If the peak voltage on a center tapped full wave rectifier circuit is 5v and diode cut in voltage is 0.7. The peak inverse voltage on diode is_______________1. A. 4.3V2. B. 9.3V3. C. 5.7V4. D. 10.7V
Question
If the peak voltage on a center tapped full wave rectifier circuit is 5v and diode cut in voltage is 0.7. The peak inverse voltage on diode is_______________1. A. 4.3V2. B. 9.3V3. C. 5.7V4. D. 10.7V
Solution
In a center-tapped full wave rectifier, the peak inverse voltage (PIV) is twice the peak voltage of the input. This is because during the negative half cycle of the AC input, the diode that was previously conducting becomes reverse biased and the voltage across it is the peak voltage of the input (Vp) plus the voltage drop across the load resistor which is also Vp. Therefore, the PIV is 2Vp.
Given that the peak voltage (Vp) is 5V, the peak inverse voltage (PIV) would be 2*5V = 10V.
However, the diode cut-in voltage is 0.7V. This is the voltage below which the diode will not conduct. But this does not affect the peak inverse voltage (PIV) of the diode.
So, the correct answer is D. 10V.
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