Chapter 2 AC to DC CONVERSION (RECTIFIER) · Power Electronics and Drives (Version 3-2003), Dr....
Transcript of Chapter 2 AC to DC CONVERSION (RECTIFIER) · Power Electronics and Drives (Version 3-2003), Dr....
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
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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
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Single-phase, half-wave, R-load
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5.02
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,
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2
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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
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i
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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R-L waveform
:i.e ,decreasing iscurrent thebecause negative is :Note
dtdi
Lv
v
L
L
=
ωt
vo
vs,
π 2π
io
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3π 4π0
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Extinction angle
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etZ
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Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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RMS current, Power
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o
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P
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)(21
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Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Half wave rectifier, R-C Load
( )
θ
ω
θ
ωθωθsin
OFF is diodewhen
ON is diodehen w)sin( /
m
RCtm
o
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eV
tVv
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iD
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π 2π 3π 4π
Vm
Vmax
vs
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Vmin
π /2
iD
3π /2
∆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 ωt=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 ωt=π/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 θ
( )
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mm
m
m
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: thenlarge, is circuits, practicalFor tantan
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11
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Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Estimation of α
αθα
θαπ
απω
ωθαπ
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for y numericall solved bemust equation This0)(sinsin(
or
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, 2tAt
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RCmm
e
eVV
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Ripple Voltage
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21 :expansoin Series Using
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) 2(
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such that large is C and 2, and If
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. 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
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Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Peak Diode Current
RV
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i
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Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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ExampleA half-wave rectifier has a 120V rms source at 60Hz. The load is =500 Ohm, C=100uF. Assume α and θ 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.
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:tageOutput vol (a)
;5.169)62.1sin(7.169sin843.048
;62.193
;7.1692120
)85.18/(62.1
/
t
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o
om
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rad
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Vm
Vmax
vs
vo
Vmin
π /2
iD
3π /2
α θ
∆Vo
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Example (cont’)
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t
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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
=+=
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ω
ωπ
ααπ
ω
ωθω
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Controlled half-wave
+vo
_
+vs
_
ig
ia
( ) [ ]
( )[ ]
( )π
απαωω
π
ωωπ
απ
ωωπ
π
α
π
α
π
α
22sin
12
]2cos(1[4
sin21
voltageRMS
cos12
sin21
: voltageAverage
2
2
,
+−=−=
=
+==
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�
�
mm
mRMSo
mmo
Vtdt
V
tdtVV
VtdtVV
ωtv
vo
α
ig
ωt
ωt
v s
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Controlled h/w, R-L load
( )
( )
( ) ( )
( ) ωτα
ωτα
ωτω
θα
θαα
α
θωωωω
−
−
−
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i
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m
m
tm
nf
sin
sin0
,0 :condition Initial
sin)()()(
+vs
_
i
+vo
_
+vR
_
+vL
_
ω t
vs
π 2π
vo
α
io
β
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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Controlled R-L load
( ) ( ) ( )
( ) ( ) ( )
( )
( ) [ ]
( )
( )
RIP
dtiI
dtiI
VtdtVV
eZ
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etZ
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o
mmo
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tm
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21
:current RMS
21
:current Average
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sin21
: voltageAverage
.angel conduction thecalled is Angle
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y numericall solved bemust angle Extinction
otherwise 0
tfor sinsin
g,simplifyin and for ngSubstituti
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θβγ
θβθββ
β
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β
α
β
α
β
α
ωτβα
ωτωα
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
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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(ωt), (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
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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
π 2π 3π 4π
iD1
io
output
Diode current
iD2
vo
0
ω t
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
25
Full wave rectifier
Center-tapped
D1is
+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
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_
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
π 2π 3π 4π
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 −
π 2π 3π 4π
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
ω tπ 2π
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
29
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 The1
11
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termsharmonics theand
2 termDC thewhere
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Series,Fourier Using
...4,2
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ω
π
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πωω
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
30
R-L load approximation
vo
vs
io
iD1 , iD2
iD3 ,iD4
is
ω tπ 2π
( )
RIP
IIII
RV
RV
I
RMSo
oRMSnoRMS
moo
2
2,
2
:load the todeliveredPower
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current eApproximat
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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=10mHa) determine the average current in the loadb) determine the first two higher order harmonics of the
load currentc) determine the power absorbed by the load
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
32
Controlled full wave, R load
( ) [ ]
( )[ ]
( )
RV
P
V
tdtVV
VtdtVV
RMSo
m
mRMSo
mmo
2
2
,
:is load R by the absorbedpower The4
2sin22
1
sin1
Voltage RMS
cos1sin1
: voltage(DC) Average
=
+−=
=
+==
�
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πα
πα
ωωπ
απ
ωωπ
π
α
π
α
+vs
_
is
i D1
+vo
_
i oT1
T4T2
T3
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
33
+vs
_
is
i D1
+
vo
_
io
+vR
_
+vL
_
Controlled, R-L load
π 2π
vo
Discontinuous mode
βα π+α
io
π 2π
vo
Continuous mode
π+α
α β
io
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
34
Discontinuous mode
[ ]
zero.an greater th bemust )(tat current operation continousFor
).( is expressioncurrent output thein when is modecurrent usdiscontino
and continousbetween boundary The
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:conditiony with numericall solved bemust and angle extinction theis that Note
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; tan and
)( for
)sin()sin()(
:load L-R with wavehalf controlled similar to Analysis
1
22
)(
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β
β
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τωθωβωα
θαθωω ωταω
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tm
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RL
RL
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Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
35
Continuous mode
[ ]
( ) απ
ωωπ
ωα
ωα
α
αθ
αθθαπ
θαπθαπαπ
πα
α
ωτπ
ωτααπ
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
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e
ei
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−
−
−
−+−
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
37
Three-phase rectifiersD1
vo =vp −vn
+vo
_
vpn
vnn
ioD3
D2
D6
+ vcn -
n+ vbn -
+ van -
D5
D4
π 2π0 4π
Vm
Vm
van vbn vcn
vn
vp
vo =vp - vn
3π
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
−−
−
−
==
=
= �
π
ωπ
ωωπ
π
ππ
π
π
vo
0
π/3 2π/3
Vm, L-L
vo
π/3
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
40
Controlled, three-phase
Vm
vanvbn vcn
T1
+vo
_
vpn
vnn
ioT3
T2
T6
+ vcn -
n+ vbn -
+ van -
T5
T4
α
vo
Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003
41
Output voltage of controlled three phase rectifier
απ
ωωπ
α
απ
απ
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.