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Power Electronics and Drives (Version 2): Dr.
Zainal Salam
1
Chapter 3
DC to DC CONVERTER(CHOPPER)
• General• Buck converter• Boost converter• Buck-Boost converter• Switched-mode power supply• Bridge converter• Notes on electromagnetic compatibility
(EMC) and solutions.
DC-DC Converter (Chopper)
• DEFINITION: Converting the unregulated DC input to a controlled DC output with a desired voltage level.
• General block diagram:
LOAD
Vcontrol(derived from
feedback circuit)
DC supply(from rectifier-filter, battery,fuel cell etc.)
DC output
• APPLICATIONS: – Switched-mode power supply (SMPS), DC
motor control, battery chargers
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
3
Linear regulator• Transistor is
operated in linear (active) mode.
• Output voltage
• The transistor can be conveniently modelled by an equivalent variable resistor, as shown.
• Power loss is high at high current due to:
TLo RIP 2=
TLo RIV =
+
−
VoRL
+ VCE − IL
MODEL OF LINEARREGULATOR
RT
EQUIVALENTCIRCUIT
Vs
RL
+ VCE −IL
VsVo
+
−
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
4
Switching Regulator• Power loss is zero
(for ideal switch):– when switch is
open, no current flow in it,
– when switch is closed no voltage drop across it.
– Since power is a product of voltage and current, no losses occurs in the switch.
– Power is 100% transferred from source to load.
• Switching regulator is the basis of all DC-DC converters
+
−
Vo
RL
+ VCE − IL
MODEL OF LINEARREGULATOR
EQUIVALENT CIRCUIT
Vs
RL
IL
VsVo
+
−
(ON)closed
(OFF)open
(ON)closed
DT T
OUTPUT VOLTAGE
Vo
SWITCH
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
5
Buck (step-down) converter
Vd
L
D C RL
S+
Vo
−
Vo
+
CIRCUIT OF BUCK CONVERTER
CIRCUIT WHEN SWITCH IS CLOSED
CIRCUIT WHEN SWITCH IS OPENED
Vo
+
−
−
iL
Vd D RL
S
Vd D RL
S
+ −vL
+ vL −
iL
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
6
Circuit operation when switch is turned on (closed)
• Diode is reversed biased. Switch conducts inductor current
• This results in positive inductor voltage, i.e:
• It causes linear increase in the inductor current
odL VVv −=
∫=⇒
=
dtvL
i
dtdiLv
LL
LL
1
Vd VD
+ vL -
C RL
+
−
Vo
Vd−Vo
−Vo
closedopened
closedopened
t
DT Tt
iLmin
iLmax
IL
vL
iL
iL
+
−
S
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
7
Operation when switch turned off (opened)
• Because of inductive energy storage, iL
continues to flow.
