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
Basic equations of an ideal
dc-dc converter:
Pin = Pout
(η = 100%)
Ig
Power
input
Vg I g = V I
+
Vg
–
I
Switching
dc-dc
converter
+
V
–
Power
output
D
V = M(D) Vg
(ideal conversion ratio)
control input
I g = M(D) I
These equations are valid in steady-state. During
transients, energy storage within filter elements may cause
Pin ≠ Pout
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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
Dc transformer
Ig
Power
input
+
Vg
–
Ig
I
M(D) I
M(D)Vg
I g = M(D) I
+
–
+
V
–
Power
input
Power
output
1 : M(D)
+
Vg
–
I
+
V
–
Power
output
D
control input
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Chapter 3: Steady-state equivalent circuit modeling, ...
The dc transformer model
Ig
Power
input
1 : M(D)
+
Vg
–
I
+
V
–
Power
output
Models basic properties of
ideal dc-dc converter:
• conversion of dc voltages
and currents, ideally with
100% efficiency
D
• 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
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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
M 2(D)R1
R1
V1
Switching
dc-dc
converter
+
Vg
–
+
–
+
V
–
+
R
M(D)V1
–
+
V
–
R
D
2. Insert dc transformer model
R1
+
V1
–
4. Solve circuit
1 : M(D)
+
Vg
–
Fundamentals of Power Electronics
+
V
–
V = M(D) V1
R
R + M 2(D) R 1
R
5
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
–
Fundamentals of Power Electronics
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Chapter 3: Steady-state equivalent circuit modeling, ...
Analysis of nonideal boost converter
L
RL
2
+
i
Vg
1
+
–
C
R
v
–
switch in position 1
i
L
RL
+ vL –
Vg
switch in position 2
+
–
i
+
iC
C
R
+ vL –
v
Vg
–
Fundamentals of Power Electronics
L
+
–
RL
+
iC
C
R
v
–
7
Chapter 3: Steady-state equivalent circuit modeling, ...
Circuit equations, switch in position 1
Inductor current and
capacitor voltage:
vL(t) = Vg – i(t) RL
iC(t) = –v(t) / R
L
i
+ vL –
Vg
+
–
RL
+
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
1
vL(t) =
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
-V/R
Average capacitor current:
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
We now have two
equations and two
unknowns:
5
RL / R = 0
4.5
RL / R = 0.01
4
0 = Vg – I RL – D'V
3.5
0 = D'I – V / R
3
Eliminate I and
solve for V:
V = 1
1
Vg D' (1 + RL / D' 2R)
V / Vg
RL / R = 0.02
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
Fundamentals of Power Electronics
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Chapter 3: Steady-state equivalent circuit modeling, ...
Inductor voltage equation
vL = 0 = Vg – I RL – D'V
L
• Derived via Kirchoff’s voltage
law, to find the inductor voltage
during each subinterval
• Average inductor voltage then
set to zero
RL
+ <vL> – + IRL –
=0
Vg
• This is a loop equation: the dc
components of voltage around
a loop containing the inductor
sum to zero
+
–
I
+
–
D' V
• IRL term: voltage across resistor
of value RL having current I
• D’V term: for now, leave as
dependent source
Fundamentals of Power Electronics
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Chapter 3: Steady-state equivalent circuit modeling, ...
Capacitor current equation
iC = 0 = D'I – V / R
node
• Derived via Kirchoff’s current
law, to find the capacitor
current during each subinterval
V/R
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
I
Vg
+
–
+
V2
D' : 1
+
–
V
R
• sources have same coefficient
• reciprocal voltage/current
dependence
–
Fundamentals of Power Electronics
I1
15
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:
R / D' 2
Solution for output voltage
using voltage divider formula:
L
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
–
I=
Fundamentals of Power Electronics
Vg
Vg
1
=
2
2
D' R + RL D' 1 + RL
D' 2 R
17
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
Fundamentals of Power Electronics
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Chapter 3: Steady-state equivalent circuit modeling, ...
Efficiency, for various values of RL
100%
1
η=
1+
0.002
90%
RL
D' 2 R
0.01
80%
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
Fundamentals of Power Electronics
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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
Fundamentals of Power Electronics
iC = 0 = I L – VC/R
20
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
+
–
IL
<iC>
=0
+
VC /R
VC
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
t
0
DTs
0
Ts
Dc component (average value) of ig(t) is
Ig = 1
Ts
Fundamentals of Power Electronics
Ts
ig(t) dt = DI L
0
22
Chapter 3: Steady-state equivalent circuit modeling, ...
Input port equivalent circuit
Ig = 1
Ts
Ts
ig(t) dt = DI L
0
Vg
Fundamentals of Power Electronics
+
–
Ig
23
D IL
Chapter 3: Steady-state equivalent circuit modeling, ...
Complete equivalent circuit, buck converter
Input and output port equivalent circuits, drawn together:
Ig
IL
RL
+
Vg
+
–
+ DV
g
–
D IL
VC
R
–
Replace dependent sources with equivalent dc transformer:
Ig
1:D
IL
RL
+
Vg
+
–
VC
R
–
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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
+
–
v
–
switch in position 1
switch in position 2
L
i
RL
+ vL –
Vg
Ts
R
+
iC
Ron
C
R
v
–
Fundamentals of Power Electronics
26
L
RL
RD
+
–
i
C
+
–
+ 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
D' Ts
DTs
t
Vg – IRL – VD – IRD – V
iC (t)
I – V/R
-V/R
vL = D(Vg – IRL – IRon) + D'(Vg – IRL – VD – IRD – V) = 0
iC = D(–V/R) + D'(I – V/R) = 0
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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
D Ron
+ IRL -
+ IDRon-
D' RD
+
–
RL
Vg
+
–
+ ID'RD +
–
I
D' V
D'I – V/R = 0
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
D Ron
D' RD
+
–
RL
Vg
+
–
+
D' V
I
+
–
D' I
V
R
–
D' VD
D Ron
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
D Ron
D' RD
D' : 1
+
–
RL
Vg
+
–
+
V
I
R
–
V= 1
D'
Vg – D'VD
V = 1
Vg
D'
1–
Fundamentals of Power Electronics
D'VD
Vg
D' 2R
D' 2R + RL + DRon + D'RD
1
R + DRon + D'RD
1+ L
D' 2R
30
Chapter 3: Steady-state equivalent circuit modeling, ...
Solution for converter efficiency
D' VD
D Ron
+
–
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:
and
Vg / D' >> VD
D' 2R >> RL + DRon + D'RD
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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
MOSFET current waveforms, for various
ripple magnitudes:
i(t)
• To correctly predict power
loss in a resistor, use rms
values
(c)
(b)
(a)
I
• Result is the same,
provided ripple is small
(a) ∆i = 0
(b) ∆i = 0.1 I
(c) ∆i = I
Fundamentals of Power Electronics
1.1 I
t
0
DTs
0
Inductor current ripple
2I
MOS FET rms current
I
D
32
Average power loss in R on
D I2 R on
(1.00167) I
(1.155) I
Ts
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, ...