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]

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
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).

[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]

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.....

[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]

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?

[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.

[bigbadcityboy]

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

[Hootenanny]

M means mega 106

[ninja_squirrel]

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.

[bhuvi]

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?[/QUOTE]

please send me the details as soon as possible please please please

[Hootenanny]

Hi bhuvi, welcome to PF. Alot of information has been given out here on the forums, I'm sure you have read some of the threads, but just incase you haven't here are links to a few;
http://www.physicsforums.com/showthread.php?t=116631

http://www.physicsforums.com/showthread.php?t=119776

http://www.physicsforums.com/showthread.php?t=119122

Basic information is available from wikipedia and other internet sites (google for photodiode/photocell/photoelectric effect etc.)

~H

[welshdragon]

Will you please help im 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

[welshdragon]

Please Will Somebody Help Me!!!

[dolarpounds]

investigation on how a photocell depends on distance using infra red radiation

[Physicsameture]

Hey people im doing this same plan. Ive 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 im 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:

[Physicsameture]

O and does any one 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).

[Hootenanny]

Hey people im doing this same plan. Ive 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 im 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;

|I| = \frac{P}{4\pi r^2}

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

~H

[Physicsameture]

Cool thanks, so does that mean that 'r' would be changing? Because im using a small led so im 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.

[Helena_88]

Hey kids,
Im 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

[Hootenanny]

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 4\pi r^2, hence the formula. Does that make sense?

[Physicsameture]

I havent got that far yet, im still stuck on the distances bit. see i actually know what to do its just this little bit tripping me up, or maybe im just trying to over complicate things.

[Physicsameture]

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 4\pi r^2, 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 wouldnt have to consider calculating the energy of each photon would i. Thanks a million :smile:

[Hootenanny]

Hey kids,
Im 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 :grumpy:

~H

[Hootenanny]

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

[Physicsameture]

lol well i think your smart. when it comes to the graph il plot an intensity current graph. what exactly wil the gradient be showing then? and what would the units be? would it be I/eV?

[welshdragon]

Just to clarify
The photocell circuit does not require a battery as the photocell produces its own current, which I will measure using an ammeter.

Im sure its been said before, but would somebody tell me if this is right?

And Im probably being stupid but would the photocell circuit work with a battery connected, or is it unnecessary?

THANX

[welshdragon]

Just to clarify
The photocell circuit does not require a battery as the photocell produces its own current, which I will measure using an ammeter.

Im sure its been said before, but would somebody tell me if this is right?

And Im probably being stupid but would the photocell circuit work with a battery connected, or is it unnecessary?

THANX

[welshdragon]

woops sorry for the double post

[Physicsameture]

Just to clarify
The photocell circuit does not require a battery as the photocell produces its own current, which I will measure using an ammeter.

Im sure its been said before, but would somebody tell me if this is right?

And Im probably being stupid but would the photocell circuit work with a battery connected, or is it unnecessary?

THANX

You would need a battery or maybe two. Because initially the photocell would have a high resistance, and you are connecting it to your circuit so it would need an initial amp reading, then you will have to investiate how the light source affects the current. hope that helps :smile:

So any one what the gradient of my graph would be if im plotting a intensity against current graph. N would the units be I/eV ?

[Hootenanny]

You would need a battery or maybe two. Because initially the photocell would have a high resistance, and you are connecting it to your circuit so it would need an initial amp reading, then you will have to investiate how the light source affects the current. hope that helps :smile:


Just to clear things up;

It depends on what your using, if you are using a photodiode or a photoemissive cell, not battery is required as it generates its own current, however if you are using a photoresistor or LDR then a circuit is required.


So any one what the gradient of my graph would be if im plotting a intensity against current graph. N would the units be I/eV ?

As for the graph, you are plotting intensity, which is measured in watts per square meter against current which is measured in amps. Therefore, your gradient would have the units of w\cdot m^{-2}\cdot A^{-1}.

~H

[welshdragon]

[QUOTE=Hootenanny]Just to clear things up;

It depends on what your using, if you are using a photodiode or a photoemissive cell, not battery is required as it generates its own current, however if you are using a photoresistor or LDR then a circuit is required.

So would a photocell require a battery?

[Hootenanny]

[QUOTE=Hootenanny]Just to clear things up;

It depends on what your using, if you are using a photodiode or a photoemissive cell, not battery is required as it generates its own current, however if you are using a photoresistor or LDR then a circuit is required.

So would a photocell require a battery?

