EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: PinheadBE on February 29, 2024, 03:53:32 pm
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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 !
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It provides you a way to control it via the isolated side or the non-isolated side. It works just like any other transistor.
But the real reason to have it is to reduce the sensitivity. You connect a resistor between the base and emitter, you can increase the current it needs on the isolated side to conduct.
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Thanks!
In practical, in what kind of applications would we want to adapt the sensitivity ?
I guess this can be done dynamically also ?
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You could also add a small capacitor across base
base and collector and turn it into a Miller integrator.
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Thanks!
In practical, in what kind of applications would we want to adapt the sensitivity ?
I guess this can be done dynamically also ?
For example if it is providing feedback in an SMPS. Or if it's an isolated digital input, and you only want to give "1" at a specific current.
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Access to the base allows the phototransistor to be used as a much faster photodiode, or tailoring of the phototransistor's response. Considerable improvements in performance are possible as shown below.
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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.
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When a phototransistor is illuminated the photo-generated electrons and holes flood the device, those that contribute to base-emitter current lead via transistor action to a much larger current through the collector.
When the illumination suddenly stops the stored charge in the base will continue this current for a while as it dissipates. Adding a resistor between emitter and base provides a faster path for this to dissipate, so the collector current can switch off much faster. But its a compromise, the sensitivity will be less.
Photo diodes are inherently much faster (no stored charge), but have no current gain, so a much larger die is needed for the same sensitivity.
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Very often with 4N25 devices we see 470K as the base resistor. Look very carefully at the published C.T.R. specs. The devices marketed as 100% C.T.R. achieve that across only a small portion of the operating specs and often at an inappropriate C. E. voltage of around 10 volts not typically found around logic circuits. With a 1K collector resistor and 5 volt VCC supply, trying to pull the collector to down to .4 volts above the logic ground would make you believe that around 5 ma. of LED current would be fine. However, you will need more like 15 ma. to get the job done. I find in most opto circuits running in the 5vdc world the true C.T.R. of a '100%' device to be more like 30-50% and I design with that expectation. Everything takes on a different meaning with photoscr's, photodarlington's or photo-logic devices. Y.M.M.V.!!
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I have seen the base used to apply a positive feedback, to make a sharp transition.
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Thank you all :-+ :-+ :-+
It's very clear, and indeed, very useful.
I have one old salvaged CNY47 laying around, and I will give it some experimentation