A full wave rectifier uses a load resistor of 1500 Ω. Assume the diodes have Rf= 10 Ω, Rr= ∞. The voltage applied to the diode 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
The AC power input; PIN=IRMS2(RF+Rr).IRMS= Im/√2 = Vm/(Rf+RL)√2 = 30/[(1500+10) x 1.414] = 13.5mA.So, PIN= (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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