- #36
Physicsameture
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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 said: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
Physicsameture said: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 said: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?
welshdragon said:Hootenanny said: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
welshdragon said: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 lolHootenanny said: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?
Physicsameture said: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.welshdragon said: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
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.
I dunno. The datasheets would say. Those may just be terms referring to the same thing. Google some datasheets to find out for sure.Physicsameture said:This has nothing to do with the plan but just out of curiosity what's the difference between an LDR and photoresistor then?
Physicsameture said:welshdragon said: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?
welshdragon said:Physicsameture said: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.
berkeman said:I dunno. The datasheets would say. Those may just be terms referring to the same thing. Google some datasheets to find out for sure.
Physicsameture said:welshdragon said: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?
welshdragon said:Physicsameture said: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.
Physicsameture said:welshdragon said: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
welshdragon said:lol I am doing the same plan.Physicsameture said: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
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 said: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
welshdragon said:Is a photoresistor the same as an LDR, and if so could I use just a normal LDR to detect infrared?
As the intensity increases, r decreases. So u should get a negative graph i think.Hootenanny said: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;
[tex]|I| = \frac{P}{4\pi r^2}[/tex]
Another quiz question for posters: What would you expect your curve of I vs. r to look like?
~H
Physicsameture said:As the intensity increases, r decreases. So u should get a negative graph i think.