Chapter 3 DC to DC Converters
Outline
3.1 Basic DC to DC converters
3.1.1 Buck converter (Step- down converter)
3.1.2 Boost converter (Step-up converter)
3.2 Composite DC/DC converters and connection of multiple DC/DC
converters
3.2.1 A current-reversible chopper
3.2.2 Bridge chopper (H-bridge DC/DC converter)
3.2.3 Multi-phase multi-channel DC/DC converters
3.1 Basic DC to DC converters
3.1.1Buck converter
1
SPDT switch changes dc
Vg
component
+
Switch output voltage waveform
Duty cycle D: 0 ≤ D ≤ 1
complement D: D´ = 1 - D
2
+
+
R
Vs(t)
-
-
Vs(t)
Vg
DTs
D Ts
0
0
switch
position:
DTs
1
t
Ts
2
V(t)
1
Dc component of switch output voltage
Vs(t)
Vg
<Vs>=DVg
area=
D Ts Vg
0
0
Ts
DTs
t
Fourier analysis:DC component =average value:
<Vs> =
<Vs> =
∫
1
Ts
Ts
Vs(t) dt
0
1 DTsVg
=DVg
Ts
Insertion of low- pass filter to remove switching harmonics and pass
only dc component
L
1
2
Vg
+
-
+
+
C
Vs(t)
V(t)
R
-
v≈<vs> =DVg
V
Vg
o
0
1
D

Basic operation principle of buck converter
L
1
Buck converter with
2
Vg
R
C
Vs(t)
+
-
ideal switch
+
+
-
+ L
Realization using
iL(t)
power MOSFET
and diode
Vg +-
+
DTs Ts
t
V(t)
VL(t)
D1
ic(t)
R
Thought process in analyzing basic DC/DC converters
1) Basic operation principle (qualitative analysis)
–How does current flows during different switching states
–How is energy transferred during different switching states
2) Verification of small ripple approximation
3) Derivation of inductor voltage waveform during different switching
states
4) Quantitative analysis according to inductor volt-second balance or
capacitor charge balance

Actual output voltage waveform of buck converter
1
iL(t)+
Buck converter
containing practical
C
+
-
+
ic(t)
VL(t)
2
Vg
L
V(t)
R
low-pass filter
-
v(t )
Actual output voltage
waveform
Actual waveform
v(t ) = V + v ripple(t)
V
DC component V
v(t ) = V + v ripple(t)
0
t

Buck converter analysis: inductor current waveform
iL(t)+
1
original
+
ic(t)
V(t)
R
C
+
-
converter
-
VL(t)
2
Vg
L
-
iL(t)+
Switch in position 1
L
-
+
-
+
ic(t)
VL(t)
Vg
Switch in position 2
L
+
-
C
R
V(t)
-
Vg
+
-
+
ic(t)
VL(t)
C
R
V(t)
iL(t)
-

Inductor voltage and current subinterval 1: switch in position 1
iL(t)+
Inductor voltage
vL=Vg - v(t)
Small ripple approximation:
L
-
Vg
+
-
+
ic(t)
VL(t)
C
R
V(t)
vL=Vg - V
-
Knowing the inductor voltage, we can now find the inductor current via
diL(t)
vL(t)=L
dt
Solve for the slope:
vL(t)
Vg - V
diL(t)
=
≈
L
L
dt
the inductor current changes with an
essentially constant slope

Inductor voltage and current subinterval 2: switch in position 2
+
Inductor voltage
vL= - v(t)
Small ripple approximation:
vL ≈ - V
L
-
Vg
+
-
+
ic(t)
VL(t)
C
R
V(t)
iL(t)
-
Knowing the inductor voltage, we can now find the inductor current via
diL (t)
dt
Solve for the slope:
vL(t)=L
V
diL(t)
≈L
dt
the inductor current changes with an
essentially constant slope
Inductor voltage and current waveforms
VL(t)
Vg -V
DTs
DTs
t
switch
position:
-V
1
iL(t)
I
iL(0)
0
2
1
iL(DTs)
Vg -V
L
vL(t)=L
ΔiL
-V
L
DTs
Ts
t
diL (t)
dt
Determination of inductor current ripple magnitude
iL(t)
I
iL(0)
iL(DTs)
Vg -V
L
0
ΔiL
-V
L
DTs
Ts
（changes in iL）=（slope）（length of subinterval）
2ΔiL =
ΔiL =
Vg -V
2L
DTs
Vg -V
L
DTs
L =
Vg -V
DTs
2ΔiL
Inductor current waveform during start-up transient
iL(t)
iL(Ts)
iL(0)=0
0 DTs Ts
Vg –v(t)
L
-v(t)
L
2Ts
iL(nTs)
iL((n+1)Ts)
nTs (n+1)Ts
When the converter operates in equilibrium:
iL((n+1)Ts)= iL(nTs)
t

The principle of inductor volt- second balance: Derivation
Inductor defining relation:
diL (t)
vL(t)=L
dt
Integrate over one complete switching period:
∫
1
iL(Ts) - iL(0)=
L
Ts
VL(t)dt
0
In periodic steady state, the net changes 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:
Ts
0 =
∫
1
Ts
0
VL(t) dt = <vL>
The average inductor voltage is zero in steady state.

