How Does Distance Affect Photocell Output?

[einstein2603]

[SOLVED] help needed on photocells please.

hi everyone.

i have been given a question to do for class. We have to investigate how the output of a photocell depends on the distance from a point source of radiation.

Now what I have tried so far is to connect the photocell in series with an ammeter. The thing is I am stuck on what to do next.

I would greatly appreciate your help.

Thanks

 

[Hootenanny]

The obvious thing to do seems to measure the distance from the source to the photocell. Vary the distance and then plot a graph.

-Hoot

 

[einstein2603]

but what method would i use? I've been told i need circuit diagrams.

 

[berkeman]

You don't need no stinking diagrams... What is the output current 1cm from the light source? 2cm? 4cm? 8cm? Be sure to do this in a dark room with only the one light source. Also measure the base current with only dark... What does the plot look like?

 

[einstein2603]

You don't need no stinking diagrams... What is the output current 1cm from the light source? 2cm? 4cm? 8cm? Be sure to do this in a dark room with only the one light source. Also measure the base current with only dark... What does the plot look like?

you are talking about current yeah. So what circuit would i construct?

This is really helping by the way. thanks

 

[einstein2603]

sorry if i am being too demanding. Its just i really want to have a thorough idea. thanks

einstein2603

 

[einstein2603]

i have the question with me and it says draw a diagram of the arrangement of your apparatus. hmmm. i am really stuck on this photocell question. If anyone can add further ideas it will be appreciated greatly.

thanks

einstein2603

 

[berkeman]

you are talking about current yeah. So what circuit would i construct?

This is really helping by the way. thanks

You already mentioned just connecting the photocell to an ammeter -- that should work fairly well for your first tests. The output of the photocell is a photocurrent that depends on the level of illumination (and the wavelengths of the incoming light). The current flows out of the cathode of the photodiode / photocell. So connect the + lead of your DVM in "mA" setting to the cathode of the photocell, and the - lead to the anode of the photocell. If your DVM's current measurements are sensitive enough, you will be able to plot a couple decades of photocurrent variation as you move the cell closer and farther away from the light source.

A better test circuit would be to use an opamp to make a current-to-voltage converter, and measure the output voltage. I'll leave that as an exercise for the reader for now...

 

[einstein2603]

You already mentioned just connecting the photocell to an ammeter -- that should work fairly well for your first tests. The output of the photocell is a photocurrent that depends on the level of illumination (and the wavelengths of the incoming light). The current flows out of the cathode of the photodiode / photocell. So connect the + lead of your DVM in "mA" setting to the cathode of the photocell, and the - lead to the anode of the photocell. If your DVM's current measurements are sensitive enough, you will be able to plot a couple decades of photocurrent variation as you move the cell closer and farther away from the light source.

A better test circuit would be to use an opamp to make a current-to-voltage converter, and measure the output voltage. I'll leave that as an exercise for the reader for now...

wow you guys know your stuff. But I am a high school student and that is all too complex. Any simpler ways? thanks

 

[Hootenanny]

As berkeman has already said you can use an ammeter. Basically, you need to connect the positive lead from your ammeter to the cathode of the photodiode and the Earth or negative lead to the anode of the photodiode. This will allow you to measure the rate of flow of electrons, which is proportional to the illuminiation of the photodiode (and dependant on the wavelength of light). If you plot current against distance from source, you should be able to determine a relationship between the two variables.

However, it is more useful to measure the voltage.

-Hoot

 

[einstein2603]

can you go through the process of measuring voltage please. thanks hootenany.

 

[Hootenanny]

Are you familiar with how to use an operational amplifier, a 741 op-amp chip for example?

 

[einstein2603]

no. i am only in high school. I am 16 and haven't come across that yet. Just simple physics (not too simple tho) please. thank you so much

 

[Hootenanny]

If you haven't used op-amps before then it is probably best to just measure current. The op-amp basically converts the current produced by the photodiode into a potential difference, a large resistance is used to prevent a significant current flowing, thus obtaining a more accurate value for the potential difference.

A basic circuit diagram can be found here ; http://www.wam.umd.edu/~toh/ElectroSim/Photodiode.html

-Hoot

 

[einstein2603]

so can this whole concept be used in the idea of automatic doors which determine whether it is safe to close the door?

