# COMMUNITY : FORUMS : GREEN

## Math help needed for converting volts/watts in solar

Here's what I know: Watts (P) = Amps (I) x Volts (E). What I don't know is if I have a solar panel that is rated at 12v, 0.125 mW I can do the above equation to get that my amps are 0.01. Does that sound right?

My real Q, is what is the best power battery to store this energy? Should I only use 8 "D" cells? Could I use a 12 v motorcycle battery?

If I use Ni-Cd batteries, is the fact that the batteries are rated at 2500 ma make any difference?
Is there a website where I can get a good feel for what batteries are good for what type of activity, solar cells, etc?

thanx, sam

verence6 years ago
Hi Sam!

> I have a solar panel that is rated at 12v, 0.125 mW I can do the above equation to get that my amps are 0.01

No, if your panel is rated 12V, 0.125mW then you will not get 0.01A, but 0.01mA which is 10µA or next to nothing.

If your panel is rated 12V, 0.125W (or 125mW) then you will get 0.01A or 10mA which is not much.

Loading a 12V accumulator will not work. First, to load a 12V accu, you need a higher voltage than 12V. Second, the rating of 12V of the panel does not mean, that it will produce 12V under any kind of lighting. Most probably, the panel will get its rating only under strong light. Otherwise the voltage will drop. Therefore
you will need a series diode to block the accumulator from discharging through the unlit panel. This diode will eat up 0.7V (silicon) or 0.3V if you can find a germanium diode. Finally, you will need any means of regulating the current into the accumulator and or the maximal voltage when the loading should stop. This depends on the type of the accumulator.

Just as a theoretical thought, ignoring all losses, loading strategy concerns etc. if you have an accumulator of 2500mAh (i guess that's what you meant by
rated at 2500 ma - please get the units right and check for correct casing) and you try to load that with a current of 10mA (using the better case of the 0.125W panel), that will take 2500mAh/10mA=250h or more than 20 days of 12 hours of bright sunlight each. And as said that is ignoring all the losses.

If you have the stuff at hand, you could try just with the series diode and a small 6V lead accumulator, but I would not guarantee for a stable operation.

Keep in mind that any accumulator loses charge, even when not in use. So it could be that more charge is lost during the night as gets recharged by day, especially in low light periods e.g. winter.
wanzie (author)  verence6 years ago
Thank you for taking the time to provide a reality check. I assume when you say "accumulator" you mean it as a general term for a battery? If so, this is helpful especially the "2500mAh/10mA=250h". And units are one of the things I haven't nailed down yet.

The panels produce 18-19 v in peak sunlight, and do pretty well, for what they are. I'll look for a germanium Shottky diode. I'm wondering if I can use the math to lower the voltage and proportionally increase the amperage? I would love to get 5-6 volts. Perhaps a circuit with a 7805 voltage regulator? would this circuit do what I would need?

http://peterflemming.ca/flemmweb_current/schem/opamp_linear/7805.pdf

Thanks for your time, sam
6 years ago
With accumulator, I just mean any rechargeable energy storage. A battery might be rechargeable, but might also just go KA-BOOM.

Yes, it is possible to convert a high voltage/low current source into a low voltage/high current source, keeping P = U1*I1 = U2*I2 constant while U1>U2 and I1
A linear regulator (78xx, 317, any OP-Amp..) will lower the output voltage, but the current is constant. The energy in Udifference*I is just convert into heat in the regulator, that is why these things often have a heat sink.

Anyway, what do you want to power with the panel?
I mean, the normal way of development is to have a problem, design the circuit to solve it and then design the power supply for the circuit. Starting with a power source for a possible power supply for an unknown circuit is hard.

Finding a problem for a solution is harder than the solution to a problem.
6 years ago
Yes, "accumulator" could be a battery, or a set of batteries, or even capacitors.