OT: PhotoCoupler Question

I am updating some controls on an older machine and have found that they are using Sharp PC829 photocouplers for stepper card control.

I noticed that they are using both 5VDC and 24VDC to "actuate" the photocoupler. The data sheet for the photocoupler shows a maximum input voltage of 3V.

I see that the collector/emitter can handle up to 35V, but how are they getting away with the higher voltage on the input side? I see no resistors in the drawings.

I confess my experience with photocouplers is minimal, and any insight would be welcome.

The input side of a photocoupler is basically a LED. Think forward biased diode. You MUST limit the current into the diode since it has no means to do it by itself. Possibly you are overlooking a resistor network on the pcb(single inline pkg, looks like an IC)which is doing the current limiting. The critical thing for the input is the current which will turn on the phototransistor fully without baking the LED. From the specs it looks like you want to limit the input current to no more than 20 mA. If you have an operating unit you could measure this. Hope this helps.

You are correct, thank you. I found a spare part here in house which the OEM made themselves which incorporated a resistor network into it.

Good call!