Two batteries in paralell with diode, common charger possible?

[Marsbar]

My battery pack did not last long, and I thought the BMS was broken, so I ordered a new BMS. When opening the 48V 15Ah pack I discovered some welded connections was broken off. I took the entire battery pack apart, and discovered 6 cells gone bad. No way to get them to life. So I re-arranged the cell pack with the remanding cells, and connected the BMS. I manually bled the cells with a lightbulb to get them the same voltage, the BMS did the rest. A test drive seemed to work well though, motor is set at 20Ah.

In order to get enough range I bought a secound battery back, 48V 10Ah.

The two battery packs were wired together with common -, and the two pluss leads were connected via a diode bride, thus on diode in series with each battery. Diodes should manage 25A and is mounted on a cooling rib.

+ from battery 1>---- o power to motor - o---< + from battery 2

- battery 1 -----o power to motor + o ---- - from battery 2

After the two diodes there is a key switch for the motor.

On other words, each pack delivers current indipendently of the other because of the diodes. Worked very well!

Setup gives me a range of approx. 20km with full power on (20A current to motor)

It must be said that battery pack only has a temperature bearly above freezing temperature, to it drains poorly...
Still cold where I live. Batteries don't like cold weather.

But problem is, I now have two chargers. The BMS breakes on the negative lead.

My question is:

Can I connect C- on each BMS in paralell, then two diodes to plus from the two chargers. Thus uising one charger to charge both batteries.

Or will it say kabong? Better use a 3A fuse?

I then get C- connected to C-, and plus to pluss via a diode, and plus to plus via a diode on the two BMS'.

Or will this setup ruin the BMS?

When I use two chargers they are isolated because of the transformer, so no worries.

[Bikemad]

I'm confused. 

I thought the 48V 15Ah battery only had 16 cells to start with. 

If there were 16 cells in series and you now only have 10 cells in series, the total battery voltage will only be 36V fully charged (10 cells @ 3.65V/cell)

If this is the case, you definitely cannot run (or charge) this battery in parallel with a 48V battery. 

If you have different voltage packs connected in parallel, but with a series diode in each positive cable, the fully charged 36V battery would not be able to provide any current/power whatsoever until the voltage at the controller's battery wires dropped below ~35.4V (36v less ~0.6V voltage drop across the diode).
The voltage at the battery needs to be at least 0.6V higher than the voltage at the controller side of the diode before any current will flow, and I suspect that the controller's undervoltage protection would be set at ~40V on the 48V battery setting.
This means that the controller would cut out completely before the 32V battery was even able to assist.

How can you safely charge a 32V battery to the required 36V maximum voltage using a charger with a regulated output of 58.4V?

If both batteries were of the same chemistry (Lead acid, LiPo, LiMn or LifePO₄ but not NiCd or NiMH) and the same voltage, they could both be used (and charged) in parallel without the need for diodes.

Charging from a single charger would require both charging plugs to be connected in parallel, but it would obviously take a lot longer to charge both batteries from a single charger than it would with 2 separate chargers.

You can connect a 48V 15Ah LiFePO₄ pack in parallel with a 48V 10Ah LiFePO₄ pack, but you cannot connect a 32V 15Ah pack in parallel (or even in series) with a 48V 10Ah pack.
The Ah rating and battery chemistry must be the same for two (or more) batteries connected in series.

Alan
 

[Marsbar]

No no, what I mean is that I took out some bad cells, then I re-arranged it to 13 cells in series like before, and 5 in paralell. Before it was 6 in paralell.

So the repaired pack is 13 cells, 48V.

The new pack is also 13 cells 48V.

The two diodes drain them both from +, but of course the battery pack with the highest voltage and power wins and will drain first.

The advantage of using diodes is that I am sure no voltage difference can go from one battery to another, thus avoiding damage to batteries.

Now for the charger:

Idea was to use two diodes to split charging current to two batteries. They would then simply get half charging current, and each BMS would shut of when the pack was full. What I am concerned about is how to arrange the diodes to avoid disaster... If ground was common, I could simply use diodes on the +, but the BMS breakes the negative, the positive is passed on straight from the battery.

Will try to make a sugested wiring diagram and post.


 

[Bikemad]

No no, what I mean is that I took out some bad cells, then I re-arranged it to 13 cells in series like before, and 5 in paralell. Before it was 6 in paralell.

That's a relief, so you're obviously not using a 48V 15Ah Golden Motor battery pack:



And you must have taken out a total of 13 x 18650 LiMn cells (6 bad and 7 good) from the old pack to convert it into a 12.5Ah pack. 

As both of your packs are now 13S LiMn you could effectively join them together in parallel to make one 22.5Ah pack.
But they can also be kept as two separate packs that could be plugged together for charging and discharging provided both packs were at similar voltages.
While they were plugged together in parallel, their voltages would always remain the same.

The only reason to use diodes in the battery power leads would be to prevent excessive current from flowing from one battery to the other if their individual voltages were substantially different when they were first connected together.

Unfortunately, if diodes are fitted, the regenerative braking function will not work, as no regen current will be able to return to either of the batteries.
The diodes could also allow an excessive voltage build up at the controller during a high speed downhill descent, as there is nowhere for the residual generated voltage to go.

Now for the charger:

Idea was to use two diodes to split charging current to two batteries. They would then simply get half charging current, and each BMS would shut off when the pack was full. What I am concerned about is how to arrange the diodes to avoid disaster...

As both packs need to be fully charged to the same 54.6V (13S x 4.2V) no diodes should be required for charging, you would simply join the charging leads together in parallel to a single charging connector to use a single charger, or keep them separate to use two individual chargers for a much quicker charge.
The voltage of both batteries should remain identical throughout the charging process. The voltage from the charger will be identical on both batteries, and the current will automatically be split between the two batteries in relation to their capacities.
The old 12.5Ah pack should draw slightly more current than the new 10Ah pack if the older cells are still in good condition, but it could draw less current if the capacity of the old cells has degraded by more than 20%.

Placing diodes in series with the charging circuit would result in a slightly lower charging voltage at the BMS (54V instead of the required 54.6V) due to the 0.6V voltage drop across the diodes. This slightly lower voltage would reduce the ability of the BMS to properly balance all of the cells at the end of the charge.

If both batteries are used every time, they should always be at an identical voltage, but if you occasionally used just one battery, you would have to make sure that both batteries were either fully charged or reading within 1 volt of each other before connecting them both together in parallel again.

Alan