How does a digital multimeter measure AC voltage? My guess would be that the AC voltage is rectified and filtered into a DC voltage which is then measured, giving the peak voltage of the AC signal. Am I correct?
A follow up question: I remember reading from somewhere that multimeters cannot very well give accurate AC voltage measurements if the signal is not a sinusoid but a square wave for example. Is this true? If so, why is this? If AC measurement works by rectification into DC as I assume, shouldn't they measure other types of waveforms just as well?
Best Answer
First, I'll note an erroneous assumption: AC meters do not display peak voltage but rather RMS voltage. This is a sort of "average" voltage that makes all the math work out right.
Now, to answer the question:
There are two main methods of doing AC measurement: the cheap way, and the accurate way.
The cheap way is indeed how you suspect: rectify it and measure the peaks, and then divide by √2 to get the RMS. This works only for sinusoids, because the factor of √2 implicitly assumed a sinusoidal signal.
The accurate way uses what's called a "true RMS converter", which is a circuit (analog in older true RMS meters, probably digital in modern ones but don't quote me on that) that actually calculates the RMS voltage with appropriate signal processing techniques.