DC average of half wave rectified AC signal - Page 1
Forum section: electronics / beginners

author: Syko_
author_rank: Contributor
timestamp: November 06, 2014, 05:39:28 am
content: Hi all,I'm kind of stuck for this question, it's asking me to compute the Vavof a half wave rectified sinusoidal AC signal. I am aware that for an ideal diode (with no voltage drop), that would be Vpk/ pi, however, the answers I get are not quite correct. For example, I get 2.99 v instead of 2.89 v, or 5.18 v instead of 5.06 v. Where is this extra voltage drop coming from? I assume that it has something to do with the fact that the diode voltage drop causes the conduction period to take slightly less time that the non-conduction period, and that causes the period to drop to just below pi, and then that causes the average voltage to be lower than what it would be in an ideal situation.Thanks

author: IanB
author_rank: Super Contributor
timestamp: November 06, 2014, 05:58:32 am
content: It's not clear what you are asking. If you compute the result it is Vpk/ pi as you said. This is exact. If you compute it that is what you will get.Are you trying to measure it? With an oscilloscope? Or something?If so, you will not get a result that exactly matches computation. Welcome to the world of measurement error.

author: Syko_
author_rank: Contributor
timestamp: November 06, 2014, 06:03:36 am
content: Nah, this is purely theoretical. Let me upload some images of the problem.

author: Syko_
author_rank: Contributor
timestamp: November 06, 2014, 06:18:28 am
content: Look at problem 2.1, b, ii. If you compute the peak value as 9.4 v (due to diode drop), then divide by pi, you receive a value of 2.99. If you see the second image, their result is 2.89 v. I thought it might have been an error with the book, but I receive the same difference if I use a different problem too.Sorry for large images, cbf scaling them.

author: Syko_
author_rank: Contributor
timestamp: November 06, 2014, 07:39:14 am
content: I just realised you might need the diagram too

author: AndreasF
author_rank: Frequent Contributor
timestamp: November 06, 2014, 07:46:35 am
content: I suspect that it has to do with the fact that your rectified half-wave is no longer a complete half-wave due to the diode drop, i.e. the (phase) length of your half wave is no longer 180 degrees (pi).Wolfram alpha integral calculationgives you an integral of 18.1511. Dividing that by the period (2*pi) to get the average gives you 2.888837...(P.S. this is completely ignoring the resistances - not sure how relevant those are)

author: Syko_
author_rank: Contributor
timestamp: November 06, 2014, 07:52:59 am
content: Well, the answer provided by the textbook is 2.89v, which happens to be the same value that you received. Great to know where it comes from, kind of a pain in the ass that it now involves an integral. I wonder if it reduces into some more helpful formula, or if I have to compute the integral every time.

author: AndreasF
author_rank: Frequent Contributor
timestamp: November 06, 2014, 08:10:36 am
content: There very well might be - I have no idea. I just knew that "9.4*sin(x)" was not going to be quite the same as "10*sin(x)-0.6"