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?

  • \$\begingroup\$ See if my answer to Average value of current or voltage helps. \$\endgroup\$
    – Transistor
    Jul 31, 2019 at 18:31
  • \$\begingroup\$ Before digital, the D'Arsonval meter movement would effectively average the rectified voltage swings. Cheap digital multimeters probably also rectify the input and attempt to find the peak value from which they can work out in a digital way the same thing. For a sine wave only, \$\overline{V}=\frac{1}{\pi}\int_0^{\pi}\,V_\text{PK}\operatorname{sin}\left(\theta\right)\,\text{d}\theta=\frac{2}{\pi}V_\text{PK}\approx 0.637\cdot V_\text{PK}\$. Other waveforms would have different results of that integral and therefore such meters would have various intrinsic errors in their measurement. \$\endgroup\$
    – jonk
    Jul 31, 2019 at 18:44

3 Answers 3


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.

  • \$\begingroup\$ If my multimeter uses the cheap way, can I measure peak voltages of non-sinusoids by simply multiplying the output reading of the MM by the square root of 2? \$\endgroup\$
    – S. Rotos
    Jul 31, 2019 at 18:14
  • 1
    \$\begingroup\$ @S.Rotos Yes, and no. You would need to multiply by a correction factor that is dependent upon the specific type of non-sinusoidal waveform you are measuring. \$ \sqrt{2} \$ is the factor that ties the peak of a sinusoid to the RMS of a sinusoid. You need other factors if you want to tie the RMS of a sinusoid to the peak of a non-sinusoid. \$\endgroup\$
    – DKNguyen
    Jul 31, 2019 at 18:15
  • 1
    \$\begingroup\$ It's not \$\frac{1}{\sqrt{2}}V_\text{peak}\$ but instead \$\frac{2}{\pi}V_\text{peak}\$ for averaging meters that assume a rectified sinusoid. \$\endgroup\$
    – jonk
    Jul 31, 2019 at 18:30
  • \$\begingroup\$ @jonk I might just be tired, but isn't \$V_{rms} = \frac{V_{pk}}{\sqrt{2}}\$? \$\endgroup\$
    – Hearth
    Jul 31, 2019 at 19:36
  • \$\begingroup\$ @Hearth I think you need more sleep. \$\endgroup\$
    – DKNguyen
    Jul 31, 2019 at 20:29

Cheap meters measure the average rectified value and then jimmy the reading up (modify the calibration - gain vs. ADC reference voltage- so that it reads higher than the actual measured voltage) by ~+11.1% to correctly read RMS for a pure sine wave. They read the incorrect RMS value of any waveform other than a sine wave (usually they are AC coupled too, so any DC component is ignored). A square wave would tend to be displayed too high because the average is equal to the RMS.

For a sine wave of peak voltage Vp:

The average over a half cycle is 2Vp/\$\pi\$ = 0.6366 Vp

The RMS is Vp/\$\sqrt{2}\$ = 0.707107 Vps

The ratio is \$\pi \over 2\sqrt{2}\$ \$\approx\$ 1.11072

More expensive meters attempt to measure the RMS value (the square root of the average of the squared voltage). Typically that's done with an analog chip and it will only handle some range of crest factor (ratio of peak to RMS) and bandwidth without losing a lot of accuracy.

  • \$\begingroup\$ Could you please clarify "jimmy" for our readers? I'm from the midwest US and I knew what you meant, but it sounds like a colloquialism. \$\endgroup\$ Aug 1, 2019 at 11:56
  • \$\begingroup\$ @ElliotAlderson Done. I do try to stick-handle those colloquialisms. 8-) \$\endgroup\$ Aug 1, 2019 at 12:01

If AC measurement works by rectification into DC as I assume, shouldn't they measure other types of waveforms just as well?

Regardless if the measurement is based on peak value or the average level of the rectified waveform, you discard information in these processes. The RMS value is related to power, so it depends on the squared value of the voltage at each instant. This squaring can be performed digitally or with analog circuits, and must be later averaged.

Also important: when you choose "AC" only, the multimeter removes any DC component in the signal before rectification. Only then it takes the peak or average value if it is not a True-RMS device.

Think what will happen with a square wave an notice that you must know how your device works to convert the value of non sinusoidal signals.


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