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I was searching a way to get the VRMS of an AC voltage.

The main transformer drops down the 230 VAC to 15 VAC and with a potentiometer I'd like to display it on a DC digital voltmeter (the AC voltmeter that I have starts from 50-80 VAC to 300-500 VAC.)

What components can be used to achieve that and what type of circuit can be built?

I have tried with an op-amp set-up as a precision full-wave rectifier and AC to DC converter, but the results don't match the multimeter VRMS.

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  • \$\begingroup\$ 50-80 is a "max" range. You can read much lower values with such a meter. When you start getting into sub-1V range, you might begin worrying about precision. \$\endgroup\$
    – Kyle B
    Commented Aug 22, 2022 at 19:27

3 Answers 3

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To display RMS voltage from the average (what you typically get from a full-wave precision rectifier followed by a low-pass filter) you need to scale the voltage by 1.11 = \$\frac {\pi}{2 \sqrt{2}}\$. That number contains an implicit assumption that you have a sinusoidal input.

If you want a general RMS-reading meter that works with different waveforms then you'll need to either take many samples and digitally process the signal or to use a RMS-to-DC converter circuit (some of which are available on a single chip- however stand-alone chips are a bit pricey). Nowadays you might be able to leverage an energy metering ASIC as used in smart meters.

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  • \$\begingroup\$ Yeah -- the issue with the chip you linked is that it's NRND -- newer parts like the AD8436 are going to be far cheaper/easier to find \$\endgroup\$ Commented Aug 23, 2022 at 3:44
  • \$\begingroup\$ @ThreePhaseEel back ordered everywhere not sketchy and pricey, I was thinking more of inexpensive chips like these. Less than $3 in 25s. \$\endgroup\$ Commented Aug 23, 2022 at 6:21
  • \$\begingroup\$ @SpehroPefhany, I am not an expert in the field, first we need a full-wave precision rectifier (that can be build with op-amps) and then link it to a low-pass filter? But how can I get the scale voltage of 1.11 (pi/2*sqrt(2))? I have to tried different configurations with an oscilloscope? \$\endgroup\$
    – philfs
    Commented Aug 23, 2022 at 17:43
  • \$\begingroup\$ Just multiply the measured voltage by 1.11 in software. You already have a scale factor due to the transformer, yes? They can be combined. \$\endgroup\$ Commented Aug 23, 2022 at 19:31
  • \$\begingroup\$ There's also the old-school method of passing the voltage through a resistor and then measuring the DC voltage that produces the same heat through an identical resistor. (Don't remember what that's called) \$\endgroup\$ Commented Aug 23, 2022 at 20:14
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I have tried with some op-amp set-up as a precision full-wave rectifier and AC to DC converter, but the results doesn't match the multimeter VRMS.

Yes, the clue is in the term RMS - you take the square of the voltage waveform, then average that modified waveform (mean) then finally take the square root to get true RMS. Starting with your suggestion of precision full-wave rectification makes life easier and doesn't affect accuracy AND, it allows a simpler multiplication algorithm or circuit.

What components can be used to achieve that and what type of circuit can be built?

You can still buy analogue multipliers such as the AD636. It is a "Low Level, True RMS-to-DC Converter": -

enter image description here

And, importantly, it's still available from digikey for example. According to their page they still have over 3,000 in stock and, it is quite expensive (circa £18). ADI no longer make the device either.

There are other options that can be pursued of course. For instance, it's quite easy to make a single-quadrant, analogue squaring circuit using op-amps, analogue-switches and pulse width modulation. This can be shown if requested but, here's a link to what I've said previously. Summary picture: -

enter image description here enter image description here

But, RMS-to-DC conversion also requires taking the square root of a number and this may sound more challenging until you realize that using a squaring circuit in the feedback loop of an op-amp does the job (albeit inverting): -

enter image description here

So, -5 volts is the input and \$\sqrt5\$ (2.236068 volts) is the output.

This would be the general form of an analogue circuit using PWM to calculate power: -

enter image description here

On the final graph above I've plotted a red horizontal line corresponding to 0.7071 volts. As you can see, the RMS output hits the target pretty accurately. In case you want to inspect the PWM on the half-wave: -

enter image description here

Hopefully, you should be able to recognize that when the squared waveform is low in value, the PWM duty cycle is nearly zero. When the squared waveform is near the peak of 1 volts, the PWM duty cycle is nearly unity.

When filtered lightly you would see a nice sine-shaped waveform of twice the frequency of the input (50 Hz) and centred at 0.5 volts. Why 0.5 volts; because, if the root-mean-square output is 0.7071 volts (as we would expect) then the mean-square value would be 0.5 volts.

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  • \$\begingroup\$ excuse me but I don't understand how LTC6992 (pulse width modulation (PWM)) can be used in the process to get the RMS value. How to use your last op-amp squaring circuit? \$\endgroup\$
    – philfs
    Commented Aug 23, 2022 at 18:03
  • \$\begingroup\$ @philfs I've added a typical circuit that can be used to generate RMS. \$\endgroup\$
    – Andy aka
    Commented Aug 23, 2022 at 18:19
  • \$\begingroup\$ @philfs PWM-ing a voltage is same as multiplying the duty cycle by that voltage. So if you use a voltage to control PWM duty cycle, and module that same voltage using PWM, you get squaring. Very accurate squaring if done correctly - can be accurate to 5-6 decimal digits easy if you wish, although the bandwidth will be low. \$\endgroup\$ Commented Aug 23, 2022 at 18:49
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\$V_{rms}\$ is presented as the equivalent dc voltage to provide the same heating.

Back in the day, true-rms meters used this principle.

An amplified or attenuated version of the signal was applied to a resistor. The temperature rise was measured as a dc voltage. The dc voltage is \$V_{rms}\$. This will work for all waveforms.

Otherwise the area under the curve must be obtained or estimated by integration.

A DSP method identifies a wave shape according to a set parameters of pre-analyzed shapes. Once the parameters identify a wave shape, then a scaling factor can be applied. If a wave shape does not have an ID, then the closest one is used as an estimate. These are also called true-rms, but I haven't tested how close to the truth they are.

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