A single phase Centre-tap transformer rectifier is required to be designed for AC-DC rectification with a transformation ratio 6:1:1 and energized from AC input of 230V 50Hz supply. with no filter capacitor. The load resistor is approximately 5.4 Ohm. The minimum PIV Rating of each of the diodes is:Select one:a. 54 Vb. 108 Vc. 38 Vd. 76 V
Question
A single phase Centre-tap transformer rectifier is required to be designed for AC-DC rectification with a transformation ratio 6:1:1 and energized from AC input of 230V 50Hz supply. with no filter capacitor. The load resistor is approximately 5.4 Ohm. The minimum PIV Rating of each of the diodes is:Select one:a. 54 Vb. 108 Vc. 38 Vd. 76 V
Solution
The Peak Inverse Voltage (PIV) of a diode in a center-tap full-wave rectifier is given by the peak secondary voltage of the transformer.
Given the transformation ratio of 6:1:1, the secondary voltage Vs can be calculated as:
Vs = Vp / a = 230V / 6 = 38.33V
Where Vp is the primary voltage and 'a' is the turns ratio.
However, the peak secondary voltage Vm is given by:
Vm = √2 * Vs = 1.414 * 38.33V = 54.18V
Therefore, the minimum PIV rating of each of the diodes should be 54.18V.
So, the closest answer is:
a. 54 V
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