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A half-wave rectifier employing a p-n junction diode is fed with an AC voltage of peak value 25 V and frequency 50 Hz. The circuit has no filter, and the load resistance is 1000 Ω. If the diode’s forward resistance R_f is 10 Ω, what is the rectification efficiency in percent?
- 40%
- 20%
- 30%
- 15%
Correct answer: 40%
Solution
Half-wave rectifier efficiency eta = 0.406/(1 + Rf/RL) = 0.406/(1 + 10/1000) = 0.402, i.e. about 40%.
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