Consider a two terminal phototransistor (collector and emitter) that is saturated and the light source is then removed.

What can one do to speed the transition from saturation to cutoff ?


This question for some reason is being interpreted as homework (I'm not in school) and this question is being interpreted as if I am new the site (lol). So let me add some additional information that I didn't think was immediately relevant to the question.

For a BJT transistor, to get the device to go transition from saturation to cutoff faster, two things would need to be done. Prevent the device from getting into saturation and apply a negative voltage at the base. Both these items are done with the base terminal.

I do not know the underlying physics of a phototransitor, but given its symbol (resembles a BJT), light hits the base, goes through some photo electric effect and we get minority carries in the base, and the operation is similar to an NPN.

With that, the phototransistor being similar to a BJT, how can I make the phototransistor (that does not have a base terminal) turn off faster than just removing the light source - since I can't add a diode between base and collector and I can't apply a negative voltage to the base ?

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    \$\begingroup\$ Your wording makes this sound like homework. What has your research revealed? \$\endgroup\$ – Transistor Dec 24 '17 at 8:19
  • \$\begingroup\$ Welcome to the site. Please quickly realise that this is not a free design house, homework-answering service or an on-line technical encyclopedia, copied out to you on demand. People will help you take the next step if your question shows that you've done as much as you possibly could on your own - which your post doesn't, I'm afraid. Please revise your question showing your work and findings so far, in considerable detail. Or delete the question if Internet searches give you your answer anyway. Again, a warm welcome to the site. \$\endgroup\$ – TonyM Dec 24 '17 at 14:04
  • \$\begingroup\$ Uh - not homework and I'm not new. Didn't see the question as homework but just a general question. I guess I'll take it as a compliment that the question was worded well that it appears to have been a copy paste from some homework problem. \$\endgroup\$ – efox29 Dec 24 '17 at 18:30

Reduce the load resistance.

For marketing reasons, datasheets typically specify a ridiculously low pull-up/-down resistor of 100 ohms or so. This alone would reduce the output voltage swing too much, but you can add more components to amplify the signal again, e.g. like this (the load is Q1's base/emitter):

phototransistor with speed-up circuit

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The phenomenon that happens to you may be due to the following phenomena:

  • The amplifier circuit that you use to detect the change of current due to the absence of light, has a very high gain, such that it requires that the phototransistor reaches a very small collector current to stop detecting the signal.

  • Ideally in the absence of light the phototransistor changes its collector current from a current Ic1 to a current Ic2 = 0, however, there will always be an offset current Ic2 a little greater than zero due to multiple phenomena such as electrical noise, static electricity from the environment , or even light frequencies that your eyes do not perceive but that the device does (infrared for example).

  • The amplifier circuit you use to detect the current change due to the absence of light has a transfer function which slows censused signal collector current of the phototransistor. Ideally, an amplifier only has A gain, but in real life this gain A is a transfer function A(f) that behaves in different ways depending on the frequency of the input signal.

The solution to this is that you use a comparator circuit, which you must calibrate to activate or deactivate with the level of light you want.

enter image description here

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