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
Analysis strategy
This converter has two
inductor currents and two
capacitor voltages, that
can be expressed as
i 1(t) = I 1 + i 1 -ripple(t)
i 2(t) = I 2 + i 2 -ripple(t)
v1(t) = V1 + v1 -ripple(t)
v2(t) = V2 + v2 -ripple(t)
To solve the converter in
steady state, we want to
find the dc components I1,
I2, V1, and V2, when the
ripples are small.
Fundamentals of Power Electronics
C1
L1
i1
Vg
L2
i2
+ v1 –
1
+
–
2
C2
+
v2
R
–
Strategy:
• Apply volt-second balance to each
inductor voltage
• Apply charge balance to each capacitor
current
• Simplify using the small ripple
approximation
• Solve the resulting four equations for the
four unknowns I1, I2, V1, and V2.
29
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
–
30
v2
R
–
L2
L1
+
+ 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
31
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
32
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
33
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
34
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
35
Chapter 2: Principles of steady-state converter analysis
Solve for steady-state inductor currents
and capacitor voltages
The four equations obtained from
volt-sec and charge balance:
Solve for the dc capacitor
voltages and inductor currents,
and express in terms of the
known Vg, D, and R:
vL1 = DVg + D' Vg – V1 = 0
vL2 = D – V1 – V2 + D' – V2 = 0
Vg
D'
V2 = – D Vg
D'
I1 = – D I2 =
D'
V
I2 = 2 = – D
R
D'
V1 =
i C1 = DI 2 + D'I 1 = 0
V
i C2 = I 2 – 2 = 0
R
Fundamentals of Power Electronics
36
D
D'
Vg
R
2
Vg
R
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
37
Chapter 2: Principles of steady-state converter analysis
Inductor current waveforms
Interval 1 slopes, using small
ripple approximation:
di 1(t) vL1(t) Vg
=
=
L1
L1
dt
di 2(t) vL2(t) – V1 – V2
=
=
L2
L2
dt
i1(t)
i1
I1
Vg
L1
DTs
Fundamentals of Power Electronics
Ts
DTs
Interval 2 slopes:
di 1(t) vL1(t) Vg – V1
=
=
L1
L1
dt
di 2(t) vL2(t) – V2
=
=
L2
L2
dt
Vg – V1
L1
I2
– V1 – V2
L2
t
t
Ts
– V2
L2
i2
i2(t)
38
Chapter 2: Principles of steady-state converter analysis
Capacitor C1 waveform
Subinterval 1:
dv1(t) i C1(t) I 2
=
=
C1
C
dt
1
v1(t)
v1
V1
I2
C1
Subinterval 2:
DTs
dv1(t) i C1(t) I 1
=
=
C1
C1
dt
Fundamentals of Power Electronics
I1
C1
39
Ts
t
Chapter 2: Principles of steady-state converter analysis
Ripple magnitudes
Analysis results
Use dc converter solution to simplify:
VgDTs
i1 =
2L 1
VgDTs
i2 =
2L 2
VgD 2Ts
v1 =
2D'RC 1
VgDTs
i1 =
2L 1
V1 + V2
i2 =
DTs
2L 2
– I 2DTs
v1 =
2C 1
Q: How large is the output voltage ripple?
Fundamentals of Power Electronics
40
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)
2
+
–
+
C
iR(t)
vC(t)
R
–
iL(t)
Inductor current
waveform.
iL(DTs)
I
iL(0)
What is the
capacitor current?
Vg – V
L
0
Fundamentals of Power Electronics
–V
L
DTs
41
iL
Ts
t
Chapter 2: Principles of steady-state converter analysis
Capacitor current and voltage, buck example
iC(t)
Total charge
q
Must not
neglect
inductor
current ripple!
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
iL
t
Ts /2
DTs
D'Ts
vC(t)
v
V
v
t
42
Chapter 2: Principles of steady-state converter analysis
Estimating capacitor voltage ripple v
iC(t)
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
Total charge
q
iL
t
Ts /2
DTs
D'Ts
vC(t)
V
v
v
t
Fundamentals of Power Electronics
43
q = C (2 v)
(change in charge) =
C (change in voltage)
Chapter 2: Principles of steady-state converter analysis
Estimating capacitor voltage ripple v
iC(t)
The total charge q is the area
of the triangle, as shown:
Total charge
q
iL
t
q=
Ts /2
DTs
D'Ts
v=
Ts
2
i L Ts
8C
v
v
t
Fundamentals of Power Electronics
iL
Eliminate q and solve for v:
vC(t)
V
1
2
44
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
L2
+
i1
Vg
Q1
+
–
vC1
C1
+
i2
D1
C2
–
vL(t)
R
v
–
Total
flux linkage
v
t
Ts /2
DTs
D'Ts
can use similar arguments, with
= L (2 i)
iL(t)
I
= inductor flux linkages
i
= inductor volt-seconds
i
t
Fundamentals of Power Electronics
45
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
46
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
47
Chapter 2: Principles of steady-state converter analysis