# Determine Vrms using OpAmps?

Discussion in 'General Electronics Chat' started by kellgine, Apr 25, 2010.

1. ### kellgine Thread Starter New Member

Feb 1, 2009
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I am wondering if it is feasible to determine the RMS voltage of a source using an integrator, whose input is a fully rectified sine wave. The output of the integrator would then be ∫|sin(ω*t)|dt ? If the DC output were checked with a voltmeter, could the RMS value of the input be determined if the voltage of the source were unknown?
I realize the resistor values in the attached diagram may be odd.

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Apr 20, 2004
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3. ### Jaguarjoe Active Member

Apr 7, 2010
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You have the inverting and non inverting inputs backwards. With R2 being such a low value you won't be doing much integrating.

4. ### kellgine Thread Starter New Member

Feb 1, 2009
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While I realize purchasing an IC that has been well tested and is known to work far better than anything I could construct, I need to do it using either an integrator or some other way. I have read that a low pass filter may be a possible route since it would eliminate the extra opamp at the output.

If the opamp inputs were connected correctly and the resistors were adjusted that they were integrating, would it be correct to say that the input Vrms can be calculated as follows : Vrms = Vout*(π√2)/4 ?

5. ### beenthere Retired Moderator

Apr 20, 2004
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Page 18 of the op amp collection shows how it's done. Forget the LM101's, of course.

6. ### ifixit Distinguished Member

Nov 20, 2008
639
110
The circiut you presented does not give an RMS output of the input (Vsin). It will, however, give you an average (.636 of Vsin[peak]) value using the simple opamp integrator. If the input is a pure sinwave, then the average reading can be simply be re-scaled to read as RMS (.707 of Vsin[peak]). I think this is what you want.

Vrms=√average[Vin$^{2}$]. This means you would need a circuit that; takes many samples of the input, squares these readings, then averages these reading over a suitable period that will cover the lowest frequency of interest, and then find the square root of the average. Doing this well with opamps is not easy... hopefully your signals are sinwaves.

Regards,
Ifixit

7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
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The circuit you posted (with the op amp inputs swapped) is not an integrator, it's a low pass filter. The true integral, ∫|sin(ω*t)|dt, will not be a DC voltage. It will go to infinity in infinite time.

8. ### bitrex Active Member

Dec 13, 2009
79
4
Hey kellgine!

Determining the RMS voltage of a sine wave source is pretty easy. RMS for a continuous function is defined as the square root of the average of the square of the function, so for a sine wave of amplitude 1 the average of the square of the function would be:
$sqrt(\frac{1}{2\pi}\int_0^{2\pi} (sin(x)^2)dx$

You can see that evaluated here, just because Wolfram Alpha is so cool :
http://www.wolframalpha.com/input/?i=sqrt(1/(2*pi)*(integrate+sin^2(x)+from+0+to+2*pi))

So it's the well known 0.707 figure. So if you want to know the RMS value of any sine wave, all you have to do is find its peak value (using a peak detector, for example) and multiply by 0.707 - you could use a precision voltage divider.

For a function that is non-sinusoidal, it's a bit more difficult. You were on the right track with the integration, but to find out the true RMS for some random waveform you first have to square the function, integrate it, and then take the square root of the result as in the equation above. As Ifixit mentioned, it isn't easy to do this in analog, though there are circuits which will do it. If you want to experiment Analog Devices makes a chip called the AD633 which is an analog multiplier - it uses the http://en.wikipedia.org/wiki/Translinear_principle
to do multiplication. It can be set up to do squares and square roots.

9. ### t06afre AAC Fanatic!

May 11, 2009
5,939
1,223
On www.edn.com you find some circuits. Search for true rms. This link Circuit measures true- rms and average value may be useful. Many digital multimeters do not use a proper "true RMS" approach. Hence they will only be accurate then measuring a low frequency sine wave.