1. Feb 19, 2005

Hi
I'm doing my physics coursework, and am rather stuck.

It's on senors, we have to choose a make a sensor and then use/test it etc -

First problem: I'm using a photodiode and, I've put it in parallel with a mulitmeter, reading it's resistance, at different light levels. i have another resistor, that is variable, so i'm going to test the sensitivity of the photodiode at different light level (baisically varying distances from a lamp, in a dark room - not i deal, i know.) Is that a good idea? I'm really not sure if thats any good?

next problem: i have to plot a graph, and i'm stumped - i was thinking about plotting distance/m against resistance, but then the gradient of the graph it redundant.
Can you suggest any ideas? I considered using the potential divider equations, but was unsure which value to put where
(my coursework plan is in for monday, so a quick reply would be greatly recieved )

2. Feb 19, 2005

Monday's in two days:

3. Feb 20, 2005

### xanthym

Your approach seems valid. Your photocell varies its Resistance depending on the incident light intensity. Vary the Distance "D" of the light source from the sensor and record the Resistance "R" for each distance. The graph would be an X-Y type graph where Resistance is the vertical axis and Distance the horizontal axis. If the sensor had a linear response to light intensity, would you expect the graph to show a straight line ("1/D") relationship with distance, or would you expect a curve (maybe "1/D^2") relationship?? (What is the relationship between light intensity received from a point source and the distance from the source??) Here are some further references to read:
http://kie.berkeley.edu/ned/data/E01-970419-003/full.html
http://science.howstuffworks.com/experiment2.htm
http://www.edinformatics.com/science_projects/labs/light_intensity.htm

~~

Last edited by a moderator: Apr 21, 2017
4. Feb 20, 2005

Thank you!!
That is so very very helpful!!
i'm expecting a 1/D^2 line, soley because it's a photodiode, and they usually have those odd curves which are like parabolas. This leads me to wonder what the area under the graph would signify, or the gradient?
Also - Can anyone think of practical use of this sensor?! I was thinking possibly a night light? but that's all i;ve got thus far
thank you for your help, xanthym

5. Feb 20, 2005

### xanthym

Photocells and sensors employing them are used for hundreds of purposes. They are used in Robotics for light & position sensing, industrial applications like monitoring certain manufacturing processes, by photographers to measure the light available for photographs, and many more. The reference below lists more applications and also presents some response characteristics of photoconductive cells (e.g., resistance versus illumination intensity).

Last edited by a moderator: May 1, 2017
6. Feb 20, 2005

### Gokul43201

Staff Emeritus
A regular room is not a great idea. You will get reflection off the walls, ceiling floor, etc, causing a large deviations from the expected behavior. If you cover all walls with some kind of black fabric, that might help. Else go outdoors (finding the right secluded spot could be tricky).

How do you plan on using the variable resistor ? You haven't explained it's role in your set-up. It looks redundant to me.

The sensitivity of the sensor is the ratio of change in resistance to change in intensity. How does intensity vary with distance for a point source. (Make sure you use a reasonable approximation to a point source, if that is the aim of this experiment. A regular incandescent bulb is pretty close; a flashlight or reflective lampshade will cause the intensity to fall away more slowly - not a point source.) Strictly from your experiment, there is no way to determine the sensitivity (you have to use additional knowledge).

As far as the plot is concerned, yes you want to plot the resistance vs. distance. A better idea would be to do a log-log plot. Plot log(Resistance) vs. log(distance). The gradient of this curve will provide useful information. (you figure out what that information is)

7. Feb 20, 2005

### Gokul43201

Staff Emeritus
If you get a 1/D^2 plot (which you should get pretty close to, if you have a "true" point source) it would not be because "it's a photodiode". It would be because that's how the intensity of light falls away from a point source. It follows direct from an Energy conservation kind of argument. Equal intnsity surfaces are spherical shells, and intensity is the power per unit area.

8. Feb 20, 2005

### xanthym

Gokul --

The situation will definitely be complicated because one "typical" photoconductive cell ("Bulk Effect Photoconductor") might be approx characterized by the following Resistance (in ohms) vs. Illumination (in foot-candles) response (approximate from manufacturer's data):

$$:(1): \color{red} \ \ \ \ Log_{10} (Ohms \ Resistance) \ = \ (-1)*Log_{10} (FootCandle \ Illumination) \ \ + \ \ 5$$

Or equivalently:

$$:(2): \color{red} \ \ \ \ (Ohms \ Resistance) \ = \ (10^{5})*(FootCandle \ Illumination)^{-1}$$

The exact response will depend on the particular manufacturer, specific photoconductive cell being used, light spectral characteristics, environmental conditions (temperature, etc.), etc.

