How do I design a simple circuit that converts a sine wave to an RMS voltage? The input will be a sine wave and the output an RMS voltage.

I know I will need an ADC, but not sure which one. I also know the analytical calculation is VRMS = Vpk × 0.7071.

I am new to electrical engineering. Any help and guidance will be appreciated.

  • 4
    \$\begingroup\$ What frequency? What sampling rate can you use? How quickly does your method need to calculate RMS? What accuracy are you looking for? What form does your input signal take i.e. where does it connect to? \$\endgroup\$
    – Andy aka
    Commented Jan 26, 2022 at 18:27
  • \$\begingroup\$ Diode bridge, RC followed by 1/1.414 divider? \$\endgroup\$
    – winny
    Commented Jan 26, 2022 at 18:27
  • \$\begingroup\$ Have you tried any of these COTS? analog.com/en/parametricsearch/11023# \$\endgroup\$
    – winny
    Commented Jan 26, 2022 at 18:28
  • \$\begingroup\$ An ADC doesn't produce a voltage, it produces a digital measurement. You know this, but I'm making the point that your requirements aren't clearly specified... \$\endgroup\$
    – TypeIA
    Commented Jan 26, 2022 at 18:29
  • 1
    \$\begingroup\$ A straight piece of wire? A sine wave already has an rms voltage. Do you actually want a DC voltage equivalent to the rms value of a sine wave? Why do you want it? \$\endgroup\$ Commented Jan 27, 2022 at 5:44

3 Answers 3


Maybe use a chip such as the AD636.

  • True rms-to-dc conversion 200 mV full scale Laser-trimmed to high accuracy
  • 0.5% maximum error (AD636K)
  • 1.0% maximum error (AD636J)
  • Wide response capability Computes rms of ac and dc signals 1 MHz, −3 dB
  • bandwidth: V rms > 100 mV Signal crest factor of 6 for 0.5% error

enter image description here

The relationship 0.707 is only valid for purely sinusoidal signals.

Cheap voltmeters display "RMS" by using a full-wave precision rectifier followed by a low-pass filter and (effectively) a DC gain of 1.11, which is the ratio of RMS to average value for a pure sine wave. It is grossly inaccurate for spike-y waveforms such as brief pulses and somewhat inaccurate for distorted sine waves.

If you want to do the above, there are three easy circuitry 'blocks' you can research and implement. True RMS (from scratch) is much more difficult because of the mathematical functions and precision required.

  • \$\begingroup\$ I cannot seem to understand how the example ircuit works, there is only 1 connection between the left and right side, you need 2 connections to measure something between them, is the left COM node supposed to go somewhere into the circuit? \$\endgroup\$
    – Ferrybig
    Commented Jan 27, 2022 at 10:24
  • \$\begingroup\$ Yes, the COM would be one test probe and Vin the other. \$\endgroup\$ Commented Jan 27, 2022 at 13:22
  • \$\begingroup\$ I mean, where if the COM node connected to in the actual schematic? there is only 1 ground symbol in the schematic. \$\endgroup\$
    – Ferrybig
    Commented Jan 27, 2022 at 13:28
  • \$\begingroup\$ Ah. The split supply is created (from the 9V battery) by the 7106. COM on the 7106 is actually an output in this configuration. So the COM terminals are connected together and could be shown grounded, but maybe they wanted to emphasize something (or maybe it's just sloppy). You might want to look at the datasheet configurations. I just wanted to show it in a complete application- but perhaps this one is not so helpful. \$\endgroup\$ Commented Jan 27, 2022 at 13:33

You could try Linear Tech's (now ADI's) frequency independent RMS detector: ∆Σ Breakthrough: LTC1966 True RMS-to-DC Converter Uses No Diodes, Heaters or Logarithms. A typical application circuit, according to the link and the datasheet, shows this much, only:

freq. indep. RMS

If you want to test it, in LTspice you can do this:

LTspice simulation

Due to the output filter it's not exactly frequency independent, but the idea is. Shown above are simulations with a 1st and a 2nd order Bessel filters for a triangular waveform (so it's not restricted to sine waves -- you may have harmonics).

  • \$\begingroup\$ Yes something like this. It's worth the $20 per chip, or the AD8436, but it's more difficult to implement. The alternative is using software. \$\endgroup\$
    – Jeroen3
    Commented Jan 27, 2022 at 14:42

If you're not wanting to use a dedicated chip to do it (as per the examples above), there's a method which was standard when I worked at a company making equipment for national grids. This was in the 1990s, and they were just in the process of changing over from analogue to software. One of my first jobs was to work through all the analogue protection circuits and turn them into design documents for software - not just the details of what they did, but the intent behind them.

Anyway, your standard method starts with a DC blocking capacitor so that you've only got the AC part. After that, the sine wave goes into a full-wave precision rectifier. This gives you a series of positive half-sine "humps". You then put that through a low-pass filter with a time constant much slower than the sine-wave frequency. The result is a DC level that's the average of the sine-wave amplitude (plus a little ripple).

Note that the DC level is the average of a rectified sine-wave (integral over time divided by time), which is (2/pi)*Vpk=0.637*Vpk. The RMS of course is (1/root2)*Vpk=0.707*Vpk. So if you want to output the actual RMS, you're going to need a non-inverting amplifier with a gain of 1.11. Or build that gain into your precision rectifier or low-pass filter.

(Thanks to @aconcernedcitizen for correcting my scaling factors!)

  • 2
    \$\begingroup\$ I hope I don't appear condescending, certainly not my intention, just want to clarify one thing: "the DC level is the average of the rectified sine-wave, which is 0.5*Vpk" -- the DC is not the average, it represents the RMS (the quantity that causes the heating, the same power); the average is 2/pi (full-wave rectification). The gain will be then pi/sqrt(2) = ~1.11. Of course, this assumes no harmonics (which can't be due to the dead-band of the bridge). \$\endgroup\$ Commented Jan 27, 2022 at 16:07
  • 2
    \$\begingroup\$ @aconcernedcitizen Oh, of course it is. My mistake - total brainfart on that! Thanks for the correction. \$\endgroup\$
    – Graham
    Commented Jan 27, 2022 at 18:15

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