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

[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

 

[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, I've been searching the net for about an hour (i know that isn't very long) but I've stubbled across you community here and I've got just about all the info i need from you! a lot of examining boards seem to like the topic of IR and distance don't they?
thanks
me

 

[i25959341]

just proposing

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 amount 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 particle 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