I need to report on it while its still fresh in my mind.

We have a client between Phalaborwa and Hoedspruit. Total offgrid setup that got hit with lightning a few weeks ago. In the meanwhile the generator also packed up. the lightning took out 2 of the 6 strings and from the time of the lightning we saw that the one battery is lagging in charge and reaching minimum SOC first resulting in the system switching off too soon.

The customer started of with one 15/12 Freedom won and my guys went up to add a second 15/12 last year. I did not accompany them and never saw what they did. They added the second and never ensured that the wires are symmetrical. I had to cut out 90 cm of cable on both the Positiva as Well as both the negative wires of the second battery to make all the wires symmetrical. Total of 3.6m of cable on the second FW.

When I arrived on site late afternoon I had 80Amps available for charge and checked how the charge got divided between the two batt. The one with the shortest wires were at 24% and took 60A of the available 80A and the one with the long wires only took 20A. I then tested them separately and both accepted the full available 80A

I decided not to mess around as we were trying to harvest as much as we could during two days of clouds.

I came back this morning and again found the batteries at 22% and 9%. I shortened the wires and restated the MPPTs. Battery one took 50 Amps and battery two took 46A of a total of 96 amps available. I think this might be due to difference in internal resistance.

Currently i am only charging the lower battery to get him up to the SOC of the leading battery. I will check again when they report the same SOC

On this one, I would really not have expected the difference to be that stark. If we assume the internal resistance of the two battery modules are the same, and that cable 1 (total length) has resistance R1 and cable 2 has resistance R2 then the current divides over the two in the ratio: Rx / (R1 + R2).

If we assume the same cable is used on both sides, we donâ€™t need to use the exact resistance values of the cable, we could even (just to make the math easy) assume it is 1â„¦/meter, ie the formula is Lx/(L1+L2), where Lx is the length of the cable on either side.

If the lengths are 2.7m and 3.6m, then division is:

2.7/6.3 vs 3.6/6.3, or 43% vs 57%.

I would have expected the 80A to split 34A vs 46A. All other things being equal (which they probably arenâ€™t).

Plus I would expect the full battery BMS to â€śslow downâ€ť and the empty battery to accept more charge as it also catches upâ€¦ and then they would both be at 100% at some stage. (Refer to Andy again here and his many batteries)

@plonkster granted itâ€™s open to interpretation, but I read this as the battery having two positive and two negative cables in parallel. And that 90cm was cut from each cable, shortening the paralleled cable loop resistance by 1.8m worth. ( Or, 1.8m x 2, thus removing 3.6m of single cable in total).

You are probably right, let me pull that part in hereâ€¦

Sounds like 1.8 meter total then. Now I just donâ€™t know if we started with 3.6 meters, or ended with 3.6 meters, but sounds like it started at 3.6 meters, which means it was halved in length, and the other battery was already at 1.8 meter total. That makes the math really easy, since it is now literally two thirds vs a third, and 80A then divides into 27A to the one, and 53A to the other. Still not quite on the 60/20 numbers, but close enough, and well in line with the final 46A vs 50A split at the end.

I donâ€™t think any BMS can do that. It can instruct the inverter to slow down (which slows down all charging, not just one battery).

But generally, this is not an issue. Once you get close to full (above ~3.45V per cell) the required charge voltage on the full pack shoots up, and it pretty much stops charging anyway, and virtually all current goes to the emptier pack.

The same happens when discharging, but the other way around.

This is a big drawback to paralleling dissimilar packs (or asymmetric wiring) - your total capacity is the sum of the two batteries, but the maximum charge/discharge current is limited to that of the lowest of the two, as towards the limits one pack will tend to 0 and the other will take all the current.

I donâ€™t think it was either. The battery has 2 negative and 2 positive cables (in parallel) so removing 0.9m from each positive and negative has shortened the loop by a total of 1.8m but required removing 3.6m of cable.

At this site, because they lost some of the charging capacity to the lightning strike, and I also heard (via my own personal back channel ) that a tree had grown enough to cause some shading. This means the batteries were not fully charged on most days, which combined with the poor cabling led to more than 10% SOC difference between the modules.

I think the moral of the story is that nothing is identical, no two crimps are the same, no two LFP cells, and we get away with a lot of these differences every time the battery is fully charged and it simply doesnâ€™t matter. The reason to keep those differences at a minimum is for those times when things go wrongâ€¦

If I may ask do both these FW units have the same shunt hardware spec (hall or prev spec)? Not sure of age / version of the first unitâ€¦
Maybe some variance between different shunt implementations?

This might be the missing link. Told Izak a few times that something just dont make sense in this installation. I think he is in exactly the same space. Cutting out 3.6 meters of cable will make a difference, but i never expected such a huge difference.

The main reason for me posting this, was to emphasize the importance of starting off correctly with the wiring. We lost one third of the PV with after a storm, resulting in the batteries never reaching 100% and the SOC of the two drifted further and further away from each other. I suspect that if the wires were correctly installed from day one, the difference would have been less over the same time period. I guess the longer cables contributed to the problem.

I still cant wrap my mind around this completely as the cables alone should not have had such a big effect, but definitely contributed.

OK, I have better numbers now While there is a whole lot of doubling up of cables and of course one has to measure the total length of the cable, all of this is just doubles the numbers in the ratio calculations, so it is quite sufficient to know what the distance from each battery to the busbar, for any one of the cables was. This distance was 1 meter for the one battery, and 1.9 meter for the other.

Then the current will divide as 1/(1+1.9) vs 1.9/(1+1.19), or 35% vs 65%.

Then 80A would have divided 28A vs 52A, which is actually quite close to what you got.

I mean, if I round that 1.9 meter to 2m, to make the math easy, the ratio is a third vs two thirds. Enough that it should make an obvious difference.

The mistake I am making here is ignoring the internal resistance of the battery. Since an LFP cell is typically around 0.3mâ„¦ per cell and you have 16 of them, ie about 4.8mâ„¦, and the cabling is about an order less (0.34mâ„¦ per meter, for 50mm^2), the internal resistance of the battery should in fact have a far larger effect on how the current divides.