[SOLVED] Solar panel and ohm's law...

Hi,

I know it's just basic ohm's law but i don't understand this:
I found a 2w solar panel on Amazon that supplies 6 volts and about 330 mA. That 330 mA is when shorted so the voltage is 0.something volts which isn't enough to power something that would require for example 6 volts and 330 mA?

I know this stuff doesn't makes any sense but it's just so you know what i'm struggling with.


Regards,
Cecemel

20-30mA at 5-6v in a sunny day is a more real value

Voltage is greatest when A=0. However that means W=0.
Amperes are greatest when V=0. However that means W=0.

You'll get maximum W at some middle value for V & A.

Amazon sells No-Name-Brand junk same as ebay. A solar panel that produces 6V with no load and 330mA when shorted is probably 4.5V at 250mA which is 1.125W, not 2W.
I have a Name-Brand 12V at 275mA solar panel rated truthfully at 3.3W. It produces about 16V with no load and 370mA when shorted.

Of course there is no sunlight when it is cloudy and at night, and your solar panel probably does not "follow" the sun. There is less solar power in the morning and late in the afternoon.

Solar panels do not obey Ohms law, or have a fixed Thevenin equivalent to calculate from.

The problem is that unlike say a battery, the output voltage and current are not linear with increasing load, and the whole thing also varies with how much sunlight there is.

You are never going to see both maximum open circuit voltage, and maximum short circuit current together.

Another problem is there are different types of solar panel technologies, from very efficient to very crappy performance. So its not really possible to even guess what to expect. You will have to actually get one and test it.

The very cheap solar panels on some of my solar garden lights are sunburned (no kidding, the plastic turned cloudy) and are destroyed in a couple of months by rain leaking under them (not sealed).
Now I buy only solar garden lights that have glass solar panels and I seal around them with caulking.

Thanks everyone for the replies!

You need to understand the basic physics: every photon produces the same voltage but the current will depend on the total number of photons falling on the panel.

This is of course the ideal case but the point I am trying to say is that the panels are constant voltage devices and the current depends on the amount of light falling on the panel.

In reality, the panels have equivalent series resistance and that appears as a load when you short the panel. See post #3 above. When you short the panel, you get zero power but the energy produced by the panel is dissipated internally and the panel gets hotter. Maximum power is available to an external load when the voltage drop is about 1/2 the open circuit voltage.

If it is really producing 2W of power internally, it can deliver max 1W to an external load. Your mileage will vary (on the lower side).