# Diode Circuits or Uncontrolled Rectifier - PowerPoint

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```					     Power Electronics
Lecture (7)
Prof. Mohammed Zeki Khedher
Electrical Engineering Department of
University of Jordan

1
Single-Phase Full-Wave Diode Rectifier
Center-Tap Diode Rectifier
                                         2 Vm
1                         2Vm               
Vdc 
       Vm sin t dt 

I dc
 R
0


 Vm sin t 
1                          Vm             Vm
Vrms                         2
dt       I rms   
                           2             2 R
0

PIV of each diode =              2Vm
Vm
IS  ID         
2R

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.
2 Vm   2 Vm
*
Pdc   Vdc * I dc               R
                                     81.05%
Pac Vrms * I rms          Vm    Vm
*
2     2R
Vm
FF 
Vrms
    2    1.11
Vdc      2 Vm 2 2

Vac
RF       FF  1  1.11  1  0.483
2          2
Vdc
The PIV is       2Vm
Single-Phase Full Bridge Diode Rectifier With Resistive Load
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.
                                                       2Vm
1                     2Vm                                       12.7324 A
Vdc 
  Vm sin t dt       
 190.956 V           I dc
 R
0
                 1/ 2
1                   
   Vm sin t  dt 
Vm
Vrms                     2
       212 .132 V
 0
                    
            2
Pdc   Vdc I dc
                81.06 %                        Vrms
Pac Vrms I rms                             FF        1.11
Vdc
Vac   Vrms  Vdc
2      2     2
Vrms
RF                      2
 1  FF 2  1  0.482           The PIV=300V
Vdc      Vdc       Vdc

Re al Power     VS I S cos 
Input power factor =                              1
Apperant Power     VS I S
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
Multi-phase Rectifier
Three-Phase Half Wave Rectifier
5 / 6
3                           3 3 Vm

            Vm sin t dt            0.827Vm            3 3 Vm 0.827 * Vm
Vdc
2                             2               I dc             
 /6                                                2 * * R   R

5 / 6

 Vm sin t 
3                                       1 3* 3
Vrms                                   2
d t        Vm  0.8407 Vm
2                                       2 8
 /6
0.8407 Vm                             08407 Vm          Vm
I rms                                Ir  IS            0.4854
R                                  R 3             R
ThePIV of the diodes is               2 VLL  3 Vm
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.
460
VS       265.58 V , Vm  265.58 * 2  375.59 V
3
3 3 Vm                      3 3 Vm 0827 Vm
Vdc            0.827 Vm   I dc            
2                          2 R      R
Vrms  0.8407 Vm                      0.8407 Vm
I rms 
R
Pdc    Vdc I dc
                 96.767 %
Pac Vrms I rms
Vrms
FF        101.657 %
Vdc
Vac   Vrms  Vdc
2      2     2
Vrms
RF                      2
 1  FF 2  1  18.28 %
Vdc      Vdc       Vdc
The PIV=       3 Vm=650.54V
Three-Phase Full Wave Rectifier With Resistive Load
IL
Ip       Is   1       3       5
VL
a
b

c
4       6       2
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 .
3 3 Vm
Vdc                1.654Vm  621.226 V

3 3 Vm 1.654Vm
I dc                   31.0613 A
 R      R
3 9* 3
Vrms          Vm  1.6554 Vm  621.752 V
2 4
1.6554 Vm
I rms               31.0876 A
R
Pdc   Vdc I dc
                99.83 %
Pac Vrms I rms
Vrms
FF        100 .08 %
Vdc

Vac   Vrms  Vdc
2      2     2
Vrms
RF                      2
 1  FF 2  1  4 %
Vdc      Vdc       Vdc

The PIV=   3 Vm=650.54V
Performance:
Single phase rectifier          3 phase rectifier
3-phase bridge rectifier with