Re: Let's build a power meter for five bucks. It's too bad TechTV is promoting non-sense like this

On Mon, 29 Mar 2004 05:15:30 -0800, "Watson A.Name - \"Watt Sun, the Dark


You could take a simultaneous voltage [Not the voltage drop across the series (shunt) resistor, but the SOURCE voltage] reading, integrate the two instantaneous values over a period of 2*pi radians, and do a vector sum. That would give you true RMS power, no?
-- -john wide-open at throttle dot info
~~~~~~~~ "The first step in intelligent tinkering is to save all the parts." - Aldo Leopold ~~~~~~~~
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
~^Johnny^~ wrote:

I'm guessing that you meant a simultaneous voltage and current measurement. You seem to be proposing an incorrect mathematical approach to a real world issue. You can't integrate two instantaneous values of two different things. Could you please re-state this to make it clear.
My guess is you really mean integrate all V readings from 0 to 2 pi radians, and also integrate the I readings for the same 360 degrees. But that mathematical solution is still incomplete - you then need to multiply. Is that what you have in mind?
Now, if you mean this to be a practical solution, what instrumentation do you buy to sample the V and I, how many samples does it take during the cycle, how are the measurements stored, and how do you integrate them?
It's much be cheaper and easier to buy a true RMS DMM and measure the V across a shunt resistor, then compute the power. Hell, I've read even have a power measurement device for about 40 bucks, which is even cheaper than a true RMS DMM, and involves no computation. I don't know if it's any good.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
snipped-for-privacy@MUNGED.OkBy.UsINVALID says...

Not vector sum, rather multiply the two, point by point and average over the cycle. BTW, it's average power (RMS power makes no sense). The vector sum would give you the power factor (or more precisely lead/lag angle). Of course the power factor is also the RMS(V) * RMS(I) / average power. Yes, this all is fairly easy to do with a microcontroller and a couple of ADCs.
--
Keith

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.