power electronics and drives (version 3-2003), dr. zainal salam, 2003 1 chapter 2 ac to dc...
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Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
1
Chapter 2AC to DC CONVERSION
(RECTIFIER)• Single-phase, half wave rectifier
– Uncontrolled: R load, R-L load, R-C load– Controlled– Free wheeling diode
• Single-phase, full wave rectifier– Uncontrolled: R load, R-L load, – Controlled – Continuous and discontinuous current mode
• Three-phase rectifier – uncontrolled – controlled
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
2
Rectifiers
• DEFINITION: Converting AC (from mains or other AC source) to DC power by using power diodes or by controlling the firing angles of thyristors/controllable switches.
• Basic block diagram
• Input can be single or multi-phase (e.g. 3-phase).
• Output can be made fixed or variable
• Applications: DC welder, DC motor drive, Battery charger,DC power supply, HVDC
AC input DC output
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
3
Single-phase, half-wave, R-load
mm
mRMSo
mm
mavgo
VV
tdtVV
VV
tdtVVV
5.02
)sin(21
,
(rms), tageOutput vol
318.0)sin(2
average),or (DC tageOutput vol
2
0
0
1
+vs
_
+vo
_
vo io
vs
t
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Half-wave with R-L load
tan
)(
:where
)sin()(
:is response forced diagram, From
response, natural response; forced:
)()()(
:Solution eqn. aldifferentiorder First
)()()sin(
:KVL
1
22
RL
LRZ
tZ
Vti
ii
tititi
tdtdi
LRtitV
vvv
mf
nf
nf
m
LRs
+vs
_
+
vo
_
+vR
_
+vL
_
i
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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R-L load
tm
mm
m
tmnf
tn
etZ
Vti
Z
V
Z
VA
AeZ
Vi
A
AetZ
Vtititi
RLAeti
tdtdi
LRti
)sin()sin()(
as,given iscurrent theTherefore
)sin()sin(
)0sin()0(
:i.e ,conducting starts diode thebefore zero is
current inductor realisingby solved becan
)sin()()()(
Hence
; )(
:in resultswhich
0)(
)(
0,source when is response Natural
0
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
6
R-L waveform
:i.e ,decreasing iscurrent thebecause negative is
:Note
dtdi
Lv
v
L
L
t
vo
vs,
io
vR
vL
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Extinction angle
otherwise
0
0for
)sin()sin(
)(
load, L-Rith rectfier w thesummarise To
and 0between conducts diode theTherefore,
y.numericall solved beonly can
0)sin()sin(
: toreduceswhich
0)sin()sin()(
. angle,tion theextincasknown ispoint This
OFF. turns whendiodeis zero reachescurrent
point when heduration)T that during negative
is source the(although radians n longer tha
biased forwardin remains diode that theNote
t
etZ
V
ti
e
eZ
Vi
tm
m
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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RMS current, Power
RMSRMSs
RMSRMSs
RMS
RMS
o
IVP
pf
IVS
S
PSP
pf
RI
tdtitdtiI
tdtitdtiI
.
.
i.e source,
by the suppliedpower apparent theis
load. by the absorbedpower the toequalwhich
source, by the suppliedpower real theis where
:definition from computed isFactor Power
P
:is load by the absorbedPower
NCALCULATIO POWER
)(21
)(21
:iscurrent RMS The
)(21
)(21
:iscurrent (DC) average The
,
,
2o
0
22
0
2
0
2
0
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Half wave rectifier, R-C Load
sin
OFF is diodewhen
ON is diodehen w)sin(
/
m
RCt
mo
Vv
eV
tVv
+vs
_
+vo
_
iD
Vm
Vmax
vs
vo
Vmin
iD
Vo
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Operation
• Let C initially uncharged. Circuit is energised at t=0
• Diode becomes forward biased as the source become positive
• When diode is ON the output is the same as source voltage. C charges until Vm
• After t=/2, C discharges into load (R).
• The source becomes less than the output voltage
• Diode reverse biased; isolating the load from source.
• The output voltage decays exponentially.