~~

Last edited: Feb 20, 2005
9. Feb 20, 2005

### Gokul43201

Staff Emeritus
Xanthym,

I thought most photodiodes have the kind of response you describe above. Up to some threshold intesity the conductivity of the diode is usually fairly linear in the incident intensity (assuming the diode doesn't get heated up in the process - heat sinking may help reduce that effect).

So, I would expect a slope of about 2 on a log-log plot, using any such typical photodiode. I don't see how this is a problem, but perhaps I'm missing the point of your post...

On the other hand, I thought photodiodes were used as sensors by using them as a current source under reverse bias, but here again, I'm a little rusty, so I wouldn't trust myself.

Last edited: Feb 20, 2005
10. Feb 21, 2005

I have posed problem to my teacher, but unfortunatly i cannot conduct the experiment outside (also it wouldn't be dark outside,) so, the room shall have to suffice, though in my final conclusive write up i shall address the issues of refection making the results not as accurate as i would have wanted them.

my explination was hurried - i intend to use the variable resistor to change the sensitivity of the photodiode - from my previous experimentation, i have found that if the resistance changes earlier in a circuit, then the photodiode becames moire sensitive (this experiment was with a thermocouple, actually, so if i'm talking rubbish, please tell me!!)

Ok, so, i think that kills my previous argument - so is there a way that i can keep the variable resistor in the circuit? because as i said earlier i NEED to sort of fill my coursework with physics (preferably advanced, but not tooo advanced, as i need to understand it), and the potential dividor equation (with the variable resistor) was what i was counting on....

Thanks, i'm going to do that - i'll do a bit of research around that topic.
thaks very much for your help.

11. Feb 21, 2005

As yet, i havn't taken any results, so i;m stil at the prediction and planning stage.

Just quickly; May i ask what a 'true point source' is, also what 'direct energy conservation' is.

12. Feb 21, 2005

### Diane_

A "point source" would be one that could be treated as a mathematical point. There is no such thing in the real world, but stars come pretty close. (Not, of course, that they put out enough light for your needs, but you get the idea.) The problem is that, with a non-point source, the light does not all come from the same place, so you get a smearing of the 1/r^2 relation.

When he says "it follows direct from an energy conservation kind of argument", he's just saying that you can derive the 1/r^2 dependence of light intensity from a point source merely by assuming energy conservation (along with a little geometry).

13. Feb 21, 2005

Cool, thank you!
Was talking to my physics teacher today, and he said that using a log graph would not be so good, this was suggested by Gokul43201, he said it would limit the accuracy of the graph; i was just wondering what the advantages of using a log graph would be?

14. Feb 21, 2005

### xanthym

The Log-Log graph's advantage in this case is that it shows "power law" relationships (like 1/D^2) with a straight line. Pretend for a moment you're trying to predict the graph's appearance if you varied the distance between the Photoconductive Cell and your point light source. Here's a table of some possibilities where {Photoconductive Cell Resistance}=R, {Light Source Distance}=D, and {Proportionality Constant}=β:

Graph Characteristics
Relationship............Regular Linear Graph.........Log-Log Graph
R = β*D....................Line (Slope β)................Line (Slope 1)
R = β*D^(-1).................Curve......................Line (Slope -1)
R = β*D^(-2).................Curve......................Line (Slope -2)
R = β*D^(-3).................Curve......................Line (Slope -3)
R = β*D^(+2)................Curve......................Line (Slope +2)
R = β*D^(+3)................Curve......................Line (Slope +3)

Now you can see the advantage: If you plot the data on a Log-Log graph and it appears to be a straight line, its slope will indicate the power law relationship and can be determined directly from the graph. (One way to approach this issue is to graph the data BOTH WAYS, first on a Regular Linear Graph and then on a Log-Log Graph.)

Now the final question. So far you've learned two basic facts about your Photoconductive Cell and light propagation physics. Both are summarized below. Can you now make a reasonable estimate of the final relationship shown by your DATA when you measure Resistance ("R") at different light source distances ("D")?? (In the following, α and β are proportionality constants.)

