Chapter 2
Principles of Steady-State Converter Analysis
2.1. Introduction
2.2. Inductor volt-second balance, capacitor charge
balance, and the small ripple approximation
2.3. Boost converter example
2.4. Cuk converter example
2.5. Estimating the ripple in converters containing twopole low-pass filters
2.6. Summary of key points
Fundamentals of Power Electronics
1
Chapter 2: Principles of steady-state converter analysis
2.1 Introduction
Buck converter
1
SPDT switch changes dc
component
+
Vg
+
2
+
–
vs(t)
R
v(t)
–
Switch output voltage
waveform
vs(t)
Vg
D' Ts
DTs
Duty cycle D:
0≤D≤1
complement D’:
D’ = 1 - D
Fundamentals of Power Electronics
–
0
0
switch
position:
2
DTs
1
Ts
2
t
1
Chapter 2: Principles of steady-state converter analysis
Dc component of switch output voltage
vs(t)
Vg
<vs> = DVg
area =
DTsVg
0
0
DTs
Ts
t
Fourier analysis: Dc component = average value
vs = 1
Ts
Ts
vs(t) dt
0
vs = 1 (DTsVg) = DVg
Ts
Fundamentals of Power Electronics
3
Chapter 2: Principles of steady-state converter analysis
Insertion of low-pass filter to remove switching
harmonics and pass only dc component
L
1
+
Vg
+
–
+
2
vs(t)
C
R
v(t)
–
–
V
Vg
v ≈ vs = DVg
0
0
Fundamentals of Power Electronics
4
1
D
Chapter 2: Principles of steady-state converter analysis
Three basic dc-dc converters
a)
+
iL(t)
Vg
M(D) = D
0.8
2
+
–
C
R
v
M(D)
Buck
1
L
1
0.6
0.4
0.2
–
0
0
0.2
0.4
0.6
0.8
1
0.6
0.8
1
0.6
0.8
1
D
b)
5
2
iL(t)
1
Vg
+
–
C
R
v
M(D) = 1 –1 D
4
+
M(D)
Boost
L
3
2
1
–
0
0
0.2
0.4
D
D
c)
0
+
–
+
2
iL(t)
C
R
v
L
–
-1
M(D)
Vg
0.4
0
1
Buck-boost
0.2
-2
-3
-4
M(D) = 1––DD
-5
Fundamentals of Power Electronics
5
Chapter 2: Principles of steady-state converter analysis
Objectives of this chapter
●
●
●
●
●
Develop techniques for easily determining output
voltage of an arbitrary converter circuit
Derive the principles of inductor volt-second balance
and capacitor charge (amp-second) balance
Introduce the key small ripple approximation
Develop simple methods for selecting filter element
values
Illustrate via examples
Fundamentals of Power Electronics
6
Chapter 2: Principles of steady-state converter analysis
2.2.
