A half-wave rectifier circuit with a 1 kΩ load operates from a 120 V (rms value), 60 Hz household supply through a 10-to-1 step-down transformer. (For sine wave, V rms= V pk/√2 .) Assume the diode voltage is 0.7 V at forward bias. (a) What is the peak voltage of the rectified output? (b) For What fraction of the cycle does the diode conduct (calculate the percentage)?

  (a) 16.27 Vpk

  (b) 48.7%

Explanation:

The transformer is assumed to be an ideal 10:1 voltage divider with no internal impedance. The diode is assumed to be modeled in the forward direction by a perfect 0.7 V voltage drop with no internal impedance. That means the frequency of the supply voltage is irrelevant.

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The peak voltage will be 0.7 V less than the transformer secondary peak voltage:

  ((120 V)√2)/10 -0.7 V ≈ 16.27 V

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The fraction of the amplitude for which the diode is non-conducting is ...

  0.7/(12√2) ≈ 0.041248

The period of conduction is symmetrical about the peak of the waveform, so it is convenient to use the arccos function to find the (half) conduction angle:

  arccos(0.041248) ≈ 87.64°

As a fraction of half the cycle, this is ...

  conduction fraction ≈ 87.64°/180° ≈ 48.7%

answer:

the answer is d

explanation: