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 ?

Edit:

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 ?