Hi Craig
On (12 Nov 97) Craig Healy wrote to Alec Cameron...
 CH> That 30 volts is open circuit, no?  So if you charge a 12v battery,
 CH> you need about 13.7v at some practical current flow.  All this current
 CH> flows through the diode.  Say you have a 5 amp current.  Five amps
 CH> times the .7 volts (Ohms Law power formula) is 3.5 watts.
This is a tricky one, a guy called Kirchoff would explain it better than me
[Kirchoff's Laws]. The 0.7 volts is an emf [electromotive force] not a
resistive drop. The diode behaves *like* a wee battery and series switch,
connected in series opposing the battery voltage. So what is "lost" is 0.7 of
the 30 volts available on open cct. .7 div by 30 =   2.33%. Less than a wee
cloud passing by.
If you find the concept of 0.7 being an emf unconvincing, then put a 
oltmeter
across the diode while increasing/ decreasing the panel output eg by shading
it. You will see that the 0.7v does not change.
 CH> If you are looking to wring every last bit out of the system, then
 CH> removing the diode and substituting some switch-type system may
 CH> help.  As I said, I'm not really satisfied with less than 101%
 CH> efficiency.  
That could make you very unhappy!! I am real happy with 50%
Cheers....ALEC
... ....Horsepower was a wonderful thing when only horses had it
--- PPoint 1.92
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