A full wave rectifier has to supply 75 mA at 250 V with a ripple that must be less than 10 V. Design the rectifier with L-filter

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written 5.6 years ago by

teamques10 ★ 65k

$I_{L}$(dc)=75 mA, $V_{L}$(dc)=250 V, $V_{r}$=ripple voltage=10 V

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I) Load resistance

$R_{L}=\frac{V_{L}(dc)}{I_{L}(dc)}=\frac{250V}{75mA}$=3.33k$\Omega \ \ \ \ \ .....(i)$

ii) The ripple factor =$\frac{V_{r}(rms)}{V_{L}(dc)}=\frac{10}{250}$

=0.04=4% ...............(ii)

iii)For inductor filtered full wave rectifier

$V_{L}(dc)=\frac{2V_{m}}{\pi}-I_{dc}\times R$

By neglecting R. i..e. transformer secondary resistance, inductor and diode resistance

$V_{L}(dc)=\frac{2V_{m}}{\pi}$

$V_{m}=\frac{-V_{L}(dc)\pi}{2}=\frac{250 \pi}{2}$=392.69 V ................(iii)

iv) Ripple …

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