
#1
Apr2806, 04:42 PM

P: 19

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



#2
Apr2806, 05:26 PM

Mentor
P: 39,683

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 



#3
May106, 11:15 AM

P: 19

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 opamp convert current to voltage? i cant find it anywhere thanks 



#4
May106, 11:25 AM

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P: 39,683

investigate how output from photocell depends on distance from infrared point sourceOn 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). 



#5
May106, 11:29 AM

P: 19

whats a decade?




#6
May106, 11:30 AM

P: 19

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




#7
May106, 11:35 AM

Mentor
P: 39,683

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. 



#8
May106, 11:53 AM

P: 19

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




#9
May106, 11:55 AM

P: 19

oh current oops cant read




#10
May106, 11:55 AM

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P: 39,683





#11
May106, 12:00 PM

P: 19

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




#12
May106, 12:01 PM

P: 19

and increase as the source gets closer




#13
May106, 12:06 PM

P: 19

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




#14
May106, 12:10 PM

Emeritus
Sci Advisor
PF Gold
P: 9,789

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 



#15
May106, 12:12 PM

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P: 39,683

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



#16
May106, 12:18 PM

P: 19

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




#17
May106, 12:26 PM

P: 19

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.




#18
May106, 12:34 PM

P: 19

does the M in 5M ohm mean 100 or 1000?



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