How Does Distance Affect Photocell Output?

[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?