide with the volt-second and charge balance equations. In converters having a pulsating input current, an additional equation is needed to model the converter input port; this equation may be obtained by averaging the converter input current. Fundamentals of Power Electronics

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Chapter 3: Steady-state equivalent circuit modeling, ...