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Diode or not diode? | 24 comments (24 topical, editorial)

Re: Diode or not diode? (3.00 / 0) (#10)
by johnlm on Thu Apr 19th, 2007 at 05:23:08 PM MST
(User Info)

Since you asked specifically how much loss would there be...

Lets assume you actually get about 4 Amps of current out of each panel. If you put another diode is series (assuming a silicon diode that has a forward drop of 0.7V) then you will be losing 0.7V X 4 amps = 2.8 watts of power that would have been going to help charge your battery. This might vary some contingent on the maximum power point, the actual drop across the diode etc, but it gives you an idea of the loss. Using a Schottky diode would cut this loss in half.

Johnlm

Re: Diode or not diode? (3.00 / 0) (#19)
by la7qz on Fri Apr 20th, 2007 at 10:41:11 PM MST
(User Info) http://home.no.net/naomij

Thanks for this.

Makes perfect sense of course, and if I had employed my brain I would have been able to figure that out for myself.

Regards

Owen Morgan
Yacht Magic
Falmouth Harbour, Antigua

If you're not living on the edge, you're taking up too much space.
[ Parent ]

Re: Diode or not diode? (3.00 / 0) (#20)
by ghurd on Sat Apr 21st, 2007 at 01:57:01 PM MST
(User Info)

Not quite right.
If the battery is charging at 13.5V, with 1V line drop, and 0.7V diode, then the PVs are at 15.2V.
The PV amps are almost the same at 15.2V or 14V. With the charging amps being the same, the power into the battery is the same.
That's why they are made for peak power at 17V, diode losses, etc.
It won't hurt.
G-

[ Parent ]

Diode or not diode? | 24 comments (24 topical, 0 editorial)

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