2
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

I have a low current current circuit and I would like to be able to measure the voltage on one point of the circuit but have electrical isolation between the low current circuit and the circuit for measuring the voltage. Needs to be impossible for higher voltage to make it to the low current circuit. Any ideas?

\$\endgroup\$
7
  • \$\begingroup\$ How "isolated" do you want this to be? Is it life-critical (someone dies if the voltage goes across), equipment-critical (your laptop or some other expensive stuff breaks if the voltage goes across), or "it's kind of annoying" critical (you don't want high voltage messing up your measurement, and failure doesn't result in any sort of monetary loss) that the voltage measuring circuit doesn't touch the low current circuit? \$\endgroup\$
    – hatsunearu
    May 8, 2018 at 13:26
  • \$\begingroup\$ Or rather, rephrasing: what is the worst that could happen if the higher voltage from the Arduino makes its way to the circuit under test? Does someone die? Does your invaluable widget break? If neither, how much risk can you tolerate? I ask because if it is non-negotiable that there be perfect/guaranteed galvanic isolation, you're looking at much higher costs, and may be even impossible. \$\endgroup\$
    – hatsunearu
    May 8, 2018 at 13:29
  • \$\begingroup\$ Potentially life critical but not really (erring on the paranoid safety side) \$\endgroup\$
    – anna
    May 8, 2018 at 13:30
  • \$\begingroup\$ You're gonna need a power supply on the other side (i.e. the side that is galvanically connected to the DUT) one way or another; is this acceptable? (perhaps using a coin cell or 9V battery) \$\endgroup\$
    – hatsunearu
    May 8, 2018 at 13:31
  • 1
    \$\begingroup\$ If your worry is too high current injected by the voltage place three 5 megaohm resistors in series to the measurement circuit. That way you are pretty safe that you will only ever inject maximum fault voltage divided by 5 megaohm. \$\endgroup\$
    – Arsenal
    May 8, 2018 at 13:37

1 Answer 1

Reset to default
1
\$\begingroup\$

Based on your answer, you're looking at a chip solution like this: ISO124. It's crazy expensive, but you can find other cheaper options with lower isolation ratings.

Either way, you definitely need a way to power the amplifier from "the other side", as in the side where the device under test is. If it's totally unworkable to have a power source on the DUT side, I don't think it's logically possible.

WARNING: I am not trained in creating life-critical/cannot fail type equipment (I don't even know the official name for this!) and my advice comes with NO WARRANTY.

\$\endgroup\$
2
  • \$\begingroup\$ thanks, don't worry this is paranoid style safe. \$\endgroup\$
    – anna
    May 8, 2018 at 13:38
  • \$\begingroup\$ if you're ok with voltages being present, you might as well just use a regular old instrumentation amplifier which presents a high enough impedance to the source that you'll get nice measurements. \$\endgroup\$
    – hatsunearu
    May 8, 2018 at 13:43

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.