I have 2 resistor configurations for an optocoupler.

The first one (on the left) is the one I've used in some previous circuits.

The second one (on the right) is something that I saw on the internet while doing my research.

If I'm not mistaken, the second one is based on a voltage divider. Though the first one works well enough for my circuit, I wanted to know which one is correct or incorrect.

If correct, which one is:

  1. More efficient
  2. More reliable
  3. Adds longevity to the optocoupler's internal LED.
  4. Finally, which one I should prefer (apart from the additional minimal cost on the second one for extra resistance.)

Schematic showing different resistor configurations on an optocoupler:

Schematic showing different resistor configuration on an optocoupler

  • \$\begingroup\$ I don't see any advantage that the right circuit would offer, it seems like it would just waste a bit of current and that's it. \$\endgroup\$
    – Drew
    May 17, 2021 at 3:11
  • \$\begingroup\$ Do you have a link to the site where you saw the version on the right? \$\endgroup\$
    – HandyHowie
    May 17, 2021 at 7:12

2 Answers 2


You are comparing two totally different versions it terms of input impedance. The one has 4.7k series resistor, while the other has 10k.

The parallel resistor provides faster turn off in case the driving circuit is open collector /or/ open drain /or/ high side P type transistor. Due to LED self capacitance (and the anti-parallel diode that is not present in your schematics), it continues to glow for a while even if was already disconnected. The resistor helps to discharge and thus the LED turns off faster.

You will find the version with parallel resistor for high speed pulse input in motor drivers (although different values than you provided, typically is 1.1k). You won't find those for low frequency PLC input signals. Don't forget to put an extra anti-parallel diode for reverse polarity protection.


simulate this circuit – Schematic created using CircuitLab

  1. What is your criteria for efficiency? Ohms law will tell you the loss of both solutions.

  2. How are you determining reliability?

  3. The first one is passing around 5mA through the led, the second around half that.

The second solution doesn’t work as a voltage divider as the led limits the voltage across it to around 1.2V. Only about 1/20th of the current flows through RP3, so it is superfluous unless there is some other design intent.

For a general input, i’d be adding a diode across the led to protect against reverse or AC voltages and ensuring the resistor is suitably rated.

  • \$\begingroup\$ Efficiency, in terms of power loss, assuming I have 50 such Opto+Resistor pairs on one PCB. Reliability, means that would filter out any unwanted spikes/current to the LED that makes it blow out. So you mean adding a diode similar to 1N4001 would be add as an advantage? \$\endgroup\$ May 17, 2021 at 3:58
  • 1
    \$\begingroup\$ For a given input voltage and led current, the loss is fixed unless you go to a more complex solution. The average industrial PLC uses a similar circuit. Your definition of reliability is EMC. Optos give you galvanic isolation but not an EMC gold pass. Define what EMC threat you want to tolerate and design the solution accordingly. \$\endgroup\$
    – Kartman
    May 17, 2021 at 5:00

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