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Estimation of
mm
m
m
RCmm
RCtm
RCtm
mm
VV
RC
RCRC
RC
RCV
V
eRC
VV
t
eRC
V
td
eVd
tVtd
tVd
sin and
Therefore wave.sine theofpeak the toclose very is 22
-tan
: thenlarge, is circuits, practicalFor
tantan
1tan
1
1sin
cos
1sincos
equal, are slopes the,At
1sin
)(
sin
and
cos)(
sin
:are functions theof slope The
11
/
/
/
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Estimation of
for y numericall solved bemust equation This
0)(sinsin(
or
)sin()2sin(
, 2tAt
)2(
)2(
RC
RCmm
e
eVV
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
13
Ripple Voltage
fRC
V
RCVV
RCe
eVeVVV
eVeVv
t
VV
VVVV
VVV
t
V
mmo
RC
RCm
RCmmo
RCm
RCmo
m
mmmm
o
2
21 :expansoin Series Using
1
:as edapproximat is voltageripple The
) 2(
:is 2at evaluated tageoutput vol The
2. then constant, is tageoutput vol DC
such that large is C and 2, and If
sin)2sin(
2at occurs tageoutput volMin
. is tageoutput volMax
2
22
2222
minmax
max
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Capacitor Current
)2(t )( i.e
OFF, is diodewhen
sin
)2(t )(2 i.e
ON, is diodewhen
)cos(
),( ngsubstituti Then,
OFF is diodewhen sin
ON is diode when )sin()(
But
)(
)(
:, of In terms
)(
)(
:as expressed becan capacitor in thecurrent The
/
/
RCtm
m
c
o
RCtm
mo
oc
oc
eR
V
tCV
ti
tv
eV
tVtv
td
tdvCti
t
td
tdvCti
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Peak Diode Current
R
VCVi
R
V
R
Vi
CVCVI
iiii
mmpeakD
mmR
mmpeakc
CRDs
sincos
:iscurrent peak diode The
sin)(2sin)(2
.
:obtained becan )(2at current Resistor
cos)(2cos
Hence. .)(2at occurscurrent diodepeak The
: thatNote
,
,
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
16
Example A half-wave rectifier has a 120V rms source at 60Hz. The
load is =500 Ohm, C=100uF. Assume and are calculated as 48 and 93 degrees respectively. Determine (a) Expression for output voltage (b) peak-to peak ripple (c) capacitor current (d) peak diode current.
(OFF) 5.169
(ON) )sin(7.169
(OFF) sin
(ON) )sin(7.169 )sin()(
:tageOutput vol (a)
;5.169)62.1sin(7.169sin
843.048
;62.193
;7.1692120
)85.18/(62.1
/
t
RCtm
mo
m
o
o
m
e
t
eV
ttVtv
VradV
rad
rad
VV
Vm
Vmax
vs
vo
Vmin
iD
Vo
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Example (cont’)
A
radradu
R
VCVi
e
t
eR
V
tCVti
VufRC
V
RCVV
VVVVVV
VVV
mmpeakD
t
RCtm
m
c
mmo
mmmmo
o
50.4)34.026.4(500
)62.1sin(7.169)843.0cos(7.169)100)(602(
sincos
:current diodePeak (d)
(OFF) A 339.0
(ON) A )cos(4.6
(OFF) )sin(
(ON) )cos(
:currentCapacitor (c)
7.5610050060
7.1692
:ionApproximat Using
43sin)2sin(
:Using
:(b)Ripple
,
)85.18/(62.1
)/(
minmax
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Controlled half-wave
+vo
_
+vs
_
ig
ia
22sin
12
]2cos(1[4
sin2
1
voltageRMS
cos12
sin21
: voltageAverage
2
2
,
mm
mRMSo
mmo
Vtdt
V
tdtVV
VtdtVV
t
v
vo
ig
t
t
v s
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Controlled h/w, R-L load
eZ
VA
AeZ
Vi
i
AetZ
Vtititi
m
m
tm
nf
sin
sin0
,0 :condition Initial
sin)()()(
+vs
_
i
+vo
_
+vR
_
+vL
_
t
vs
vo
io
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Controlled R-L load
RIP
dtiI
dtiI
VtdtVV
eZ
Vi
etZ
V
ti
A
RMSo
RMS
o
mmo
m
tm
2
2
)(
)(
:load by the absorbedpower The
2
1
:current RMS
2
1
:current Average
coscos2
sin2
1
: voltageAverage
.angel conduction thecalled is Angle
sinsin0
y numericall solved bemust angle Extinction
otherwise 0
tfor sinsin
g,simplifyin and for ngSubstituti
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
21
Examples
1. A half wave rectifier has a source of 120V RMS at 60Hz. R=20 ohm, L=0.04H, and the delay angle is 45 degrees. Determine: (a) the expression for i(t), (b) average current, (c) the power absorbed by the load.
