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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 converterSPDT switch changes dc
component
Switch output voltage waveform
Duty cycle D: 0 ≤ D ≤ 1
complement D: D´ = 1 - D
+ -
+
-
V(t)
R
Vg
+
-
Vs(t)
Vs(t) Vg
switch position:
DTs
D Ts
0
1
1
2
t
1
2
0
DTs
Ts
Dc component of switch output voltage
Vs(t) Vg
<Vs>=DVg
DTs
Ts
0
t
Fourier analysis:DC component =average value:
0
area= D Ts Vg
<Vs>
=
∫
0
Ts Vs(t)
dt
Ts
1
<Vs>
=
=DVg
1
Ts DTsVg
Insertion of low- pass filter to remove switching harmonics and pass only dc component
+ -
+
-
V(t)
R
Vg
+
-
Vs(t)
1
2
L
C
v≈<vs>
=DVg
V
Vg
o
0 1 D
Basic operation principle of buck converter
+ -
+
-
V(t)
R
Vg
+
-
Vs(t)
1
2
L
C
Buck converter with
ideal switch
Realization using
power MOSFET
and diode
+ -
+
-
VL(t)
ic(t)
Vg
iL(t)
t
DTs
Ts
+
L
D1
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
+ -
+
-
V(t)
R
Vg
+
-
VL(t)
1
2
L
C
Buck converter
containing practical
low-pass filter
ic(t)
iL(t)
Actual output voltage
waveform
v(t ) = V + v ripple(t)
v(t )
V
0
t
Actual waveform
v(t ) = V + v ripple(t)
DC component V
Buck converter analysis: inductor current waveform
+ -
+
-
V(t)
R
Vg
+
-
VL(t)
1
2
L
C
original
converter
ic(t)
iL(t)
Switch in position 1
Switch in position 2
+ -
+
-
V(t)
R
Vg
+
-
VL(t)
L
C
ic(t)
iL(t)
+ -
+
-
V(t)
R
Vg
+
-
VL(t)
L
C
ic(t)
iL(t)
Inductor voltage and current subinterval 1: switch in position 1
Inductor voltage
vL=Vg - v(t) Small ripple approximation:
vL=Vg - V
Knowing the inductor voltage, we can now find the inductor current via
+ -
+
-
V(t)
R
Vg
+
-
VL(t)
L
C
ic(t)
iL(t)
vL(t)=L diL(t)
dt Solve for the slope:
diL(t)
dt =
vL(t)
L ≈ Vg - V
L
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
Knowing the inductor voltage, we can now find the inductor current via
vL(t)=L diL (t)
dt Solve for the slope:
diL(t)
dt ≈
V L
the inductor current changes with an
essentially constant slope -
+ -
+
-
V(t)
R
Vg
+
-
VL(t)
L
C
ic(t)
iL(t)
Inductor voltage and current waveforms
VL(t)
Vg -V
switch position:
DTs
DTs
-V
1
1
2
t
vL(t)=L diL (t)
dt iL(t)
t
DTs
Ts
0
I
iL(0)
iL(DTs)
Vg -V
L
-V
L
Δ iL
Determination of inductor current ripple magnitude
(changes in iL)=(slope)(length of subinterval)
Vg -V
L
2Δ iL
DTs
=
Δ iL
=
Vg -V
2L DTs
L
=
Vg -V
2Δ iL
DTs
iL(t)
DTs
Ts
0
I
iL(0)
iL(DTs)
Vg -V
L
-V
L
Δ iL
Inductor current waveform during start-up transient
iL(t)
t
DTs
Ts
0
When the converter operates in equilibrium:
iL(0)=0
iL(Ts)
Vg –v(t)
L
-v(t)
L
2Ts
iL(nTs)
nTs
(n+1)Ts
iL((n+1)Ts)
iL((n+1)Ts)= iL(nTs)
The principle of inductor volt- second balance: DerivationInductor defining relation:
Integrate over one complete switching period:
In periodic steady state, the net changes in inductor current is zero:
Hence, the total area(or volt-seconds)under the inductor voltage waveform is zero whenever the converter operates in steady state.