• Diode is forward biased
• Current now flows through the diode and
oL Vv −=
Vd
+ vL -
C RL
+
−
Vo
Vd−Vo
−Vo
closedopened
closedopened
t
DT Tt
iLmin
iLmax
IL
vL
iL
iL
S
D
(1-D)T
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
8
Analysis for switch closed
( ) DTL
VVi
LVV
DTi
ti
dtdi
i
i
LVV
dtdi
dtdiL
VVv
odclosedL
odLLL
L
L
odL
LodL
⋅
−
=∆
−=
∆=
∆∆
=
−=⇒
=
−=
Figure From
linearly. increase must Therefore
constant. tive-posi a is of vative
-deri thesince :Note
oltage,inductor v The
IL
iL max
DT T
iL
Vd− Vo
vL
t
t
iL min
closed
∆iL
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
9
Analysis for switch opened
( ) TDLVi
LV
TDi
ti
dtdi
ii
LV
dtdi
dtdiL
Vv
oopenedL
oLLL
LL
oL
LoL
)1(
)1(
Figure From
linearly. decreasemust constant, tive
-nega a is of vative-deri thesince :Note
opened,switch For
−⋅
−=∆
−=
−∆
=∆∆
=
−=⇒
=
−=
IL
iL max
DT T
iL
Vd− Vo
vL
t
t
iL min
opened
∆iL
(1− D)T
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
10
Steady-state operation
( ) ( )
do
so
sod
openedLclosedL
L
L
DVV
TDLVDT
LVV
ii
i
i
=⇒
=−⋅
−−⋅
−
=∆+∆
0)1(
0
:i.e zero, is period oneover of change theisThat cycle.next theof begining
at the same theis cycle switching of endat the that requiresoperation state-Steady
iL Unstable current
Decaying current
Steady-state current
t
t
t
iL
iL
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
11
−−=
∆−=
−+=
−+=∆+=
==⇒
=
LfD
RViII
LfD
RV
TDL
VR
ViII
RVII
oL
L
o
ooLL
oRL
2)1(1
2
:current Minimum
2)1(1
)1(21
2
:current Maximum
Rin current Average currentinductor Average
min
max
L
Average, Maximum and Minimum inductor current
IL
Imax
Imin
iL
∆iL
t
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
12
Continuous current operationiL
Imax
Imint
0
min
min
min
min
be bechosen is Normally operation. of mode continous ensure
current toinductor minimum theis This2
)1(
02
)1(1
,0 operation, continuousFor
2)1(1
2
analysis, previous From
LL
RfDLL
LfD
RV
I
LfD
RViII
o
oL
L
>>
⋅−
=≥⇒
≥
−−⇒
≥
−−=
∆−=
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
13
Output voltage ripple
2
2
8)1(
factor, ripple theSo,8
)1(8
82221
:formula area triangleuse figure, From
LCfD
VVr
LCfD
CiTV
iTiTQ
CQVVCQCVQ
iii
o
o
Lo
LL
ooo
RLc
−=
∆=
−=
∆=∆∴
∆=
∆
=∆
∆=∆⇒∆=∆⇒=
+=
iRiL
LiC
iL
iL=IR
imax
imin
0
0Vo
Vo/R+
−
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
14
Design procedures for Buck
Vd(inputspec.)
SWITCH
f = ?D = ?TYPE ?
D
L
Lmin= ?L = 10Lmin
Cripple ?
RLPo = ?Io = ?
• Calculate D to obtain required output voltage.
• Select a particular switching frequency:– preferably >20KHz for negligible acoustic
noise– higher fs results in smaller L, but higher device
losses. Thus lowering efficiency and larger heat sink. Also C is reduced.
– Possible devices: MOSFET, IGBT and BJT. Low power MOSFET can reach MHz range.
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
15
Design procedures for Buck
• Determine Lmin. Increase Lmin by about 10 times to ensure full continuos mode.
• Calculate C for ripple factor requirement.
• Capacitor ratings:– must withstand peak output voltage– must carry required RMS current. Note RMS
current for triangular w/f is Ip/3, where Ip is the peak capacitor current given by ∆iL/2
• Wire size consideration:– Normally rated in RMS. But iL is known as
peak. RMS value for iL is given as:
22
, 32
∆+= L
LRMSLiII
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
16
Examples of Buck converter
• A buck converter is supplied from a 50V battery source. Given L=400uH, C=100uF, R=20 Ohm, f=20KHz and D=0.4. Calculate: (a) output voltage (b) maximum and minimum inductor current, (c) output voltage ripple.
• A buck converter has an input voltage of 50V and output of 25V. The switching frequency is 10KHz. The power output is 125W. (a) Determine the duty cycle, (b) value of L to limit the peak inductor current to 6.25A, (c) value of capacitance to limit the output voltage ripple factor to 0.5%.
• Design a buck converter such that the output voltage is 28V when the input is 48V. The load is 8Ohm. Design the converter such that it will be in continuous current mode. The output voltage ripple must not be more than 0.5%. Specify the frequency and the values of each component. Suggest the power switch also.
