Solar Pannel Power as a function of light intensity

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Solar Pannel Power as a function of light intensity

Postby jmki » Mon Dec 24, 2012 6:51 am

Hello,
I have just conducted an experiment involving 5 light bulbs ( of different intensities), a solar panel and a 50 Kilo-Ohm (ohmic) resistor in order to find out the relation of light intensity and the solar panel's output power. I hooked the resistor to the solar panel and I put a lamp (whose light bulb I changed several times) at a constant distance from the panel. According to what I have read, since the frequency of the incident light does not change, the voltage should stay the same but the current should change. However, what I got was that both. The current and voltage would increase, which makes sense since in order for the current to become bigger, bigger voltage is needed. However, my REAL PROBLEM is that the solar panel's power output was NOT directly proportional to the intensity of the light. The graph of power vs light intensity looks like a parabola instead of a strait line. Shouldn't the power be directly proportional to the light intensity?? Why am I getting these results? Have I done anything wrong?

PLEASE HELP ME as fast as possible.

Thank you for your time.
jmki
 
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Project Question: solar panel power output as a function of the intensity of the incident light.
Project Due Date: January 7th 2013
Project Status: I am finished with my experiment and analyzing the data

Re: Solar Pannel Power as a function of light intensity

Postby rmarz » Mon Dec 24, 2012 2:18 pm

jmki - I'm trying to understand how you were trying to measure power output from the solar cell when you had a 50KΩ resistor placed somehow in the circuit. If your 'load' was 50KΩ across the panel, you were reading virtually open circuit voltage (no load) across the cell. You didn't mention the nominal voltage output of the solar panel. But let's say it was 12 volts. If that's how the resistor load was connected, the current flowing in the 50KΩ resistor load would be 240 microamps, and the power would be about 3 milliwatts. The impedence of your source (the solar panel) and your load should be closely matched for peak power transfer. I don't think your setup for measuring or calculating power from your measurements is correct. Did you measure any current through various loads?

Let us know more about how you made your measurements, what loads you used, and how you calculated your results. If your panel was specified as a 12 volt unit capable of producing 100 milliamps in bright sunlight (the seller would want to advertise the best performance) you could infer that a realistic load resistance might be in the range of 50Ω to 200Ω (certainly not 50KΩ). You could easily calculate an optimized load value through experimentation. Good luck, we'll be happy to help when we know more.

Rick Marz
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Re: Solar Pannel Power as a function of light intensity

Postby jmki » Tue Dec 25, 2012 8:14 am

In the experiment, my goal was to find some relation between the intensity of the incident light and the solar panel's output power. Since I was just trying to find a relation between the two, I thought that it would make no difference if I used a different resistor. However my measurements are:

Lumens __ Voltage (V)__ Current (mA) __ Power (W)
170________11,42_______ 0,227________ 0,00259
405________13,23_______ 0,263_________ 0,00348
915________15,29_______ 0,304__________0,00465
1320_______15,9________ 0,318 _________ 0,00506
1980 ______ 16,6 _______ 0,331 _________ 0,00549

I would expect a linear relation between the power and the light intensity. However the results were not what I expected.

Please tell me if I have made any wrong measurements or if I should expect these results.

Thank you.
jmki
 
Posts: 2
Joined: Mon Dec 24, 2012 5:48 am
Occupation: student
Project Question: solar panel power output as a function of the intensity of the incident light.
Project Due Date: January 7th 2013
Project Status: I am finished with my experiment and analyzing the data

Re: Solar Pannel Power as a function of light intensity

Postby rmarz » Tue Dec 25, 2012 11:15 am

jmki - Your measurements and calculations all appear to be within range and accurate as to power dissipated by the 50KΩ load. The solar panel is really not working to produce power, however, and leads to a flawed experiment. It needs an appropriate working load from which to take readings and make your calculations. You are essentially just making 'no load' open circuit voltage measurements. I don't know how much current your solar panel is capable of producing, but it seems to have been manufactured to produce an operating voltage of 12 VDC (you measured over 16 VDC with your highest intensity light source). I would try various loads using, say, 68Ω resistors. Take three of them and take measurements as a single resistor, 68Ω, then add an additional resistor, in series, one at a time, to give you several different loads of 68Ω, 136Ω, and 204Ω. This should put your panel into an active working zone, producing real power, so that you can make your measurements and calculate power. Use at least a 2 watt rated resistor for the first, single resistor test. The rest can be 1 watt. The experiment will also demonstrate the fact that your source and load impedence selection, or matching, will affect optimum power transfer. For example, each different bulb will produce a different power output with different loads. Good luck, report your results if you can. As an aside, you expected a linear relationship between illumination and voltage/power. There are few purely linear outcomes in the real world, so expect some non-linear readings in your experiment.

Rick Marz
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