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Mixing old and new batteries is bad.

But there are various options how they can be mixed, and I guess there are use cases in which certain ways of mixing can be justified.

In my off-grid solar system I am using quite good lead-carbon batteries capable of surviving 4000 cycles when discharged up to 50%. Their Partial State of Charge (PSOC) wear is minimal. Also they can be charged 2.5 times as fast as traditional lead acid types. At the moment I am running 12 of them — connected in 3 parallel quartet series. I am giving them quite hard times — discharging down to 50% every night and topping up to 100% during daylight.

Now I want to scale the array up — add another quartet or two. Say if I do it this way it perhaps would be quite bad:

enter image description here

But a gut feeling tells me I might get away with this:

enter image description here

I am sort of lazy to do the calculations vs ask this question. Is my gut feeling wrong or right? Is there any difference between the two options above? How much will the new batteries suffer if the old ones are only say 20% worn? Should I even think of scaling the array up, or should I just wear it out completely and buy all new batteries thereafter?

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    \$\begingroup\$ Please don’t mix SoC and SoH. They are completely different and are mitigated in different manners. Your first suggestion is correct given that you refer to SoH. \$\endgroup\$
    – winny
    Commented May 21, 2019 at 11:10
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    \$\begingroup\$ @winny Do I mix SoC and SoH? I only refer to SoC to show how hard the batteries work for me. I expect that the existing batteries are 20% worn given their life expectancy, use pattern and time in use (which is 10 months now). \$\endgroup\$
    – Greendrake
    Commented May 21, 2019 at 11:53
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    \$\begingroup\$ You write “State of Charge (PSOC)“, “discharging down to 50%“, both refer to SoC. \$\endgroup\$
    – winny
    Commented May 21, 2019 at 12:03
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    \$\begingroup\$ once the old batteries die, it renders the good batteries in series with it useless unable to use their remaining charge. only as strong as the weakest link. all your parallel legs would be balanced at least though \$\endgroup\$
    – DKNguyen
    Commented Jun 1, 2019 at 21:32
  • \$\begingroup\$ @DKNguyen That would be fine. The goal is to scale the array up and keep using old batteries until they pass away without hurting new ones too much (vs replacing the whole array with all new batteries and discarding old ones while they still can work). \$\endgroup\$
    – Greendrake
    Commented Jun 1, 2019 at 21:43

2 Answers 2

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There is a great deal of literature in this space that is becoming more and more important as life extension and secondary markets for these systems grow.

But of course, generally it depends.

If the chemistry provides a stable voltage-SoC relationship, then this mixing can work without extra electronics. In that case, your gut is wrong and the all-in-one-stack option is best. That's because you have two scenarios to consider: charge and discharge.

During charge, the weaker batteries will reach higher states of charge quicker. Provided the voltage-SoC relationship is stable, that means they will automatically accept a smaller proportion of the charge current as the stronger batteries remain at a lower internal voltage for longer. If this were not to happen the weaker batteries would overcharge.

During discharge the opposite happens. The weaker batteries will reach lower state of charge quicker but the arrangement will automatically compensate by the stronger batteries taking a greater share of the load.

If instead you were to go with your gut, you'd find the weaker batteries in the string are both overcharged and undercharged as they are forced to take the current that the stronger batteries are capable of.

When it comes to scenarios where you don't/can't know which batteries are the weaker ones, then you have to turn to individual compensation electronics. There are various systems for bypassing individual batteries in a string to compensate for their more rapid changes in State of Charge. Again, they generally rely on the Voltage-SoC relationship being stable. If it is not (eg. Lithium based chemistries), then generally a coulomb-counting approach has to be used.

