I've replaced the original battery in a Peg Perego Polaris 600 with a higher capacity SLA 12V battery (9AH). I also added a second SLA 6V (7.5AH) battery in series for increased speed. I've read that connecting batteries of different capacities can lead to issues during charging and discharging. Since I am charging each battery individually I don't believe there should be any issue with charge balancing. My questions are regarding discharge:

1) It has been mentioned online that in such a series configuration with mismatched capacities one battery can discharge completely before the other at which time the larger capacity battery can cause the weak battery to go into reverse polarity. If this is the case then would it be as though the 6V is now in parallel with the 12V with the load between the negative terminals of the battery as though the 6V is being charged by the 12V?
(see diagram)

2) If the above is correct, does this condition of reverse polarity/charging by the larger cell only occur when the lower capacity battery is in a completely drained state? What are the ramifications of this state -- will the 6V SLA be permenantly damaged or will it lead to catastrophic failure such as catching fire/explode/leak?

3) From what I understand this complete drained state is 1.75v/cell = 5.25V for a 6V SLA. When charged the 6V is at 6.74V. After the toy is used for about an hour (at this time the toys performance appeared slower as though it only had the 12V) the voltage of the 6V SLA is around 5.74V and can go down to 5.24V if stressed (trying to hold the vehicle in one spot to put a greater load on the motor to simulate going up hill). After about 30min of non-use the voltage recovers to 5.94V and after several hours of non-use it recovers to 6V. In this case I am thinking that a battery cutoff circuit may not work well as the voltage under load will not represent the true state of the battery which would be its resting voltage after several hours. Would it be suggested then to simply monitor the voltage during use. Perhaps periodically during use leave the toy idle for a minute or so and once the voltage is reading around 5.7V it would be time to recharge?

 

Why is + to + and both - to the load in your diagram? Diodes in reverse to each battery with a fuse for each could help make it a little safer. But is there any reason you cannot just use two batteries of the same capacity? This is a much safer option.

 

Why is + to + and both - to the load in your diagram? Diodes in reverse to each battery with a fuse for each could help make it a little safer. But is there any reason you cannot just use two batteries of the same capacity? This is a much safer option.

I was trying to illustrate my understanding of what happens after by the lower capacity cell reverses polarity as I have read that the cell then begins to get charged by the larger capacity cell. Is my diagram correct in terms of what happens at that stage?

Yes I can get a 9AH 6V battery if this setup is not safe. Is this setup dangerous (i.e. Fire/explosion/etc)? I've not found much material online which speaks of the discharge dangers in such a setup and what exactly could happen. One piece of advice was to ensure the lower capacity cell does not discharge completely and it would be ok.

If I do get a 9AH 6V it may be from another manufacturer and therefore may not be perfectly matched with the existing 12v battery (plus I've used the 12v a couple of times already). In this case would I still face the same balancing issues?

 

It isn't going to catch fire or explode.
The smaller battery can be reverse charged in this situation but with lead acid batteries, the battery can be recovered by normal charging though there may be some reduction in capacity.
Read this thread.

 

Batteries in series always take the lower capacity value, so yours is 18V 7.5aH.

 

Here's a sim showing the battery voltages for a steady 10A discharge. With an initial state-of-charge (SOC) of 100% each, polarity reversal of the 6V battery occurs at ~2.7k sec.

 

It isn't going to catch fire or explode.
The smaller battery can be reverse charged in this situation but with lead acid batteries, the battery can be recovered by normal charging though there may be some reduction in capacity.
Read this thread.

I was reviewing the linked thread and also the article referenced in it entitled Cell balancing and battery replacement. In that article the author uses an example of alkaline batteries where one cell in a series of cells is driven to a discharged state. The author then goes on to say:

"... If one is fortunate, all that will happen is that the weak cell will be irreparably damaged and the assembly will become useless, with no further drama. The unfortunate may find this situation to result in the release of caustic, corrosive, toxic, and/or flammable materials, cell rupture due to internal pressure buildup, spewing of flame, and perhaps a combination of these and other inconveniences…"​

Is the above catastrophic failure such as catching fire or exploding for discharged cells (in a series connection with other good cells) applicable for non-lead acid batteries such as alkaline or some other type such as lithium cells?

For my application since it won't catch fire or explode then I'll forget a battery cutoff circuit and instead just monitor the voltage to ensure the cell is not completely discharged to prevent damage to it.

Here's a sim showing the battery voltages for a steady 10A discharge. With an initial state-of-charge (SOC) of 100% each, polarity reversal of the 6V battery occurs at ~2.7k sec.

Thank you for this simulation. Does this mean that when U2 becomes completely discharged the polarity of the battery is reversed and then begins to get reverse charged by U1? Is this behavior similar to my original diagram (below) where the
6V 7.5AH battery becomes discharged and then its polarity reverses. The 6V battery would then appear to be in parallel with the 12V and would start to get reverse charged by it?

 

Does this mean that when U2 becomes completely discharged the polarity of the battery is reversed and then begins to get reverse charged by U1?

Yes.