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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. |
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Answer» (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` |
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