Inductor volt-second balance:Buck converter example
VL(t)
inductor voltage waveform
total area λ
Vg -V
previously derived:
t
DTs
Integral of voltage waveform is area of rectangles:
∫
Ts
λ =
VL(t) dt = (Vg –V)( DTs)+( -V) ( DTs)
0
average voltage is
<vL> =
-V
λ
=D (Vg –V) +D'( -V)
Ts
Equate to zero and solve for V:
0=D Vg –(D+D')V= D Vg –V
V=D Vg
3.1.2Boost converter

Boost converter example
2
L
Boost converter
with ideal switch
iL(t) +
Vg
-
+
iC(t)
vL(t)
1
+
-
v
R
C
+ L
Realization using
iL(t)
power MOSFET
and diode
-
D1
ic(t)
VL(t)
Q1
Vg +DTs Ts
t
C
+
R
v
+
-
-
Boost converter analysis

2
L
iL(t) +
original
Vg
vL(t)
1
+
-
converter
+
iC(t)
v
R
C
Switch in position 1
Switch in position 2
L
L
iL(t) +
Vg
+
-
vL(t)
+
iC(t)
C
R
iL(t) +
v Vg +
vL(t)
+
iC(t)
C
R
v
-
-
-

Subinterval 1: switch in position 1
Inductor voltage and capacitor current
vL=Vg
iC= - v/R
Small ripple approximation:
vL=Vg
iC= - V/R
iL(t) +
Vg
+
-
L
vL(t)
+
iC(t)
C
R
v
-

Subinterval 2: switch in position 2
Inductor voltage and capacitor current
L
vL=Vg -v
iL(t) +
iC=iL - v/R
Small ripple approximation:
vL=Vg -V
iC= I - V/R
Vg
+
-
vL(t)
+
iC(t)
C
R
v
-

Inductor voltage and capacitor current waveforms
VL(t)
Vg
DTs
D'Ts
t
Vg -V
iC(t)
1 –V/R
DTs
D'Ts
t
-V/R

Inductor volt- second balance
Net volt-seconds applied to inductor
VL(t)
over one switching period：
∫
Ts
Vg
DTs
D'Ts
t
VL(t) dt = ( Vg) DTs+(Vg –V) D'Ts
0
Vg -V
Equate to zero and collect terms：
Vg（D+ D'）-VD'=0
Solve for V：
V=
Vg
D'
The voltage conversion ratio is therefore：
1
V
1
M（D）=
= D' =
1-D
Vg
Conversion ratio M(D) of the boost converter
5
1
1
M（D）=
D' = 1-D
4
M（D） 3
2
1
0
0
0.2
0.4
0.6
D
0.8
1

Determination of inductor current dc component
Capacitor charge balance：
∫
Ts
iC(t) dt =（-
0
V
R
）DTs +（I-
iC(t)
V
R
DTs
I –V/R
D'Ts
）D'Ts
t
-V/R
Collect terms and equate to zero：
V
- （D+D'）+I D'=0
R
Solve for I：
I=
V
I
Vg/R
8
6
4
D'R
Eliminate V to express in terms of Vg：
Vg
I= 2
D' R
0
2
0
0.2 0.4 0.6 0.8
D
1
Continuous- Conduction- Mode (CCM) and Discontinuous
Conduction-Mode (DCM) of boost
L
M
VD
EM
V
a)
uo
E
3.2 Composite DC/DC converters and connection of multiple DC/DC
converters
3.2.1 A current reversible chopper
uo
V1
VD2
E
L
R
io
V2
O
VD1
uo
M
EM
io iV1
iD1
O
Can be considered as a
combination of a Buck and a Boost
Can realize two- quadrant (I & II)
operation of DC motor:
forward motoring,
forward braking
t
t
uo
t
O
io
O
iV2 iD2
t
3.2.2Bridge chopper (H-bridge chopper)
V1
VD2
E
V2
VD1
V3
uo
L
R io
+
VD4
M
EM
V4
VD3
3.2.3Multi-phase multi-channel DC/DC converter
L
V1
i1
V2
i 2 L2
i 3 L3
V3
E C
L1
VD1 VD2 VD3
u1
i0
M
u
O
u2 u 3
Current output capability is increased due
to multi- channel paralleling.
Ripple in the output voltage and current is
reduced due to multi-channel paralleling.
Ripple in the input current is reduced due to
multi- phase paralleling.