# Help needed on photocells please.

1. Apr 5, 2006

### 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

2. Apr 5, 2006

### Hootenanny

Staff Emeritus
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

3. Apr 5, 2006

### einstein2603

but what method would i use? ive been told i need circuit diagrams.

4. Apr 5, 2006

### Staff: Mentor

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?

5. Apr 5, 2006

### einstein2603

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

This is really helping by the way. thanks

6. Apr 5, 2006

### einstein2603

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

einstein2603

7. Apr 5, 2006

### 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

8. Apr 5, 2006

### Staff: Mentor

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....

9. Apr 6, 2006

### einstein2603

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

10. Apr 6, 2006

### Hootenanny

Staff Emeritus
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

11. Apr 6, 2006

### einstein2603

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

12. Apr 6, 2006

### Hootenanny

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

13. Apr 6, 2006

### einstein2603

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

14. Apr 6, 2006

### Hootenanny

Staff Emeritus
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

15. Apr 6, 2006

### einstein2603

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

16. Apr 6, 2006

### Staff: Mentor

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.

17. Apr 7, 2006

### 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!

18. Apr 7, 2006

### einstein2603

also would i need to mention the inverse square law?

19. Apr 7, 2006

### einstein2603

can you guys help me asap please. thanks

20. Apr 7, 2006

### Staff: Mentor

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!

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