How Does Distance Affect Photocell Output?

[Zalaras]

About the reply earlier. Say I was to set up a circuit with a photocell and ammeter in a dark room and shot IR radiation from a remote control at the photocell. Would this give a good set of results? How would I be able to see the readings in the dark ?

I have not seen those sites but I will make sure to have a browse through. Thank You .

 

[berkeman]

i couldn't find any data sheets for solar cells on the digikey thing. I am assumin' itl output something in the mA range

Well, I just now googled solar cell datasheet, and got a bzillion hits. I went down into one, and it had the J = xxxx mA/cm^2 type number that you are looking for. Check the hits out to see if you find what you need:

http://www.google.com/search?hl=en&q=solar+cell+datasheet

 

[Rufio]

Ive already completed this exercise and handed it in and I'd just thought I'd tell all those still doing this that this is a "planning exercise". Therefore, they are testing your ability to plan an experiment, not necessarily your ability to understand theory. A little theory may be required to explain why you are doing something, but a lot of information on this thread would gain you no extra marks. Remember, its only 2 marks per point (as Einstein wrote earlier) and this is only worth about 5% of your total grade! A thesis is not required or expected!

 

[Cheery Blossom]

Thanks for the consolation Rufio-Chan

 

[Rufio]

No problem! All you've got to remember is to not leave out the simple physics. They are the bits that'll give you marks!
And a quick tip: An empty table of results is a very handy way of showing what you are going to measure, in what scale, and does not count in the word count! (I don't think there is a specified word count but the recommendation from the exam board is between 500 and 1000 words)

 

[PenguinBoy]

if i put a resistor into stop the components from overloading, would that stop the ammeter readings from being inversely proportional to the square of the distance?

 

[Hootenanny]

if i put a resistor into stop the components from overloading, would that stop the ammeter readings from being inversely proportional to the square of the distance?

No. You wouldn't change the relationship, you would just 'shift' it, similar to translating a curve.

~H

 

[duck86]

Hahah poor little einstein, like many others i also have the misfortune of doing this planning exercise, but then again i also have the misfortune of re-doing my a-levels when I am 20 (Buggery). Its quite a tasking one, but with some thought its easily done. Some of the points raised on this discussion are far beyond what is needed for such a simplistic experiment. I can only suggest to think of the ideas you have been taught this year and not look for answers which you can't comprehend. It gave me a huge headache for a few days, but I am now smiling again.

 

[testament]

hey, i was wondering if u think it would be sufficient to make a potential divider with a pull up resistor and measure the output voltage across the photo detector and if I should use it to work out the current if that would be more accurate, thanks

 

[berkeman]

hey, i was wondering if u think it would be sufficient to make a potential divider with a pull up resistor and measure the output voltage across the photo detector and if I should use it to work out the current if that would be more accurate, thanks

See posts #28 and #62.

 

[testament]

oh, sorry, luckily I'm not being tested on my observational skills, having now read the whole thread you've pretty much done the work for us, reading wot everyone's posted on here i don't think there's a single brain cell between us!

 

[evrox]

ok so I am going to have a photocell in series with a resistor and I am going to mesure the pd( voltage) of the resistor. ok this is really stuid question but I am confused. when the photocell receives more light its resistance drops so does that mean that the pd across the resistor will increase?

 

[evrox]

so I am therefore going to have a negative gradient pd vs distance graph? please help! I am having a temporary momeny of insanity

 

[berkeman]

A photocell or photodiode is not a "light dependent resistor". An LDR is constructed differently, and is probably based on more of a CMOS structure. A photocell or photodiode is a bipolar PN junction, where incident photons liberate electrons that form the photocurrent. The fundamental "thing" that you get from a photocell is current, not resistance or voltage. The fundamental thing that will drop off as the receiving photocell is moved farther from the light source is the photocurrent.

There are different ways to measure the photocurrent, and you should take a look at a typical DVM to figure out whether it will give you the best range of readings when it is set to a current measuring scale and connected directly to the photocell, or if it will work better if you connect a resistor across the anode and cathode of the photocell and measure the voltage generated by the photocurrent passing through the resistor. I honestly don't know which would be better with a standard DVM. You need to do the math to figure that out.

