Half-Wave Diode Rectification

[lion]

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 subject

I have a simple Lab work with the following circuit:



a)My Vs is 14.5VAC RMS -> 20.5Vp
b)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 DMM http://img.dxcdn.com/productimages/sku_224265_3.jpg), which I got 7.5V
d)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 rectified
3. How can I respond to d).


Thanks

[lion]

"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

[Andy Watson]

My questions are:
1. Can I measure Vaverage with a DMM?

Depends on the DMM. Many DMMs do measure Vaverage.

2. If so why Vaverage must be measured with VAC and not VDC, since the wave is half rectified

Are you sure? I think you should be using the DC range to measure the average DC level.

3. How can I respond to d).

What 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.

[danadak]

Google "vave vs vrms".

DMMs come in various flavors, some computing, some RMS......Consult
manual.

For a sinewave there is a relationship between Vave and Vrms. Also true
for half wave.



[url=http://site.iugaza.edu.ps/abdos/files/exp3.pdf]http://site.iugaza.edu.ps/abdos/files/exp3.pdf[/url]



Regards, Dana.

[Z80]

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 rectified
Your 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 maths
For a sinewave
VRMS = 1/(sqroot 2) * Vpeak
Or approx 0.707 * Vpeak
You are only using half the waveform so divide this by 2.

[Ammar]

Strictly speaking this is RMS the average is a slightly different thing.

Very 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.

[WaveyDipole]

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.html

After 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.