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Old 01-02-2012, 02:59 AM
TheLaw TheLaw is offline
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Default Finding Peak-to-Peak Voltage of a Square Wave Using a Multimeter?

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.
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Old 01-02-2012, 03:06 AM
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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
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Old 01-02-2012, 03:13 AM
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and hence the peak-to-peak voltage would be twice the RMS reading.
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Old 01-02-2012, 03:17 AM
TheLaw TheLaw is offline
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Quote:
Originally Posted by thatoneguy View Post
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.
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Old 01-02-2012, 03:19 AM
TheLaw TheLaw is offline
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Quote:
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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?
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Old 01-02-2012, 03:19 AM
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Yes, T is usually the period, or \frac{1}{f} where f is frequency.
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Old 01-02-2012, 03:22 AM
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Quote:
Originally Posted by TheLaw View Post
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.
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Old 01-02-2012, 03:23 AM
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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.
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Old 01-02-2012, 03:24 AM
TheLaw TheLaw is offline
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Quote:
Originally Posted by thatoneguy View Post
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/....htm?PID=73125

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

Thank you.
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Old 01-02-2012, 03:27 AM
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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.
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