 

[evrox]

ive spoken to my physics teacher and this is definitely the way to do it. I am using a cds photocell.

 

[super_swifty]

for i LDR i would use a multimeter and measure the resistance. (would i need a current for this?)

 

[Hootenanny]

for i LDR i would use a multimeter and measure the resistance. (would i need a current for this?)

As far as I know you would be required to connect your LDR to a power source before measuring the resistance. I could not say what would happen to the relationship, whether or not it would be altered.

~H

 

[Iamphoebe]

I've been reading through this thread over the last few weeks, and it has provided me with really valuabe information - just about the only info I could find. I can see that, although i am confused myself, it is frustrating for the people offering advice. thank you all so much for your help and patience. (and good luck everyone)

 

[Hootenanny]

I've been reading through this thread over the last few weeks, and it has provided me with really valuabe information - just about the only info I could find. I can see that, although i am confused myself, it is frustrating for the people offering advice. thank you all so much for your help and patience. (and good luck everyone)

Thank-you phoebe, it is nice to be appreciated . It was a pleasure. I also echo phoebe's best wishes for anyone involved in this planning examination.

~H

 

[lining515]

i hve read through all the threads...
it seems quite useful...but it really confused me ~
coz we hve been doing some preliminary work in class, and all we did is to connect a radient heater to a power supply...and it emitts out infrared and we connected da photocell with a voltage sensor and a data logger and measured with different distance and figured out da inverse square law relationship..i just don't understand why you can't not connect a voltagmeter directly across da photocell..because our physics teacher suggested us to do it..so ?
and all we need now is da various distance?and homework to produce da point source of infrared...im thinking of conver da radient hearter with sum sort of card and make a hole underneath it so it emitting out infrared frm tat pinhole~~

 

[berkeman]

Maybe the photocell had a resistor connected between the anode and cathode, to convert the photocurrent into an output voltage. Just read a datasheet on a typical photocell (or better yet, the one you used), and you should see that it's a diode that generates a reverse photocurrent when illuminated.

 

[lining515]

ok thankx
bt what do u suggest we could use for a source of infrared?is it ok...if we cover da radient heater with some sort of things and put a pinhole through?would it work?

 

[lining515]

oh and~
any suggestion of da improvement on this experiment and da safety things...

 

[berkeman]

ok thankx
bt what do u suggest we could use for a source of infrared?is it ok...if we cover da radient heater with some sort of things and put a pinhole through?would it work?

This will only work if da cover is flammable. Or inflammable might work too.

 

[imogen89]

Or inflammable might work too.

haha

what exactly is photocurrent. this is probably really obvious, but i am terrible at electronics, and really not understanding any of this.

 

[imogen89]

and if anyone has a circuit diagram [a simple one, all of the ones on the links you have put up make me want to cry] of what this should look like, that would be lovely:)

 

[imogen89]

i am so going to fail this...

 

[berkeman]

haha

what exactly is photocurrent. this is probably really obvious, but i am terrible at electronics, and really not understanding any of this.

The explanation at wikipedia should be straighforward enough. Try there, and list that reference on your paper.

 

[imogen89]

yes, i read that already [all of my references seem to be from wikipedia:)], but i don't understand, "photocurrent is the current that flows through a photosensitive device"---does this mean that it is just normal current when it flows out of the photocell and into the wires, because it is no longer in a photosensitive device?

 

[berkeman]

Okay, here's a present for you and all your classmates (wherever the heck you guys and gals are):

http://www.makingthings.com/teleo/cookbook/photocell.htm

I found it just now when I searched on photocell at HowStuffWorks.com trying to find you another explanation of photocurrent. Please be sure to list that web page as a reference in your report if you use the info from it (and I'm sure you will).

 

[imogen89]

does that question even make sense?

also, it says in a [much] earlier post that the current flows <i>out</i> of the cathode. doesn't current usually go <i>into</i> the cathode?

 

[imogen89]

oof. that's the most helpful thing i have seen all day:):):)

 

[imogen89]

ok i just reposted something and it didnt come up.

i said...

oof, that's the most helpful thing I have seen all day:):):)

 

[imogen89]

ok now it did come up. I'm goign to be quiet and do my physics work now.