Inductor volt-second balance, capacitor charge
balance, and the small ripple approximation
Actual output voltage waveform, buck converter
iL(t)
1
Buck converter
containing practical
low-pass filter
L
+ vL(t) –
2
+
–
Vg
+
iC(t)
C
R
v(t)
–
Actual output voltage
waveform
actual waveform
v(t) = V + vripple(t)
v(t)
V
v(t) = V + vripple(t)
Dc component V
0
Fundamentals of Power Electronics
t
7
Chapter 2: Principles of steady-state converter analysis
The small ripple approximation
actual waveform
v(t) = V + vripple(t)
v(t)
v(t) = V + vripple(t)
V
Dc component V
0
t
In a well-designed converter, the output voltage ripple is small. Hence,
the waveforms can be easily determined by ignoring the ripple:
vripple << V
v(t) ≈ V
Fundamentals of Power Electronics
8
Chapter 2: Principles of steady-state converter analysis
Buck converter analysis:
inductor current waveform
iL(t)
1
L
+ vL(t) –
original
converter
2
+
–
Vg
+
iC(t)
C
R
v(t)
–
switch in position 1
iL(t)
L
L
+ vL(t) –
Vg
+
–
switch in position 2
C
+
+ vL(t) –
iC(t)
R
Vg
v(t)
iL(t)
C
R
v(t)
–
–
Fundamentals of Power Electronics
+
–
+
iC(t)
9
Chapter 2: Principles of steady-state converter analysis
Inductor voltage and current
Subinterval 1: switch in position 1
iL(t)
Inductor voltage
L
+ vL(t) –
vL = Vg – v(t)
Vg
Small ripple approximation:
+
–
C
+
iC(t)
R
v(t)
–
vL ≈ Vg – V
Knowing the inductor voltage, we can now find the inductor current via
vL(t) = L
diL(t)
dt
Solve for the slope:
diL(t) vL(t) Vg – V
=
≈
L
L
dt
Fundamentals of Power Electronics
⇒ The inductor current changes with an
essentially constant slope
10
Chapter 2: Principles of steady-state converter analysis
Inductor voltage and current
Subinterval 2: switch in position 2
L
Inductor voltage
+ vL(t) –
vL(t) = – v(t)
Vg
Small ripple approximation:
+
–
iL(t)
C
+
iC(t)
R
v(t)
–
vL(t) ≈ – V
Knowing the inductor voltage, we can again find the inductor current via
vL(t) = L
diL(t)
dt
Solve for the slope:
diL(t)
≈– V
L
dt
Fundamentals of Power Electronics
⇒ The inductor current changes with an
essentially constant slope
11
Chapter 2: Principles of steady-state converter analysis
Inductor voltage and current waveforms
vL(t)
Vg – V
D'Ts
DTs
t
–V
switch
position:
1
2
1
vL(t) = L
iL(t)
iL(DTs)
I
iL(0)
0
Fundamentals of Power Electronics
diL(t)
dt
∆iL
Vg – V
L
–V
L
DTs
12
Ts
t
Chapter 2: Principles of steady-state converter analysis
Determination of inductor current ripple magnitude
iL(t)
iL(DTs)
I
iL(0)
Vg – V
L
0
∆iL
–V
L
Ts
DTs
t
(change in iL) = (slope)(length of subinterval)
Vg – V
DTs
2∆iL =
L
⇒
V –V
∆iL = g
DTs
2L
Fundamentals of Power Electronics
L=
13
Vg – V
DTs
2∆iL
Chapter 2: Principles of steady-state converter analysis
Inductor current waveform
during turn-on transient
iL(t)
iL(Ts)
iL(0)=0
Vg – v(t)
L
– v(t)
L
0 DTs Ts
iL(nTs)
2Ts
nTs
iL((n+1)Ts)
(n+1)Ts
t
When the converter operates in equilibrium:
iL((n + 1)Ts) = iL(nTs)
Fundamentals of Power Electronics
14
Chapter 2: Principles of steady-state converter analysis
The principle of inductor volt-second balance:
Derivation
Inductor defining relation:
di (t)
vL(t) = L L
dt
Integrate over one complete switching period:
iL(Ts) – iL(0) = 1
L
Ts
vL(t) dt
0
In periodic steady state, the net change in inductor current is zero:
Ts
0=
vL(t) dt
0
Hence, the total area (or volt-seconds) under the inductor voltage
waveform is zero whenever the converter operates in steady state.
An equivalent form:
T
s
1
0=
v (t) dt = vL
Ts 0 L
The average inductor voltage is zero in steady state.
