investigate how output from photocell depends on distance from infrared point source

bigbadcityboy

I am required to design a labratory experiment to investigate how the output from a photocell depends on its distance from a point source of infra-red radiation.

wat procedure could i follow?
how would i measure the output of the photocell?
what other measuring instruments should i use?

berkeman

Welcome to the PF. Here's a recent thread with lots of info to get you started. The OP (original poster) didn't get all that he wanted out of the thread, but there's lots of good info in it. Post more questions here if the thread doesn't get you going.

http://www.physicsforums.com/showthread.php?t=116631

bigbadcityboy

to measure the current of the photocell can i just make a circuit with a photocell and an ammeter? this seems to be a pretty simple circuit to me and i think maybe it should be more complicated.

also, just out of curiosity how does an op-amp convert current to voltage? i cant find it anywhere

thanks

berkeman

Yes, as stated in that other thread, with a reasonably sensitive current setting on your DVM, you will be able to measure about a decade of photocurrent variation, starting very close to the source. But to get much of a plot of photocurrent over several decades, you will want to make a simple current to voltage converter circuit using a CMOS opamp. The linearity of the photodiode is also improved by placing a reverse bias across it of several volts, which you can do if you take the anode of the photodiode to V- instead of ground.

On your last question, I just googled current to voltage converter photodiode, and got lots of good hits. Here's one of the first ones:

http://www.ecircuitcenter.com/Circui...tov/opitov.htm

and as I just mentioned, consider taking the anode of the photodiode to V- instead of Ground (the first figure in that link shows it connected to Ground).

bigbadcityboy

whats a decade?

bigbadcityboy

im using a photocell, is that the same as a photodiode?

berkeman

Um, a decade is a change by a decimal order of magnitude. Like if you measure a photocurrent of 1uA, then a decade larger would be 10uA, and a decade smaller would be 100pA. Just like a decade change in frequency (from 1kHz to 10kHz) or in any other quantity. Another standard ratio change is an octave, which is a doubling or halving of something.

A photocell is a photodiode. Typicaly photocells have large surface areas to generate a larger photocurrent than a typical photodiode. The larger the surface area, the higher the photocurrent, but the slower the response time to changes in light intensity. Photodiodes that are optimized for fiber optic communication have *very* small photoreceptor areas.

bigbadcityboy

so did you mean a decade of distance or of current?

bigbadcityboy

oh current oops cant read

berkeman

Whichever. BTW, how would you expect the received photocurrent to vary with distance? What is your initial guess as to what you will observe in the experiment? You should always have an initial guess based on the equations and the setup.....

bigbadcityboy

to decrease a lot as it gets further away, if u see what i mean

bigbadcityboy

and increase as the source gets closer

bigbadcityboy

how big a resistor will i need for the opamp thing? 5 ohm??

Hootenanny

I would recommend something along the lines of a 5M ohm resistor, you want the biggest resistance possible, to prevent a a large current flowing, thus obtaining a more accurate measure of the potential difference.

~H

berkeman

Nope. What value of Rf did they start with in the SPICE simulation in the link I posted above? How would you tune the value of your Rs resistor based on the expected photocurrent?

BTW, what grade are you in? What is your background so far in electronics? How much help is your teacher giving you on this?

bigbadcityboy

im from england, doing physics A level im 17. i only know basic electronics, as i havent finished learning my electronics module.

bigbadcityboy

im in year 12 doing A level. im 17. iv only just started learning about electronics. i was off school ill the lesson we had about this so i've had no help.