I'll probably be back though. i am not getting what analog in means...

 

[berkeman]

it says in a [much] earlier post that the current flows <i>out</i> of the cathode. doesn't current usually go <i>into</i> the cathode?

Depends on the situation, now, doesn't it? What's special about a photodiode/photocell? How do the photons of the light that hit the diode near the junction get turned into mobile electrons? What happens to those electrons?

Also think about it from this angle -- Why do you get net power out of a battery? Because the current flows out of the + terminal, right? If current flows into the + terminal of something (like a resistor) and there is a voltage drop to the - terminal, that thing is a sink of power. Look at the previous reference and the polarity of the photocurrent and diode voltage -- is a photocell a source or sink of energy?

 

[Hootenanny]

Analogue means that the data can take any of an infitie number of values. Digitial can only take two values, 1 or 0 on / off. An example of an analogue wave would be a sine or cosine curve.

~H

 

[imogen89]

now, i am doing the same planning excersise as einstein was--->

you are required to design a laboratory experiment to investigate how the output from a photocell depends on its distance from a point source of infrared radiation.
pay particular attention to:
1. the procedure to be followed
2. how the output of the photocell would be measured
3. range of distances to be used
4. ranges of measuring distances uesd
5. any relevant circuit diagrams
6. any safety precautions
7. particular features which ensure the accuracy and reliability of the results
8. diagram of apperatus

and i was wondering, do you think that it would be enough to do the experiment you mentioned earlier where you just attatch a DVM on mA setting to the photocell, or whether it would give more of a valid conclusion to do the thing where you put a resistor and a voltage divider [?] into the circuit and measure the voltage?

 

[imogen89]

Analogue means that the data can take any of an infitie number of values. Digitial can only take two values, 1 or 0 on / off. An example of an analogue wave would be a sine or cosine curve.

~H

yea that's what i thought, but it says to plug the circuit into analogue in on the circuit diagram [half way down the page that the other guy gave me a link for]...where is that wire going. or do i not even need to know...

 

[razor12]

One thing to watch out for is the example posted on the last page is for a light dependent resistor, a standard LDR will not react to IR as its wavelength is too long, this means the only thing it will do is react to visable light,
You can in theory get LDR's that respond to IR (think they are, Ge:Cu)but you need to use components you can source

 

[welshdragon]

Will you please help I am confused
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

 

[dolarpounds]

could you help me with the apparatud that will be used for this experiment

 

[rich_b62003]

Ok here's my take on things.
I am also doing this experiment and whil it seemed a bit strange at first its not so bad. Really just asking berkemen and hootenanny whether what I've done is the right sort of lines. A few people have mentioned batteries. however having done a bit of reading on photocells i thought they generated current themselves, so I've left out a battery. By the way not sure if anybody has acctually clarified this. We don't acctually do the experiment, just plan it. my plan is to have the photocell connected to a fixed resistor with a large resistance, and to measure the voltage acroos that large resistance. From that i could work out the current using simple maths.
think that's right but just a yes or no would help enormously

 

[Hootenanny]

Will you please help I am confused
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

Yes, this should work but you may only obtain a limited range of data.

~H

 

[Hootenanny]

Ok here's my take on things.
I am also doing this experiment and whil it seemed a bit strange at first its not so bad. Really just asking berkemen and hootenanny whether what I've done is the right sort of lines. A few people have mentioned batteries. however having done a bit of reading on photocells i thought they generated current themselves, so I've left out a battery. By the way not sure if anybody has acctually clarified this. We don't acctually do the experiment, just plan it. my plan is to have the photocell connected to a fixed resistor with a large resistance, and to measure the voltage acroos that large resistance. From that i could work out the current using simple maths.
think that's right but just a yes or no would help enormously

This would probably work, but it would depend on how sensitive your voltmeter is and it is likely your obtained curve will not be nice . If you want simplicity a better method to use would be to connect your photocell directly to an ammeter and measure the current.

~H

 

[me2]

hmmmm i believe I am doing the exact same question! I am just as stuck, i was under the impresion that you simply hook up a curcit with a powersource, the photocell and reviver thing (is that the diode?) and an ammeter, then alter the distances and ,easure the current.
am i any where near right?