Do you know if it would be acceptable to build another vtc6 pack with new cells and then connect those in parallel? The current pack is a bit over 1 year old with around 50-70 cycles. No crazy loads only 40-60A maximum for a sort time.
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I have a theory on how they actually distribute the load but I have yet to test it. There was a time where I wanted to combine 95C lipo packs with slow discharge laptop cells. When I was doing research one person wrote out a large post discussing internal resistance, and it seemed accepted. I did see issues with it as did some other users.
(theory part dont accept as fact, I donāt want to spread misinformation)
The problem I saw was that amp load on a battery should be proportional to its voltage dropā¦ For example, if battery A at 1v sag produces 10A, and battery B at 1v sag also produces 10A, then as a pair at 1v sag they should produce 20A.
If battery A produces 10A at 1v sag, and battery B produces 1A at 1v sag, then as a pair at 1v sag we should see 11A.
The problem I believe we could run into is the discharge ratings and continuous draw. For example if Pack A is a 1Ah 100C discharge pack, and Pack B is a 1Ah 1C discharge pack. At first when we draw lets say 100A, everything is okay. The voltage sag is within limits and working like the above cases. Eventually though, the pack as a whole will begin to drop in voltage as we are consuming all of its power. When the discharging voltage gets to a certain point, the voltage drop from initial voltage will be okay with Pack A, but definitely not okay with pack B. This is because we are treating pack B like it can discharge at 100C and thus its voltage drops too quick and too low in this short time. In this case Pack B would also increase its amp contribution proportional to the amount of amps it would use at a voltage sag that of the overall pack, but its voltage sag will most definitely be below what its voltage should be.
Recharging in this theory could work like this using the same high discharge example above. Lets say we drain the battery pack to 20% or something like 3.3v per cell. The high discharge pack A is 100% okay with this, and its voltage drop was honest to its power consumption. Removing its load would reveal the pack A to go up by 1v lets say. pack B however was way beyond its limits, and its voltage was not honest. A very large majority of it was all voltage sag that was hurting the battery, not actual voltage drop due to using up its energy. If we were disconnect pack B and check its voltage, it could go up by .7v lets say! Here is where it gets interesting in this theory.
I use the āhonest voltageā to define the voltage of a pack once the load is removed and each pack is disconnected separately. With a load as a pair the packs will both be the same, and when you remove the load and keep them connected it will still be the same. (this is because they are connected, realistically you wont see a voltage gradient on their parallel lines)
How to find out how much charging will occur?
Since Pack A has an honest voltage of 3.3+.1=3.4, and Pack B has an honest voltage of 3.3+.7=4.0, the amount of charging should in theory be equal to the amp draw of (4.0-3.4)v worth of sag in pack B. In this case, if Pack B supplies 2A at .5v sag, then it will charge the other pack at 2A, which will slowly decrease in Amp flow as it gets more and more charged, until both packs honest voltage is the same.
My whole point here is that it should be okay as long as batteries are pretty similar, but once we cross the line of over discharging not by amp draw, but just by overall rate of overall pack voltage, we will see issues. These include over discharging the weaker pack or having a recharging effect. As long as you set your limits to double the C rating of the weaker pack (to account for the discharge of the stronger pack) you will be okay. According to the theory you could have a pack discharge higher than this and still be safe, but you would always have to have the overall voltage drop across time not be steeper than what the weaker pack can handle.
(theory part dont accept as fact, I donāt want to spread misinformation)
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I will try to test this this winter with some amp flow meters, RC 95C packs and the laptop cells I have lying around that can barely do discharge of 1C. I can try to post my findings as accurately as I can and recreate my theory as stated above. Although I wan to believe in the resistance theory, I still believe that the weaker battery if treated like a 100C battery will be damaged, and attempt to charge the higher discharge pack with load removed. There is no way you could have a combined pack go from 4.2v to 3v in 2 minutes, and not damage the weaker half that at most could only do that in an hour. Its current combined voltage of 3v is a lie, and when disconnecting the packs from each other you will see that it was all voltage sag in the weaker pack and the stronger pack will still be at 3v(or lower, because in this theory the weaker pack was charging it and it would have had a small positive sag)
(Again some of this message contains theory, do not use it as fact in your builds)
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One last thing, I do believe that resistance theory plays a role in this. Like in my example where if battery A at 1v sag produces 10A, and battery B at 1v sag also produces 10A, then as a pair at 1v sag they should produce 20A. In this case if their resistance values are different, I think we could see slightly different values here in terms of amp contribution, however functional it should still work like the theory I presented still. Just with a few differences in numbers that should be minor but still accounted for.
Perhaps when I go through the crazy testing and figure out what actually is happening, I can write up a program that does all this calculating for you.
(Again some of this message contains untested theory, do not use it as fact in your builds)
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Should I be worried about a spot weld that blasted through the nickel? There was a bad smell for around 10-20 seconds; Iām thinking it could be the smell of pvc wrap burning?
(my welds are so far apart just to avoid the middle as much as possible)
My guess is I should probably scrap it?
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If you think it punctured the can, you should probably scrap it. Likely this happened because you were not keeping enough pressure on your probes, which caused the current to arc.
It also looks like all your welds are a little too hot.
Maybe burn a bit of pvc shrink wrap to compare the smell to what you smelled from the cell?
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Could be a couple things. Most likely the PVC, because if you follow that circle of PVC its definitely under that weld. It could also be the nickel vaporizing since it couldnāt spot weld properly. In general I would just worry about the battery having a hole in it, definitely look around and make sure this isnāt the case. Looks like your welds are a little hot too but if its working well then there isnāt too much to be concerned about.
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Yeah, my hand slipped a bit right as I hit the pedal 
Iāll try burning some wrap and see what happens
Yeah thats what I was thinking too. Iāll get a peek at it later to make sure
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Weāve all been there.
My understanding is that if you actually punctured the can, the smell that would be released would be persistent, and very strong. Perhaps @Battery_Mooch can chime in on this when he has a minute.
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Itās unlikely that you punctured the can, but if you have a sharp pointy object like the probe of a multimeter you can poke it to see if its still solid. Wear glasses and do it outside though
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Yeah thatās what i recall reading too, so i thought its probably the pvc burning; and after burning some outside it smells pretty similar to what i smelled earlier ā just without the welded nickel smell
poke the weld?
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Poke the spot where is blasted through, a structurally intact can is strong enough to stab a few times with no damage but if you damaged the steel then the needle will go in or at least deform and you need to dispose of the cell before it ignites
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ahh I see, Iāll do that
Iām glad I bought a fire extinguisher then
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Fire extinguisher is not enough, a lithium fire is really hard to extingish, just let it burn out in the open where it wonāt damage anything
Edit: sand or dirt is good as long as itās dry, that should keep the flames down, but any water could make it explode
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Bucket of sand and bucket of water should be on hand always
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I keep a metal bucket around at my table.
Water is super dangerous as it only accelerates the burn especially with lipos. Just let it burn off, try not to breath any of the fumes.
I believe it was Mooch who said that water helps to cool down the reaction and extinguish the fire. Dont quote me on that though.
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Depends on the water container, if you only have a small volume then it will just vaporize the water and splash boiling lithium water, but if you have a pool or something then it will have enough thermal mass that the vapor dissipates back into the water
Yea, if the can was punctured there would be a persistent chemical odor near that end of the battery. Maybe even a film of liquid on the battery.
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@YeetMeat I urge you not to poke a battery can that could potentially be damaged. If the can is weakened or opened then poking could push around the active layers of stuff inside and possibly cause a short circuit.
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