# Hm. What is this resistor for?

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#### Nrets

##### New Member
So I wanted to procrastinate studying for an exam last night and whipped out some components to make a simple light sensor circuit that would output analog between 0-5v. I had an IR phototransistor laying around that, so I used that instead of a photoresistor. At first I did not have R2 in the circuit, and although I was able to measure a signal, the data was very noisy when no light was on, or if the light level was low. I tried changing some resistors, but it didn't help anything. I then realized I had nothing connected to ground, but couldn't really see where anything should be connected to ground. I finally ended up putting a 100kohm resistor on the emitter of the transistor, and that solved all my problems. I think I understand vaguely how this works, but I don't fully understand the purpose of R2. Can anyone explain?

I drew the schematic in power point, please forgive me.

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#### Reikitronic

##### New Member
Stability

Hello Guys, i think it is important to mention what this r2 resistor contributes with the stability of this circuits and dont allow noises.

Have a nice day.

#### crutschow

##### Well-Known Member
Without the resistor, there's nowhere for the transistor emitter current to go. The transistor couldn't work and was not doing anything.

#### Nrets

##### New Member
THe BJT is behaving as a current source. R2 converts the output current into an output voltage. To get a proper reading without R2, you would need to measure the current (rather than voltage) between emitter and ground, then you would probably get a better reading.

Common collector - Wikipedia, the free encyclopedia
Common emitter - Wikipedia, the free encyclopedia
I see. Actually reading those links almost confused me with the usage of impedence and voltage buffers, but I think I get the basic jist. So all the resistor is doing is creating a voltage drop across the resistor so that a voltage can be measured on V_out? So if I used a smaller resistor, such as 500 ohms, the voltage drop would be small, and the maximum reading on V_out would be less than 5v?