Example:

The input to the half-wave rectifier of given Figure is 120 Vac. If the resistance R_d of the diode during conduction is 20 Ω and the load resistance R_L = 1000 Ω , determine the peak, DC and rms load currents and the power dissipated in the load and diode.

Solution

The peak load current is I_p      = V_p /(R _L+ R_d  ) = 120(√2 /(1000 + 20)) = 0.166 A . The DC

current is I_DC    =I_p/ π = 0.053 A . The rms current, is, hence,

The entire input power to the circuit is P₁ = P_L + P_d. the power dissipated in the load is P_L =(I_rms)² R_L  = 0.0832  . 1000 = 6.92 W, and the power dissipated in the diode is  P^d=        (I_rms  )²R_d = 0.14 W . The total power supplied by the source is, hence, 7.06 W

In the above calculations, we have neglected the 0.7 V voltage drop across the conducting diode. Also note down that the peak reverse voltage across the diode during the non-conducting phase is 120 √2 = 170 V , which the diode should be capable of withstanding.