Hi Craig
On (23 Oct 97) Craig Healy wrote to Alec Cameron...
-> panel output.
-> So I connected a permanent diode in a lead from each of the panels.
-> Problem solved.
 CH> But you lost the .7 volt drop across the diode in the process...
That is NO great "loss". Solar panels operate at about 50% efficiency, to
charge a 12v battery you use a panel with an output of about 30v. To lose 
.7v
from a 30v source, is to "lose" just 3 percent.
In a practical set up this is negligible compared with the 50% loss 
rmitted.
The intentional waste of 50% of the generated energy, is in the interests of
maximum power flow. For every watt of electricity produced by the sunlight,
half a watt is lost as resistive losses and half a watt is sent to the
battery. You could run your solar panels at [say] 80% efficiency but the
output watts would be lower and more panels [and dollars] would have to be
committed to the project.
Cheers......ALEC
... Wunce i coodn even spll ingineer. Now i are wun!!
--- PPoint 1.92
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