2. Design a circuit to produce an average voltage of 40V across a 100 ohm load from a 120V RMS, 60Hz supply. Determine the power factor absorbed by the resistance.
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
22
Freewheeling diode (FWD)• Note that for single-phase, half wave rectifier
with R-L load, the load (output) current is NOT continuos.
• A FWD (sometimes known as commutation diode) can be placed as shown below to make it continuos
+vs
_
io
+vo
_
+vR
_
+vL
_
+vs
_
+vo
_
D1 is on, D2 is off
io
vo= vs
io
+vo
_
io
D2 is on, D1 is off
vo= 0
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Operation of FWD
• Note that both D1 and D2 cannot be turned on at the same time.
• For a positive cycle voltage source,– D1 is on, D2 is off
– The equivalent circuit is shown in Figure (b)
– The voltage across the R-L load is the same as the source voltage.
• For a negative cycle voltage source,– D1 is off, D2 is on
– The equivalent circuit is shown in Figure (c)
– The voltage across the R-L load is zero.
– However, the inductor contains energy from positive cycle. The load current still circulates through the R-L path.
– But in contrast with the normal half wave rectifier, the circuit in Figure (c) does not consist of supply voltage in its loop.
– Hence the “negative part” of vo as shown in the normal half-wave disappear.
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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• The inclusion of FWD results in continuos load current, as shown below.
• Note also the output voltage has no negative part.
FWD- Continuous load current
iD1
io
output
Diode current
iD2
vo
t
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Full wave rectifier
Center-tapped
D1
is
+vs
_
vo +
iD1
iD2
io
+vs1
_
+vs2
_
D2
+ vD1
+ vD2
• Center-tapped (CT) rectifier requires center-tap transformer. Full Bridge (FB) does not.
• CT: 2 diodes
• FB: 4 diodes. Hence, CT experienced only one diode volt-drop per half-cycle
• Conduction losses for CT is half.
• Diodes ratings for CT is twice than FB m
mmo
m
mo
VV
tdtVV
ttV
ttVv
637.02
sin1
: voltage(DC) Average
2 sin
0 sin
circuits,both For
0
+vs
_
is
i D1
+vo
_
i o
Full Bridge
D1
D2
D4
D3
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Bridge waveforms
+vs
_
is
i D1
+vo
_
i o
Full Bridge
D1
D2
D4
D3
Vm
Vm
-Vm
-Vm
vs
vo
vD1 vD2
vD3 vD4
io
iD1 iD2
iD3 iD4
is
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Center-tapped waveforms
Center-tapped
D1is
+vs
_
vo +
iD1
iD2
io
+vs1
_
+vs2
_
D2
+ vD1
+ vD2
Vm
Vm
-2Vm
-2Vm
vs
vo
vD1
vD2
io
iD1
iD2
is
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Full wave bridge, R-L load
+vs
_
is
i D1
+
vo
_
io
+vR
_+vL
_
vo
vs
io
iD1 , iD2
iD3 ,iD4
is
t
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Approximation with large L
,for ,2
:i.e. terms,harmonic the
all drop topossible isit enough, large is If
. increasingry rapidly ve decreases Thus
decreases. harmonic increases, As
:currents harmonic The
curent DC The
11
112
termsharmonics theand
2
termDC thewhere
)cos()(
Series,Fourier Using
...4,2
RLR
V
R
VIti
L
nI
Vn
LjnR
V
Z
VI
R
VI
nn
VV
VV
tnVVtv
moo
n
n
n
n
nn
oo
mn
mo
nnoo
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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R-L load approximation
vo
vs
io
iD1 , iD2
iD3 ,iD4
is
t
RIP
IIII
R
V
R
VI
RMSo
oRMSnoRMS
moo
2
2,
2
:load the todeliveredPower
,2
current eApproximat
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
31
Examples
Given a bridge rectifier has an AC source Vm=100V at
50Hz, and R-L load with R=100ohm, L=10mH
a) determine the average current in the load
b) determine the first two higher order harmonics of the load current
