I would like to increase my RV marine deep cycle battery capacity. I currently have a Group 24 battery with 64 amp hours. I would like to add a second battery to increase my capacity to the max. As an example, I could add a group 27 battery with 88 amp hours. That would give me 152 combined amp hours. What is the safe 50% level for depletion? 50% of 152 would be 76. However, I think 88 is the correct value as I do not want to go below 44 amp hours on the capacity battery.
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1\$\begingroup\$ When batteries are combined in parallel, they do not necessarily discharge with equal currents. However, if their chemistries are well-matched, they will combine well and deplete at roughly the same rate. Do you have more details on the battery specifications? (alternatively you can use the two batteries separately one at a time to power the RV while tracking their discharge independently, but that's more work on your end) \$\endgroup\$– nanofaradCommented Jan 6, 2020 at 20:16
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1\$\begingroup\$ On my calculator, half of 152 is 76, the sum of 50% capacity of each battery. However, batteries of different ages may not share well, even if they are exactly the same chemistry. 50% is not a 'safe' level for depletion, there is no cliff edge, 45% is better, 55% is not much worse. \$\endgroup\$– Neil_UKCommented Jan 6, 2020 at 20:17
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\$\begingroup\$ What everyone is trying to tell you is that this is generally a bad, and possibly dangerous, thing to do. You would need some kind of battery management system (BMS) to safely combine the batteries. \$\endgroup\$– Elliot AldersonCommented Jan 6, 2020 at 20:22
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2 Answers
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Assuming that you are directly connecting the batteries in parallel, and the cables you use are adequate, the two batteries will always have pretty much the same voltage, and if they're the same chemistry they will discharge to the same level, and basically the whole combined capacity will be usable. If you take 50% of the combined capacity, each battery will end up at approximately a 50% state-of-charge.
Yes, the first approximation is that if the batteries have the same voltage, then they will provide equal current to the load, and that will cause the smaller battery to discharge more quickly. But, a battery's voltage lowers as it discharges, and any miniscule difference in voltage between the two batteries will cause a current to flow from the higher-voltage battery to the lower-voltage battery, through a very small resistance.
When you add the current from these two effects, what you see in the steady-state is that each battery will provide current to the load proportional to its capacity, because they're all trying to have the same voltage droop, and that any imbalance between the batteries will only be short-lived.
Here is an article from a battery manufacturer on "parallel operation of lead-acid batteries". (Actually series-parallel, but there's nothing to prevent a "string" of one battery). They state
However, there is no technical reason that the individual parallel strings must all be of the same capacity. When battery strings of unequal capacity are connected in parallel each string will deliver power and accept recharge current proportional to its percentage of the total combined capacity as shown in Figure 3.
Their only recommendation is that each string (each battery, in your case) has an individual circuit breaker / fuse to protect against overcurrent conditions.
There is also a question on this SE from 2011 about charging lead-acid batteries in parallel, although it pertains to SLAs, and not marine wet cells.
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In addition my concurrence with @hobbs
I support your endeavor to reduce DoD to 50% but also prevent overvoltage.
The stronger battery has lower ESR and will provide most of the surge currents and possibly the older battery will age slower as a result of tandem connections with a new battery. (unless it develops a cell short and drains the other battery)
I suggest you consult Mfr.
50% SoC between 12.05V & 12.11V @ 25'C with light or no load.
AGM batteries have a small negative temperature coefficient on voltage & ESR but also age much faster with elevated temperatures.
Internal battery resistance,Ri ought to be < 5 Ohms so current ΔV=I*Ri should indicate the change in voltage after some settling time.
