Hi,
What's the purpose of the connection to the base of the photo-transistor in an opto-coupler (see pin 6 on the pin diagram of a DIP-6 opto) ?
And how do we use it ?
Thanks !
Hello there,
The most important reason for that base lead is so that you can reduce the turn off time of that internal transistor.
To turn a bipolar off fast you have to get the charge out of the base region. The faster you can do that, the faster the transistor turns off. To get maximum turn off time, you would use a resistor and negative voltage power source usually less than 5 volts. That however would require adding a negative power rail to most projects so usually just a resistor is used. For a common 4N35 opto coupler, even a 100k resistor from base to emitter can speed it up significantly, although some data sheets do not reflect that functionality correctly. This allows a much higher frequency signal to be passed.
The lower the base emitter resistor, the faster the turn off, but the downside is you lose some of the CTR of the device so you have to check that after adding a resistor. 100k does not seem to bother it too much, but it will mean you don't get the 100 percent CRT anymore it may go down to 50 percent. It's still a very workable solution though. If with no resistor you get 2ma out with 2ma in, then with 100k you may get 1ma out with 2ma in, but I would advise you to check that once you build the circuit.
Long ago I don't think they even mentioned this on the data sheets but it looks like they include that too now, but the speed up is better than what some data sheets indicate.