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Ralph Mowery

In article ,
says...
Speaking of solar chargers, I have one charging the 6 volt battery in
my Ford 9N tractor. The panel is rated at 7 watts max. Here in the
great Pacific Northwest, with the cloudy days and all, The thing is
charging at about 17 milliamps. Open circuit voltage is 18 volts. But
when the thing is connected the battery voltage measures at 6.5 volts.
So I don't think there is any danger of overcharging unless the sun
goes nova and really lights the panel up. But I was wondering how to
calculate the wattage. is at 18 volts times 17 mA or 6.5 volts at 17
mA? I think I should use the open circuit voltage but I don't know.
Eric

YOu really need to load the panel with a resitive load. If it really is
putting out 7 watts,that resistor should be around 5 to 10 ohms. Then
measure the voltage and current.

You are just barley putting into the battery (when mostly charged) any
power at all. It would be the 17 ma times what ever the open circuit of
the battery is ( say 6.3 maybe) minus the charging voltage (6,5 in your
case) or about .2 volts. So .2 x .017 equals almost nothing.

If the battery was mostly dead you would have a much different ammount
of current if that solar cell is relly 7 watts. The cell is in very
bright sun is it not ?

I have a lawn tractor with a 12 volt battery and a panel hooked to it.
It has ran 2 winters hooked up for about 5 months at a time and no
problem so far and the tractor starts right up with a good spin the
first time of the mowing season. Water does not boil out over the
winter.