Again, it depends on terminology. I would consider a photocell to be a photoemissive cell and not an LDR but I have heard of LDR being called photocells so I don't know I'm afraid. If you wait until Berkeman comes online, he'll probably know.

~H

[Physicsameture]

[QUOTE=Hootenanny]Just to clear things up;

It depends on what your using, if you are using a photodiode or a photoemissive cell, not battery is required as it generates its own current, however if you are using a photoresistor or LDR then a circuit is required.

So would a photocell require a battery?
Im using a photo resistor which i think is a kind of photocell, n the way i think this works is that the more light that is shone on it the less resistance it produces therefore allowing more current to pass through which will lead to your increase in current on your ammeter. this requires a battery. :smile: am i rite anyone who actually knows lol

[berkeman]

[QUOTE=welshdragon]
Im using a photo resistor which i think is a kind of photocell, n the way i think this works is that the more light that is shone on it the less resistance it produces therefore allowing more current to pass through which will lead to your increase in current on your ammeter. this requires a battery. :smile: am i rite anyone who actually knows lol
That sounds right. But as Hoot points out, terminology can get mixed up sometimes. I would expect an LDR to be a CMOS structure, where the light helps to liberate electrons into the conduction band, where they are moved along by the external battery. More light liberates more electrons, hence you measure a lower resistance. Photodiodes are different. They are a PN junction structure, and the electrons that are liberated by photons are swept across the junction and have to return by an external connection. That current is called the photocurrent. You don't need a battery in this measurement.

Quiz question for those of you who have paid attention all through this long thread(s): Would the photodiode's photocurrent that you measure be affected by a battery that is connected + to cathode and - to anode? That is, your circuit is a series connection of cathode, ammeter -, ammeter +, battery +, battery -, anode.

[Physicsameture]

This has nothing to do with the plan but just out of curiosity whats the difference between an LDR and photoresistor then?

[berkeman]

This has nothing to do with the plan but just out of curiosity whats the difference between an LDR and photoresistor then?
I dunno. The datasheets would say. Those may just be terms referring to the same thing. Google some datasheets to find out for sure.

[welshdragon]

[QUOTE=welshdragon]
Im using a photo resistor which i think is a kind of photocell, n the way i think this works is that the more light that is shone on it the less resistance it produces therefore allowing more current to pass through which will lead to your increase in current on your ammeter. this requires a battery. :smile: am i rite anyone who actually knows lol

So for a photoresistor you just require a battery, photoresistor and a ammeter in a circuit?

What would the measurement on the ammeter be for this circuit?

[Physicsameture]

[QUOTE=Physicsameture]

So for a photoresistor you just require a battery, photoresistor and a ammeter in a circuit?

What would the measurement on the ammeter be for this circuit?

mA meaning miliamps. or it depends on the voltage your using mate. Do a quick calculation V=IR and see what kind of figures you get for I when you use different voltages.

[Physicsameture]

I dunno. The datasheets would say. Those may just be terms referring to the same thing. Google some datasheets to find out for sure.

lol i was guessing one in the same, just a fancier name i suppose

[welshdragon]

[QUOTE=welshdragon]

mA meaning miliamps. or it depends on the voltage your using mate. Do a quick calculation V=IR and see what kind of figures you get for I when you use different voltages.

What is the range of resistances for the photoresistor then?

[Physicsameture]

[QUOTE=Physicsameture]

What is the range of resistances for the photoresistor then?

lol depends on the light source you are using seeing as some can go on for meters, and others not very far.

[welshdragon]

[QUOTE=welshdragon]

lol depends on the light source you are using seeing as some can go on for meters, and others not very far.

Im using a IR source that will travel about a meter. Do u have any idea what sort of range the resistance would be for this source?

If u can answer that then ur a legend:biggrin:

THANX

[Physicsameture]

[QUOTE=Physicsameture]

Im using a IR source that will travel about a meter. Do u have any idea what sort of range the resistance would be for this source?

If u can answer that then ur a legend:biggrin:

THANX
lol im doing the same plan.
Err the resistance again depends on what kinda photocell ur using in my case ldr/photoresistor. If u search on the net u should find some, and they shoul come with like a resistance already on it.:smile:

[welshdragon]

Just out of interest do I need to include a graph of what i expect to see?

Thanks Physicsameture you've been a great help

[Physicsameture]

Just out of interest do I need to include a graph of what i expect to see?

Thanks Physicsameture you've been a great help

Well im adding a graph, but im not actually gona draw it im just gona say what the person should plot if they wer to draw 1. But if ur using an ldr u should expect to see that as the distance increases, the current drops.

