Three Phase Controlled Rectifiers2003
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Transcript of Three Phase Controlled Rectifiers2003
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VAN
VCN
VBN
1200
1200
1200 RN AN
YN BN
BN CN
v v
v v
v v
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T1
T2
T3
Rload
3-supply
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PhaseVoltages
0 30 60 90 120 150 180 210 240 270 300 330 360
Vo(=0 )
Vo(30+
)
Va Vb Vc
0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360
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PhaseVoltages
0 30 60 90 120 150 180 210 240 270 300 330 360
Vo(=0 )
Vo(30+
)
Va Vb Vc
0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360
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01
0
2
0
3
0
306
5
1506
7 270
62
Each thytistor conducts for 120 or radians3
T is triggered at t
T is triggered at t
T is triggered at t
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5
6
6
If the reference phase voltage is
sin , the average or dc output
voltage for continuous load current is calculated
using the equation
3
sin .2
RN an m
dc m
v v V t
V V t d t
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5
6
6
56
6
3 sin .2
3cos
2
3 5cos cos
2 6 6
mdc
mdc
mdc
VV t d t
VV t
VV
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0 0
0
Note from the trigonometric relationship
cos cos .cos sin .sin
5 5cos cos sin sin
6 63
2co
cos 150 cos sin 150 sin32 cos 30
s .cos sin sin6 6
.cos
mdc
m
dc
A
VV
B A B A B
V
V
0sin 30 sin
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0 0
0 0 0 0
0 0
0 0
0
0
0
0
0 0
Note: cos 1
cos 180 30 cos sin 180 30 sin3
2 cos 30 .cos sin 30 sin
cos 30 cos sin 30 sin3
2 cos 30 .cos sin 30 s
80 30 cos 30
sin 180 30 sin 30
in
m
dc
m
dc
VV
V
V
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03 2cos 30 cos2
3 32 cos
2 2
3 3 33 cos cos
2 2
3 cos2
Where 3 Max. line to line supply voltage
m
dc
m
dc
m m
dc
Lm
dc
Lm m
VV
VV
V VV
VV
V V
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15 26
2 2
6
12
The rms value of output voltage is found by
using the equation
3sin .
2
and we obtain
1 33 cos 2
6 8
mO RMS
mO RMS
V V t d t
V V
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R-Lload
T1 T3 T5
T4 T2T6
A
B
C
6 - pulse combinations are:
(6,1) (1,2) (2,3) (3,4) (4,5) (5,6)
Io
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R-Lload
T1 T3 T5
T4 T6
A
B
C
When thyristors T1& T6conducts
T2
A T1 RL T6 B
Io
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R-Lload
T1 T3 T5
T4 T6
A
B
C
When thyristors T1& T2conducts
T2
A T1 RL T2 C
Io
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R-Lload
T1 T3 T5
T4 T6
A
B
C
When thyristors T2& T3conducts
T2
B T3 RL T2 C
Io
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R-Lload
T1 T3 T5
T4 T6
A
B
C
When thyristors T3& T4conducts
T2
B T3 RL T4 A
Io
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R-Lload
T1 T3 T5
T4 T6
A
B
C
When thyristors T4& T5conducts
T2
C T5 RL T4 A
Io
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R-Lload
T1 T3 T5
T4 T6
A
B
C
When thyristors T5& T6conducts
T2
C T5 RL T6 B
Io
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Line
Voltages
PhaseVoltages
0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360
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LineVoltages
Vo
(=0 )
AB AC BC BA CA CB
0 30 60 90 120 150 180 210 240 270 300 330 360VCBVCAVAB VBAVBCVAC
0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360Vo(=60 )
Io
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LineVoltages
Vo
(=0 )
AB AC BC BA CA CB
0 30 60 90 120 150 180 210 240 270 300 330 360VCBVCAVAB VBAVBCVAC
0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360
Vo
(=60 )
Io
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0 30 60 90 120 150 180 210 240 270 300 330 360
LineVoltages
Vo
(=90 )
AB AC BC BA CA CB
0 30 60 90 120 150 180 210 240 270 300 330 360
Vo
(=120 )0 30 60 90 120 150 180 210 240 270 300 330 360
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0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360
LineVoltages
Vo
(=90 )
AB AC BC BA CA CB
0 30 60 90 120 150 180 210 240 270 300 330 360
Vo
(=120 )
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The output load voltage consists of 6 voltage
pulses over a period of 2radians, Hence the
average output voltage is calculated as
2
6
6 . ;2
3 sin6
dc OO dc
O ab m
V V v d t
v v V t
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2
6
mL
max
3
3 sin .6
3 3 3cos cos
Where V 3 Max. line-to-line supply vo
The maximum average dc output voltage is
obtained for a delay angle
ltage
3 3 0,
3
dc m
m mL
dc
m
m m
dmdc
V V t d t
V VV
V
V VV V
L
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1
22
2
6
The normalized average dc output voltage is
cos
The rms value of the output voltage is found from
6.2
dc
dcn n
dm
OO rms
VV V
V
V v d t
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1
22
2
6
1
22
2 2
6
1
2
6 .2
33 sin .
