Dc Converter
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Transcript of Dc Converter
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DC TO DC CONVERTER
PRESENTED BY
ANKAN BANDYOPADHYAY
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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 multipleDC/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
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Basic DC to DC converters
Buck converter
SPDT 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
osition:
DTs DTs
0
1 12
t
12
0 DTs Ts
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Dc component of switch output voltage
Vs(t)
Vg=DVg
DTs Ts0
t
Fourier analysis:DC component =average value:
0
area=
D Ts Vg
=
0
TsVs(t) tTs
1
= =DVg1
TsTsVg
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Insertion of low- pass filter to remove switching
harmonics and pass only dc component
+-
+
-
V(t)RVg
+
-
Vs(t)
1
2
L
C
v =DVgV
Vg
o0 1 D
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Basic operation principle of buck converter
+-
+
-
V(t)RVg
+
-
Vs(t)
1
2
L
C
Buck converter with
ideal switch
Realization using
power MOSFET
and diode+-
+ -
VL(t) ic(t)
Vg
iL(t)
tDTs Ts
+
L
D1 R
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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
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Actual output voltage waveform of buck converter
+-
+
-
V(t)RVg
+ -
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
0t
Actual waveform
v(t) = V+ v ripple(t)
DC component V
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Buck converter analysis: inductor current waveform
+
-
+
-
V(t)RVg
+ -
VL(t)
1
2
L
C
original
converter
ic(t)
iL(t)
Switch in position 1 Switch in position 2
+-
+
-
V(t)RVg
+ -
VL(t)
L
C
ic(t)
iL(t)
+-
+
-
V(t)RVg
+ -
VL(t)
L
C
ic(t)
iL(t)
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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)RVg
+ -
VL(t)
L
C
ic(t)
iL(t)
vL(t)=LdiL(t)
dtSolve for the slope:
diL(t)
dt=
vL(t)
L
Vg - V
L
the inductor current changes with an
essentially constant slope
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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)=LdiL (t)
dtSolve for the slope:
diL(t)
dt
V
L
the inductor current changes with an
essentially constant slope-
+-
+
-
V(t)RVg
+ -
VL(t)
L
C
ic(t)
iL t)
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Inductor voltage and current waveforms
VL(t)
Vg -V
switch
osition:
DTs DTs
-V
1 12
t
vL(t)=LdiL (t)
dtiL(t)
tDTs Ts0
I
iL(0)
iL(DTs)
Vg -VL
-VL
iL
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Determination of inductor current ripple magnitude
changes in iL=slopelength of subinterval
Vg -V
L2iL DTs=
iL=Vg -V
2LDTs L =
Vg -V
2iLDTs
iL(t)
DTs Ts0
IiL(0)
iL(DTs)
Vg -V
L
-V
L
iL
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Inductor current waveform during start-up transient
iL(t)
tDTsTs0
When the converter operates in equilibrium:
iL 0 =0iL(Ts)
Vgv(t)
L-v(t)
L
2Ts
iL(nTs)
nTs n+1 Ts
iL((n+1)Ts)
iL((n+1)Ts)=iL(nTs)
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The principle of inductor volt- second balance:Derivation
Inductor 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 waveformis zero whenever the converter operates in steady state.
An equivalent form:
The average inductor voltage is zero in steady state.
vL(t)=LdiL (t)
dt
0 TsVL(t)tL1iL(Ts) -iL(0)=0TsVL(t)t=0
=
0Ts
VL(t)dtTs
1 =0
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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
0Ts
VL(t)dt= = (VgV)( DTs)+( -V) ( DTs)
=Ts
=D (VgV) +D'( -V)
0=D Vg(D+D')V= D VgV V=D Vg
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3.1.2Boost converter
Boost converter example
+-
+
-
vR
Vg
+ -vL(t)
1
2L
C
Boost converter
with ideal switch
iL(t) iC(t)
Realization using
power MOSFET
and diode+-
ic(t)
Vg
iL(t)
tDTs Ts
+ -
VL(t)
L D1
R
+-
Q1
+
-
vC
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Boost converter analysis
original
converter
Switch in position 1 Switch in position 2
+-
+
-
vRVg
+ -
vL(t)1
2L
C
iL(t)
+-
+
-
vRVg
+ -
vL(t)
L
C
iL(t) iC(t)
iC(t)
+-
+
-
vRVg
+ -
vL(t)
L
C
iL(t) iC(t)
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Subinterval 1: switch in position 1
Inductor voltage and capacitor current
vL=Vg
Small ripple approximation:
iC= - v/R+-
+
-
vRVg
+ -vL(t)
L
C
iL(t) iC(t)
vL=Vg
iC= - V/R
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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
+-
+
-
vRVg
+ -
vL(t)
L
C
iL(t) iC(t)
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Inductor voltage and capacitor current waveforms
VL(t)
VgDTs D'Ts
Vg -V
t
iC(t)
-V/R
DTs D'Ts
1V/R
t
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Inductor volt- second balance
VL(t)
Vg
DTs D'Ts
Vg -V
t
0Ts
VL(t)dt = ( Vg) DTs+(VgV) D'Ts
Net volt-seconds applied to inductor
over one switching period
Equate to zero and collect terms
VgD+ D'-VD'=0
Solve for V
V= VgD'
The voltage conversion ratio is therefore
V
Vg D'MD= = =
1
1-D
1
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Conversion ratioM(D) of the boost converter
D'
MD= =1
1-D
1
D
MD
0
01
2
3
4
5
0.2 0.4 0.6 0.8 1
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Determination of inductor current dc component
Vg/R
I
D0
02
46
8
0.2 0.4 0.6 0.8 1
iC(t)
-V/R
DTs D'Ts
IV/R
t
Capacitor charge balance
0
TsiC(t)dt =- D'Ts
V
RDTs+I-
V
R
Collect terms and equate to zero
-V
RD+D'+I D'=0
Solve for I
V
D'RI=
Eliminate V to express in terms of Vg
Vg
D'I= 2
R
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Continuous- Conduction- Mode (CCM) and Discontinuous Conduction-
Mode (DCM) of boost
M E
VDL
V uoEM
a)
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3.2 Composite DC/DC converters and connection of multiple DC/DC
converters
3.2.1 A current reversible chopper
E L
V1
VD1 uo
ioV2
VD2
EMM
R
t
tO
O
uo
io iV1 iD1
t
tO
O
uo
io
iV2 iD2
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
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3.2.2Bridge chopper (H-bridge chopper)
E L R
+ -
V1
VD1
uo
V3
EM
V2
VD2 io
V4
VD3
VD4
M
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3.2.3Multi-phase multi-channel DC/DC converter
C
L
E M
V1
VD1
L1
i0
uO
V2
V3
i1
i2
i3
VD2VD3
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.