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
17
Boost (step-up) converter
Vd
L D
C
RL
S
Vd
L D
CRL
S
Vd
LD
C RLS
+ vL −
+
Vo
−
+ vL -
Vo
+
CIRCUIT OF BOOST CONVERTER
CIRCUIT WHEN SWITCH IS CLOSED
CIRCUIT WHEN SWITCH IS OPENED
Vo
+
−
−
iL
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
18
Boost analysis:switch closed
Vd
L D
CS
+ vL −
iL
+vo−
( )LDTVi
LV
dtdi
DTi
ti
dtdi
LV
dtdi
dtdiL
Vv
dclosedL
dL
LLL
dL
LdL
=∆
=⇒
∆=
∆∆
=
=⇒
=
=
DT T
iL
vL
CLOSED
t
t
Vd
Vd− Vo
∆iL
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
19
Switch opened
( ) ( )L
DTVViL
VVdtdi
TDi
ti
dtdi
LVV
dtdi
dtdiL
VVv
odopenedL
odL
L
LL
odL
LodL
)1(
)1(
−−=∆⇒
−=⇒
−∆
=∆∆=
−=⇒
=
−=
DT T
( 1-D )T
iL
vL
OPENED
t
t
Vd
Vd− Vo
∆iL
Vd
D
CS
+ vL -
iL
+vo-
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
20
Steady-state operation( ) ( )
( )
DVV
LTDVV
LDTV
ii
do
odd
openedLclosedL
−=⇒
=−−
−
=∆+∆
1
0)1(
0
• Boost converter produces output voltage that is greater or equal to the input voltage.
• Alternative explanation:– when switch is closed, diode is reversed. Thus
output is isolated. The input supplies energy to inductor.
– When switch is opened, the output stage receives energy from the input as well as from the inductor. Hence output is large.
– Output voltage is maintained constant by virtue of large C.
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
21
Average, Maximum, Minimum inductor current
LDTV
RDViII
LDTV
RDViII
RDVI
RDV
RD
V
IV
RVIV
ddLL
ddLL
dL
d
d
Ld
odd
2)1(
2
2)1(
2
currentinductor min Max,
)1(
currentinductor Average
)1()1(
powerOutput powerInput
2min
2max
2
2
2
2
2
−−
=∆
−=
+−
=∆
+=
−=
−=
−
=
=
=
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
22
L and C values
( )
( )
RCfD
VVr
RCfDV
RCfDTVV
VCDTR
VQ
fRDD
TRDDL
LDTV
RDV
I
o
o
ooo
oo
dd
=∆
=
==∆
∆=
=∆
−=
−=
≥−−
≥
factor Ripple
21
21
02)1(
0
operation, continousFor
2
2
min
2
min
������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������
DT T
imax
imin
imin
imax
ic
iD
iL
Vd
vL
∆Q
Vd−Vo
Io=Vo / R
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
23
Examples
• The boost converter has the following parameters: Vd=20V, D=0.6, R=12.5ohm, L=65uH, C=200uF, fs=40KHz. Determine (a) output voltage, (b) average, maximum and minimum inductor current, (c) output voltage ripple.
• Design a boost converter to provide an output voltage of 36V from a 24V source. The load is 50W. The voltage ripple factor must be less than 0.5%. Specify the duty cycle ratio, switching frequency, inductor and capacitor size, and power device.
