Industrial Electronics - Module 1: Steady State Devices

Industrial Electronics - Module 1: Steady State Devices

1. 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

2. 3.1. The dc transformer model
Basic equations of an ideal
dc-dc converter:
P
in
=P
ou
t
V
g
I
g
=V
I
(η = 100%)
V=M(D)V
g
(ideal conversion ratio)
I
g
=M(D)
I
These equations are valid in steady-state. During
transients, energy storage within filter elements may cause
P
in
≠ P
out
Switching
dc-dc
converter
D
Control input
Power
input
Power
output
I
g
I
+
V
–
+
V
g
–

3. Equivalent circuits corresponding to
ideal dc-dc converter equations
P
in
=P
ou
t
V
g
I
g
=V
I
V=M(D)V
g
I
g
=M(D)
I
Dependent sources DC transformer
Power
output
+
V
–
I
+
–
M(D)V
g
Power
input
+
V
g
–
I
g
M(D)I
D
Control input
Power
input
Power
output
+
V
–
+
V
g
–
I
g
I
1: M(D)

4. The DC transformer model
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
• 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
D
Control input
Power
input
Power
output
+
V
–
+
V
g
–
I
g
I
1: M(D)

5. Example: use of the DC transformer model
1. Original system
2. Insert dc transformer model
3. Push source through transformer
4. Solve circuit
V=M(D)V
1
R
R+M 2
(D)R
1
D
R
V
1
R
1
+
–
+
V
g
–
+
V
–
Switching
dc-dc
converter
1 : M(D)
R V
1
R
1
+
–
+
V
g
–
+
V
–
R M(D)V
1
M
2
(D)R
1
+
–
+
V
–

6. 3.2. Inclusion of inductor copper loss
Dc transformer model can be extended, to include converter nonidealities.
Example: inductor copper loss (resistance of winding):
Insert this inductor model into boost converter circuit:
LR
L
L
+
–
CR
+
v
–
1
2
i
V
g
R
L

7. Analysis of nonideal boost converter
switch in position 1 switch in position 2
L
+
–
CR
+
v
–
1
2
i
V
g
R
L
LR
L
+
–
i
CR
+
v
–
+ v
L
–
i
C
V
g
LR
L
+
–
i
CR
+
v
–
+ v
L
–
i
C
V
g

8. Circuit equations, switch in position 1
Inductor current and
capacitor voltage:
v
L
(t)=V
g
–i(t)R
L
i
C
(t)=–v(t)/R
Small ripple approximation:
v
L
(t)=V
g
–IR
L
i
C
(t)=–V/R
LR
L
+
–
i
CR
+
v
–
+ v
L
–
i
C
V
g

9. Circuit equations, switch in position 2
v
L
(t)=V
g
–i(t)R
L
–v(t)5V
g
–IR
L
–V
i
C
(t)=i(t)–v(t)/R5I–V/R
LR
L
+
–
i
CR
+
v
–
+ v
L
–
i
C
V
g

10. Inductor voltage and capacitor current waveforms
Average inductor voltage:
v
L
(t)
=
1
T
s
v
L
(t)dt
0
T
s
=D(V
g
–IR
L
)+D'(V
g
–IR
L
–V)
Inductor volt-second balance:
0=V
g
–IR
L
–D'V
Average capacitor current:
i
C
(t)
=D(–V/R)+D'(I–V/R)
Capacitor charge balance:
0=D'I–V/R
v
L
(t)
t
V
g
– IR
L
DT
s
D'T
s
V
g
– IR
L
– V
i
C
(t)
–V/R
I – V/R
t

11. Solution for output voltage
We now have two
equations and two
unknowns: 0=V
g
–IR
L
–D'V
0=D'I–V/R
EliminateI and
solve for V:
V
V
g
=
1
D'
1
(1+R L
/D'
2
R)
D
V/V
g
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
R
L
/R = 0
R
L
/R = 0.01
R
L
/R = 0.02
R
L
/R = 0.05
R
L
/R = 0.1

12. 3.3. Construction of equivalent circuit model Results of previous section (derived via inductor volt-sec balance and
capacitor charge balance):
v
L
=0=V
g
–IR
L
–D'V
i
C
=0=D'I–V/R
View these as loop and node equations of the equivalent circuit.
Reconstruct an equivalent circuit satisfying these equations

13. Inductor voltage equation
v
L
=0=V
g
–IR
L
–D'V
• Derived via Kirchhoff’s voltage
law, to find the inductor voltage
during each subinterval
• 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
•IR
L
term: voltage across resistor
of value R
L
having current I
•D’V term: for now, leave as
dependent source
+
–
LR
L
+
–
+v
L
œ –
= 0
+ IR
L
–
I
D'V V
g

14. Capacitor current equation
• Derived via Kirchoff’s current
law, to find the capacitor
current during each subinterval
• Average capacitor current then
set to zero
• 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
i
C
=0=D'I–V/R
R
+
V
–
C
i
C
œ
= 0
Node
V/R
D'I

15. Complete equivalent circuit
The two circuits, drawn together:
The dependent sources are equivalent
to a D
′
: 1 transformer:
Dependent sources and transformers
+
–
+
V
2
–
nV
2
nI
1
I
1
n : 1
+
V
2
–
I
1
•
sources have same coefficient
•
reciprocal voltage/current
dependence
+
–
+
–
D'V
R
L
I
D'I
R
+
V
–
V
g
+
–
R
L
I
R
+
V
–
D' : 1
V
g

16. Solution of equivalent circuit
Converter equivalent circuit
Refer all elements to transformer
secondary:
Solution for output voltage
using voltage divider formula:
V=
V
g
D'
R
R+
R
L
D'
2
=
V
g
D'
1
1+
R
L
D'
2
R
+
–
R
L
I
R
+
V
–
D' : 1
V
g
+
–
D'I
R
+
V
–
V
g
/D'
R
L
/D'
2

