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. 38 Vb. 76 Vc. 108 Vd. 54 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. 38 Vb. 76 Vc. 108 Vd. 54 V
Solution
The Peak Inverse Voltage (PIV) of a diode in a center-tap full-wave rectifier can be calculated using the formula:
PIV = 2 * Vm
Where Vm is the peak voltage of the secondary winding of the transformer.
Given that the transformation ratio is 6:1:1, the voltage on the secondary winding (Vs) is:
Vs = Vin / transformation ratio = 230V / 6 = 38.33V
The peak voltage (Vm) is √2 times the RMS voltage (Vs), so:
Vm = √2 * Vs = √2 * 38.33V = 54.19V
Therefore, the PIV of each diode is:
PIV = 2 * Vm = 2 * 54.19V = 108.38V
So, the minimum PIV rating of each of the diodes should be 108V. Therefore, the correct answer is c. 108 V.
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Controlled Rectifier
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