Power Electronics and Drives (Version 2) Dr. Zainal Salam 1 Chapter 3 DC to DC CONVERTER (CHOPPER) • Basic non-isolated DC-DC converter topologies: Buck, Boost, Buck-Boost, Cuk in CCM and DCM mode • Non-ideal effects on converter performance • Isolated DC-DC converters, switched- mode power supply • Control of DC-DC converters • High frequency transformer and inductor design • Notes on electromagnetic compatibility (EMC) and solutions.
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Power Electronics and Drives (Version 2) Dr. Zainal Salam
1
Chapter 3DC to DC CONVERTER
(CHOPPER)• Basic non-isolated DC-DC converter
topologies: Buck, Boost, Buck-Boost, Cuk in CCM and DCM mode
• Non-ideal effects on converter performance
• Isolated DC-DC converters, switched-mode power supply
• Control of DC-DC converters• High frequency transformer and inductor
design• 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
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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
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−−=
∆−=
−+=
−+=∆+=
==⇒
=
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
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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
Continuous Current Mode (CCM)
( )
( )
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
Cuk Converter
CIRCUIT OF CUK CONVERTER
Vd
L2
D C2 RLS+
Vo
−
L1 C1
iL1 iL2+ vc1-+
−vc2
Vd
L2
DC2 RL
+
Vo
−
L1
C1iL1 iL2
+
−vc2
S
Vd
L2
DC2 RLS
+
Vo
−
L1
C1iL1 iL2
+
−vc2
CIRCUIT WHEN SWITCH IS CLOSED
21 LC ii −=
11 LC ii =
Power Electronics and Drives (Version 2) Dr. Zainal Salam
30
Cuk analysis: from capacitor current point of view
iC1
OPENED
t
−iL2
iL1
CLOSED
( )
( )
( )[ ] ( )[ ]
)1(
0)1(ng,Substituti
0)1(zero, iscurrent average theoperation, periodicFor
i.e. source, by the suppliedpower the toequal is load by the absorbedpower The
:is C1in current The on. diode theforce L2 and L1in current theopened, isswitch When the
:is C1in current theand off is diodeclosed, isswitch When the
KVL,by computed is C1 across voltageaverage The
2
112
11
12
11
21
1
DD
II
TDIDTI
TDiDTi
IVIV
ii
ii
VVV
L
LLL
openCclosedC
LsLo
LopenC
LclosedC
odC
−=
=−+−
=−+
=−
−=
−=
−=
Power Electronics and Drives (Version 2) Dr. Zainal Salam
31
Cuk analysis
fLDV
LDTVi
orLV
DTi
dtdi
Lvv
fCLD
VV
DD
VV
VV
II
IVIV
dd
d
dL
o
o
s
o
s
o
L
LLsLo
L
L
L
11
1
11
222
2
112
1
1
1
closed, isswitch when DT interval In time8
1Hence, converter.buck theasion configurat same the
in are adnR) C2 (L2, stageoutput that theNote
)1(
:as written becan tageoutput vol Combining,
i.e. source, by the suppliedpower the toequal is load by the absorbedpower The
==∆
=∆
⇒
==
−=
∆
−
−=
−=
=−
Power Electronics and Drives (Version 2) Dr. Zainal Salam
32
Cuk design parameters
fRDL
DfRDL
fLDV
LDTVi
or
dtdiLv
VVVVVvvVvv
dd
LL
doodocLoLc
L
2)1(
2)1(
operation,current continuousFor
)(0
closed, isswitch when DT interval in time L2,For
min,2
min,1
2
