I am setting up a solar panel to charge a 12 VDC lead/acid battery to run a trolling motor for my dinghy. Does anyone have an idea as to how much leakage current there is through a 12VDC 8.5W panel when dark? Thank you.

  • \$\begingroup\$ Next to none if you include the diode that you should have in there to stop leakage current... \$\endgroup\$
    – Majenko
    Commented Jan 16, 2015 at 20:11

1 Answer 1


Solar cells are diodes. They produce voltage when illuminated such that the current is flowing backwards thru the diode. This means when dark, a cell just looks like a forward biased diode to any voltage applied with the same polarity as the diode produces when illuminated.

This is also why there is usually a Shottky diode in opposite polarity across each series cell of a solar panel. When dark, a cell will act like a diode, which will block the light-caused current of a panel. A single dark cell in a series string is enough to kill all the current. If enough cells are in series, the remaining cells can cause enough voltage to exceed the breakdown of the dark cell and cause damage. The Schottky diode across it also prevents this from happening.

Solar panels therefore must have a diode in series with them if directly connected to charge a battery. This is just normal operation of a solar panel, and any real installation will include this diode.

For low voltages like a 12 V battery, the diode can be Schottky, which has a lower forward drop and therefore wastes less power when the battery is being charged. Schottky diodes have higher reverse leakage current the full junction diodes, but that will still be small compared to the self-discharge current of a typical "car" lead-acid battery.

  • \$\begingroup\$ FYI only. My "model" at very low light is still incomplete. Consider PV panel directly connected to a battery, with ample PV voltage for charging in daylight conditions. Under very low levels of illumination (say moonlight) the N forward PV potentials will be << less than the typical battery voltage and you will get reverse panel current at a low level.This reverse current may be >> the Isc of the panel under those conditions. eg 30V, 250 W poly Si panel. Torchlight - a few lux. Voc ~= 1V. Isc ~= 0.25 mA. I reverse at Vbat ~= 28V = 85 mA. Reduce illum to ~= 0 lux and Ireverse about the same. \$\endgroup\$
    – Russell McMahon
    Commented Jan 17, 2015 at 10:43
  • \$\begingroup\$ ie Series diode protects against low but non trivial battery discharge. In this case about perhaps 1 Ah overnight. Behaviour without diode as light rises uncertain. Must check. \$\endgroup\$
    – Russell McMahon
    Commented Jan 17, 2015 at 10:45
  • \$\begingroup\$ Blip from torch used to induce Voc = 1V - plus 24 hours of other stuff. \$\endgroup\$
    – Russell McMahon
    Commented Jan 17, 2015 at 11:16

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