Peak rectifier - conduction interval

[theBEAST]

Homework Statement

I am confused about how they got the expression:
V_Pcos(ωΔt)=V_P-V_r

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:
V_Pcos(ωT)-V_Pcos(ωt_d)=V_P-V_r

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

 

[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

 

[BVR]

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

 

[berkeman]

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

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.

 

[BvU]

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

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) .