Fact #1 (from manufacturer's data referenced earlier in this thread):
{Photocell Resistance ("R") } = α*{Illumination From Light Source}^(-1)
Fact #2 (from physics of point light source propagation):
{Illumination From Point Light Source} = β*{Dist From Point Light ("D") }^(-2)

What is your predicted relationship between "R" and "D"??

~~

Last edited: Feb 21, 2005
15. Feb 24, 2005

Wow! And thank you very much.
Ok, I have done some preliminary investigating, and have took the output voltage around the photodiode, this gave me a graph that went up rapidly, then started to curve off - as you suggested, xanthym, I then plotted a log distance vs. log voltage, this is contrary to what you suggest, as you said that I should plot log distance vs. log resistance.
I talked to my teacher (though I have little faith in him...) about the idea of using a log/log graph, and he said it wouldn't be amiable - I don't understand why. He also said that plotting log distance vs. log resistance would mean it wouldn't be very good for a sensor, as they are supposed to measure output voltage.

In regards to the slope, gradient of the graph, I believe it shall be able to account for the sensitivity of the sensor. But more usefully does it give me the Proportionality Constant??

Can I ask the benefit of using both the log/log graph and the regular linear graph?
(If someone tells me how, i can upload my graph so you can see where i;ve screwed up! it didn’t produce a straight line, which is most confusing!)

Oh, and the links you sent, Xanthym are invaluable!

16. Feb 25, 2005

### xanthym

First, and MOST IMPORTANT, what type of photocell do you have?? There are 2 basic types. One type acts like a variable resistor in which the resistance decreases with increasing light intensity. The other acts like a small battery, producing a voltage (or current) which increases with light intensity.

From your previous msgs, it seemed like yours was the variable resistance type because you mentioned measuring its resistance. For this type, you would measure resistance and graph Resistance versus Distance on a graph for your experiment. However ....

However, from your most recent msg, apparently the photocell is the "small battery" type, which you called a "photodiode." If it's the "small battery" type, the VOLTAGE would be measured, and you would graph the VOLTAGE versus Distance for your experiment.

For the Voltage type of photocell, the Voltage it produces (when measured with a standard voltmeter) typically has the following approximate relationship to the light intensity illuminating its surface:

$$:(1): \ \ \ \ V \ = \ c*LOG(L) \ + \ d$$

where "V" is the measured Voltage, "L" is the Light Intensity illuminating its surface, "c" is a Proportionality Constant, "d" is an additive Constant, and "LOG" is the standard Base 10 Logarithm.

Furthermore, we know that Light Intensity "L" varies with distance "D" from a Point Light Source according to:

$$:(2): \ \ \ \ L \ = \ k*D^{-2}$$

where "k" is a Proportionality Constant. Combining equations #1 and #2, we get:

$$:(3): \ \ \ \ V \ = \ c*LOG(k*D^{-2}) \ + \ d$$

Finally, we can rearrange terms in Eq #3 and express "V" using new Constants "a" and "b":

$$:(4): \ \ \ \ V \ = \ a*LOG(D) \ + \ b$$

On a standard Linear-Linear graph, this would produce a curve which increased rapidly at first and then "curved off":

http://www.ltcconline.net/greenl/courses/103b/premid1/log2.gif

For the relationship between "V" and "D" shown above, it does NOT make sense to graph it on a Log-Log graph since only 1 of the variables ("D") is involved with the Log function . (Both variables would need Logs in order to make sense using the Log-Log graph.) So stay with the ordinary Linear-Linear graph of Voltage "V" versus Distance "D".

In order to attach files to your msg, go below the msg entry area to the "Additonal Options" panel and use "Attach Files".

~~

Last edited: Feb 25, 2005
17. Feb 26, 2005

Right, it has been my fault i didn;t realise that there were two kinds of photocells, (in fact now i realse there are many!) I've attached my graphs, and i think it's the second one you mentioned, though i found that the voltage got higher as the light intensity got lower...

The resistance was just me attempting to get the physics into my coursework, and i would do that by utilising the potential dividor equations, relating R1 and R2 togethrt with the output voltage. But the graph's suggest later look fansastic in regards to physics.

this is really great, thank you very much, but i'm unsure how i'd find both constants, i would be able to find one through putting lOG distance and output voltage in, but not sure how the 2nd one would turn up - would it be the gradient of the straight line i plot?

