How does photocell output change with distance from infrared point source?

In summary, the photocell should be close to the IR emmiter to generate the most photocurrent. To measure the photocurrent, you will need a current to voltage converter circuit and a DVM. Other measuring instruments may be used to determine the output of the photocell.
  • #1
bigbadcityboy
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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?
 
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  • #2
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.

https://www.physicsforums.com/showthread.php?t=116631
 
  • #3
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 can't find it anywhere

thanks
 
  • #4
bigbadcityboy said:
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 can't find it anywhere
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/Circuits/opitov/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
whats a decade?
 
  • #6
im using a photocell, is that the same as a photodiode?
 
  • #7
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
so did you mean a decade of distance or of current?
 
  • #9
oh current oops can't read
 
  • #10
bigbadcityboy said:
so did you mean a decade of distance or of current?
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...
 
  • #11
to decrease a lot as it gets further away, if u see what i mean
 
  • #12
and increase as the source gets closer
 
  • #13
how big a resistor will i need for the opamp thing? 5 ohm??
 
  • #14
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
bigbadcityboy said:
how big a resistor will i need for the opamp thing? 5 ohm??
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?
 
  • #16
im from england, doing physics A level I am 17. i only know basic electronics, as i haven't finished learning my electronics module.
 
  • #17
im in year 12 doing A level. I am 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
does the M in 5M ohm mean 100 or 1000?
 
  • #19
M means mega 106
 
  • #20
well i know bugger all about electrionics. but the procedure is pretty simple, but iv had to go out and find the exact IR emmiter i want to use, because they all vary on things like the power needed to work, and also the range they will work over. its all about doors, so the emmiter will need a range of c.2-3 metres, and a lot of them only have about 25cm.
 
  • #21
help

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?[/QUOTE]

please send me the details as soon as possible please please please
 
  • #22
  • #23
Will you please help I am confused :cry:
Ive been told by my physics tutor to connect the photocell to a battery, and take a reading of the current.
Would this work?

THANX
 
  • #24
Please Will Somebody Help Me!
 
  • #25
investigation on how a photocell depends on distance using infra red radiation
 
  • #26
Hey people I am doing this same plan. I've managed to do most of it by myself, but i was speaking to my teacher today and she told me that i would need to calculate the energy of each photon using E=hf obviously. Alos i know that the intensity of the light depends on how much photons are realeased and not the energy of the photons. But now my problem is how would i link the two ideas together. Also the intensity of the light will not be changing (because I am not changing the power and the area of the bulb is still the same), but i think that the futher away i move my bulb the less current i will get on my ammeter. Is this changing the intensity in anyway? and ifso can someone please explain how it is. Thanks :smile:
 
  • #27
O and does anyone know a site where i can get a good infrared light source from. Preferably an IR led, and not just some lamp thank you. Sorry for double posting.

Erm how do you calculate the energy of a single photon when u are using an led? would i just use E=hf (after i have found the intensity of the light).
 
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  • #28
Physicsameture said:
Hey people I am doing this same plan. I've managed to do most of it by myself, but i was speaking to my teacher today and she told me that i would need to calculate the energy of each photon using E=hf obviously. Alos i know that the intensity of the light depends on how much photons are realeased and not the energy of the photons. But now my problem is how would i link the two ideas together. Also the intensity of the light will not be changing (because I am not changing the power and the area of the bulb is still the same), but i think that the futher away i move my bulb the less current i will get on my ammeter. Is this changing the intensity in anyway? and ifso can someone please explain how it is. Thanks :smile:

Your almost there, however, intensity is also dependant upon the distance from the source and is defined mathematically (for a point source, radiating equally in all directions) as;

[tex]|I| = \frac{P}{4\pi r^2}[/tex]

Note; this is in an undamped medium. I.e. perfect vacumm

~H
 
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  • #29
Cool thanks, so does that mean that 'r' would be changing? Because I am using a small led so I am think i should change the distance of the light source in groups of 5cm. So i was wondering how this would affect r (radius) in the intensity formula.
 
  • #30
Hey kids,
I am new and would like to thank you for all the help you have given me with this! and would like to say i simpathise with you with the persistat nagging and total want to be spoon fed the answers to this planning exercise. BUT, (lol, hypocrite!) i'd like to know what safety precautions pple are putting down. (seriously, if you need to follow safety precautions for this type of experiment you shouldn't be allowed out of the safety of your own home) anyway, thanks
Helena
 
  • #31
r is the distance between the source and the detector. Consider the source as a point source, imagine creating a sphere of radius r around the source. Intensity can be considered power per unit area. The surface area of a sphere is given by [itex]4\pi r^2[/itex], hence the formula. Does that make sense?
 
  • #32
I haven't got that far yet, I am still stuck on the distances bit. see i actually know what to do its just this little bit tripping me up, or maybe I am just trying to over complicate things.
 
  • #33
Hootenanny said:
r is the distance between the source and the detector. Consider the source as a point source, imagine creating a sphere of radius r around the source. Intensity can be considered power per unit area. The surface area of a sphere is given by [itex]4\pi r^2[/itex], hence the formula. Does that make sense?

LOl thank you so much you are a genius. Its all clicked into place now. So i will be increasing r by 5cm each time i change the length. So then i wouldn't have to consider calculating the energy of each photon would i. Thanks a million :smile:
 
  • #34
Helena_88 said:
Hey kids,
I am new and would like to thank you for all the help you have given me with this! and would like to say i simpathise with you with the persistat nagging and total want to be spoon fed the answers to this planning exercise. BUT, (lol, hypocrite!) i'd like to know what safety precautions pple are putting down. (seriously, if you need to follow safety precautions for this type of experiment you shouldn't be allowed out of the safety of your own home) anyway, thanks
Helena

I agree with you totally there! I once had to write safety precautions when measuring the time period of a pendulum? :confused: For your risk assesment, just use simply common sense, what don't you do around electricity? Basic lab safety, trailing leads etc. ,not shining your source at someone and/or their eyes. Real basic stuff, but if its a requirement, you need to put it in. Just jump through the hoops

~H
 
  • #35
Physicsameture said:
LOl thank you so much you are a genius.

I think you're being over generous there! :redface: Anyway, I forgot to define the symbols in the equation I gave you, just incase you didn't know P is the power of the source, e.g. a 60w light bulb.

~H
 
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