17. Solution for input (inductor) current
I=
V
g
D'
2
R+R
L
=
V
g
D'
2
1
1+
R
L
D'
2
R
+
–
R
L
I
R
+
V
–
D' : 1
V
g

18. Solution for converter efficiency
P
in
=(V
g
)(I)
P
out
=(V)(D'I)
d
=
P
out
P
in
=
(
V
)(
D
'I
)
(V
g
)(I)
=
V
V
g
D'
d
=
1
1+
R
L
D'
2
R
+
–
R
L
I
R
+
V
–
D' : 1
V
g

19. Efficiency, for various values of R
L
d
=
1
1+
R
L
D'
2
R
D
d
R
L
/R= 0.1
0.02
0.01
0.05
0.002
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%

20. 3.4. How to obtain the input port of the model
Buck converter example —use procedure of previous section to
derive equivalent circuit
Average inductor voltage and capacitor current:
v
L
=0=DV
g
–I
L
R
L
–V
C
i
C
=0=I
L
–V
C
/R
+
–
CR
+
v
C
–
LR
L
i
L
i
g
1
2
+ v
L
–
V
g

21. Construct equivalent circuit as usual
v
L
=0=DV
g
–I
L
R
L
–V
C
i
C
=0=I
L
–V
C
/R
What happened to the transformer?
• Need another equation
i
C
œ
= 0
R
+
V
C
–
R
L
+
–
DV
g
+v
L
œ –
= 0
I
L
V
C
/R

22. Modeling the converter input port
Input current waveform i
g
(t):
Dc component (average value) of i
g
(t) is
I
g
=
1
T
s
i
g
(t)dt
0
T
s
=DI
L
i
g
(t)
t
i
L
(t)5 I
L
0
DT
s
T
s
0
area =
DT
s
I
L

23. Input port equivalent circuit
I
g
=
1
T
s
i
g
(t)dt
0
T
s
=DI
L
+
–
DI
L
I
g
V
g

24. Complete equivalent circuit, buck converter
Input and output port equivalent circuits, drawn together:
Replace dependent sources with equivalent dc transformer:
R
+
V
C
–
R
L
+
–
DV
g
I
L
+
–
DI
L
I
g
V
g
R
+
V
C
–
R
L
I
L
+
–
I
g
1 : D
V
g

25. 3.5. Example: inclusion of semiconductor
conduction losses in the boost converter model
Boost converter example
Models of on-state semiconductor devices:
MOSFET: on-resistance R
on
Diode: constant forward voltage V
D
plus on-resistance R
D
Insert these models into subinterval circuits
+
–
DT
s
T
s
+
–
i
L
CR
+
v
–
i
C
V
g

26. Boost converter example: circuits during
subintervals 1 and 2
switch in position 1 switch in position 2
+
–
DT
s
T
s
+
–
i
L
CR
+
v
–
i
C
V
g
LR
L
+
–
i
CR
+
v
–
+ v
L
–
i
C
R
on
V
g
LR
L
+
–
i
CR
+
v
–
+ v
L
–
i
C
R
D
+
–
V
D
V
g

27. Average inductor voltage and capacitor current
v
L
=D(V
g
–IR
L
–IR
on
)+D'(V
g
–IR
L
–V
D
–IR
D
–V)=0
i
C
=D(–V/R)+D'(I–V/R)=0
v
L
(t)
t
V
g
– IR
L
– IR
on
DT
s
D'T
s
V
g
– IR
L
– V
D
– IR
D
– V
i
C
(t)
–V/R
I – V/R
t

28. Construction of equivalent circuits
V
g
–IR
L
–IDR
on
–D'V
D
–ID'R
D
–D'V=0
D
'I–V/R=0
R
L
+
–
D'R
D
+
–
D'V
D
DR
on
+ IR
L
– + IDR
on
– + ID'R
D
–
+
–
D'V
I
V
g
R
+
V
–
V/R
D'I

29. Complete equivalent circuit R
L
+
–
D'R
D
+
–
D'V
D
DR
on
+
–
D'V
R
+
V
–
D'I
I
V
g
R
L
+
–
D'R
D
+
–
D'V
D
DR
on
R
+
V
–
I
D' : 1
V
g

30. Solution for output voltage
V=
1
D'
V
g
–D'V
D
D'
2
R
D'
2
R+R
L
+DR
on
+D'R
D
V
V
g
=
1
D'
1–
D'V
D
V
g
1
1+
R
L
+DR
on
+D'R
D
D'
2
R
R
L
+
–
D'R
D
+
–
D'V
D
DR
on
R
+
V
–
I
D' : 1
V
g

31. Solution for converter efficiency
P
in
=(V
g
)(I)
P
out
=(V)(D'I)
d
=D'
V
V
g
=
1–
D'V
D
V
g
1+
R
L
+DR
on
+D'R
D
D'
2
R
Conditions for high efficiency:
R
L
+
–
D'R
D
+
–
D'V
D
DR
on
R
+
V
–
I
D' : 1
V
g
V
g
/D'>V
D
D'
2
R>R
L
+DR
on
+D'R
D

32. 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:
Inductor current ripple MOSFET rms current Average power loss in R
on
(a) ∆i= 0 ID
D I
2
R
on
(b) ∆i= 0.1 I (1.00167)ID
(1.0033) D I
2
R
on
(c) ∆i = I (1.155)ID
(1.3333) D I
2
R
on
i(t)
t
0
DT
s
T
s
0
I
2 I
1.1 I
(a)
(c)
(b)

33. 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.