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

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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.

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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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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.

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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

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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.

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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.

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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.

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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 ! :-(

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

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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, Iload = Vload/Rload.
Then, the internal cell resistance, Rint, = (Voc-Vload)/Iload.
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.

 

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

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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.

 

well at least I can its accomplishing something

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Never buy "Trust" products, all mine broke !!!