Fundamentals of Power Electronics
15
Chapter 2: Principles of steady-state converter analysis
Inductor volt-second balance:
Buck converter example
Inductor voltage waveform,
previously derived:
vL(t)
total area λ
Vg – V
t
DTs
–V
Integral of voltage waveform is area of rectangles:
Ts
λ=
vL(t) dt = (Vg – V)(DTs) + ( – V)(D'Ts)
0
Average voltage is
vL = λ = D(Vg – V) + D'( – V)
Ts
Equate to zero and solve for V:
0 = DVg – (D + D')V = DVg – V
Fundamentals of Power Electronics
16
⇒
V = DVg
Chapter 2: Principles of steady-state converter analysis
The principle of capacitor charge balance:
Derivation
Capacitor defining relation:
dv (t)
iC(t) = C C
dt
Integrate over one complete switching period:
vC(Ts) – vC(0) = 1
C
Ts
iC(t) dt
0
In periodic steady state, the net change in capacitor voltage is zero:
0= 1
Ts
Ts
iC(t) dt = iC
0
Hence, the total area (or charge) under the capacitor current
waveform is zero whenever the converter operates in steady state.
The average capacitor current is then zero.
Fundamentals of Power Electronics
17
Chapter 2: Principles of steady-state converter analysis
2.3 Boost converter example
L
Boost converter
with ideal switch
iL(t)
2
+ vL(t) –
1
Vg
+
iC(t)
+
–
C
R
v
–
D1
L
Realization using
power MOSFET
and diode
iL(t)
Vg
+
+ vL(t) –
Q1
+
–
DTs
Ts
+
–
iC(t)
C
R
v
–
Fundamentals of Power Electronics
18
Chapter 2: Principles of steady-state converter analysis
Boost converter analysis
L
iL(t)
original
converter
2
+
+ vL(t) –
iC(t)
1
Vg
+
–
C
R
v
–
switch in position 1
switch in position 2
L
L
iL(t)
Vg
+ vL(t) –
+
–
iL(t)
+
iC(t)
C
R
Vg
v
+
iC(t)
C
R
v
–
–
Fundamentals of Power Electronics
+
–
+ vL(t) –
19
Chapter 2: Principles of steady-state converter analysis
Subinterval 1: switch in position 1
Inductor voltage and capacitor current
L
vL = Vg
iC = – v / R
iL(t)
Vg
Small ripple approximation:
+
–
+ vL(t) –
+
iC(t)
C
R
v
–
vL = Vg
iC = – V / R
Fundamentals of Power Electronics
20
Chapter 2: Principles of steady-state converter analysis
Subinterval 2: switch in position 2
Inductor voltage and capacitor current
L
vL = Vg – v
iC = iL – v / R
iL(t)
Vg
Small ripple approximation:
+
–
+ vL(t) –
+
iC(t)
C
R
v
–
vL = Vg – V
iC = I – V / R
Fundamentals of Power Electronics
21
Chapter 2: Principles of steady-state converter analysis
Inductor voltage and capacitor current waveforms
vL(t)
Vg
DTs
D'Ts
t
Vg – V
iC(t)
I – V/R
D'Ts
DTs
t
– V/R
Fundamentals of Power Electronics
22
Chapter 2: Principles of steady-state converter analysis
Inductor volt-second balance
Net volt-seconds applied to inductor
over one switching period:
vL(t)
Vg
DTs
D'Ts
Ts
t
vL(t) dt = (Vg) DTs + (Vg – V) D'Ts
0
Vg – V
Equate to zero and collect terms:
Vg (D + D') – V D' = 0
Solve for V:
Vg
D'
The voltage conversion ratio is therefore
V =
M(D) = V = 1 = 1
Vg D' 1 – D
Fundamentals of Power Electronics
23
Chapter 2: Principles of steady-state converter analysis
Conversion ratio M(D) of the boost converter
5
1 = 1
M(D) = D'
1– D
M(D)
4
3
2
1
0
0
0.2
0.4
0.6
0.8
1
D
Fundamentals of Power Electronics
24
Chapter 2: Principles of steady-state converter analysis
Determination of inductor current dc component
Capacitor charge balance:
iC(t)
I – V/R
D'Ts
DTs
Ts
0
iC(t) dt = ( – V ) DTs + (I – V ) D'Ts
R
R
Collect terms and equate to zero:
– V (D + D') + I D' = 0
R
t
– V/R
I
(V g / R)
8
Solve for I:
