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
         RL load
Condition for continuous load current

				
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posted:2/28/2012
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