I try to find max (useful) input voltage on the PC817 general purpose photocoupler.
I can only find "reverse voltage" and "forward voltage" on the input stage, and the forward voltage is max 1.4V so I guess it could not be it, surely it can withstand higher voltages than that? Or am I mistaken? It is max voltage on pin #1 I'm asking for. Or can it withstand any voltage as long is under 5KV and under max forward current (50mA)?
Datasheet is here.
As with any LED you choose a current and calculate the series resistance from R=(Vcc-Vf)/If. Usually the input current is determined from the desired output current and transmission ratio, but this optocoupler seems to have a very weak current dependency.
Use a resistor to limit the current into the input LED, just like you would for any LED. 10 mA is more than sufficient (330 Ohms for a 5V output, 220 for 3.3V).
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In a series circuit the voltage gets divided between the series components in proportion their resistance.
LEDs are non-linear and their resistance goes dramatically down as the voltage goes up. But if you supply the right current the voltage magically "falls into place".
We consider LEDs to be "current operated devices".
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So, in this case voltage is not that important for the optocoupler as long as the input current is well under the max values? ,and this would be the "forward current" in the datasheet i presume? Thank you everybody for helping me understand!
And the opto LED current doesn't have to be much more than the opto transistor current has to switch. This spec for this is "Current Transfer Ratio". If the transistor only has to switch 1mA, then an opto LED current of 1-2mA should be enough. Anything more than needed is wasted power and possibly a shorter LED life.
Leo..
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