Half-Wave Diode Rectification  - Page 1
Forum section: electronics / beginners

author: lion
author_rank: Contributor
timestamp: October 23, 2016, 09:23:28 pm
content: Hello guys.I've been on chat asking about the Half-Wave Diode subject. Some of the questions Im asking are more clear now, but I still want your opinion on the subjectI have a simple Lab work with the following circuit:a)My Vs is 14.5VAC RMS -> 20.5Vpb)With my analog scope I visually measured Vout =19.5Vp which is almost correct according to "20.5Vp - 0.7V from diode".c)The teacher asked me to get the Vaverage by measuring VAC on the load with a DMM (link of the DMMhttp://img.dxcdn.com/productimages/sku_224265_3.jpg), which I got 7.5Vd)Then asked me to justify that Vaverage on DMM with my measurement on the scope.PS: my teacher is still new to this kind of subjects.My questions are:1. Can I measure Vaverage with a DMM?2. If so why Vaverage must be measured with VAC and not VDC, since the wave is half rectified3. How can I respond to d).Thanks

author: lion
author_rank: Contributor
timestamp: October 23, 2016, 10:13:07 pm
content: "2. If so why Vaverage must be measured with VAC and not VDC, since the wave is half rectified"Does this answers my question? VDC = VAverage?http://nrredc.blogspot.pt/2016/05/average-or-dc-output-voltage-vav-or-vdc-derivation.html

author: Andy Watson
author_rank: Super Contributor
timestamp: October 23, 2016, 11:51:24 pm
content: Quote from: lion on October 23, 2016, 09:23:28 pmDepends on the DMM. Many DMMs do measure Vaverage.QuoteAre you sure? I think you should be using the DC range to measure the average DC level.QuoteWhat average value would you expect for a half-sine wave? Now include the portion of the wave for which the value is zero and take the mean.

author: danadak
author_rank: Super Contributor
timestamp: October 23, 2016, 11:51:39 pm
content: Google "vave vs vrms".DMMs come in various flavors, some computing, some RMS......Consultmanual.For a sinewave there is a relationship between Vave and Vrms. Also truefor half wave.[url=http://site.iugaza.edu.ps/abdos/files/exp3.pdf]http://site.iugaza.edu.ps/abdos/files/exp3.pdfRegards, Dana.

author: Unknown
author_rank: Unknown
timestamp: October 24, 2016, 09:22:20 am
content: 1. Can I measure Vaverage with a DMM?Yes you can, if your DMM is a trueRMS type (the manual will tell you this & it is often printed on the meter). Strictly speaking this is RMS the average is a slightly different thing.2. If so why Vaverage must be measured with VAC and not VDC, since the wave is half rectifiedYour circuit doesn't produce DC as there is no smoothing capacitor. The diode just passes one half of the AC wave.For DC the voltage doesn't vary so the average and peak values are the same.3. How can I respond to d).Use mathsFor a sinewaveVRMS = 1/(sqroot 2) * VpeakOr approx 0.707 * VpeakYou are only using half the waveform so divide this by 2.

author: Ammar
author_rank: Regular Contributor
timestamp: October 24, 2016, 10:53:16 pm
content: Quote from: Z80 on October 24, 2016, 09:22:20 amVery important point. Make sure you bring this up with your lecturer. I once marked some exams where the lecturer made this error in their solutions.Vrms is the equivalent DC voltage that delivers the same average power.Vavg is the average voltage over a period.

author: WaveyDipole
author_rank: Frequent Contributor
timestamp: October 25, 2016, 12:22:16 am
content: You don't mention the value of R, but you state that your Vs is 14.5Vrms. You visually get a reading on your scope of the Vp of 19.5v, which, as you say, taking into account the voltage drop across the diode is not far off the 19.8v expected. The formula Vrms = Vp * 0.707 applies to a full sine-wave only. Other waveforms will have other proportions and hence a different formula. See here:http://meettechniek.info/compendium/average-effective.htmlAfter the diode, you will have a half-wave rectified sine wave. The formula that applies here is Vrms = Vp/2. The RMS voltage therefore is 19.5/2 = 9.75v. The formula for the average voltage in this circumstance is Vavg = Vp/PI (or can be expressed as Vp * 0.318), so Vaverage would be 19.5/3.141 ~ 6.201v.Maybe the point of (D), is to demonstrate that what you might get in practice is not always the same as what the theory would suggest, in which case its important to understand the reason for the difference. In this case, the DMM you are using does not appear to show True RMS, so will not display accurate AC voltage information under these circumstances, particularly when wave-forms more complex than a symmetrical sine-wave are involved.