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A diode having $R_{1}=40 \Omega$ is used for half wave rectification. If V=20sin$\omega$ t and $R_{L}$=800$\Omega$

Calculate :

(i) $I_{m}$,$I_{dc}$,$I_{rms}$

ii) Efficiency

iii) DC output voltage

1 Answer
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$R_{L}=800\Omega , R_{f}=40\Omega$

i) Secondary voltage is given by

$V_{secondary}=V_{ra} sin \omega t$

=20 sin $\omega t$

$v_{m}$=20V

$I_{m}=\frac{V_{m}}{R_{L}+R_{F}}=\frac{20V}{800+40}$=23.8mA ........(i)

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$I_{dc}=\frac{I_{m}}{\pi}$=7.57mA.............(ii)

$I_{rms}=\frac{I_{m}}{2}$=11.9mA..........(iii)

ii)Efficieny $\eta-\frac{P_{dc}}{P_{ac}}=\frac{I^{2}_{dc}R_{L}}{I^{2}rms(R_{L}+R_{f})}$

=$\frac{I^{2}m R_{L}}{\pi^{2}}\times \frac{4}{I_{m}^{2}(R_{L}+R_{f})}$

$\frac{4}{\pi^{2}}.\frac{R_{L}}{R_{L}+R_{f}}=\frac{4}{\pi^{2}}\times\frac{800}{800+40}$

38.59%.......(iv)

iii)DC output voltage $V_{dc}=I_{dc}\times R_{L}=6.06V \ \ \ ........(iv)$

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