# Finding Peak-to-Peak Voltage of a Square Wave Using a Multimeter?

#### TheLaw

Joined Sep 2, 2010
228
Hello,

How can I find the peak-to-peak voltage of a square wave produced by a function generator when I only have a True-RMS multimeter?

I'm new at this so if someone could dumb it down for me, I'd appreciate it.

Is it possible? Thanks.

#### thatoneguy

Joined Feb 19, 2009
6,359
• jaygatsby

#### MrChips

Joined Oct 2, 2009
21,823
and hence the peak-to-peak voltage would be twice the RMS reading.

#### TheLaw

Joined Sep 2, 2010
228
For a square-wave, RMS is the peak voltage.

Or do you need it for duty cycles other than 50%?

ETA: RMS - Peak Calculator for different waves
No 50% duty is all I need. It says positive square wave. I forgot to mention that it is a +/- square wave equally bipolar. So if it was 10Vpp, it would be +5, -5V...if that makes sense.

Is duration (T) one period?

Thank you.

#### TheLaw

Joined Sep 2, 2010
228
and hence the peak-to-peak voltage would be twice the RMS reading.
The reading of a multimeter is RMS, but since it's meant to give the RMS value of a sine wave, wouldn't it screw something up when its a square wave?

#### thatoneguy

Joined Feb 19, 2009
6,359
Yes, T is usually the period, or $$\frac{1}{f}$$ where f is frequency.

#### thatoneguy

Joined Feb 19, 2009
6,359
The reading of a multimeter is RMS, but since it's meant to give the RMS value of a sine wave, wouldn't it screw something up when its a square wave?

If the meter is TRUE - RMS and marked that way, then it will give you the correct RMS voltage of any waveform (up to the frequency ability of the meter)

If the meter cost under $40-$50 or so, it is probably only True RMS for sinewaves, and the actual voltage of a different waveform is anybody's guess in the case of an Sine RMS only meter.

#### MrChips

Joined Oct 2, 2009
21,823
The RMS value is what constant DC voltage will give the same heating effect as the signal of arbitrary wave form. Hence mathematically, you have to compute the integral of the square of the amplitude and then take the square root. So when you square the negative portion, it is the same value as the positive part. Hence the RMS is the same as if the square wave was a constant DC signal of half the peak-to-peak voltage.

#### TheLaw

Joined Sep 2, 2010
228
If the meter is TRUE - RMS and marked that way, then it will give you the correct RMS voltage of any waveform (up to the frequency ability of the meter)

If the meter cost under $40-$50 or so, it is probably only True RMS for sinewaves, and the actual voltage of a different waveform is anybody's guess in the case of an Sine RMS only meter.
http://www.amprobe.com/amprobe/usen/Multimeters/Industrial-Multimeters-/AM-270.htm?PID=73125

I'll have to look at the datasheet for my meter then.

Thank you.

#### thatoneguy

Joined Feb 19, 2009
6,359
That meter is a True RMS meter, meaning it will give you the RMS value of any waveform.

It may also give you the peak voltage if you set it to peak record.

#### TheLaw

Joined Sep 2, 2010
228
The RMS value is what constant DC voltage will give the same heating effect as the signal of arbitrary wave form. Hence mathematically, you have to compute the integral of the square of the amplitude and then take the square root. So when you square the negative portion, it is the same value as the positive part. Hence the RMS is the same as if the square wave was a constant DC signal.
I got the first part about the heating. (Actually learned that in school!). Second part not so much..Let's see. I'll read it again.

#### MrChips

Joined Oct 2, 2009
21,823
Draw a square wave centered above and below 0V. So in your example of 10V p-p, you end up with 5V above and 5V below.
Now flip the negative portion to the positive side. You end up with a DC voltage equal to 5V, half the peak-to-peak value.

A true RMS reading voltmeter would give the same reading as the 5V DC voltage if they were to have the same heating effect.

#### bountyhunter

Joined Sep 7, 2009
2,512
Hello,

How can I find the peak-to-peak voltage of a square wave produced by a function generator when I only have a True-RMS multimeter?

I'm new at this so if someone could dumb it down for me, I'd appreciate it.

Is it possible? Thanks.
Be careful. Most modern "true RMS meters" do some kind of integral approxination and are only accurate on signals with a crest factor less than 3. The phrase "True RMS" must be taken with a grain of salt, unless it is a thermal meter and they are rare and very expensive.

#### crutschow

Joined Mar 14, 2008
25,389
Be careful. Most modern "true RMS meters" do some kind of integral approxination and are only accurate on signals with a crest factor less than 3. The phrase "True RMS" must be taken with a grain of salt, unless it is a thermal meter and they are rare and very expensive.
The crest factor of a square-wave is 1, so that should not be a problem for the meter.

#### bountyhunter

Joined Sep 7, 2009
2,512
The other problem is bandwidth limitation. The leading edges of a square wave have very high frequency content. It would be like using a scope with a bandwidth of about 50 kHz to look at a square wave. The vertical edges would be "leaning" more like a triangle wave.

Joined Apr 30, 2011
1,569
I'm going to have to try that with my meter sometime. It's a true RMS with min, max and average (Fluke 89 IV). I know that mathematically a 10Vpp square wave of 50% duty cycle should read 5Vrms, min = -5V, max = 5V and average = 0V but it would be interesting to verify.