How do I design a new battery system using two old but large AGM batteries in series?

I realize a load test must be done to determine capacity differences.
AGM batteries are also sensitive to overcharging, and voltage is temperature sensitive.

What are the key critical design specs to define? I can follow this to implement a reliable design.

I have a floor scrubber that has four 6V flooded lead acid Trojan T-125 batteries. Problem is after using it to scrub the floor, it takes about 4 hours to recharge using the onboard 21A max charger. Using a more powerful charger on them is not recommended as the batteries will heat up excessively and likely shorten their life. I have a candidate solution though and wanted to know if it will work and what problems may arise. At home I have 2 large (but identical) Odyssey PC2250 12V AGM batteries that are quite old (at least 10 years) but have a rated capacity of 126Ah. They do not hold identical charges and one has slightly higher actual capacity than the other but they have not been in service for many years but I maintained them. They usually hold 12.7V OCV (Open Circuit Voltage) which tells me they are still in decent condition.

So here is my plan and I would like to know if this will work or if I am headed down the wrong path. I will do a load test on each of these batteries using my 12V power inverter and 500 watts of incandescent lights as the actual load. I will simply measure the duration of the test. The inverter will beep when the load voltage drops to about 10.5V. At that point I will shut off the inverter and put the battery back on the charger for a full charge. I will do this same test for each battery solo and probably twice for each battery to confirm the results. So assuming they are both similar Ah capacity (let's say the test lasts 1 hour each which would be about 42Ah each since 500W = about 42A at 12V), would it be ok to put these back into service but with a new task as floor scrubber batteries? To get a quicker charge, I would use an external charger with 55A charge capability which will far exceed the onboard chargers maximum 21A rate. I expect the actual charge time to go from about 4 hours on the old system to maybe 1.5 hours on the new system.

  • 1
    \$\begingroup\$ I must be missing something. The set of T-125 batteries has a nominal capacity of about 200 A-h. Although for some reason, you seem to be recharging them with only about 80 A-h. Are they old, too? What would be the point of replacing them with a set of old batteries (nominal capacity 126 A-h) that has maybe 75 A-h of capacity left? \$\endgroup\$
    – Dave Tweed
    Sep 12, 2016 at 15:37
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    \$\begingroup\$ Can you use two 12 V chargers, one for each battery? That will help balancing the batteries and you can push more power into it in two hours. \$\endgroup\$
    – winny
    Sep 12, 2016 at 17:38
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    \$\begingroup\$ @Passerby - It would be too much work for the people that use the floor scrubber to swap out the drained batteries and swap in the fully charged ones. The main advantage of AGM technology is they can be fully charged within 1 hour (2 hours is better). I actually tested this one time. I drained a 12V 14Ah AGM battery down to 10.5V under a reasonable load which is considered nearly full drained. I put it on a 25A charger for 1 hour and although it was warm to the touch afterwards, it run the complete charge curve to 100% SoC. I was impressed. 2 hours at 40A for the big batteries might work. \$\endgroup\$
    – David
    Sep 12, 2016 at 17:38
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    \$\begingroup\$ @winny - I considered this idea such as using a pair of 12V 25A chargers but a problem there is extra complexity for whoever uses the floor scrubber. They would have to be keen enough to connect to the proper terminals to get that to work. I do use this technique on a 36V golf cart because it seems to charge better than the 36V charger but I have experience and know what I am doing (somewhat). I still think the pair of large AGM batteries with a 24V 40A charger is a good candidate solution. This is for indoor use so AGM is a better choice cuz FLA releases fumes and has higher maintenance. \$\endgroup\$
    – David
    Sep 12, 2016 at 17:42
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    \$\begingroup\$ Good. You seem to know what you are doing and the associated trade offs. Having worked with battery chargers for the majority of my career, I highly recommend fully float charging each battery individually every one in a while to equalize them. \$\endgroup\$
    – winny
    Sep 12, 2016 at 18:05

1 Answer 1


I think this will work. My best candidate solution so far is to first load test the pair of Odyssey PC2250 12V 126Ah rated AGM batteries solo and record the actual Ah rating of each. If they are both at least 75Ah for example, they still may be usable. To charge them, I would use 2 separate 12V 55A chargers that work on standard 120VAC (60Hz) here in USA. I would assume the 2 batteries could remain connected in series (24VDC) even while they are being charged separately by separate chargers. At 55A max, assuming the batteries only hold a true 75Ah max, the maximum charge time should be no more than 2 hours, likely less (perhaps 1.5 hours). I do not know the actual load of the floor scrubber but if I did, it would likely be best to simulate that load. However, using my 500 watt load might suffice to determine the relative performance of the 2 batteries. For example, if one battery lasts 1.5 hours at 500 watts and the other only 1 hour, I might then have a problem.

