Question

A sinusoidal voltage of amplitude 25 volts and frequency 50 Hz is applied to a half wave rectifier using PN diode. No filter is used and the load resistor is `1000 Omega`. The forward resistance `R_(f)` ideal diode is `10 Omega`. Calculate.
(i) Peak, average and rms values of load current
(ii) d.c power output
(ii) a.c power input
(iv) % Rectifier efficiency
(v) Ripple factor.

Answer 1

(i) `I_("in")=(V_(m))/(R_(f)+R_(L))=(25)/((10+1000))=24.75A`,
`I_(dc)=(I_(m))/(pi)=(24.75)/(3.14)=7.88mA`,
`I_(max)=(I_(m))/(pi)=(24.75)/(2)=12.38mA`
(ii) `P_(dc)=I_(dc)^(2)xxR_(L)=(7.88xx10^(-3))^(2)xx10^(3)~~63mW`
(iii) `P_(dc)=I_("rms")^(2)(R_(f)+R_(L))=(12.38xx10^(-3))^(2)xx(10+1000)~~155mW`
(iv) Rectifier efficiency `eta=(P_(dc))/(P_(ac))xx100=(62)/(155)xx100=40%`
(v) Ripple factor `= [((I_("rms"))/(I_(ac)))^(2)-1]^(1//2)=[((12.38)/(7.88))^(2)-1]=1.21`