2
22
222
1221
−=
−=
==∆
=
=+−=+=+−+=
Power Electronics and Drives (Version 2) Dr. Zainal Salam
33
Cuk analysis from inductor current point of view
decrease to causes This .n larger tha is since negative iswhich
:as written becan oltageinductor v The L1. and Vdfromenergy by diode he through tcharged C1isCapacitor
diode. he through tflow and off, isswitch When the
and n larger tha is seen that becan alsoIt
). ofpolarity the(Note ,
zero. are and thatassumed becan it state,steady In
11
11
21
01
1
21
LdC
CdL
LL
dC
oodC
LL
iVV
VVv
ii
VVV
VVVV
V V
+=
+=
Vd
L2
DC2 RLS
+
Vo
−
L1
C1iL1 iL2
+
−vc2
SWITCH IS OPENED
Closed
Open
t
DT T
t
vL1
iL1
vd
-vo
iL1
Closed
Open
t
DT T
t
vd
-vo
IL1
vL2
iL2
iL1
Power Electronics and Drives (Version 2) Dr. Zainal Salam
34
Cuk analysis
Vd
L2
DC2 RL
+
Vo
−
L1
C1iL1 iL2
+
−vc2
S
CIRCUIT WHEN SWITCH IS CLOSED
21 LC ii −=
increase. to causing L1, energy to feedsinput The
increases. reL2.Therefonoutput theenergy to ngtransferriswitch, he through tdischarged C1capacitor Since
switch. he through tflow andcurrent inductor The diode. thebiased-reverse on, isswitch When the
decrease. to causesh which whic
side,output on theSimilarly
1
21
221
1
L
LoC
LLC
Ldo
i
iVV
iiV
iVV
>
−=
Closed
Open
t
DT T
t
vL1
iL1
vd
-vo
iL1
Closed
Open
t
DT T
t
vd
-vo
IL1
vL2
iL2
Power Electronics and Drives (Version 2) Dr. Zainal Salam
35
Cuk Analysis
output theofpolarity theNote
1
:Combining
10)1)(()( :L2
1
10)1)(( :L1
L2, and L1 across voltages theof integral theEquating
1
1
1
1
DD
VV
VD
V
TDVDTVV
VD
V
DVVDTV
d
o
oc
ooc
dc
cdd
−=
=⇒
=−−+−
−=⇒
=−−+
Power Electronics and Drives (Version 2) Dr. Zainal Salam
36
Converters in CCM: Summary
Vd
L2
D C2 RLS+Vo
−
L1 C1
iL1 iL2+ vc1-+
−vc2
Vd
L D
CRL
S+Vo
−
Vd
L
DC RL
S
+Vo
−
Vd
L
D C RL
S+Vo
− fRDL
LCfDVV
DVVBuck
do
do
2)1(
81
min
2
−=
−=∆
=
fRDDL
RCfDVVD
VVBoost
do
do
2)1(
11
2
min−
=
=∆−
=
fRDL
RCfDVV
DDVV
BoostBuck
do
do
2)1(
1
2
min−
=
=∆−
−=
−
fRDL
DfRDL
LCfDVV
DDVV
Cuk
do
do
2)1(
2)1(81
1
2
2
1
2
−=
−=
−=∆
−−=
Power Electronics and Drives (Version 2) Dr. Zainal Salam
37
Buck in discontinuous current mode (DCM)
Vd VD
+ vL -
C RL
+
−
Vo
Vd−Vo
−Vo
closedopened
closedopened
t
tDT
is
vL
iL
iL
+
−
S
Imax
Imax
BUCK CONVERTER
D1T
T
( )
DTI
DTi
ti
LVV
dtdi
VVv
TDDI
TDIDTIT
I
RVII
DDD
VV
DVDVVTDVDTVV
LL
odL
odL
L
oRL
d
oood
ood
max
1max
1maxmax
1
11
inductor, across Voltage21
21
211
figure, From
zero) iscurrent capacitor average (because
current resistor equalscurrentinductor Average
)(0)(
zero, isoltageinductor v Average
=∆
=∆∆
⇒
−=
−=
+=
+=
==
+
=
=−⇒=−−
Power Electronics and Drives (Version 2) Dr. Zainal Salam
38
Buck in DCM
( ) ( )
++=
+
=
+−=
=−+
=+
=+
=
−
=∆=
−
RTLDD
DVDD
DVV
RTLDD
D
D
RTLDDD
RVDD
LTDVDDI
LTDVDT
LVViI
DVVI
ddo
oo
oosL
os
82
Hence,
2
8,for Solving
02gives,Which
21
21
,Substitute
,)( using and for Solving
21
2
1
1
12
1
11
1max
1max
max
Power Electronics and Drives (Version 2) Dr. Zainal Salam
39
Example
example. thisof parameters for the ratioduty and voltage