Choosing a resistor for balancing lead acid batteries

Question

I have a solar-panel-and-battery kit with a built-in inverter and what not built into the battery box. (Goal Zero Yeti 400) The battery box has Anderson Powerpole connectors for chaining it to external batteries. As far as I know, these are connected directly to the battery inside.

The battery built-in is a 33 Ah 12V sealed lead acid battery.

Suppose I'm out camping, and someone joins me and brings another 12V, perhaps of different capacity, and at an unknown charge level. How can I safely connect these batteries together in parallel?

I assume that if the voltage difference between them is too great, the current flow between them will be a lot and potentially damaging to one or both of the batteries. Therefore, I'm thinking that some sort of resistor is called for, even if just until they're balanced.

Is this a good idea? And if so, how do I choose what resistor to use in this case? I suppose I could use something of very high resistance and let the battery trickle up to the right voltage over a long time, but it would be nice if it didn't take days. What sort of current can these batteries take before they become damaged?

Answer 1

The ideal way to transfer charge between two batteries with substantial voltage differences is with a buck converter, with efficiency able to be in excess of 90%.
However, at low voltage differentials the use of a resistor may be more efficient!
Power input = Vin x I. Power Output = Vout x I.
Efficiency = Pout/Pin = Vout/Vin.
For eg 12.5V in and 12V out, efficiency = 12/12.5 = 96% !

Use of a resistor is better than "just a thick wire" for safety purposes but regardless of what form of resistor you use you will dissipate energy as heat in the process.
The resistor value could be set based on known conditions, or a worst case value could be used.

Rminimum = (Vsource - Vtarget) / Imax_allowed.

If you have two batteries with say 12.8V (about fully charged in rest state) and 12V terminal voltages and can tolerate say 10A current then R = V/I = (12.8-12)/10 = 0.08 Ohms.

ie enough "thinnish" wire may suffice.
Some prior experiment (or measurement) could allow you to size a wire that limits current to some maximum value and does not get too hot (or glow in the dark) in the process.
Something around a few 10ths of an Ohm all up (wire, clip connections, contact resistance) would probably suffice.

The danger is in using eg jumper leads with batteries of very different charge levels.
Many 10's of amps could flow and, very worst case - but unlikely, 100 + A.

One possibility is a low voltage light bulb - say a 3V bulb (although where you'd get one of adequate current rating is hard to imagine). As current increases bulb resistance will rise and maintain "a degree of" current limiting.