# measuring power of solar panels

Discussion in 'Renewable Energy' started by Thunderchild, Aug 24, 2008.

1. ### ThunderchildNew Member

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how would I determine the power output of an unknown solar panel ? if i just measure the voltage its going to be higher than normal but when measuring current the voltage will drop to nout !

so what do I do ?

one clue may be that I can count the number of "junctions" in the panel or cells

2. ### ericgibbsWell-Known MemberMost Helpful Member

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hi,
If its a 12V panel, do you have a 120R 2Watt resistor [ thats 100mA]

or a 12V cycle lamp, say 3Watt, thats about 250mA.

Problem today is the lack of Sun.

3. ### ThunderchildNew Member

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yes on this particular day sun is hard to come by well i have many resistors I'm assuming I'm dealing with 6 V panels

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5. ### ericgibbsWell-Known MemberMost Helpful Member

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hi,
It should be easy to see if its a 6V or 12V nominal panel.

Choose a resistor that will pass about 100mA for starters and measure the output voltage at the same time.

6. ### ThunderchildNew Member

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ok I'll do that I have 2 panels with 15 strips/junctions rated at 7 volts and these have 10 junctions so i presume I'm looking at about 5 volts and I think the output is approximately 100 mA I'll give them all a go when the sun comes out

7. ### ericgibbsWell-Known MemberMost Helpful Member

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A quick test would be a tungsten table lamp, you might wait a long time for some Sun today.

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Corrected.

9. ### Nigel GoodwinSuper ModeratorMost Helpful Member

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But bear in mind the output from a solar panel under those circumstances is only a tiny fraction of what you get from the sun.

Do you remember the robot school olympics programme that used to be on TV?, and how massive the lights were for the solar car event.

10. ### ericgibbsWell-Known MemberMost Helpful Member

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hi,
I was suggesting a way on how to determine if they are 6V or 12V panels, the OP isnt sure.
Of course I agree, ref the high intensity of artificial light versus current output.

11. ### ThunderchildNew Member

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well eric its more of a case that they are "custom" panels they come from equipment like i said one is probably 6 volts but the other compared to that must be 5 ! :-(

12. ### ThunderchildNew Member

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corrected again we are having some sun here (just a bit don't blink or you will miss it)

13. ### WillbeNew Member

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". . . determine the power output. . ."

You measure the open circuit panel voltage and call it Voc.
You put a load resistor, Rload, onto the panel with a resistance low enough to drop the voltage 10% to 50%, and you measure the voltage, Vload, across the resistor.

Then, the maximum power obtainable from the cell is achieved when it is loaded down by a load resistor of value equal to Rint.

(Voc and Rint equals the Thevenin equivalent voltage & Thevenin equivalent resistance).

If the panel doesn't mind being shorted, just measure the Voc and short circuit current Isc, then Rint = Voc/Isc.

Rint will probably vary with incident sunlight intensity; for sure Voc will.

Last edited: Aug 24, 2008
14. ### ThunderchildNew Member

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well then Rint is 87.5 ohms for the smaller panels and 115 for the slightly bigger ones but I'm not sure i can calculate Rint as Voc/Isc I think I'd best go the longer way round

thanks for the calculations

15. ### WillbeNew Member

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So with Voc = 6v and a 100 Ω load resistor you'd get (3^2)/100 = 90 mW, enough to power an LED.

16. ### ThunderchildNew Member

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well at least I can its accomplishing something

Last edited: Aug 25, 2008
17. ### awrightNew Member

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I agree with Willbe's comments, but a little more intuitive and simplistic approach is available.

The easiest way to evaluate panel power is, of course, to measure it directly with a wattmeter as you vary the load with a potentiometer or resistor load bank having sufficient power/current rating. Observe the load current or the load voltage or the load resistance at which the displayed power is maximized as you vary the power resistor.

Although I have been lucky enough to pick up a few wattmeters surplus from a local university, unfortunately very few of us have wattmeters, the passive, analog form of which are instruments with meters that resemble the familiar d'Arsonval (pointer) type analog meter except that the magnetic field is created not from a permanent magnet, but from a fixed coil that carries the load current. The moving coil is energized by the load voltage. The result is that the pointer displacement indicates the PRODUCT of the load voltage and the load current - i.e., the power. The scale will typically be very non-linear. There are also wattmeters that employ active analog or digital circuitry to compute and display the power.

Lacking a wattmeter, you can determine the peak power point by measuring the load voltage and the load current and plotting the product of the two vs. load resistance on a chart. At zero load resistance (short circuit) the current will be maximum, the voltage will be zero, and the product (power) will be zero. At infinite load resistance, the current will be zero, the voltage will be maximum, and the product (power) will again be zero. As you vary the load resistance and plot the product of voltage and current at several points, you will find a maximum in the power output curve somewhere near the mid point of voltage and current. Ideally you would let the panel reach steady state thermal conditions at each new point but getting data at nearly constant solar irradiance requires that you take data quickly.

One way to obtain a first-order correction factor for varying solar irradiance during your measurements would be to plot simultaneous readings of voltage or current from a properly loaded separate solar panel or even a tiny solar cell like you can (or used to be able to) buy at Radio Shack that is held at constant conditions and orientation.

Have fun.

awright

18. ### ThunderchildNew Member

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thanks awright that's a very good idea

19. ### rajieNew Member

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I am currently involved in project of Comparison study of Sensor and using Genetic Algorithm for Optimum Power for Solar Panel. I’ve developed a mini prototype, which includes a mini solar panel and PIC16f877a microcontroller and a motor as part of the tracking system.
For the first part of my project using sensors, the panel moves accordingly to the sensor values. (determine which is higher)

For the second part of my project using Genetic Algorithm, I would need a method to measure the voltage/power output so that the Genetic Algoritm could compute the accurate tilt of the panel, and move accordingly.

My problem now is, is there any method for me to measure the voltage/power output?
If P = V*I and
P=V2 / R

How could I measure voltage/power from mini panel? Let’s say we need 4 points from the panel. Is there any method?

20. ### awrightNew Member

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Unless you have a very constant load resistance, your second formula will be of limited value in calculating power due to variations in load resistance. Measurement of both current and voltage and computation of the product will give you the instantaneous power, but that seems like a poor parameter upon which to make the decision of panel orientation. Variations in actual power output not due to orientation, but rather due fluctuations in loading or insolation, could lead to the wrong decision regarding optimum orientation.

awright

21. ### ronp849New Member

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Maybe the apparent internal resistance could be used as a Sun Seeking parameter. This resistance would appear to increase as the panel went off track.

This could be measured by rapidly changing the load applied to the panel, (switch in a known resistor value) and measure the new terminal voltage. The voltage difference could give an indication of the panel internal resistance.

Ron