# Peak rectifier - conduction interval

1. Jan 25, 2014

### theBEAST

1. The problem statement, all variables and given/known data
I am confused about how they got the expression:
VPcos(ωΔt)=VP-Vr

Specifically you can find the notes in page 5 of this article:
http://whites.sdsmt.edu/classes/ee320/notes/320Lecture8.pdf

You may also find it useful to refer to the graphs as a reference.

Shouldn't the expression be:
VPcos(ωT)-VPcos(ωtd)=VP-Vr

Since we want the voltage difference between the points td and T.

2. Jan 26, 2014

### BvU

You've had little response because you make it a little hard to help you.
I agree with your conclusion. Didn't look into it any further to see how it influences the results

3. Dec 31, 2016

### BVR

Hello, can someone please explain me why does the diode ONLY conduct in that interval? Thanks

4. Dec 31, 2016

### Staff: Mentor

Welcome to the PF.

I didn't look much at the article, but the diodes in the rectifier only conduct when their cathode voltage drops to about 1.4V (twice the 0.7V diode drop, because there are two conducting diodes in series with the input waveform) below their anode voltage. That only happens for the part of the AC cycle near the peak voltages, because that's the only time that the input voltage value is above the voltage value stored on the top of the smoothing capacitor.

If there is no output current into a load, there will be no conduction at the peaks of the input waveform because the storage cap will just stay charged up to the peak input voltage minus about 1.4V.

5. Dec 31, 2016

### BvU

An ideal diode only conducts if the forward voltage difference is greater than zero, i.e. if the blue line threatens to go above the red line.
Berk refers to a more realistic case where a diode conducts but with a voltage drop of approximately 0.7 V (google diode characteristics).
(And he also appears to think of a bridge rectifier) .