2 6
1 3 33 cos 2
2 4
abO rms
mO rms
mO rms
V v d t
V V t d t
V V
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For four quadrant operation in many industrial variable speed dcdrives , 3 phase dual converters are used.
Used for applications up to 2 mega watt output power level.
Dual converter consists of two 3 phase full converters which areconnected in parallel & in opposite directions across a common load.
+I0-I0
+V0
-V0
Conv. 1
Inverting
1 > 900
Conv. 1
Rectifying
1 < 900
Conv. 2
Rectifying
2 < 900
Conv. 2Inverting
2 > 900
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The Dual Converter is operated such that:
One conv. should be Converter One conv. should be Inverter
{1 < 90}{2 > 90}
i.e, 1+2= 180
Conditions:
1. O/P voltage mag. of 2 converters should be same.
2. Polarities should be same.
i.e,Vo = Vo1 = -Vo2
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Non Circulating current mode:
only one converter is in operation at a time, and it alonecarries the entire load current.
No Reactor is used between two converters.
Circulating current mode:
A reactor is inserted between two reactors. Thisreactor limits the magnitude of circulating current
to a reasonable value.
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load
V0
T11 T13 T15
T14 T12T16 T21T23T25
T24T22 T26
A
B
C C
B
A
V01 V02
Lr2
Lr2
Vr Vr
2 2
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LineVoltages
Vo1
(=60 )
AB AC BC BA CA CB
0 30 60 90 120 150 180 210 240 270 300 330 360VCB V
CA
VAB
VBA
VBC
VAC
0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360Vo2(=120 )
Io
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LineVoltages
Vo1
(=60 )
AB AC BC BA CA CB
0 30 60 90 120 150 180 210 240 270 300 330 360VCB V
CA
VAB
VBA
VBC
VAC
0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360Vo2(=120 )
Vo=Vo1+Vo2
Vo2 (=120 )
2
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LineVoltages
Vr
AB AC BC BA CA CB
0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360
Ic =
Vr dt
Ic
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1 2
3 sin sin6 2
3 cos6
The circulating current can be calculated by
using the equation
r O O ab bc
r m
r m
v v v v v
v V t t
v V t
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1
1
6
6
1
max
1.
13 cos .
6
3sin sin
63
t
r r
r
t
r m
r
m
r
r
m
r
r
i t v d t
L
i t V t d t L
Vi t t
L
Vi
L
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Because of Source Inductance effect, overlap conductionof thyristors takes place.
At an instant, 3 thyristors will conduct
i.e, 2 thyristors from +ve group& 1 thyristor from -ve group
(or)
2 thyristors from -ve group& 1 thyristor from +ve group
So, .output voltage decreases.
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Decrease in output voltage due to overlap angle:
Vo = Area of the triangle
Vo = 6x (1/2)0Vl d(t)VO = 3Ls
Io
Io
For 3- full bridge conv. With R-L Load without Ls
VO= 33 Vm cos
Therefore, Avg. o/p voltage with overlap angle (or) Source Inductance
VO(avg)= 33 Vm cos
3Ls
Io
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3 Phase semiconverters are used in Industrial dc driveapplications upto 120kW power output.
Single quadrant operation is possible.
Power factor decreases as the delay angle increases.
Power factor is better than that of 3 phase half waveconverter.
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R-Lload
T1 T3 T5
D4 D 2D6
A
B
C
Io
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R-Lload
T1 T3 T5
D4 D6
A
B
C
When thyristors T1& T6conducts
D2
A T1 RL T6 B
Io
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LineVoltages
Vo
(=30 )
AB AC BC BA CA CB
0 30 60 90 120 150 180 210 240 270 300 330 360
VCBVCAVAB VBAVBCVAC
0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360Vo(=60 )
Io
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LineVoltages
Vo
(=30 )
AB AC BC BA CA CB
0 30 60 90 120 150 180 210 240 270 300 330 360
VCBVCAVAB VBAVBCVAC
0 30 60 90 120 150 180 210 240 270 300 330 360
0 30 60 90 120 150 180 210 240 270 300 330 360
Vo(=60 )
IphA
1,2 3,4 5,6
Phase A For Inductive LOAD
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To derive an Expression for the
Average Output Voltage of 3 PhaseSemiconverter for > / 3and Discontinuous Output Voltage
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76
6
76
6
For and discontinuous output voltage:
3the Average output voltage is found from
3 .2
33 sin
2 6
dc ac
dc m
V v d t
V V t d t
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max
3 31 cos
23
1 cos2
3 Max. value of line-to-line supply voltage
The maximum average output voltage that occurs at
a delay angle of 0 is
3 3
m
dc
mL
dc
mL m
m
dmdc
VV
VV
V V
VV V
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17 2
6
2
6
The normalized average output voltage is
0.5 1 cos
The rms output voltage is found from
3.
2
dcn
dm
acO rms
VV
V
V v d t
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