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

[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 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.

 

[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 wouldn't have to consider calculating the energy of each photon would i. Thanks a million

 

[Hootenanny]

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? 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

 

[Hootenanny]

LOl thank you so much you are a genius.

I think you're being over generous there! 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 I am 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 I am 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 I am 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

So anyone what the gradient of my graph would be if I am 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

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 anyone what the gradient of my graph would be if I am 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]

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]

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]

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. am i rite anyone who actually knows lol

 

[berkeman]

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. 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 what's the difference between an LDR and photoresistor then?

 

[berkeman]

This has nothing to do with the plan but just out of curiosity what's 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]

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. 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]

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]

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]

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]

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

THANX

 

[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

THANX

lol I am 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.

 

[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 I am adding a graph, but I am not actually gona draw it I am 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