Voltage-Current Characteristics of a Solar Cell

[iwfc87]

Hi, I'm doing an experimental write up report based on an experiment I did at school and the title is Voltage-Current Characteristics of a Solar Cell.

 

What we did was to use different resistors of different resistors and put in a circuit powered by a Solar cell and then record the voltage and current of the circuit.

 

What I need to know is:

1. What is the power delivered to the resistor and how can that be calculated?

2. How would you calculate the internal resistance if I graphed it out?

3. If i drew a graph of power delvired to the resistor (y-axis) and Resistance (x-axis), how would I estimate the resistance for which the power is delivered is the maximum?

 

Thanks a lot.

[Kedas]

1) Power = current * voltage.

 

2) theoretically if you make a graph voltage-current and you put all your resistor measuring points on it then you draw a line trough those point. that line will cross the voltage and current axis.

if you take tose two values then you have Rinternal= voltage/current

The voltage should be the voltage that you measure without resistor attached, the current should be the current when you short-circuit the the power-cell.

 

 

3) The Rload power is at his highest when Rload = Rinternal.

I can prove that but then you will have to pay me. LOL

 

I have a feeling I'm doing your homework ?

[YT2095]

also bear in mind that solar cells are more current stable than they are voltage stable!

 

so you will also need a graph for your cell responses at a constant light over load in order to work it out accurately

[Kedas]

of course, everything I said is only valid when the energy source (light) is constant.

[iwfc87]

Lol..yes and no.

 

I've been trying to think about it but I can't seem to figure it out, and sometimes I don't understand the question/s.

 

Sorry bout that!

But thanks a heap though. Makes things a lot easier to understand!

[iwfc87]

Hey, I'm back again.

 

I've done as Kedas recommended in his reply to my 2nd question, and some how my calculated internal resistance is very high.

 

If I then tried to apply this internal resistance to V t= EMF-Ir, where Vt is terminal voltage and Ir is internal resistance, I get a negative terminal voltage because my Ir is very large.

I've taken the EMF as the voltage intercept of the graph.

 

Thanks.

[YT2095]

ok, across the solar cell, put a voltage meter across the pos and neg terminals, leave it there, that will read the voltage suplied.

 

now, from your pos terminal attatch a current meter, then the other wire from the current meter to a known value of resistor (1 Ohm) would be a good place to start, the neg lead from your cell attatches to the other side of the resistor, making a circuit.

 

from here you will be able to see both voltage and current. take these results and plot it on your graph. then use a different resistor value, maybe 2 ohms or 5 ohms, and plot that on your graph, do this with as many different values as you like, then join the dots

[Kedas]

You will also have to bring the internal resistance of the ampere-meter into account. To check measure the voltage before and after the ampere-meter.

 

Also try to be clear if you give a formula.

use an . or * like Vt=EMF-I.r

what are the values that you have to determine the internal resistance? (values that cross the axis)

[YT2095]

or the voltage across the current meter and do a simple PD calc between that and the suplly voltage, either way will tell you the meters internal resistance if it doesn`t have documents with it

[Sayonara]

BTW, I changed the title of this thread to make it more user- and search engine-friendly.

 

Hope nobody minds

[iwfc87]

Sorry for being so unclear in writing formulas etc.:S. I'm pretty much a newbie

 

I think I've solved the error now. I just got confused with the formula because the information sheet I got made it look confusing! WHoops >_<.

 

THanks a whole lot for everyone's help but!

Your all legends!

[iwfc87]

Oh..my values were 23.54 when crossing the Voltage axis, and 0.0817 when crossing the Current axis.

 

Hence using the formula as advised, my Rinternal= 288.13 (this is pretty large if u asked me)

 

Now if I use VT=EMF-I*r,

VT= 23.54- (0.0817 * 288.13)

VT= -2.21 EXP-4

 

Would this be right? It doesn't seem so to me, but it could be :S.

 

Thanks a lot

(Sorry for bugging you guys so much. I enjoy physics, just takes me a while to understand it!:S)

[Kedas]

VT = -2.21 EXP-4 = -0.000221Volt it isn't zero because you rounded the value of the resistor.

 

Are you sure you had 23.54V what was it when you just measured the voltage on the solar-cell without resistor.

 

BTW the equation VT=EMF-I.r is the mathematical representation of that line you drew. Y=EMF-X.r

[iwfc87]

So if my line drawn has the equation of y=-287.59+23.547x, what does that mean?? *confused*

 

Thanks.

[Kedas]

with Y=EMF-X.r I mean

X is the current through the resistor.

Y is the voltage over it.

 

VT=23.547 - 287.59 * I

VT = voltage over the resistor

I = current through the resistor.

 

example: your current is 0.02A

Then your voltage is about 17.8V

(assuming that the values that you give are correct)

 

 

your equation y=-287.59+23.547x doesn't make much sense.

[iwfc87]

So the current can be different values?

 

So if my current is 0.063A (it's one of my values when a 12 ohm resistor is used), my terminal voltage would be

 

VT=23.547 - 287.59 * 0.063

VT=5.43 V

 

I think I understand it now!

 

Thanks a lot for all your help!

[Kedas]

Yes, but 5.43V/12ohm = 0.4525A not 0.063A

meaning that that line you drew isn't going through that point ??

how many points did you have and could you more or less connect them with a straight line?

 

maybe could you just give me all your measuring points?

R= V= I=

R= V= I=

 

etc.

[iwfc87]

yeah i know..

 

I was getting the results off the digital multimeter, and it was weird.

 

Though there could of been various errors that caused this, such as the resistor's tolerance of 5% or the sunlight's intensity and colour not being constant causing the results to be different.

 

R= (Ohm) all ±5% V= v Current=mA

12 0.76 63.9

33 2.11 64.3

100 6.22 63.3

220 12.95 59.8

470 16.06 32.2

1000 16.82 17.1

 

So yeah..I got the voltage and current readings off the multimeter. I don't know whether I should use the multimeter's reading or my calculated reading, but I have a feeling I should use the multimeter's.

 

Thanks.

[iwfc87]

whoa..that didn't come out as i expected.

but the 1st column is Resistance

2nd is voltage and 3rd is current in mA

[Dave]

I would say those results are pretty much screwed up. I just plotted them and they really suck, I have to say. You need to ensure that the intensity of the light being shone on the cell is constant or the entire results set don't come out right.

[Kedas]

iwfc87 said in post #18 :

I don't know whether I should use the multimeter's reading or my calculated reading, but I have a feeling I should use the multimeter's.

Thanks.

 

What you calculate is only based a mathematical representation, you should useally always believe what you read on your measuring device.

 

I did draw your graph and I see that the internal resistance of the solar-cell isn't constant. I did some checking because I have no experiance with solar-cells and your graph is correct.

 

Now the question is which resistor does have the max power well that is where Rload=Rinternal and that is where the curve/line changes direction that is in your case around 220ohm.

 

for Rloads above 220ohm the solar cell has a small internal resistance, for values below 220ohm the solar cell will have a high internal resistance.

 

The slope of your curve defines the value of the internal resistance.

 

So forget about the values that cross the axis because that is only valid when the internal resistance is constant. (you can use that methode if you don't mix the values above and below 220ohm)