Electronic – Amplifying photodiode current with BJT transistor

amplifiercurrentphotodiodetransistors

I'm trying to develop a kind of light detector. To do that, I'm going to use a photodiode and a bjt transistor to amplify its current. I found a circuit in this tutorial:

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which explains how the phototransistors work. It is pretty simple, but I'm not able to understand two things:

  1. It is said that Rb is optional. But the photodiode needs bias to work in that way. How on Earth it will be biased? Maybe due to base-emitter junction, but how exactly does it work?
  2. What if the Rl resistor would be connected to collector? I tried such circuit with this photodiode and this transistor. The Rb and Rl was both 100K. I expected Rb voltage to be about 1,2V because of the 12uA current produced by the diode, but actually it was about 0.7V and the circuit was extremely unstable – it turns off and on for unexplainable reason.

I guess I'm missing something obvious, but just can't get, what exactly.

Best Answer

  1. A photodiode doesn’t have to be biased; it will try and force a photo current into whatever circuit it is connected to. In your emitter follower circuit if you removed Rb, there will be a path for photo current to flow through the base emitter junction, through the emitter resistor and back to the negative terminal on the supply. The base emitter junction behaves like a forward biased diode and drops around 0.7 volts in a typical circuit but, this can range from a couple of hundred milli volts to around 1 volt (depending on the circuit).
  2. You can put a resistor in the collector and short emitter to ground and you get a very high gain amplifier. Because base and emitter conduct at typically around 0.7 volts (as per (1) explanation), you won’t ever see 1.2 volts. Yes it can be unstable and be influenced by temperature and light changes (sometimes) and, that is why it tends not to get used like that in engineered circuits.
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