Question

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)?

Answer 1

Answer:

  (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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(a)

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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(b)

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%