c) determine the power absorbed by the load
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Controlled full wave, R load
R
VP
V
tdtVV
VtdtVV
RMSo
m
mRMSo
mmo
2
2
,
:is load R by the absorbedpower The4
2sin
22
1
sin1
Voltage RMS
cos1sin1
: voltage(DC) Average
+vs
_
is
i D1
+vo
_
i oT1
T4T2
T3
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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+vs
_
is
i D1
+
vo
_
io
+vR
_
+vL
_
Controlled, R-L load
vo
Discontinuous mode
io
vo
Continuous mode
+
io
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Discontinuous mode
zero.an greater th bemust )(tat current operation continousFor
).( is expressioncurrent output thein when is modecurrent usdiscontino
and continousbetween boundary The
0)(
:conditiony with numericall solved bemust and angle extinction theis that Note
)(
:ensure toneed mode, usdiscontinoFor
; tan and
)( for
)sin()sin()(
:load L-R with wavehalf controlled similar to Analysis
1
22
)(
o
tm
i
RL
RL
LRZt
etZ
Vti
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Continuous mode
cos2
sin1
:asgiven is tageoutput vol (DC) Average
tan
mode,current continuousfor Thus
tan
for Solving
,01)sin(
),sin()sin(
:identityry Trigonomet Using
0)sin()sin(0)(
1
1
)(
)(
mmo
VtdtVV
RL
RL
e
ei
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
36
Single-phase diode groups
• In the top group (D1, D3), the cathodes (-) of the two diodes are at a common potential. Therefore, the diode with its anode (+) at the highest potential will conduct (carry) id.
• For example, when vs is ( +), D1 conducts id and D3 reverses (by taking loop around vs, D1 and D3). When vs is (-), D3 conducts, D1 reverses.
• In the bottom group, the anodes of the two diodes are at common potential. Therefore the diode with its cathode at the lowest potential conducts id.
• For example, when vs (+), D2 carry id. D4 reverses. When vs is (-), D4 carry id. D2 reverses.
+vs
_+vo
_
vp
vn
io
D1
D3
D4
D2
vo =vp vn
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Three-phase rectifiersD1
vo =vp vn
+vo
_
vpn
vnn
ioD3
D2
D6
+ vcn -
n+ vbn -
+ van -
D5
D4
Vm
Vm
van vbn vcn
vn
vp
vo =vp - vn
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
38
Three-phase waveforms• Top group: diode with its anode at the
highest potential will conduct. The other two will be reversed.
• Bottom group: diode with the its cathode at the lowest potential will conduct. The other two will be reversed.
• For example, if D1 (of the top group) conducts, vp is connected to van.. If D6 (of the bottom group) conducts, vn connects to vbn . All other diodes are off.
• The resulting output waveform is given as: vo=vp-vn
• For peak of the output voltage is equal to the peak of the line to line voltage vab .
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
39
Three-phase, average voltage
phase.-single a ofn higher thamuch isrectifier phase- threea
ofcomponent voltageDCoutput that theNote
955.03
)cos(3
)sin(3
1
: voltageAverage
radians.3or degrees 60over average itsObtain segments.six theof oneonly Considers
,,
323
,
32
3,
LLmLLm
LLm
LLmo
VV
tV
tdtVV
vo
Vm, L-L
vo
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Controlled, three-phase
Vm
vanvbn vcn
T1
+vo
_
vpn
vnn
ioT3
T2
T6
+ vcn -
n+ vbn -
+ van -
T5
T4
vo
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Output voltage of controlled three phase rectifier
cos3
)sin(3
1
:as computed becan voltageAverage
SCR. theof angledelay thebe let Figure, previous theFrom
,
32
3,
LLm
LLmo
V
tdtVV
• EXAMPLE: A three-phase controlled rectifier has an input voltage of 415V RMS at 50Hz. The load R=10 ohm. Determine the delay angle required to
produce current of 50A.