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  • \$\begingroup\$ So, in a nut shell, mixing old and new in parallel is no problem provided that voltage-SoC relationship is stable? \$\endgroup\$
    – Greendrake
    Commented Jun 2, 2019 at 2:25
  • \$\begingroup\$ Pretty good summary. In other words, keep all the batteries in a string similar. \$\endgroup\$
    – Heath Raftery
    Commented Jun 2, 2019 at 7:19
  • \$\begingroup\$ Most of this answer is just completely wrong. If you tie the batteries together then the terminal voltage of the series/parallel array is ABSOLUTELY the same. The only difference is that older battery's have a lower CAPACITY. The older (lower capacity) batteries will NOT overcharge or undercharge compared to the newer (higher capacity) batteries. The side effect of putting older (lower capacity) batteries in parallel with newer batteries is shown clearly in the graph Laptop2d showed. Since the terminal voltage is related to SoC you want to reduce your discharge terminus in a mixed pack. \$\endgroup\$
    – Jack Creasey
    Commented Jun 7, 2019 at 18:33
  • \$\begingroup\$ Thanks @JackCreasey, I added the word "internal" to try to clarify. I didn't want to get too complex but as you suggest it is confusing to imply that the batteries will be at different voltages. You might want to have another look at the graph - it shows battery life (number of cycles) vs how deep each cycle is (DoD), which is not related to the matter at hand. On the other hand, over/undercharging of weaker batteries in a series string is a well known phenomena, as is the self balancing effect of dissimilar batteries in parallel. mpoweruk.com/balancing.htm \$\endgroup\$
    – Heath Raftery
    Commented Jun 8, 2019 at 0:28
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    \$\begingroup\$ Alright, this explanation is not working for you. Let's leave it at that. Hopefully someone can provide an alternate explanation to help more people. Perhaps this would be a good place to start: neuralfibre.com/paul/wp-content/uploads/2007/05/… \$\endgroup\$
    – Heath Raftery
    Commented Jun 8, 2019 at 9:57
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I am sort of lazy to do the calculations vs ask this question. Is my gut feeling wrong or right?

It's wrong, the problem is if the cells do not match, more current (usually) gets dissipated in the unmatched cells.

Is there any difference between the two options above? How much will the new batteries suffer if the old ones are only say 20% worn?

The problem is in the Depth of Discharge, it goes down over time. In either configuration the batteries that still need to be charged will show a lower voltage then the ones that don't if the batteries are not matched. This will at minimum waste energy as heat, at worst it will lead to failures in the battery.

enter image description here

Should I even think of scaling the array up, or should I just wear it out completely and buy all new batteries thereafter?

I think the best thing to do would to keep the old batteries on their own string with their own charger and leave them matched to each other.

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  • \$\begingroup\$ The answer from Heath Raftery suggests that paralleling new and old batteries is fine given that voltage-SoC relationship does not depend much on SoH: although older batteries will have lower voltage, this will grow up quicker on charging; on discharging, although the voltage will go down quicker the new batteries will balance it, effectively charging the old ones. Do you still say that either configuration is equally bad? \$\endgroup\$
    – Greendrake
    Commented Jun 2, 2019 at 4:12
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    \$\begingroup\$ All I know is when we installed batteries on satellites we had to match them, and if you don't match batteries its bad and it degrades them and wastes energy. There is plenty of stuff you can get away with, depends on what your risk tolerance is. I would say no to either configuration, but if I had to pick one I'd pick the parallel one \$\endgroup\$
    – Voltage Spike ♦
    Commented Jun 2, 2019 at 4:53
  • \$\begingroup\$ The complication is that if you charge the older batteries separately then the terminal voltage of old vs new becomes a management problem. You could provide a relay to disconnect the older batteries when they reach the lower SOC, but again the variation in terminal voltage becomes a significant issue. I'd suggest that the OP pick a target depth of discharge (and therefore pack terminal voltage) that uses as much as possible from the older batteries. The newer batteries are therefore not being discharged as deeply and will provide longer life (cycles). \$\endgroup\$
    – Jack Creasey
    Commented Jun 7, 2019 at 18:47

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