Results of the load test were as follows:

800 watt max rated non pure sine wave output 12V power inverter was used. This is low quality output with much distortion but adequate for load testing.

Load was a pair of 300 watt incandescent bulbs estimated to draw 250 watts at the lower voltage (estimated 110V) output of the inverter.

Battery # 1 (the expected weaker one from previous other load testing)

Part 1: 48 minutes at an estimated 500 watts.

Part 2: Load was then reduced to approximately 250 watts for another 0.5 hours (one bulb was turned off with no interruption of load).

Total calculated watt hours is therefore (0.8 hours * 0.5 kW) + (0.5 * 0.25 kW) = 0.4 KWh + 0.125 Kwh = 0.525 KWh There was supposed to be a part 3 with an even lower load but I cancelled it cuz the load is not representative of a floor scrubber so it is not really relevant but I would call the battery about 0.6KWh total had I let part 3 conclude. Remember the KWh rating of a battery decreases when the average load increases. If the floor scrubber is something like 50A at 24VDC, this first battery wont do well cuz it only lasted 0.8 hours at an estimated 42A.

Part 1 of the test is the most significant since the load of the floor scrubber will likely be constant. Part 2 is just to better inform me how much total capacity the battery has but is not really relevant to high loads.

The battery is now slow charging overnight at 2A setting unattended but when I get my new charger I will blast it with 55A for about 1 hour or however long it takes to get a 100% charge. Since the battery is so old, I am not sure if it will "like" 55A of charge current and I may have to use the slower 20A charge mode. Perhaps doing many charge/discharge cycles will improve the Ah capacity of it since it has been sitting so long out of service.

I will slightly "overcharge" it and try this test again to see if I can get close to 1 hour at the 500 watt load. Even if I get that, it is only about half as long as a healthy battery should last so something is wrong with my battery, likely it is way past it's usable service life and should be "retired" to a much easier life such as UPS backup duty which is rarely needed here.

  • \$\begingroup\$ The weakest of any cell determines the Ah capacity. the cells are sealed which makes it more difficult. to know you are overcharging any 1 cell. THis can lead to accelerated aging. I suggest you overcharge them 5% to attempt to equalize the cells and check the ESR but the step load Vrise during static load test. ESR will rise as SoC drops depending on % of cell mismatch. For charging to prevent unbalance use an OVP active Zener FET transistor to bypass battery charge current thru a dummy 1V/55A wire with a lower RdsOn than the wire heater. 55W bypass is equiv to 10% mismatch batteries. \$\endgroup\$ Sep 12, 2016 at 22:05
  • \$\begingroup\$ plot Vbat VS time and no load and measure Vbat no load for a few seconds then resume load . From this compute in a spreadsheet V and Delta V vs T,,which is a strong indicator of approaching 10% SoC and dead battery from rise is delta V as well as Vmin at 11.5V \$\endgroup\$ Sep 12, 2016 at 22:52
  • \$\begingroup\$ @TonyStewart - can you give me more details on how to do your test? I did the load test and didn't get encouraging results. The true Ah capacity of the first battery (and expected weaker of the 2 batteries) at high load (close to 50A load) seems to be slightly less than 1 hour. \$\endgroup\$
    – David
    Sep 13, 2016 at 4:15
  • \$\begingroup\$ Like I said measure V and Delta V and preferably I during discharge test where Delta V/I= ESR and charge capacity =I*t... try burning off some sulphation with 5~8% over voltage after full charge with a large series R overnight and see if capacity improves. \$\endgroup\$ Sep 13, 2016 at 4:29
  • \$\begingroup\$ I only got 0.8 hours at about 42A load which is about 33.6 Ah before my power inverter started to beep to warn me of low input voltage. The attached chart indicates a healthy PC-2250 battery should yield about 2 hours of output with an approximate 500W load on it. I got less than half of this but maybe running the test twice will yield better results the 2nd time. I will try it and I will try to measure the actual load use a Kill-A-Watt meter but I don't think that will be accurate with non pure sine wave AC output from the inverter.odysseybatteries.com/batteries/pc2250.htm \$\endgroup\$
    – David
    Sep 13, 2016 at 5:08

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