A full wave rectifier uses a load resistor of 1500 ohm. Assume the diodes have Rf = 10 ohm, and Rr = infinity. The voltage applied to the diode is 30V with a frequency of 50Hz. Calculate AC power input.
Correct answer: B. 275.2 mW
- A. 358.98 mW
- B. 275.2 mW
- C. 145.76 mW
- D. 456.78 mW
Explanation
The AC P(in)=Irms^2(RF+Rr).Irms=Vm/(Rf+RL)√2=30/(1500+10) x 1.414=13.5mASo, P(in)=(13.5 x 10^-3)^2 x (1500+10)=275.2mW.
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A p-n junction forms when p-type and n-type semiconductor regions meet, creating a depletion layer, barrier potential and rectifying action. The work covers forward and reverse bias, junction current, diode characteristics, and how a diode converts alternating current into pulsating direct current through half-wave and full-wave rectification.
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