 

[berkeman]

so can this whole concept be used in the idea of automatic doors which determine whether it is safe to close the door?

Absolutely. Light is used in many applications like that. You typically have a light source as the sending element and a photodiode of some sort as the receiving element. You can have the light source on one side of something and the receiver on the other side, or you can have them both together pointing at a reflecting element on the other side of whatever you want to sense. The term for this type of arrangement is "photointerruptor", and it is used in everything from computer mice to factory automation to automatic doors.

 

[einstein2603]

lastly, does anyone know what distances to vary and use because i heard that photocells use very small distances. Can anyone find that out for me.

also how could i improve the experiment?safety precautions? any more diagrams you would recommend me to use?the ranges of any measuring instruments used?the list of apparatus needed?

The help i have been given so far from Hootenanny and Berkeman has been unbelievable. Physics Forums is a must for anyone needing help. I will recommend to everyone I know.

If you could help me with those questions, that would be great. Thanks guys!

 

[einstein2603]

also would i need to mention the inverse square law?

 

[einstein2603]

can you guys help me asap please. thanks

 

[berkeman]

can you guys help me asap please. thanks

Asap? We're still waking up here in Cali.

I think I already suggested some starting distances to try. The distances you use for your plots will depend some on how sensitive your DVM's current measurement ranges are. With no amplification (using the opamp current-to-voltage converter that Hoot and I were mentioning), you're probably only going to get measurable current on your DVM out to 10cm or so.

Yes, the inverse square law plays a part in your experiment -- do you know why? As for safety considerations, nah. You're working with only a volt or two output from the photocell. When you start working with voltages over about 40-60V, that's when you have to start being more careful (and that's when Underwriters Laboratories safety regulations start to kick in...) Have fun with the experiment!

 

[einstein2603]

i have no idea about the inverse square law and how it plays a part in the experiment?could you kindly tell me?

 

[berkeman]

I thought you mentioned it first...or did one of us bring it up? I forget.

The inverse square law relates the intensity of radiation to the distance from the source. Think of it in terms of the surface area of a sphere, spaced at some distance from the source. A sphere of radius 1cm has a certain surface area (what is it?), and a sphere of radius 2cm has 4x the surface area of the 1cm sphere (right?). The area if these spheres varies as the square of the radius. If the same amount of radiation goes through each sphere, then the intensity at any distance from the source will be dropping off at a rate of 1/r^2. Hence the name.

You can also google "inverse square law" for more info.

 

[einstein2603]

so how could i incorporate that into into what i am doing? thanks berkeman

einstein2603

 

[Hootenanny]

berkeman explained it very well. I would just like to add a specific example. If we consider your light source a 'point source' and we assume that no energy is lost to their air (no damping) then we can say that the intensity of light landing on a sphere at a distance r from a source of power P is given by;

\left| I \right| = \frac{P}{4\pi r^2}

This assumes that the source radiates energy equal in all directions.

So, if you had a 60 watt light blub and a distance of 10cm from the source you would expect the intensity to be;

\left| I \right| = \frac{60}{4\pi \cdot 0.1^2}

-Hoot

 

[einstein2603]

could this stuff be used in high school?

also how could i incorporate the inverse square law into what i am doing?

thanks guys

einstein2603

 

[Hootenanny]

I suppose you could plot a graph of your results. The current produced by a photodioide is proportional to the itensity, therefore if you plotted a graph of current against \frac{1}{r^2} you should get a straight line. You should disccuss any theory you use in your report. Also I would speak to your tutor about the theory.

-Hoot

 

[einstein2603]

ok cool.

also, how do you get a measure out of a photodiode?it might have been talked about before but i just want to make sure.

thanks

 

[Hootenanny]

The most sensitive way to connect the photodiode to an ammeter directly. You need to connect the positive lead from your ammeter to the cathode of the photodiode and the Earth or negative lead to the anode of the photodiode.

Hope this helps you

-Hoot

 

[einstein2603]

can you tell me anything else to put to get extra marks?

 

[berkeman]

can you tell me anything else to put to get extra marks?

Include the PF in your bibliography...