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
24
Buck-Boost converter
Vd L
D
C RL
S+
Vo
−
Vo
+
CIRCUIT OF BUCK-BOOST CONVERTER
CIRCUIT WHEN SWITCH IS CLOSED
CIRCUIT WHEN SWITCH IS OPENED
Vo
+
−
−
iLVd vL
+
−
iLVd vL
+
−
D
DS
S
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
25
Buck-boost analysis
������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������
DT T
imax
imin
imin
imax
ic
iD
iL
Vd
vL
∆Q
Vd−Vo
Io=Vo / R
LTDVi
LV
TDi
ti
LV
dtdi
dtdiLVv
LDTVi
LV
DTi
ti
LV
dtdi
dtdiLVdv
oopenedL
oLL
oL
LoL
dclosedL
dLL
dL
LL
)1()(
)1(
openedSwitch
)(
closedSwitch
−=∆
=−∆
=∆∆
=⇒
==
=∆
=∆
=∆∆
=⇒
==
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
26
Output voltage
−
−=⇒
=−
+
DDV
LTDV
LDTV
s
od
1V
0)1(
:operation stateSteady
o
• NOTE: Output of a buck-boost converter either be higher or lower than the source voltage.– If D>0.5, output is higher– If D<0.5, output is lower
• Output voltage is always negative• Note that output is never directly • connected to load. Energy is stored in
inductor when switch is closed and transferred to load when switch is opened.
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
27
Average inductor current
2
2
2
2
)1(
,for ngSubstituti
:ascurrent inductor average torelated iscurrent source averageBut
i.e. source,by thesuppliedpower equalmust load by the absorbedpower
converter, in the losspower no Assuming
DRDV
DVP
RDVVI
V
DIVR
V
DII
IVR
V
PP
d
d
o
d
oL
o
Ldo
Ls
sdo
so
−===⇒
=⇒
=
=
=
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
28
L and C values
RCfD
VVr
RCfDV
RCDTVV
VCDTR
V
fRDL
LDTV
DRD
LDTV
DRDViII
LDTV
DRDViII
o
o
ooo
oo
d
ddLL
ddLL
=∆
=
==∆
∆=
=∆
−=⇒
=+−
−−
=∆
−=
+−
=∆
+=
Q
ripple, tageOutput vol
2)1(
02)1(
Vcurrent, continuousFor
2)1(2
2)1(2
current,inductor min andMax
2
min
2d
2min
2max
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
29
Control of DC-DC converterusing pulse width modulation-
PWM
Comparator
Vcontrol
SawtoothWaveform
Vo (desired)
Vo (actual)
+
-
Switch control signal
SawtoothWaveform
Vcontrol 1
Switchcontrolsignalton 2
T
Vcontrol 2
ton 1
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
30
Switch-mode power supply (SMPS)
• Advantages over linear power-Efficient (70-95%)-Weight and size reduction
• Disadvantages -Complex design-EMI problems
• However above certain ratings,SMPS is the only feasible choice
• Types of SMPS-Flyback-forward-Push-pull-Bridge (half and full)
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
31
Linear and switched mode power supplies block diagram
Basic Block diagram of linear power supply
Base/gateDrive
ErrorAmp.
