I am using TPS26601 for my next design. There is a pin IMON which outputs current proportional to current flowing throwing the output pin. The gain ranges from 72 to 85 µA/A (Page 7).

In reference schematics (Page 27), they have used a resistor at the IMON pin to convert current to voltage so that it can be sampled by an ADC downstream.

However, I am planning to use an optocoupler. The current that would be drawn from OUT pin of the IC ranges from 0.1 A to 2 A (or even more in case of short-circuit). So, the current that would appear from IMON pin ranges from 8.5 µA to 170 µA.

I am not sure about optocouplers which can transfer signal if If ranges from 8.5 µA to 170 µA. For example, I was considering EL3H7(B)(TA)-VG and found out they don't show graph of Current Transfer Ratio in the range from 8.5 µA to 170 µA.

Is there any optocoupler which can be activated with the current in µA ranges?

  • \$\begingroup\$ Why are you using an opto? \$\endgroup\$
    – Andy aka
    Dec 11, 2018 at 13:20
  • \$\begingroup\$ For isolation, IC have it's own Device Reference Pin (RTN) and Imon's resistor is connected across Imon and RTN. I want to add isolation between MCU and TPS26601. \$\endgroup\$
    – abhiarora
    Dec 11, 2018 at 13:21
  • 2
    \$\begingroup\$ You may need to pass a larger signal and then process it first with an op-amp? Also that is a fairly large dynamic range for an opto. \$\endgroup\$
    – MadHatter
    Dec 11, 2018 at 13:28
  • 1
    \$\begingroup\$ Use a differential amplifier instead. \$\endgroup\$
    – Andy aka
    Dec 11, 2018 at 13:35
  • \$\begingroup\$ What is the point of the isolation? It makes no sense in the context of an e-fuse. \$\endgroup\$
    – Matt Young
    Dec 11, 2018 at 15:28

1 Answer 1


So, first of all, typically optocouplers are not meant to be linear measurement devices – as you can see by the fact that you need a graph to represent your current transfer ratio (instead of a single value).

Now, you ask whether there's an optocoupler that has a current transfer for a input current of 8.5 µA.

Probably there's no such device. Remember that this current is just used to power an LED in the optocoupler – and single-digit microamperes aren't really much! You need to put this into perspective with the thermally-induced noise in the receiving end of the optocoupler.

A simple solution would be to just amplify the current – even if it's just with a single BJT – and use that increased current. Sure, the nonlinearity of your transistor makes calibration of the system necessary, but it's already necessary because of the non-constant current transfer ratio!

Or, you just use an isolated ADC (e.g. AMC1035) that includes optical isolation in the package.

Other than that, you could simply use a galvanically isolated method of measuring current and get rid of the optocoupler; measurement transformers (for AC) or hall-effect based measurement comes to mind.

  • \$\begingroup\$ Do you think isolation is required at all? \$\endgroup\$
    – abhiarora
    Dec 11, 2018 at 13:53
  • \$\begingroup\$ how should I know that? It's your system! You don't mention any isolation requirements, so we can't make a statement on that. \$\endgroup\$ Dec 11, 2018 at 14:40

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