# True RMS explaned

in power supply by | 9 comments

Not everyone understands how multimeters measure RMS voltage. Doug Criner has put a good explanation.

Can there be more than one RMS value of a voltage waveform, but only one of them is true?  Apparently so.To use simple Ohm’s law and power calculations for a periodic waveform, we must know its RMS, or root-mean-square value–which is the effective power-producing value.  For a sinusoidal waveform, the RMS value is equal to the peak voltage divided by √2.  What we call 120-VAC power is the RMS value of a sinusoid with positive and negative peaks of 170 V.

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1. rsdio says:

The linked article may explain what Fluke’s “True RMS” (TM) feature means, but it really misses an important point: The RMS value of a waveform is different for each waveform shape. The article only mentions that a sine wave has an RMS value of √2 of its peak value, but then mistakenly looks at a square wave as an example. That’s very misleading because the RMS value of a square is the same as its peak value!

A good resource to learn about “real” RMS is http://designtools.analog.com/dt/dbconvert/dbconvert.html

A triangular waveform has even less than √2 of the peak as its RMS, and pure noise is only 15% of peak.

2. I agree with rsdio. I read the article and this Doug Criner guy is clueless Funny Fluke didn’t set him straight.

• I take back my statment. I did that math as well as contacted Fluke. Doug Criner is 100% correct about this. Strange o_0

• rsdio says:

Are you saying that you no longer agree with me (aw, my feelings are hurt), or that you take back your statement that Criner is clueless?

I still believe that Doug is not 100% correct, at least not with his square wave example.

• @rsdio, That Doug Criner is clueless. He is not. I contacted Fluke and he is right about how the meters work like the 87-V and most others. Only higher modules like the Fluke 289 have a setting for AC+DC when working with waves that have a DC offset. And his math on the square wave is correct as well. I checked that too. He should have used a sin wave because that would have looked normal to me and would have been a better point to make but still his clipped square wave example is correct.

3. SQKYbeaver says:

for a simple sin wave he is right, for anything else you really need to use a scope and a calculator.

the history of the “true RMS” meters is the interesting part.

• rsdio says:

You don’t always need a scope and calculator. If the waveform is well-defined, then you can use calculus to determine the RMS. I guess if your calculator does calculus then you’re set!

Distorted waveforms would pose serious issues. You’d need a sampling scope which calculates the actual RMS, and hope that the sample rate is high enough that the scope is measuring something close to the original waveform.

4. hyper says:

I think you mean ‘explained’.

5. SQKYbeaver says:

for a pure sine wave it is much easier, just multiply the peak to peak voltage to 0.707.