Physics Coursework: Sensors, Photodiodes and Graphs

[Padford]

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 that's 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 received )
Thanks in advance

-Padford

 

[Padford]

Monday's in two days:
Please help! :D

 

[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




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[Padford]

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

-Padford

 

[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).
http://optoelectronics.perkinelmer.com/content/RelatedLinks/photocell%20introduction.pdf

 

[Gokul43201]

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)

 

[Gokul43201]

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.

-Padford

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.

 

[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.


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[Gokul43201]

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.

 

[Padford]

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).

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.

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.

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!)

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).

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...

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)

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

Padford

 

[Padford]

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.

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.
Padford

 

[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).

 

[Padford]

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?
Thanks in advance
Padford

 

[xanthym]

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?
Thanks in advance
Padford&

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"??


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[Padford]

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!)

Thanks in advance
Oh, and the links you sent, Xanthym are invaluable!
Padford.

 

[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".


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[Padford]

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.

Right, it has been my fault i didn;t realize 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...

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.

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.

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":

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

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?

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

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.

Thanks in advance

Padford.

 

[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.



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[Padford]

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

 

[Padford]

Ok, I've graphed the line, and although it still seems to be curved, it resembles your graph in the URL (also, which graphing program did you use - we have autograph here, it's not a very good version.)
With the formula i now have for the line of best fit, i can find the gradient, but what else useful can i do with this formula? Is it possible to differentiate it to find the gradient? Also, what use would the gradient be?
and finally, does '*' mean mulitbly by, or to the power of?! (it's new notation to me!)
Thanks again

-Padford

 

[xanthym]

.
NOTE:
The symbol "*" means "multiply" and the symbol "^" means "to the power of".

The resulting "Best-Fit Line" from your first example experiment was (from previous msg):
y = (29.625)*x + (-24.0475)
which is in standard slope-intercept form. Thus, from algebra, the "gradient" or "slope" is (29.625) and the "y-intercept" is (-24.0475). Of course, the final equation form is the following (where we substitute the measured and computed quantities for "x" and "y" in the above formula):

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

The "gradient" is still (29.625) and the "intercept" is still (-24.0475). However, it's difficult to interpret these quantities physically since we don't have a good physical interpretation of the basic equation of "V" versus "{Log D}^0.05" above.

The most important point is that the above procedures are EXAMPLES of mathematical techniques which can be used in conjuction with experimental studies. Each case will be different.

The graphing program used for the previous msg's graph was called "Sci-Graph". You should be able to use the graphing program you already have. Here one site that lists some others:
http://directory.google.com/Top/Science/Math/Software/Graphing/
(NOTE: Always be careful if you download and use software off the network.)


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[Padford]

Results have been gathered

Hey,
I have gathered all my results, now, the experiments are over! I have to start the write up now, i have put my results into tables, and graphed a few (not attached as they are too big ), but you can see from the results table that they are similar to the prediction graphs; I'm not going to start to analise them, but am unsure where to start. I was thinking about starting with talking about the sentitivity of each the graphs with their varying set resistance, but then after that I'm at a loss.
I think that I'm going to do what Xanthym suggested a few posts ago and find the straight line for the diode (i have mastered the graphing program on excel, so that's good.) but I'm unsure what the graoh shall show me!

Thanks in advance

-Padford

 

[xanthym]

Determining your graph's PHYSICAL interpretation will be difficult because they are not entirely based on (or derived from) physics theories. Such graphs display what are called "Empirical Relationships" because they describe certain phenomena based on experimental observations which are not fully understood yet. Later, you may explain them and win the Nobel Prize!

Nevertheless, these graphs are VERY USEFUL. They describe how to use your device. In particular, they show how to translate your device's Voltage reading into the light source Distance. (That's also a purpose of your "Best-Fit Line" formula.) With that information, your device can be utilized in various applications, such as robotics, automated sensing systems, etc.

Incidentally, methods for obtaining "best-fit" curves are called Regression techniques. When you have some spare time, the following site describes and allows you to explore Regression methods. (You'll need to scroll all the way down the page to reach the interactive portion.) Good luck!
http://www.arachnoid.com/polysolve/


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[Padford]

Many thanks again!

 

[Padford]

Hi, again!
This is just a quick question; I was wondering why 'd' is an additive constant, and indeed what an additive constant it -
Also, we have yet to cover the maths by which you derived the last equation regarding the last equation: V = a*LOG(D) + b, if it not too much of a hassle, could you explain it to me, soley for my inquisitive mind!
thanks in advace

-Padford

 

[Padford]

results

I'm just graphing my results at the moment; if you have a look at two sets of the, you'll see that the normal either LOG distance or LOD distnace^0.05 doesn';t produce a straight line.
If it;s possible could you suggest a factor by which i should multiply LOG(D) by to get a relevtivally straight line.
Attached are the two tables of results, otu of 8, that I'm having trouble with;
thanks in advance

-Padford.