I have played around with my graphs, and have produces 3, a log/log graph (now realising that it's a bad idea!) distance vs. V^2, and the normal D vs. V.
these are not my actual results for my coursework, only preliminary experiments as class work, if you could possibly show, or prompt me how the straight line would be formed, the'd be excellent, the attachements are in three word documents, as i wasn't allowed to upload all three graphs and results table on one document.

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18. Feb 26, 2005

### xanthym

Your data is certainly intersting since Voltage (V) appears to increase with Distance (D). However, there are many types of Photocells, and so we'll try to study (or "model") this data with whatever methods appear to produce the best results (without trying to find the detailed physics of the results).

A good method to study this data is to apply a mathematical formula (or "transformation") to make it graph approximately in a straight line. Such a formula is not always obvious, sometimes takes a bit of experience to find, and you should be prepared to obtain only an approximately straight line.

The following formula appears to graph your data with an approximately straight line:

$$:(1): \ \ \ \ V \ = \ a*(Log \ D)^{0.05} \ + \ b$$

where "V" is measured Voltage, "D" is measured Distance, and both "a" and "b" are Constants.

The procedures for graphing this data and for finding values for "a" and "b" are given below. The indicated procedures have been somewhat customized to your situation, but the techniques shown are quite general and can be applied to similar problems.

Step 1 ---> Make a table of your measured data (similar to those you've already made) with the following columns:
Measured Distance (D)..........{Log D}^(0.05)..........Measured Voltage (V)
Complete the table by copying your measured data into it (both "D" and "V"). Then compute the entry for Column #2 with the formula "{Log D}^(0.05)" for each value of "D" in your table. (Do NOT worry about the constants "a" and "b" at this stage. We'll determine them later.)

Step 2 ---> Next, take an ordinary piece of graph paper and label the Horizontal Axis with "{Log D}^(0.05)" and the Vertical Axis with "Voltage (V)". Then graph your data by graphing the SECOND (2nd) Column data on the Horizontal Axis and the THIRD (3rd) Column data on the Vertical Axis, one pair of points at a time until all pairs have been graphed. (Finish the graph by connecting the successive points with line segments.)

Step 3 ---> If the proper formula has been applied (like Eq #1 above), the points should form an approximately straight line. Next, take a RULER and draw a "Best-Fit Line" through the points. Try to get this straight line to contain the data points. However, this is usually impossible. Thus, an approximate fit is obtained by drawing the straight line such that the straight line has an equal area above it and below it of the space between the straight line and your graphed data point line segments. If everything has proceeded correctly, you should now have the graphs shown here:
http://img206.exs.cx/img206/3458/photocellgraph8qj.jpg

Step 4 ---> The last step is obtaining the values for Constants "a" and "b". This is done by reading TWO (2) POINTS FROM THE STRAIGHT LINE (preferrably near the straight line's endpoints). For each point, we'll call the Horizontal Axis value "x" and the Vertical Axis value "y".
Using algebra techniques, derive the line equation {y = a*x + b} from these 2 "{x,y}" points. (Later, the "x" will be replaced by "{Log D}^(0.05)" and the "y" will be replaced by "V" when we're finished). For the "Best-Fit Line" shown in the above graph for your data, we have:
Point #1: {x=(0.94), y=(3.8)}
Point #2: {x=(1.02), y=(6.17)}
From algebra, the straight line thru these 2 points is given by:
(y - 3.8) = {(6.17 - 3.8)/(1.02 - 0.94)}*(x - 0.94) (<--- From Algebra)
Or rearranging terms and writing the above in {y = a*x + b} format:
y = (29.625)*x + (-24.0475)
Thus, a=(29.625) and b=(-24.0475), and the "Best-Fit Line" (or "Best-Fit Model") for your data is given by:

$$:(2): \ \ \ \ \color{red} V \ = \ (29.625)*(Log \ D)^{0.05} \ + \ (-24.0475)$$

Of course, this was just one example, and every situation will be slightly different. However, the above example illustrates the general methods used for this type of study.

~~

Last edited: Feb 27, 2005
19. Feb 27, 2005

thank you so much, really you're so helpful, i'm gonna do that this evening, i'll get back to you with results etc how it went
Thanks again