Given that I'm using the following circuit (adapted from this forum post):


simulate this circuit – Schematic created using CircuitLab

Which takes a 24V (Nominal) input and converts it over to 3.3V for an input on the Raspberry Pi Compute module. I'd like to have an LED present on the board to signify that the input is on.

The issue is that I'm using the LM2676SX-3.3 to power the 3.3V, which provides a maximum of 3A, and I need to have 128 of these LEDs in the system. At the above values, each LED will consume 8.7mA of power which is 1.3A.

I'd like to power the LED from its own 3.3V source (probably another LM2676SX-3.3), but I'm not sure how to modify the above schematic to use a separate supply?

The other option is to power the LED on the 24V side, but it needs to support a voltage range from 12V to 32V as the input. I'm not concerned with a particular brightness as long as the LED is viewable.

So how can I modify this to use a separate source for D3 (considering I need to duplicate this 128 times, so low part count) or how can I run D3 off of the 24V side with a wide input range and not turn the enclosure into a small heater?


You can put an LED in series with the opto-coupler LED, no other components required. It won't increase the power consumption at all.

It will increase the minimum voltage slightly but would still be within your 12v minimum voltage requirement.

From your resistor values it looks like the current through the opto-coupler LED is compatible with the required current through the illuminating LED. What value is the zener diode?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ The zener diode specifically is this one which is a 6.2V. So what you are saying is place it in series, either on the anode side or the cathode side? \$\endgroup\$ – Ron Beyer Mar 20 '17 at 19:33
  • \$\begingroup\$ I've added a schematic to my answer. The LED can be connected as shown or on the Q1 side of the optocoupler. This will give a current of about 4mA into the indicator LED - that should be quite adequate for a modern LED. \$\endgroup\$ – Kevin White Mar 20 '17 at 19:44
  • \$\begingroup\$ Thank you, that is incredibly helpful and actually reduces my part count. \$\endgroup\$ – Ron Beyer Mar 20 '17 at 19:49

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