On 2014-10-26 02:05, jdebert wrote:
Because all batteries are not the same. When discharging deeply, one will go flat before the other and may be reverse-charged.
Huh. Reverse charging happens when connected in series, not in parallel. (-) bat 1 (+) (-) bat 2 (+) 50% -------- 80 % ---> current flow, 1 A on all cells, in series. later: bat 1 bat 2 0% ---- 30 % ---> current flow, 1 A on all cells, in series. bat 1 bat 2 -10% ---- 20 % ---> current flow, 1 A on all cells, in series. This is the normal configuration, by the way. Unavoidable.
And it is not resistance but difference in potential that causes one to discharge through another.
When in parallel, one cell discharges on the other, charging it, till they are at the same potential, and, probably, at the same charge level. Energy is not "lost", just redistributed, save losses: (-) bat 1 (+) +--- 50% -----+------ --> external current flow = 0 | | (standby) | | | bat 1 | internal current flow +--- 80% -----+----- anti-clockwise (-) (+) later: (-) bat 1 (+) +--- 60% -----+------ --> external current flow = 0 | | (standby) | | | bat 1 | +--- 68% -----+----- (-) (+) later: (-) bat 1 (+) +--- 62% -----+------ --> external current flow = 0 | | (standby) | | | bat 1 | +--- 62% -----+----- (-) (+) (invented loss figure) What happens if you charge them and they are slightly different, is that one will charge earlier than the other, and then start to overcharge, if you charge fast enough. If you trickle charge, nothing (bad) happens. However, classic lead acid batteries could stand some overcharging. They just heated a bit, and gassed. You just replaced the water lost and done. With extra-fast charging you could destroy them instead. -- Cheers / Saludos, Carlos E. R. (from 13.1 x86_64 "Bottle" at Telcontar)