 

[Padford]

the wall i have hit is lead. I've derived the formula, much like you showed me, Xanthym, for the 100kOHM fixed resistance, (shown in the previous attachement), i got the formula v =25.1(LOG D)^0.05 - 24.04, which, when i put it next to the measued voltage, produced gaps of 1v, which is huge, compared to the other formulae i worked out - and i have checked the maths twice! please can you help, the graphs also look pretty awful too. it's in for monday :D
thanks in advance

Padford

 

[Padford]

lastly, i hope!
to find the sensitvity of the sensor, one must use the gradient - i think that the units for that would be Vcm^-1, am i correct?

 

[Padford]

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:

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



~~

where'd you get these formulae from too?
thanks
Padford

 

[xanthym]

Unfortunately, there are some complicating factors involved in answering your questions:
(1): The attachment (your document) titled "Table of Results for 100k.doc" will NOT download. Thus, it's currently impossible to review your data. Above the attachment box it states "Attachments Pending Approval". Is it waiting for something from you?? Maybe that's why it won't download.
(2): There's additional info needed to properly review your data: the exact electrical circuit used, exactly what was measured in the circuit, values of the various resistors used, exact specifications of your light source (flashlight??, light bulb??, etc.), and other items.
(3): Time is very short if you need answers before Monday (3/14). An entire week went by between your Msg #24 and this current series. There's just not enough time to review everything and also determine solutions to the latest problems (given the slow turn-around time between forum msgs).
(4): At this point, the best solution may be to NOT determine a Straight Line fit for your data. This objective is currently causing you MAJOR problems, and it'll take too much time to consider all the details required for its solution.
(5): Instead, just graph your data on a standard Linear-Linear scale graph. (In other words, use the values you measured without applying any formulas, such as "LOG", to them.) Use the Horizontal Axis for Distance "D" between Light Source and Photocell, and use the Vertical Axis for your measured quantity, in this case Voltage "V". The result will be a curve (which is OK).
(6): Then determine the best fit POLYNOMIAL to this curve from the Web Site given below. Scroll down the web page to the program (give it time to load), enter each complete data set into the entry box in pairs "Distance Voltage <return>" like the example shows, and then determine the Polynomial of least degree (say 2 or 3 or 4) that gives a reasonable fit. Read the Polynomial's equation from the area below the graph.
(7): This is all you need to make your device useful (You do NOT need a straight line). With the Polynomial equation, you can translate any voltage reading into the distance between Light Source and Photocell.
(8): If you have several different data sets (e.g., for different circuit resistors), determine the best fit Polynomial for each one separately.
(9): Of course, each curve's gradient (slope of the tangent line) will be continually changing. However, you can compare sensitivities between curves by choosing the center point of your Distance data (maybe 15 cm) and determining the tangent line slope for each curve at that Distance value. Units for the gradient will be Volts/cm.
(10): Good Luck!
http://www.arachnoid.com/polysolve/



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[Padford]

damn, well, let's cut the problem down, a hellovalot
in regards to the working-out equations, i managed to work out them for the 1st four fixed resistance values. but that last two, which i shall attach again seemed not to work, but i believe i justifed these well:

"This sensor was the one I took man more values for, it seemed, while I was setting the sensors up, to be the most sensitive, though having done the result analysis and gathering, I see I was incorrect, but I think that taking all those results would give a more accurate graph – you can see, however how many anomalies there are, this, shall be due to either unsystematic errors, systematic errors, or the fact, again the fluctuations are caused by the sensitivity of this particular sensor. The sensitivity of this sensor was 25.1Vcm-1, which is good, but the great thing about this sensor is that it worked effectively over a comparatively long range, even after I stopped the voltage had not reached 4.5V (the maximum battery output).
"The formula tells us that, V = 25.1(LOG D) 0.05 – 24.04, this is actually quite incorrect, as you can see. It seems that the idea behind the relationship may only function correctly for those sensors with a relatively low fixed resistance. Though, if a decent relationship was found between the two values, then this sensor would be very effective, and efficient."
This set of results, and indeed all the results and tables are prime example of information displaying an “Empirical Relationship” they describe certain occurrences based on experimental observations, which are not fully understood yet. However, as you can see, these graphs and results can show us the relationship between voltage and distance from the point source quite effectively, though lacking is the understanding of regression techniques ."
is that ok, so i don't have to go into details about rectifying the formula with the polynomial regression techniques. i think i understand wat I'm talking about, so that's useful...
also, that foumula relating light inensity and voltage output together, where'd ou get it from?!
i used a 12V light buld 24W, and fixed resistances at 1kOHM 2,2, 22, 47, 100 and 470 - i think i WAY over-did it on the whole result taking thing, hence my finalk write up being 18 pages...

 

[Padford]

i have sorted my problem! which i good! Thanks very much for your help in this piece, is have referenced this thread in my acknowledgments

Padfrd

 

[xanthym]

Excellent! (Msg #30 was meant to provide an alternative in case you were having major problems. However, if you indeed wrote 18 pages, then you obviously have lots of good results to talk about and have done a great job!)

The logarithmic relationship between Photodiode output Voltage and incident Light Intensity can be found here:
http://www.centrovision.com/tech2.htm
----------> Scroll all the way down the page to Section called "Photovoltaic Operation - Rl>>Rd, load line (a)"


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