I know the normal way of optically isolating an analog signal is to use an analog optocoupler with two detectors, and using an op-amp to servo the emitter voltage until the detector output matches the input, like this:


simulate this circuit – Schematic created using CircuitLab

However, there are some issues with this, particularly the need to have some kind of supply voltage on the input side to power the op-amp and provide a voltage reference for the detector to use.

Would it be possible to build a circuit that didn't need an input-side supply voltage?

One idea I had is to use two separate optocouplers, driving one directly from the input and driving the second from the output of an op-amp, servoing until the second optocoupler is driven identically to the first, like this:


simulate this circuit

But I haven't ever seen a schematic like this in any appnotes or documentation. Would this actually work?

Also, how would I find such a "matched" pair of optocouplers that could be used in this way? Potentially, the two couplers in a single-package dual channel optocoupler would be reasonably similar to each other, but I've yet to see a datasheet that actually gives any kind of specification for that beyond the general tolerances for the individual elements.

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    \$\begingroup\$ One potential problem with your proposal is that input impedance is relatively low and nonlinear, because it's an LED. Depending on what's driving it, that may or may not matter. \$\endgroup\$ Jan 28, 2020 at 16:21
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    \$\begingroup\$ It's a cool idea (kudos) but it's really quite easy to generate isolated power supplies these days and get a more accurate coupling either digitally or using conventional analogue isolation methods. \$\endgroup\$
    – Andy aka
    Jan 28, 2020 at 16:25
  • \$\begingroup\$ @pericynthion That is something I'd considered, but in this case, it's being driven by an buffer amplifier at the other end of a cable, so the impedance isn't too much of an issue. \$\endgroup\$ Jan 28, 2020 at 16:29
  • \$\begingroup\$ @Andyaka if you want to elaborate on how to do that, I'd be open to that as an answer. In the instant project I'm working on I don't actually need the full 5kV isolation guarantees of an optocoupler, although I'd also be interested to see how you'd do it when you do need isolation guarantees that strong. As far as I know the only way to generate a power supply with that kind of isolation is to AC couple the power over a transformer, and that's a lot of extra expense. \$\endgroup\$ Jan 28, 2020 at 16:50
  • \$\begingroup\$ Look up Traco as a supplier of isolated dc to dc converters. If you require many channels, the price is shared per channel sometimes. \$\endgroup\$
    – Andy aka
    Jan 28, 2020 at 17:12

1 Answer 1


Linearity will be not very good in the second circuit.

In the first circuit you get quite good proportionality of photo currents at both detectors (even if they don't match very well).
This is not the case in the second circuit.

Also note that the first circuit is a linear (input) voltage to (photo) current converter. It works correctly also for input voltages that are smaller than the forward voltage of the LED.

Not so the second circuit. It doesn't work below forward voltage of the LEDs and if even if input voltage is above forward voltage, if the forward voltages of both LEDs differ only a little bit the input to output linearity will suffer a lot.


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