simulate this circuit – Schematic created using CircuitLab
I am using the emitter voltage to drive another transistor. But voltage is zero.
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7\$\begingroup\$ Did you really connect 5V directly to the input diode? If so, you've ruined the part. \$\endgroup\$– Alfred CentauriCommented Oct 16, 2013 at 11:32
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\$\begingroup\$ This cannot be answered unless we know the current flowing through D1. Do you know the current flowing through D1? \$\endgroup\$– Connor WolfCommented Oct 16, 2013 at 11:36
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\$\begingroup\$ I am using new one, this time it is 4n35, and current through d1 is approx. 8mA \$\endgroup\$– bharath thinkCommented Oct 16, 2013 at 11:51
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\$\begingroup\$ Not the problem here, but you should connect a resistor from base to emitter (maybe 56K) to avoid humidity causing a partial turn-on of your opto. Refer to my answer in why mosfet is getting very hot? for more details. \$\endgroup\$– TutCommented Oct 16, 2013 at 14:54
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\$\begingroup\$ Have you tried disconnecting Vo from the transistor it is driving? \$\endgroup\$– TutCommented Oct 16, 2013 at 16:28
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2 Answers
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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.
answered Oct 16, 2013 at 14:46
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If you implemented this very schematic, then your opto-isolator need to be replaced, because the LED inside is already burned.
And before switching ON the schematic next time, put a current limiting resistor in series with the input.
