0
\$\begingroup\$

I'm looking for a circuit to determine the RMS voltage of the AC mains supply regardless of its frequency.

The monitoring circuit needs to be isolated from the AC mains, and I can't use a transformer. I've seen some circuits using optocouplers, but they seem to be frequency-dependent.

Update

Below is an example that ive found that would seemingly work well for my application, but changing the mains frequency changes the frequency of the pulses. How to measure AC voltage using non-linear opto-couplers?

\$\endgroup\$
10
  • 3
    \$\begingroup\$ so AC mains frequency puts this in the range of 45 to maybe 75 Hertz. I don't think that's going to be hard for any optoisolated measurement IC. Can you refer us to one of your optocoupler-based circuits that are so frequency-dependent that they don't work at these low frequencies? \$\endgroup\$ Feb 2, 2023 at 21:49
  • 1
    \$\begingroup\$ ”seem to be frequency dependent” Please show! \$\endgroup\$
    – winny
    Feb 2, 2023 at 21:51
  • 1
    \$\begingroup\$ also, can you explain why you can't use a transformer? I think this sounds a bit like you're thinking of big bulky transformers, while I think of things like measurement clamps, which are also transformers. So, I'm not sure I understand the restriction well enough to base recommendations of it. \$\endgroup\$ Feb 2, 2023 at 21:51
  • 4
    \$\begingroup\$ Use an incandescent lamp and photovoltaic cell -- good isolation and frequency independence to RF, though with limited range and requiring calibration, since light output is highly nonlinear. Make a paper scale for a d'Arsonval mA (or µA) meter for the photocell output. \$\endgroup\$ Feb 2, 2023 at 23:32
  • 1
    \$\begingroup\$ If you truly need RMS, and not just an assumed value, note that RMS can be done with random (specifically, uncorrelated and evenly distributed) sampling of any waveform. I'm not sure this helps much with respect to isolation, but it is frequency independent. \$\endgroup\$ Feb 6, 2023 at 22:34

0

Your Answer

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

Browse other questions tagged or ask your own question.