I= V
D' R
6
4
2
Eliminate V to express in terms of Vg:
Vg
I= 2
D' R
Fundamentals of Power Electronics
0
0
0.2
0.4
0.6
0.8
1
D
25
Chapter 2: Principles of steady-state converter analysis
Determination of inductor current ripple
Inductor current slope during
subinterval 1:
diL(t) vL(t) Vg
=
=
L
L
dt
Inductor current slope during
subinterval 2:
diL(t) vL(t) Vg – V
=
=
L
L
dt
iL(t)
∆iL
I
Vg
L
0
Vg – V
L
DTs
Ts
t
Change in inductor current during subinterval 1 is (slope) (length of subinterval):
Vg
2∆iL =
DTs
L
Solve for peak ripple:
∆iL =
Vg
DTs
2L
Fundamentals of Power Electronics
• Choose L such that desired ripple magnitude
is obtained
26
Chapter 2: Principles of steady-state converter analysis
Determination of capacitor voltage ripple
Capacitor voltage slope during
subinterval 1:
dvC(t) iC(t) – V
=
=
C
RC
dt
v(t)
∆v
V
–V
RC
Capacitor voltage slope during
subinterval 2:
dvC(t) iC(t) I
=
= – V
C
C RC
dt
0
I – V
C RC
DTs
Ts
t
Change in capacitor voltage during subinterval 1 is (slope) (length of subinterval):
– 2∆v = – V DTs
RC
Solve for peak ripple:
∆v = V DTs
2RC
Fundamentals of Power Electronics
• Choose C such that desired voltage ripple
magnitude is obtained
• In practice, capacitor equivalent series
resistance (esr) leads to increased voltage ripple
27
Chapter 2: Principles of steady-state converter analysis
2.4 Cuk converter example
C1
L1
Cuk converter,
with ideal switch
i1
Vg
L2
i2
+ v1 –
1
+
–
2
C2
+
v2
R
–
Cuk converter:
practical realization
using MOSFET and
diode
C1
L1
i1
Vg
L2
i2
+ v1 –
+
–
Q1
D1
C2
+
v2
R
–
Fundamentals of Power Electronics
28
Chapter 2: Principles of steady-state converter analysis
Cuk converter circuit
with switch in positions 1 and 2
L1
Switch in position 1:
MOSFET conducts
Capacitor C1 releases
energy to output
L2
i1 + vL1 –
Vg
–
+
–
v1
i2
+
iC1 + vL2 –
C1
iC2
C2
+
i1
Switch in position 2:
diode conducts
Capacitor C1 is
charged from input
Fundamentals of Power Electronics
iC1
+ vL1 –
Vg
+
–
+
C1
v1
–
29
R
–
L2
L1
v2
+ vL2 –
i2
iC2
C2
+
v2
R
–
Chapter 2: Principles of steady-state converter analysis
Waveforms during subinterval 1
MOSFET conduction interval
Inductor voltages and
capacitor currents:
vL1 = Vg
vL2 = – v1 – v2
i C1 = i 2
v
i C2 = i 2 – 2
R
L1
i1 + vL1 –
Vg
+
–
L2
–
v1
+
iC1 + vL2 –
C1
i2
iC2
C2
+
v2
R
–
Small ripple approximation for subinterval 1:
vL1 = Vg
vL2 = – V1 – V2
i C1 = I 2
V
i C2 = I 2 – 2
R
Fundamentals of Power Electronics
30
Chapter 2: Principles of steady-state converter analysis
Waveforms during subinterval 2
Diode conduction interval
Inductor voltages and
capacitor currents:
vL1 = Vg – v1
vL2 = – v2
i C1 = i 1
v
i C2 = i 2 – 2
R
i1
L2
L1
iC1
+ vL1 –
Vg
+
–
+
C1
v1
–
+ vL2 –
i2
iC2
C2
+
v2
R
–
Small ripple approximation for subinterval 2:
vL1 = Vg – V1
vL2 = – V2
i C1 = I 1
V
i C2 = I 2 – 2
R
Fundamentals of Power Electronics
31
Chapter 2: Principles of steady-state converter analysis
Equate average values to zero
The principles of inductor volt-second and capacitor charge balance
state that the average values of the periodic inductor voltage and
capacitor current waveforms are zero, when the converter operates in
steady state. Hence, to determine the steady-state conditions in the
converter, let us sketch the inductor voltage and capacitor current
waveforms, and equate their average values to zero.
