Page 34 - Electrician - TT (Volume 2)
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ELECTRICIAN - CITS




           Peak inverse voltage - Bridge rectifier: Fig 7  shows a bridge rectifier at the instant the secondary voltage has
           reached its maximum value.
           Diode D4 is ideally short (as it is conducting) and D1 is ideally open.  summing the voltages around the outside
           loop and applying Kirchhoff’s law,
           V s(peak)  - PIV across D  + 0 = 0
                              1
           or        PIV across D  = V s(peak)
                             1
           Therefore, the peak inverse voltage across D  is equal to the peak secondary voltage V
                                                   1                                    s(peak)
           In a similar way, the peak inverse voltage across each diode will be equal to the peak secondary voltage V s(peak)  of
           the transformer secondary.  Hence the PIV ratings of the diodes used should be greater than V s(peak)
           Example
           In Fig 7 if the transformer secondary voltage V s(rms)  is 24 volts, find the minimum PIV of diodes used.  In a bridge
           rectifier PIV across the diodes is same and is equal to V s(peak)
           Therefore, in the given example,







             Fig 6

























               Fig 7



























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                                    CITS : Power - Electrician & Wireman - Lesson 60-69
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