The peak detector circuit based on a precision rectifier with buffer allows to obtain the peak value.

My peak detector circuit

The RMS value in an AC circuit is obtained from:

RMS = V peak / sqrt (2) = V peak * 0.7071

AC source 15 VAC 50 Hz (image)

If the amplitude of the alternating input signal increases, the capacitor charges to the new value, if it decreases the value does not change.

How to continuously update the "stored" peak value in the capacitor? Avoiding manually discharging the capacitor (reset). Maybe adding a 10-100K resistor in parallel with the capacitor? Without discharging the capacitor too much before the positive semi-sinusoid.

What other circuits could be used to get the accurate and updated peak value and then get the RMS value?

I've already asked a similar question in the past and I thank everyone for the replies but sadly I'm not an electronic engineer and my knowledge is mediocre, furthermore the required ICs are too expensive for Italy (shipping), and I'm not able (I don't understand) how to make the circuit.

Thanks in advance.

EDIT 04/03/2023

Thank you all for your valuable advice.

The final circuit is this:

full wave prec. rectifier with AVG

It should be correct, can you check it and let me know?

VAC input 7 V

V_RMS (V_peak * 0.70711) = 4.9477 V (4.93 V sim.)

V_AVG (V_peak * 0.637) = 4.459 V (4.44 V sim.)

V_RMS circuit (V_AVG * 1.11) = 4.93 V

Thanks again.

  • 2
    \$\begingroup\$ You need a resistor in parallel with the capacitor, otherwise that output node will float. The bigger it is, the least droop it will have (meaning, less peak decay), but it'll also introduce an offset due to the input bias current developing a voltage drop across such a resistor. \$\endgroup\$
    – Designalog
    Jan 31, 2023 at 20:47
  • 1
    \$\begingroup\$ The RMS formula is only valid if the waveform is a sinewave. Ditto the above. \$\endgroup\$
    – Andy aka
    Jan 31, 2023 at 21:06
  • \$\begingroup\$ Almost any commonly available op-amp is suitable for a circuit such as this. You might be able to salvage some usable components from old junk equipment such as audio amplifiers and tape recorders/players. \$\endgroup\$
    – PStechPaul
    Jan 31, 2023 at 22:13
  • \$\begingroup\$ If frequency does not change, use a sampler. \$\endgroup\$
    – Antonio51
    Feb 1, 2023 at 15:42

2 Answers 2


Spehro gave a good answer in your question from August. AC voltmeter circuit

Finding the peak is not the best method.

The way this is normally done: assume that the signal is a sine, rectify, average, apply correction factor.

If you want to use a half-wave rectifier, then the correction factor is twice 1.11. It is easier to filter a full-wave signal, so that is preferred. You can find full-wave precision rectifiers if you search for them.

This circuit has a 1 Vrms input. If you simulate this long enough for the signal to stabilize, you get 0.451 V out.

0.451 * 2.22 = 1.00 V

You could use another opamp to apply the correction factor.


simulate this circuit – Schematic created using CircuitLab

Note that R1 must be small compared to R2. You want a low source impedance for the filter (R2 & C1).

enter image description here

  • \$\begingroup\$ A clever trick is to connect the discharging resistor R1 to the rectifier's output and not in parallel with the capacitor... \$\endgroup\$ Jan 31, 2023 at 22:04
  • \$\begingroup\$ @Circuitfantasist how do you mean? I see R1 already connected at v_rectified to ground. \$\endgroup\$
    – Designalog
    Feb 1, 2023 at 19:34
  • 1
    \$\begingroup\$ @ErnestoG, I mean the conventional connection of the discharging resistor is in parallel with the capacitor but that would affect the charging as well. Here the resistor does not affect the charging because it is connected directly to the output of the rectifier, which has a low resistance. I liked the idea of this connection; that is why I was delighted. ErnestoG, I also like your answers because they are written in detail and with great enthusiasm! \$\endgroup\$ Feb 1, 2023 at 19:47
  • \$\begingroup\$ Thanks for the reply mr. @Mattman944. I modified the circuit with your advice. I wanted to ask if I could change the values of the filter RC (with a smaller capacitor to charge it faster) and keep the same result without change the AVG value. Please, could you check the new circuit and let me know? \$\endgroup\$
    – philfs
    Feb 4, 2023 at 9:47

What other circuits could be used to get the accurate and updated peak value and then get the RMS value?

If RMS is the final goal, you could build a full-wave rectifier with gain, as mentioned in the answer to your previous question, so the average of the output is the RMS of the sine wave input value:

enter image description here

The input opamp acts only on the positive input semi-cycle, with the gain -0.5, determined by R1 and R2. This semi-cycle is amplified by -2.22, by the output opamp, according go R3 and R4.

At the negative semi-cycle, the input is just amplified by -1.11, since the diode is not conducting, and this gain is given by: \$\frac{R4}{(R1+R2+R3)}\$

As mentioned by Spehro in his answer, this 1.11 factor comes from the scaling factor of the average of the full-wave rectified signal:

\$\dfrac {\pi}{2 \sqrt{2}} \approx 1.11\$


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