Waveforms:
Inductor voltage vL1(t)
vL1(t)
Volt-second balance on L1:
Vg
DTs
vL1 = DVg + D'(Vg – V1) = 0
D'Ts
t
Vg – V1
Fundamentals of Power Electronics
32
Chapter 2: Principles of steady-state converter analysis
Equate average values to zero
Inductor L2 voltage
vL2(t)
– V2
DTs
D'Ts
t
– V1 – V2
vL2 = D( – V1 – V2) + D'( – V2) = 0
Capacitor C1 current
iC1(t)
Average the waveforms:
i C1 = DI 2 + D'I 1 = 0
I1
DTs
D'Ts
t
I2
Fundamentals of Power Electronics
33
Chapter 2: Principles of steady-state converter analysis
Equate average values to zero
Capacitor current iC2(t) waveform
iC2(t)
I2 – V2 / R (= 0)
DTs
D'Ts
t
i C2 = I 2 –
V2
=0
R
Note: during both subintervals, the
capacitor current iC2 is equal to the
difference between the inductor current
i2 and the load current V2/R. When
ripple is neglected, iC2 is constant and
equal to zero.
Fundamentals of Power Electronics
34
Chapter 2: Principles of steady-state converter analysis
Cuk converter conversion ratio M = V/Vg
D
0
0.2
0.4
0.6
0.8
1
0
M(D)
-1
-2
-3
-4
V2
M(D) =
=– D
Vg
1–D
-5
Fundamentals of Power Electronics
35
Chapter 2: Principles of steady-state converter analysis
Inductor current waveforms
Interval 1 slopes, using small
ripple approximation:
i1(t)
∆i1
I1
Vg
L1
di 1(t) vL1(t) Vg
=
=
L1
L1
dt
di 2(t) vL2(t) – V1 – V2
=
=
L2
L2
dt
Vg – V1
L1
DTs
Ts
Interval 2 slopes:
di 1(t) vL1(t) Vg – V1
=
=
L1
L1
dt
di 2(t) vL2(t) – V2
=
=
L2
L2
dt
Fundamentals of Power Electronics
I2
– V1 – V2
L2
t
Ts
DTs
t
– V2
L2
∆i2
i2(t)
36
Chapter 2: Principles of steady-state converter analysis
Capacitor C1 waveform
Subinterval 1:
dv1(t) i C1(t) I 2
=
=
C1
C1
dt
v1(t)
∆v1
V1
I2
C1
I1
C1
Subinterval 2:
DTs
Ts
t
dv1(t) i C1(t) I 1
=
=
C1
C1
dt
Fundamentals of Power Electronics
37
Chapter 2: Principles of steady-state converter analysis
Ripple magnitudes
Analysis results
VgDTs
∆i 1 =
2L 1
V + V2
∆i 2 = 1
DTs
2L 2
– I DT
∆v1 = 2 s
2C 1
Use dc converter solution to simplify:
VgDTs
∆i 1 =
2L 1
VgDTs
∆i 2 =
2L 2
VgD 2Ts
∆v1 =
2D'RC 1
Q: How large is the output voltage ripple?
Fundamentals of Power Electronics
38
Chapter 2: Principles of steady-state converter analysis
2.5 Estimating ripple in converters
containing two-pole low-pass filters
Buck converter example: Determine output voltage ripple
L
1
iL(t)
Vg
iC(t)
+
iR(t)
2
+
–
C
vC(t)
R
–
Inductor current
waveform.
