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Ok, so wiring batteries in parallel of different voltage is bad because the batteries try to equalize voltage with high currents essentially by discharging. The thing is, even among batteries of the same type, the voltage is slightly different. How far apart do the voltages have to be that I should consider not paralleling them?

So, an example: I have 3 lead acid batteries (fully charged) that are from the same place, same make, model, and voltage (kind of, they're supposed to be 12V each). They have: 12.75V, 12.78V, and 12.85V. This is clearly a range of voltages, but does the .03 drop from one or the .1 drop in the most extreme case matter much?

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  • \$\begingroup\$ Normally SoC levels are 12.5 to 11.5 after a brief load . Did you load them at all to discharge the absorption voltage during charging. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 26 '16 at 2:50
  • \$\begingroup\$ @TonyStewart No, so far they're fresh off of the charger. I don't even really know about absorption voltage. I'm guessing it's a temporary heightened state that dissipates quickly? \$\endgroup\$ – Sarah Szabo Aug 26 '16 at 18:51
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That small difference should be fine.

I don't think I'd want to charge a gel-cell or AGM in parallel with a flooded battery, but if yours are same make and type there should be no problem.

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When you connect two batteries with different voltages, you'll have a current \$I=\frac{\Delta V}{2*R_{int}}\$, assuming both batteries have the same internal resistance. If this current value is under the maximum charge/discharge current, it's OK to connect the batteries.

As a rule of thumb, \${\Delta V}\$ should be less than the voltage drop (\$V_{unloaded}-V_{loaded}\$) under nominal load, minus ~50% safety margin.

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Consider using big powerful diodes at the output of each battery. This way you drop the output voltage of each battery by about 0.7V, but only the diode leakage current (usually in microamps) will flow from one battery to another...

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  • \$\begingroup\$ Care to explain the downvote? \$\endgroup\$ – nurchi Aug 29 '16 at 16:27
  • \$\begingroup\$ I see two issues here: losing ~5% efficiency for no good reason, and potential difficulty to find the right diodes (~500A in case of car batteries) \$\endgroup\$ – Dmitry Grigoryev Feb 7 '18 at 13:38
  • \$\begingroup\$ A downvote is perhaps a little harsh, but this doesn't really answer the question, and the diodes will waste a decent amount of power. \$\endgroup\$ – pipe Feb 7 '18 at 13:38

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