A diode with internal resistance rf=20 X in half-wave rectifier, is applied V = 50 sin xt and RL =800 X. Calculate (a) Im, Idc and Irms, (b) ac power input and dc power output, (c) dc output voltages and (d) efficiency of rectifier.

Question

A diode with internal resistance [latex]r_f=20 \Omega [/latex] in half-wave rectifier, is applied [latex]V = 50 \sin \omega t[/latex] and [latex]R_L =800 \Omega[/latex]. Calculate (a) [latex]I_m[/latex], [latex]I_{dc}[/latex] and [latex]I_{rms}[/latex], (b) ac power input and dc power output, (c) dc output voltages and (d) efficiency of rectifier.

Accepted Answer

[latex]V=50\sin\omega t[/latex]
[latex]V_m=50[/latex]volt
(a) [latex]I_m=\frac{V_m}{r_f+R_L}=\frac{50}{20+800}=0.061A =61mA[/latex]
[latex]I_{dc}=\frac{I_m}{\pi}=\frac{61}{\pi}=19.4mA[/latex]
[latex]I_{rms}=\frac{I_m}{2}=\frac{61}{2}=30.5mA[/latex]
(b) ac power input[latex]=I_{rms}^2(r_f+R_L)[/latex]
[latex]=(30.5 imes 10^{-3})^2(20+800)[/latex]
[latex]=0.763W[/latex]
d.c. power output[latex]=(I_{dc})^2 R_L[/latex]
[latex]=(19.4 imes 10^{-3})^2 imes 800[/latex]
[latex]=0.301W[/latex]
(c) d.c. output voltage [latex]=I_{dc}R_L=19.4mA imes 800\Omega[/latex]
[latex]=15.2V[/latex]
(d) Efficiency[latex]=\frac{P_{dc}}{P_{ac}} imes 100[/latex]%[latex]=\frac{0.301}{0.763} imes 100[/latex]%=39.5%

Question 2.12: A diode with internal resistance rf=20 X in half-wave rectif...

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