LineInput
φφ 3/1 50/60 HzIsolation
Transformer
Rectifier+
Vd
-
Vce=Vd-VoC E
B
Vo
Vref
RL
+Vo
+
Vo
-
Basic Block diagram of SMPS
EMIFILTER
RECTIFIERAND
FILTER
HighFrequency
rectifierandfilter
Base/gatedrive
PWMController
errorAmp
Vo
Vref
DCRegulated
DC-DC CONVERSITION + ISOLATION
DCUnregulated
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
32
High frequency transformer
: Models
;
iprelationshoutput -input Basic voltage varying- meup/down ti step ii)
isolation electricaloutput -Input i)
:function Basic
1
2
2
1
2
1
2
1NN
ii
NN
vv
==
V1V2
+
−
+
−
i1 i2N1 N2
Ideal model
V1V2
+
−
+
−
i1 i2N1 N2
Model used formost PE application
Lm
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
33
Flyback Converter
Vs
+
−Vo
Vs
Vs
Vs
N1 N2i1
i2
+
-v2v1
+
-
iLM
iD
+ -vD
iCiR
C R
vSW+ −
+
−
Vo
iS
N1 N2
+
-
0
0
v1
v1=Vs
iLM
is=iLM
+=
2
1NNVVV ossw
+ −
−
2
11 N
NVv o
v1
+
−iLM
N1 N2
v2= -VS
+
− +
−Vo
iD
LM
vSW
Vo
+
−
Flyback converter circuit
Model with magnetisinginductance
Switch closed
Voltage and currentconditions when switchopened
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
34
Flyback waveforms
DT T
iLm
DT
T
t
t
is
DT
T
t
t
t
iD
iC
T
v1
-V(N1/N2)
Vo/ R
Vs
DT
DT T
∆iLM
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
35
Analysis: switched closed
( )
00
Therefore,
0
er, transform theof side load On the
21
1
2
1
2
1
212
1
==
<
−−=
=
=
=∆⇒
=∆
=∆
=
==
ii
NNVVv
NNV
NNvv
LDTVi
LV
DTiL
dtiL
dtdiL
dtdiLVv
doD
d
m
dclosedL
m
dmmm
Lmmd
m
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
36
Analysis: switch opened
( )
( )
( ) ( )
( )
−
=⇒
=
−+⇒
=∆+∆
−−=∆⇒
−=
−∆
=∆
=
−==
−=
=⇒
−=
−=
2
10
2
10
2
10
2
10
2
101
2
10
2
121
022
101
)1(
01
0
operation, state-steadyFor
)1(
1
;
NN
DDVV
NN
LTDV
LDTV
ii
NN
LTDVi
NN
LV
TDi
dti
dtdi
NNVv
dtdi
L
NNV
NNvv
VvNNVv
d
mm
d
openedLclosedL
mopenmL
m
mLmLmL
mLm
mm
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
37
Output voltage
• Input output relationship is similar to buck-boost converter.
• Output can be greater of less than input,depending upon D.
• Additional term, i.e. transformer ratio is present.
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
38
Average inductor current
( )
( )
−
=
−
=
=⇒
=
==
=
=
1
202
1
22
20
20
20
0
)1()1(
:as written also iscurrent inductor average The
for solving and Substitute
:as torelated is
NN
RDV
NN
RDDVI
DRVVI
RVDIV
I
DITDTI
I
II
RVIV
PP
dL
dL
Ld
L
LL
s
Ls
sd
s
m
m
m
m
mm
m
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
39
Max, Min inductor current, Lmin, C values
( )
RCfD
VVr
NN
fRDVL
fLDV
LDTV
NN
RDDV
I
LDTV
NN
RDDVi
II
LDTV
NN
RDDVi
II
dm
m
d
m
dd
L
m
ddLLL
m
ddLLL
m
mmm
mmm
=∆
=
−=
==
−
=
−
−
=∆
−=
+
−
=∆
+=
0
0
2
2
12
min
2
1
22
min
2
1
22min,
2
1
22max,
converter,boost similar to isn calculatio ripple The
2)1(
22)1(
0, operation, continuosFor
2)1(2
2)1(2
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
40
Full-bridge converter
SW2SW4
VS
NS
NS
+
−vx C R
+
−Vo
+
−
vp
SW1,SW2
SW3,SW4 DT T
2T DTT
+2VP
VS
-VS
Vx
P
SS N
NV
DT T T T
SW3Lx
SW1
2DT+
2
Power Electronics and Drives (Version 2): Dr.
Zainal Salam
41
Full bridge: basic operation
• Switch “pair”: [S1 & S2];[S3 & S4].
• Each switch pair turn on at a time as shown. The other pair is off.
• “AC voltage” is developed across the primary. Then transferred to secondary via high frequency transformers.
• On secondary side, diode pair is “high frequency full wave rectification”.
• The choke (L) and © acts like the “buck converter” circuit.
• Output Voltage DNNVV
p
sso ⋅
= 2
Recommended