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Power ElectronicsLecture (7)
Prof. Mohammed Zeki Khedher
Department of Electrical Engineering
University of Jordan
1
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Single-Phase Full-Wave Diode Rectifier Center-Tap Diode Rectifier
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m
mdcV
tdtVV2
sin1
0
R
VI mdc
2
2
sin1
0
2 mmrms
VtdtVV
R
VI mrms
2
PIV of each diode = mV2
R
VII mDS 2
Example 3. The rectifier in Fig.2.8 has a purely resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple
factor (d) TUF (e) Peak inverse voltage (PIV) of diode D1 and(f) Crest factor of transformer secondary current.
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%05.81
2*
2
2*
2
*
*
R
VVR
VV
IV
IV
P
P
mm
mm
rmsrms
dcdc
ac
dc
11.1222
2
m
m
dc
rmsV
V
V
VFF
483.0111.11 22 FFV
VRF
dc
ac
The PIV is mV2
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Single-Phase Full Bridge Diode Rectifier With Resistive Load
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Example 4 single-phase diode bridge rectfier has a purely resistive load of R=15 ohms and, VS=300 sin 314 t and unity transformer ratio.
Determine (a) The efficiency, (b) Form factor, (c) Ripple factor, (d) The peak inverse voltage, (PIV) of each diode, , and, (e) Input power factor.
VV
tdtVV mmdc 956.190
2sin
1
0
A
R
VI mdc 7324.12
2
VV
tdtVV mmrms 132.212
2sin
12/1
0
2
%06.81rmsrms
dcdc
ac
dc
IV
IV
P
P11.1
dc
rms
V
VFF
482.011 22
222
FFV
V
V
VV
V
VRF
dc
rms
dc
dcrms
dc
ac The PIV=300V
Input power factor = 1cosRe
SS
SS
IV
IV
PowerApperant
Poweral
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Comparision between Single Phase Rectifiers
Half wave Full wave Fullwave• center-tap bridge• Peak repetitive reverse voltage VRRM 3.14Vdc 3.14 Vdc 1.57
Vdc• Rms input voltage per transformer leg Vs 2.22 Vdc 1.11 Vdc 1.11
Vdc• Diode average current IF.AV. 1.00 Idc 0.50 Idc 0.50
Idc• Diode rms current IF.RMS. 1.57 Idc 0.785 Idc 0.785 Idc• Form factor of diode current IF.RMS. 1.57 1.57 1.57• Rectification ratio 0.405 0.81 0.81• Form factor 1.57 1.11 1.11• Ripple factor 1.21 0.482 0.482• Transformer rating primary VA 2.69 Pdc 1.23 Pdc 1.23 Pdc• Transformer rating secondary VA 3.49 Pdc 1.75 Pdc 1.23 Pdc• Output ripple frequency fr 1 fi 2 fi 2 fi
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Multi-phase Rectifier
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Three-Phase Half Wave Rectifier
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mm
mdc VV
tdtVV 827.02
33sin
2
36/5
6/
R
V
R
VI mmdc
*827.0
**2
33
mmmrms VVtdtVV 8407.08
3*3
2
1sin
2
36/5
6/
2
R
VI mrms
8407.0
R
V
R
VII mmSr 4854.0
3
08407
ThePIV of the diodes is mLL VV 32
Example A 3-phase star rectifier is operated from 460 V 50 Hz supply at secondary side and the load
resistance is R=20. If the source inductance is negligible, determine (a) Rectification efficiency, (b)
Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of each diode.
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VVVV mS 59.3752*58.265,58.2653
460
mm
dc VV
V 827.02
33
R
V
R
VI mmdc
0827
2
33
mrms VV 8407.0R
VI mrms
8407.0
%767.96rmsrms
dcdc
ac
dc
IV
IV
P
P
%657.101dc
rms
V
VFF
%28.1811 22
222
FFV
V
V
VV
V
VRF
dc
rms
dc
dcrms
dc
ac
The PIV= 3 Vm=650.54V
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Three-Phase Full Wave Rectifier With Resistive Load
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Example 10 The 3-phase bridge rectifier is operated from 460 V 50 Hz supply and the load resistance is R=20ohms. If the
source inductance is negligible, determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage
(PIV) of each diode .
VVV
V mm
dc 226.621654.133
AR
V
R
VI mmdc 0613.31
654.133
VVVV mmrms 752.6216554.14
3*9
2
3
AR
VI mrms 0876.31
6554.1
%83.99rmsrms
dcdc
ac
dc
IV
IV
P
P
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%08.100dc
rms
V
VFF
%411 22
222
FFV
V
V
VV
V
VRF
dc
rms
dc
dcrms
dc
ac
The PIV= 3 Vm=650.54V
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Single phase rectifier 3 phase rectifier
Performance:
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3-phase bridge rectifier with RL load
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Condition for continuous load current