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I am using the emitter voltage to drive another transistor. But voltage is zero.
Electronic – what is the voltage at the emitter of optocoupler 4N25
opto-isolator
Related Solutions
Considering that particular type of optocoupler is effectively just a light controlled switch with no need of it's own supply or ground, it's conceptually like a relay, so you can place it anywhere in the circuit and it should still work.
However, if the opto has any driver electronics (latch, logic gate, fet driver etc.), something that requires it's own supply, then things would be different. But a phototransistor by itself (which it looks like here) can go anywhere, the light provides the base drive.
where does the base current go in this closed circuit?
Photoelectric current current flows from the Collector to the Base. If the Base is left open then it also flows into the Emitter, so Collector and Emitter currents are identical. The equivalent circuit looks like this:-
simulate this circuit – Schematic created using CircuitLab
Best Answer
4N25 and quite a few other opto devices have what is known as "current transfer ratio". For the 4N25 this is specified as being greater than 20%.
This means if you stuff 20mA into the diode, you should expect to see at least 4mA through the opto-transistor and your emitter resistor is 1k so you should see something like 4V or possibly a little greater.
If you are putting 10mA into it don't expect much above 2V on the emitter.
This is the table from the data sheet - note that there testing conditions are with collector-emitter voltage at 10V.
If you are not registering any voltage at the emitter, the diode could be broken - try measuring its volt drop after the resistor in series with it. Alternatively you may have connected it up incorrectly. Also check your 1k emitter resistor isn't something like 1 ohm and check your collector is at 5v.