In a half-wave rectifier using a PN junction | vedclass

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(B) The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L} = \frac{25}{10 + 1000} = 24.75 \ mA$.The $d.c.$ current is $I_{dc} = \frac{I_m}{\p

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$62 \ mW, 155 \ mW$

Vedclass · Answer

(B) The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L} = \frac{25}{10 + 1000} = 24.75 \ mA$.The $d.c.$ current is $I_{dc} = \frac{I_m}{\pi} = \frac{24.75}{3.14} \approx 7.88 \ mA$.The $r.m.s.$ current is $I_{rms} = \frac{I_m}{2} = \frac{24.75}{2} = 12.38 \ mA$.The $d.c.$ power output is $P_{dc} = I_{dc}^2 	imes R_L = (7.88 	imes 10^{-3})^2 	imes 1000 \approx 62 \ mW$.The $a.c.$ power input is $P_{ac} = I_{rms}^2 	imes (R_f + R_L) = (12.38 	imes 10^{-3})^2 	imes (10 + 1000) \approx 155 \ mW$.

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