A full wave rectifier uses a load resistor of 15000. Assume the diodes have Rf=l0 Ohms and Rr= infinite. The voltage applied to the diodes is 30V with a frequency of 50Hz. Calculate the 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
A.c power input = Irms^2 (RF+Rr) Irms = Imax/ √2 = Vmax/ (Rf+Rr) √2 [ as V=IR and I=V/R] = 30/ (1500+10) *1.414 = 13.5 mASo, power input = (13.5*10^-3)^2 + ( 1500+10) = 275.2 mW
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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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