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A `p-n` diode is used in a half wave rectifier with a load resistance of `1000 Omega`. If the forward resistance `(r_(f))` of diode is `10 Omega`, calculate the efficiency of this half wave rectifier. |
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Answer» Load resistance `R_(L) = 1000 Omega` Forward resistance of the diode `= r_(f) = 10 Omega` Efficiency of half wave rectifier `[(0.406 R_(L))/(r_(f)+R_(L))] =(0.406 xx 1000)/(1010) = 0.4019` The percentage efficiency of the half wave rectifier `eta = 40.19 %`. |
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