I have an existing circuit design needs a simple indicator LED which is then going ON when the output of the ULN2803 has output respect to the +VS.

Initially I ran a test just hooking up a LED with a current limiting resistor at the output of the ULN2803. It seems to be OK. It turns ON and OFF as per the input from the optocoupler. But, I don't have a test load with me so, I am not sure what concerns could happen if I connect one. The load is essentially won't be more than a few hundreds mA.

Here is the schematic piece;

enter image description here

My main concern is the load's self. Maybe it is not in my case but I have to think about it anyway. I have two concerns come to my mind (you can add more if you see more);

Concern #1. How can I prevent a current leakage when the load is connected? If this happens then the LED will lid at least a bit so it would be annoying. If so, what would be the solution for that? Perhaps, increasing the resistor value according to the load's effect? or placing the LED somewhere else, maybe at the input of the ULN2803?

Concern #2. Should I worry about a possible impact paralleling two resistors (LED circuit and load)? If so, again, I should maybe move the LED somewhere else, but, it will be then the same effect goes at the new place like the input of the ULN2803, won't it?

+VS = +24V LED: 0402 smd package, 20mA, 2V Forward voltage

Any input greatly appreciated.

  • \$\begingroup\$ What is your load and how it is connected to the output? \$\endgroup\$
    – Justme
    Commented Aug 31, 2022 at 10:58
  • \$\begingroup\$ I don't have the specifications of the load yet as this is some generic design ideally can drive low current loads. \$\endgroup\$
    – Sener
    Commented Aug 31, 2022 at 11:00
  • 1
    \$\begingroup\$ How would current leakage result in a voltage higher than 2V? \$\endgroup\$
    – Lundin
    Commented Aug 31, 2022 at 11:10
  • 1
    \$\begingroup\$ Have you considered adding the indicator LED in series with the opto-LED? With a 5 V supply you would have enough headroom for a red, yellow or green. That tells you that the output is enabled but doesn't confirm that the +Vs supply is on. A separate LED could be used to confirm that, if required. \$\endgroup\$
    – Transistor
    Commented Aug 31, 2022 at 12:38
  • 1
    \$\begingroup\$ Question is clear. Closevoter may wish to explain their concern. \$\endgroup\$
    – Russell McMahon
    Commented Sep 8, 2022 at 11:51

1 Answer 1


The ULN datasheet is clear. There is no issue with the LED drawing < 20 mA when the driver can sink 350 mA at 1.6V max. Since the voltage drop on the current limiting resistor only reduces by 1.6V from something > 20V , the LED current change is insignificant. However if the 2V LED is dissipating 65 mW, the 20V drop on the resistor will be 650 mW.

A smarter solution is to use ultrabright LEDs with > 10k mcd intensity@ 30 deg. and run them at 2 or 3 mA with a 10x bigger current limiting R.

Also, keep in mind, do not exceed the recommended safe maximum case and junction temperature with multiple drivers dissipating heat (sum {Vol * Io = P * Thermal resistance 'C/W }


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