# Homework Help: Electrical Circuity - Phototransistors

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1. Jan 1, 2018

### connormmoore

• Please post homework questions in the homework section using the template.
The following circuit was created. I would be very grateful if you could check that the what I have said about it is correct.

"A phototransistor was connected in a series circuit as shown below. It is used like a resister in this case with a battery and multimeter measuring current. When the light increases, resistance decreases. As the voltage remains constant, the current must then increase as the resistance decreases. We know this from the equation V = IR."

2. Jan 1, 2018

### Staff: Mentor

Yes. I looks correct to me. Reduce the resistance and current will increase.

3. Jan 1, 2018

### Staff: Mentor

I think I would prefer the $I = \frac V R$ form of the equation (even if they are equivalent).

4. Jan 1, 2018

### Baluncore

Connormmorre, welcome to PF.

The photocurrent that flows will be proportional to the number of photons of sufficient energy that reach the light sensitive base, multiplied by the current gain of the transistor.

I would expect the transistor to source or sink photocurrent, independent of the circuit voltage. The voltage across the transistor will be the battery voltage and will not change significantly with changes to the illumination.

Resistance and ohms law will not come into play, except for the internal resistance of the current meter.

5. Jan 2, 2018

### connormmoore

I have just created the circuit and the current measured with the ammeter varied with the lux incident on the phototransistor.

I would be very grateful if you could explain what's happening here and why the phototransistor does not act like a variable resistor.

Many thanks,

Connor

6. Jan 2, 2018

### Baluncore

You have a fixed supply voltage, the meter has a very low resistance, so almost all the supply voltage will appear across the collector of the transistor. As the illumination and photocurrent changes there can be little change in voltages.

Most introductions to electronics study voltage drops across resistors or potentiometers. But photons that release electrons generate a pure source of current, that will not appear as a proportional voltage without a resistor. Ohms law requires there to be resistance. The collector voltage of a BJT is quite independent of the collector current flowing. Any change in BJT collector voltage is due to the external collector load resistance.

There are many possible modes of operation of a photodiode or phototransistor.
https://en.wikipedia.org/wiki/Photodiode#Principle_of_operation

7. Jan 2, 2018

### connormmoore

So would the photo transistor act like a variable resistor in the following circuit:
d

8. Jan 2, 2018

### Baluncore

No. A BJT collector is always going to be a simple current source. The collector current is a function of illumination. The collector voltage is a function of the external collector resistor and supply voltage. As current rises VCE falls due to the external load resistance. If the collector was a variable resistor it would have negative resistance.

9. Jan 2, 2018

### Staff: Mentor

To rephrase what @Baluncore wrote: it is quite commendable that you are trying to understand the circuit in terms of what you already know, but it is one of these cases where you have to accept there is more to it

10. Jan 2, 2018

### connormmoore

Ah, but in this instance, for this experiment (bearing in my mind this is as secondary school standard), can it be used as a variable resister?

11. Jan 2, 2018

### connormmoore

Could the same idea be applied if I were to swap the photo transistor for an LDR?

12. Jan 2, 2018

### connormmoore

If I were to swap the photo transistor for an LDR, could I then say that the current will increase as the light incident on the LDR increases in a series circuit?

13. Jan 2, 2018

### connormmoore

If I were to swap the photo transistor for an LDR, could I then say that the current will increase as the light incident on the LDR increases in a series circuit?

14. Jan 2, 2018

### Baluncore

You need to specify exactly what you mean by "the same idea".

An LDR is a light-variable resistor, ohms law applies. Increasing the supply voltage will increase the meter current.
A phototransistor is quite different, it is a current source, significantly independent of the supply voltage.

You might imagine the phototransistor collector to be an “intelligent self-adjusting variable negative resistor”, but it is much more convenient to avoid that quite misleading concept, and to model a collector as a simple "floating current source".

Edit: Snap, we crossed posts.

15. Jan 2, 2018

### Tom.G

To rephrase what @Baluncore was stating:

Recall that current is a flow of electrons
The Phototransistor, when dark is mostly an insulator, very few free electrons to move around and carry a current
When Light hits the Phototransistor, it knocks some electrons loose, allowing them to move
This is the Base current of the Phototransistor
There is a bias voltage, applied by the battery, across the Phototransistor Collector to Emitter
Being a Transistor, this Base current is of course amplified by the transistor β (Beta) and appears as the Collector current

Note that the Collector Current has little to no dependence on the Collector Voltage (battery voltage), it is dependent on the Base current, which is depentent on how many electrons are knocked loose by the light.

The current being independent of the applied voltage is pretty much the definition of a Current Source.

So, yes, as a first-order approximation the Phototransistor behaves in a somewhat similiar manner as a variable resistor. But as you can see from the above, "There is a bit more to it!"

Happy New Year,
Tom