-> -> 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
-> from a 30v source, is to "lose" just 3 percent.
That 30 volts is open circuit, no?  So if you charge a 12v battery,
you need about 13.7v at some practical current flow.  All this current
flows through the diode.  Say you have a 5 amp current.  Five amps
times the .7 volts (Ohms Law power formula) is 3.5 watts.  If that is
negligible depends on what you consider negligible.  I will be the
first to admit it isn't a major factor.
If you are looking to wring every last bit out of the system, then
removing the diode and substituting some switch-type system may
help.  As I said, I'm not really satisfied with less than 101%
efficiency.  
-c-
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