Giles,

This is what I figured, but what would be the "peak" value?
In an AC sinewave the peak value is half of the peak to peak
value and the RMS is the peak x 0.707. Is their a math formula
for finding the peak value?

Thanks for your help!
Steve

-----Original Message-----
From: Giles Honeycutt [mailto:giles-pl@AMTECH-ENG.COM]
Sent: Monday, September 16, 2002 9:06 AM
To: PICLIST@MITVMA.MIT.EDU
Subject: Re: [EE]: Measuring the peak value of an AC squarewave.


Steve,
The formula for the RMS of a AC square wave is Divide by 2
Or you can just flip the negative, and you have DC and DC is RMS

Best regards,
Giles

----- Original Message -----
Subject: [EE]: Measuring the peak value of an AC squarewave.
Friends,

I am having a problem that I really need help with.
I am trying to measure an AC squarewave at 800Hz with a
dutycycle of 50%. I am using a Keithley 2000 multimeter
set to read ACV. I need to measure the volts peak, the meter
gives me the volts RMS. I figure I could do the formula but
nothing matches.

I am measuring 5.6Vpp and 2.8Vrms. This doesn't make any sense to
me. When I measure the sinewave of an equivelant circuit everything
works out mathematically. Could someone point me to a possible solution.
I have a feeling I am missing / forgetting something trivial.

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