[Hootenanny]

If I were doing this I would plot a graph of Current on the x axis vs. Displacement from source on the y axis, if you equipment is sensitive enough, you should be able to obtain a nice curve that will confirm the equation;

|I| = \frac{P}{4\pi r^2}

Another quiz question for posters: What would you expect your curve of I vs. r to look like?

~H

[welshdragon]

In my plan should I say that I was using a photocell, or would I have to be more clear and say that I will use a photoresistor.

[welshdragon]

Is a photoresistor the same as an LDR, and if so could I use just a normal LDR to detect infrared?

[Hootenanny]

Is a photoresistor the same as an LDR, and if so could I use just a normal LDR to detect infrared?

Yes, a photoresistor is the same as an LDR. I think you would be able to detect an IR source, I know that Cadmium Sulphide Cells can detect IR radiation, although I'm not sure how sensitive they are, on would have to look at the data sheet for such information. The data sheet should provide a range of detectable wavelengths, anything longer than 750mm is infrared, but you might want to look at what wavelength your transmitter transmits and source your LDR acordingly. CdS cells are inexpensive.

Hope this is useful.

~H

[Physicsameture]

If I were doing this I would plot a graph of Current on the x axis vs. Displacement from source on the y axis, if you equipment is sensitive enough, you should be able to obtain a nice curve that will confirm the equation;

|I| = \frac{P}{4\pi r^2}

Another quiz question for posters: What would you expect your curve of I vs. r to look like?

~H
As the intensity increases, r decreases. So u should get a negative graph i think.

[Hootenanny]

As the intensity increases, r decreases. So u should get a negative graph i think.

Yes, it would be a curve because I \propto r^2 with the gradient being \frac{P}{4\pi}.

~H

[jerromes8]

i'm doing the same thing and you don't know how much help you've given me

[me2]

sam ur 1000% percent right! you people are a god send, ive been searching the net for about an hour (i know that isnt very long) but ive stubbled across you community here and ive got just about all the info i need from you! a lot of examining boards seem to like the topic of IR and distance dont they?
thanks
me

[i25959341]

hey i am doing the same experiment and i am proposing my idea and i want ur comments on it
ermm i wud connect a photocell in series with an ammeter with a voltage meter. The photocell/ LDR/ Photodiode watever u call it, wud be directly pointed toward my light source. Therefore my circuit wud have a change in resistance wen my light source hit the LDR or a photocell. Using that information i cud vary the distance and obsevre the difference in resistance and therefore the strenght of my signal. i wonder if it s possible to calculate number of photons hitting my LDR???
The experiment wud be performed in a dark room and with safety google due to the dange of infrared light.
my thought on this experiment wud be since light diffract sometimes therefore the amout of photons hitting my LDR wud varied with the distances. Yet the question said how the output from a photo cell depends on its distance from a POINT SOUCE of infre-red radiation, so i have doubt whether it wud diffract or not due the confusion caused by the wording of the question. Nevertheless, that s my simple plan at the moment and i wud be really gratefull if some1 cud spot any wrong doing in my ignorant planning

[i25959341]

i also wonder how cud i use the information provided from my experiement to calculate the number photons hitting my LDR??
i meant shud i use the power equations like P=VI cos this is like joules per second and divided it by E=hf(frequency of infrared ) so i can get number photons hittin the LDR. Therefore i cud make a even more brilliant graph like how the number of photons decrease as the distance get longer due to diffraction.
does that sound right ??

[i25959341]

'A photoresistor is made of a high resistance semiconductor. If light falling on the device is of high enough frequency, photons absorbed by the semiconductor give bound electrons enough energy to jump into the conduction band. The resulting free electron (and its hole partner) conduct electricity, thereby lowering resistance.'
therefore does that mean wen more photon hitting the LDR then the Amp increase since it allowed more charged particle to go around the circuit??? like photo electric effect??

[i25959341]

THEREFORE, since amp is number of charged partical per second or something. the current of an infrared hitting my resistor and the current of my resistor without any light, the differences is the number of electrons released by the infrared. and since according to photoelectric effect each electron cud only be released by one photon then the number of electrons= the number photon hitting my LDR???

[i25959341]

assuming each photon has archeieved the threshold frequency or something??? but i think that wud depend on the sensitivity of my LDR right???

[i25959341]

mannn i think i am wrong here but i just guessing

[i25959341]

hello hello reply plz some1 plz wake up or something loool