An equivalent form:
The average inductor voltage is zero in steady state.
vL(t)=L diL (t)
dt
∫ 0
Ts VL(t)
dt
L 1
iL(Ts) -
iL(0)=
∫ 0
Ts VL(t)
dt
=
0
<vL>
=
∫ 0
Ts VL(t)
dt
Ts 1
=
0
Inductor volt-second balance:Buck converter example
Integral of voltage waveform is area of rectangles:
average voltage is
Equate to zero and solve for V:
inductor voltage waveform
previously derived:
VL(t)
Vg -V
DTs
-V
t
total area λ
∫ 0
Ts VL(t)
dt
=
λ
= (Vg –V)( DTs)+( -V) ( DTs)
<vL>
=
Ts λ
=D (Vg –V) +D'( -V)
0=D Vg –(D+D')V= D Vg –V
V=D Vg
3.1.2Boost converter Boost converter example
+ -
+
-
v
R
Vg
+
-
vL(t)
1
2
L
C
Boost converter
with ideal switch
iL(t)
iC(t)
Realization using
power MOSFET
and diode
+ -
ic(t)
Vg
iL(t)
t
DTs
Ts
+
-
VL(t)
L
D1
R
+ -
Q1
+
-
v
C
Boost converter analysis
original
converter
Switch in position 1
Switch in position 2
+ -
+
-
v
R
Vg
+
-
vL(t)
1
2
L
C
iL(t)
+ -
+
-
v
R
Vg
+
-
vL(t)
L
C
iL(t)
iC(t)
iC(t)
+ -
+
-
v
R
Vg
+
-
vL(t)
L
C
iL(t)
iC(t)
Subinterval 1: switch in position 1
Inductor voltage and capacitor current
vL=Vg
Small ripple approximation:
iC= - v/R + -
+
-
v
R
Vg
+
-
vL(t)
L
C
iL(t)
iC(t)
vL=Vg
iC= - V/R
Subinterval 2: switch in position 2
Inductor voltage and capacitor current
vL=Vg -v
Small ripple approximation:
iC=iL - v/R
vL=Vg -V
iC= I - V/R
+ -
+
-
v
R
Vg
+
-
vL(t)
L
C
iL(t)
iC(t)
Inductor voltage and capacitor current waveforms
VL(t)
Vg
DTs
D'Ts
Vg -V
t
iC(t)
-V/R
DTs
D'Ts
1 –V/R
t
Inductor volt- second balance
VL(t)
Vg
DTs
D'Ts
Vg -V
t
∫
0
Ts VL(t)
dt
= ( Vg) DTs+(Vg –V) D'Ts
Net volt-seconds applied to inductor
over one switching period:
Equate to zero and collect terms:
Vg(D+ D')-VD'=0
Solve for V:
V=
Vg
D'
The voltage conversion ratio is therefore:
V
Vg
D'
M(D)=
=
=
1
1-D
1
Conversion ratio M(D) of the boost converter
D'
M(D)=
=
1
1-D
1
D
M(D)
0
0
1
2
3
4
5
0.2
0.4
0.6
0.8
1
Determination of inductor current dc component
Vg/R
I
D
0
0
2
4
6
8
0.2
0.4
0.6
0.8
1
iC(t)
-V/R
DTs
D'Ts
I –V/R
t
Capacitor charge balance:
∫
0
Ts iC(t)
dt
=(-
)D'Ts
V
R
)DTs
+(I-
V
R
Collect terms and equate to zero:
-
V
R
(D+D')+I D'=0
Solve for I:
V
D'R
I=
Eliminate V to express in terms of Vg:
Vg
D' I=
2
R
Continuous- Conduction- Mode (CCM) and Discontinuous Conduction-Mode (DCM) of boost
M E
VD L
V uo EM
a)
3.2 Composite DC/DC converters and connection of multiple DC/DC
converters 3.2.1 A current reversible chopper
E L
V 1
VD 1 u o
i o
V 2
VD 2
E M M
R
t
t O
O
u o
i o i V1 i D1
t
t O
O
u o
i o
i V2 i D2
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
3.2.2Bridge chopper (H-bridge chopper)
E L R
+ -
V 1
VD 1
u oV 3
E M
V 2
VD 2 io
V 4
VD 3
VD 4
M
3.2.3Multi-phase multi-channel DC/DC converter
C
L
E M
V1
VD1
L1
i0
uO
V2
V3
i1
i2
i3
VD2 VD3
u1 u2 u3
L2
L3
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.