Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value Vp = 100 V. Let the load resistance R = 10 kΩ. Calculate the fraction of the cycle during which the diode is conducting2.12 %

The diode in a peak rectifier conducts when the input voltage is greater than the voltage across the load resistance. For a sinusoidal input, this happens during the positive half cycle.

1. The period of the 60 Hz sinusoid is T = 1/60 = 0.01667 seconds.

2. The diode conducts during the positive half cycle of the sinusoid, which is half the period, or T/2 = 0.00833 seconds.

3. To find the fraction of the cycle during which the diode is conducting, we divide the conduction time by the period. This gives us (0.00833 seconds) / (0.01667 seconds) = 0.5, or 50%.

4. However, the diode does not conduct for the entire positive half cycle. It stops conducting when the input voltage drops below the peak voltage across the load resistance. This happens at the peak of the sinusoid, or π/2 radians into the cycle.

5. The time at which this happens is (π/2) / (2π) = 1/4 of the cycle, or 0.25.

6. Therefore, the diode conducts for 0.5 - 0.25 = 0.25, or 25% of the cycle.

7. However, the question asks for the fraction of the cycle during which the diode is conducting in percentage. Therefore, the diode conducts for 25% of the cycle.

Please note that the value you provided (2.12%) seems to be incorrect based on the given information. The calculation above is based on the assumption that the diode is ideal and there is no forward voltage drop. If there is a forward voltage drop across the diode, the conduction angle would be slightly less than 25%.