0
\$\begingroup\$

I found this https://www.edn.com/isolated-circuit-monitors-ac-line/ circuit in my quest for a low cost isolated AC line voltage monitor and constructed it. enter image description here

On testing however, I am not getting the correct voltage as expected from the formula provided. Ve is supposed to be 48V but my calculation by rearranging the second formula to get Ve, is giving around 133V.

Is the formula wrong or have I made a mistake somewhere?

Additional information: I changed the values of R1 and R2 to 100k each as I found the 10k 0.5W resistors were getting quite hot and the 200k obtained should provide enough current to turn on the optocoupler. I used EL817 as I had a few and the diode is 1N4007. AC RMS voltage is 230V for which I got a low pulse width (Ttotal in the formula) of 0.0073s (it should be about 0.0091s if Ve is 48V). I used an STM8S microcontroller's input capture for measurement. Unfortunately I do not have a DSO to verify the pulse width. I did verify that the pulse width was changing using a variac and an old analog scope.

Also are there design errors in the circuit which should be corrected?

\$\endgroup\$
3
  • \$\begingroup\$ I did use a 47V zener diode. The 1N4007 was used instead of 1N4004 in the diagram. \$\endgroup\$
    – Subin Roy
    Commented Dec 17, 2020 at 15:37
  • \$\begingroup\$ Why are you commenting on your own question? There's an edit link below the question if you need to add in additional information. Can you add the schematic into your question so we don't have to follow a link to understand it and so the question still makes sense if the link dies? Thanks. \$\endgroup\$
    – Transistor
    Commented Apr 6, 2021 at 21:12
  • 1
    \$\begingroup\$ I recommend you calibrate your setup. Reality trumps theory. \$\endgroup\$
    – Abel
    Commented Apr 7, 2022 at 5:41

3 Answers 3

1
\$\begingroup\$

Use a very small power transformer.

Nothing else will give you a signal that is directly proportional to the input amplitude (and frequency!), fully-isolated, with complete safety certifications, for such a low cost and zero circuit design time.

NOTE: The EDN circuit does not produce an output signal amplitude that is related to the input AC amplitude. It produces a constant amplitude square wave with a duty cycle that varies with the input amplitude.

\$\endgroup\$
0
\$\begingroup\$

There are known problems with the EDN circuit. See https://sound-au.com/appnotes/an005.htm , read near the end for Rod's experiences. There are plenty of working circuits available on the ESP page.

\$\endgroup\$
2
  • \$\begingroup\$ Rod's experiences were with a different circuit. \$\endgroup\$
    – AnalogKid
    Commented Feb 26, 2022 at 16:10
  • 3
    \$\begingroup\$ None of the circuits on that page do what the TS wants. All of the circuits are zero-crossing detectors, which do not convey any information about the amplitude of the input. The output of the EDN circuit is a waveform that is proportional to the amplitude of the input AC. The implementation might be too simplified for precision use, but the concept is sound. \$\endgroup\$
    – AnalogKid
    Commented Feb 26, 2022 at 16:11
0
\$\begingroup\$

The best solution I have found is to use a transformer. Once the primary is connected all of the voltages are low and isolated. Transformers have a fixed ratio between the primary and secondary, this will enter into your calculations. You can power a voltmeter or add other electronics and convert it to what you want. Here is a link but you will need to replace the front end of the AC circuit with the above. https://simple-circuit.com/measure-ac-voltage-arduino-ac-voltmeter/

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.