# Understanding the phototransistor

I am trying to learn how a phototransistor works, and I have made a schematic to show how I understand the photo-diode and NPN transistor to be linked in a phototransistor.

I think that when light intensity reaches a certain level, the photodiode should conduct current and the bulb should light up. Is this correct? I am also wondering whether there would need to be a connection between the photodiode and the emitter, as shown with a question mark in the image below.

simulate this circuit – Schematic created using CircuitLab

The image below is how I image the electron flow should be when the photodiode conducts:

For reference, this is a less detailed depiction of the design I would like to achieve with the above schematic:

• This is so big bunch of radical misconceptions that the full corrective explanation would be long. Are you sure this is not a bad joke?
– user136077
May 16, 2021 at 13:21
• In the first figure, both the ends of the photo diode are shorted. This means that any circulating current through that loop cannot contribute any current to the base of the transistor. Secondly, the transistor junctions may not be properly biased to the required voltages. For an NPN transistor, the emitter is usually connected to a lower potential than the collector.
– AJN
May 16, 2021 at 13:30
• There is no photodiode "connected" to a phototransistor, the base-emitter and base-collector junctions of the transistor form photodiodes. May 16, 2021 at 13:30
• Like I said, there is no photodiode involved, the transistor itself is photosensitive. May 16, 2021 at 14:55
• Also, @user287001, let's try to be a bit more friendly to newcomers who are interested in actual understanding? If they didn't have some kind of misconception or misunderstanding, they wouldn't be asking the question in the first place. May 16, 2021 at 15:14

This image presents normal useful voltage polarities and current directions for NPN transistor in a circuit:

I guess you know that base current can be increased by reducing the resistance R1 or by increasing the voltage of BAT1.

The collector current in a loop which contains a voltage source BAT2, a load and the BE-section of transistor Q1, can be controlled by changing the current in the base loop.

Normally the collector current can be tens or hundreds times bigger than the base current due the current gain in the transistor, but not higher than the voltage of BAT2 and the resistance of the load allow by Ohm's law.

Think that in phototransistor light (actually carrier generation in the semiconductor molecular structure caused by photons) replaces the base current, there's no need for the base current loop.

I do not recommend a beginner starts to keep a circuit which has photodiode and ordinary transistor as an exact equivalent of phototransistor. The latter is probably designed for certain performance which is difficult to achieve by connecting a photodiode and ordinary transistor and simply the light affects to the parts of a specially shaped transistor, there's no separate photodiode in a phototransistor.

In addition too well adopted not so accurate equivalent circuit can harm learning the real semiconductor physics of the operation of a component, if he one day starts semiconductor engineering studies.

I think this is the correct schematic and diagram of electron/conventional current flow according to the advice others have given. If it is not, I will delete this answer.

simulate this circuit – Schematic created using CircuitLab

• Photodiode should go to the collector, not the emitter. May 16, 2021 at 15:05
• @KD9PDP I have edited the schematic - is that correct now? May 16, 2021 at 15:08
• I can't really say this is wrong, but I imagine it's likely to be misleading if you're interested in the physics of the device beyond just using it as a switch. In particular, I think this misrepresents how nonlinear phototransistors are when compared to photodiodes. May 16, 2021 at 15:08
• If, however, you're just interested in operation as a light-sensitive switch, this is likely a good enough understanding. Just be aware that the actual physics are more complicated! (and an area of relatively specialized study that I never studied, so I can't explain exactly how. Might be worth looking into if you're interested, though!) May 16, 2021 at 15:16
• @Hearth OK sure! Thanks so much for your help!! May 16, 2021 at 15:21