# PIV for Full Wave Rectifier (Quick/Simple)

Discussion in 'Homework Help' started by jegues, Apr 21, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
I've got a quick question about the PIV.

Shouldn't it be,

$PIV = 2v_{S} - v_{D}$

?

This is a full wave rectifier, what he has got looks like what it should be for a Bridge rectifier.

Which one is correct?

EDIT: I've got it figured out I was misreading something in the original question.

It also asks about what fraction of a cycle each diode conducts.

On any half cycle the supply must overcome a voltage drop of 0.7V.

$wt = sin_{-1}(\frac{0.7}{12\sqrt{2}})$

So the fraction of a cycle that each diode conducts is,

$\frac{\pi - 2wt}{2\pi} * 100 = 12.38%$

Is this correct?

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Last edited: Apr 21, 2011

Dec 26, 2010
2,147
300
NB indeed this is not a bridge rectifier: the lost voltage is one diode drop, not two. The calculation in your attachment shows two diode drops, which is wrong.

Each diode therefore conducts for half a cycle, less the time that the 12V rms AC is less than one diode drop. You seem to have taken the "lost" conduction angle due to the diode drop to be the total. Think again.

3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Let $wt = \theta$

$\theta = 0.0412rad$

$\frac{\pi - 2\theta}{2\pi} * 100 = 48.69% \text{ for EACH diode}$