What is the
capacitor current?
Fundamentals of Power Electronics
iL(t)
iL(DTs)
I
iL(0)
∆iL
Vg – V
L
–V
L
0
DTs
39
Ts
t
Chapter 2: Principles of steady-state converter analysis
Capacitor current and voltage, buck example
iC(t)
Must not
neglect
inductor
current ripple!
total charge
q
∆iL
Ts / 2
DTs
If the capacitor
voltage ripple is
small, then
essentially all of
the ac component
of inductor current
flows through the
capacitor.
Fundamentals of Power Electronics
t
D'Ts
vC(t)
V
∆v
∆v
t
40
Chapter 2: Principles of steady-state converter analysis
Estimating capacitor voltage ripple ∆v
iC(t)
total charge
q
∆iL
t
Ts / 2
DTs
D'Ts
vC(t)
V
∆v
∆v
t
Fundamentals of Power Electronics
41
Current iC(t) is positive for half
of the switching period. This
positive current causes the
capacitor voltage vC(t) to
increase between its minimum
and maximum extrema.
During this time, the total
charge q is deposited on the
capacitor plates, where
q = C (2∆v)
(change in charge) =
C (change in voltage)
Chapter 2: Principles of steady-state converter analysis
Estimating capacitor voltage ripple ∆v
The total charge q is the area
of the triangle, as shown:
iC(t)
total charge
q
∆iL
t
Ts / 2
DTs
q = 12 ∆iL
Eliminate q and solve for ∆v:
D'Ts
∆v =
vC(t)
V
∆v
∆v
t
Fundamentals of Power Electronics
Ts
2
42
∆iL Ts
8C
Note: in practice, capacitor
equivalent series resistance
(esr) further increases ∆v.
Chapter 2: Principles of steady-state converter analysis
Inductor current ripple in two-pole filters
L1
Example:
problem 2.9
iT
+
i1
Vg
+
–
L2
Q1
C1
+
i2
vC1
D1
C2
R
–
vL(t)
total
flux linkage
λ
∆v
v
–
t
Ts / 2
DTs
D'Ts
can use similar arguments, with
λ = L ∆i
λ = inductor flux linkages
iL(t)
I
∆i
∆i
= inductor volt-seconds
t
Fundamentals of Power Electronics
43
Chapter 2: Principles of steady-state converter analysis
2.6 Summary of Key Points
1. The dc component of a converter waveform is given by its average
value, or the integral over one switching period, divided by the
switching period. Solution of a dc-dc converter to find its dc, or steadystate, voltages and currents therefore involves averaging the
waveforms.
2. The linear ripple approximation greatly simplifies the analysis. In a welldesigned converter, the switching ripples in the inductor currents and
capacitor voltages are small compared to the respective dc
components, and can be neglected.
3. The principle of inductor volt-second balance allows determination of the
dc voltage components in any switching converter. In steady-state, the
average voltage applied to an inductor must be zero.
Fundamentals of Power Electronics
44
Chapter 2: Principles of steady-state converter analysis
Summary of Chapter 2
4. The principle of capacitor charge balance allows determination of the dc
components of the inductor currents in a switching converter. In steadystate, the average current applied to a capacitor must be zero.
5. By knowledge of the slopes of the inductor current and capacitor voltage
waveforms, the ac switching ripple magnitudes may be computed.
Inductance and capacitance values can then be chosen to obtain
desired ripple magnitudes.
6. In converters containing multiple-pole filters, continuous (nonpulsating)
voltages and currents are applied to one or more of the inductors or
capacitors. Computation of the ac switching ripple in these elements
can be done using capacitor charge and/or inductor flux-linkage
arguments, without use of the small-ripple approximation.
7. Converters capable of increasing (boost), decreasing (buck), and
inverting the voltage polarity (buck-boost and Cuk) have been
described. Converter circuits are explored more fully in a later chapter.
Fundamentals of Power Electronics
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Chapter 2: Principles of steady-state converter analysis