Split battery into two packs?

carboncanyon

1 µW
Joined
Jan 6, 2014
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4
I'm building a custom electric motorcycle and looking for guidance on the battery pack construction. Here's what I have so far:

  • Motor (already have this):
    Motenergy ME1507 (600A peak, 157A cont)
  • Battery (planning to buy):
    Molicel P42A 21700 cells 4200mAh 45A
    28s8p (224 cells), approx 100V nominal, 360A discharge, 3.39 kWh
    Layout is 14 wide by 8 long by 2 tall (cells are vertical)

I don't need much range as this will be more of a show piece, but I would like it to have decent acceleration and freeway-capable speed.

What I'm wondering is:

  • Can I trust the Molicel discharge number?
  • Is 360 amps enough, or do I need more to cover the motor's peak amps?

If this isn't sufficient, I do have space in the bike to do another battery pack. To simplify, I would build two identical packs (two 14x8x2 packs). I'd rather stick with one (cost, simplicity, weight, effort, etc.), but if my setup needs it I will go with two. Would it then be better to build each pack 14s16p and wire them in series to make a 28s16p battery?

I have a ton of experience with ICE bikes, but this is my first EV project, so any advice is appreciated!
 
:D
carboncanyon said:
What I'm wondering is:

  • Can I trust the Molicel discharge number?
  • Is 360 amps enough, or do I need more to cover the motor's peak amps?
..
Would it then be better to build each pack 14s16p and wire them in series to make a 28s16p battery?

-Yes, but not really.. At 30A the battery is already at 50degC in tests for longer duration. Wouldn’t use it at more than short 45A peaks due to this even if it’s possible. (=active driving, not for longer time at 45A on the highway)
100W per cell seems doable from this diagram which fits with the 30A assumption (got it from molicel site):
F0B9D5C0-E830-47AF-88D6-D67AB76A679E.jpeg
-good question, depends on your goals. If 600A is the motor peak then i’d build a pack also with this peak. (600/45)*30 gives 400A max for a longer duration, this would be my minimum goal for the current output and the needed parallel strings would be 400/30=13.3 so 14P seems like a good estimate either way you connect it.
 
Thank you for the reply!

Would this battery need copper, or can I get away with nickel strips? If it requires copper, would going from 28s to 24s alleviate that need? Thanks again!
 
carboncanyon said:
Would this battery need copper, or can I get away with nickel strips? If it requires copper, would going from 28s to 24s alleviate that need? Thanks again!

600A/8P = 75A

That will take a chunky bit of nickel. The nickel strips most of us are accustomed to are .015" x .3 to .4 inches, and are doing well to transmit 10A without problems.
 
Most people will be surprised how much power is lost in the battery pack. Motor efficiency or reducing weight, drivetrain or rolling resistance is most often the focus of efficiency discussions. It’s as easy as to calculate the resistance of nickel vs battery cells to make a judgement if copper will give a large efficiency improvement or significantly less heating.

Let’s look at one battery series string during full load
A 28s battery will have a length of about 22(mm)*28(cells) so 616mm. This will be the length of the series connection nickel carrying the current. Assume nickel strip 0.2mm, 10mm wide. It is 2mm2 in area.

Resistance calculator gives resistance of 21mohm for the nickel strip alone

Assume that each series string will output 45A current at peak power.

Power lost in the nickel=ri2=0.021*45*45=42W
Same copper strip would have about a quarter of the losses, 0.005*45*45=11W

Efficiency of the battery
Power total p=u*i=28*3.6*45=4536w

Cell losses: internal resistance from the cells 0.0185*28=0.518ohm, which gives 1049w losses at 45A

Already here we can see that there isn’t a huge improvement to be found in the connection resistance relative to the cell resistance.

Battery efficiency: power out/power in
(4536-1049-42)/4536=75.9% with nickel strip
(4536-1049-11)/4536=76.6% with copper strip
(4536-1049)/4536=76.9% with no connection resistance at all

Voltage drop in nickel 0.021*45=0.945V
Voltage drop in copper 0.005*45=0.225V
Voltage drop in cells 0.518*45=23.3V

So, no, i wouldn’t say that copper is ’needed’. it will give below 1% battery efficiency increase, this is small in relation to the total losses caused by cell resistance. Just to give some perspective, an added parallel string from 14 to 15 will give 2% battery efficiency increase at the same total current - and more range, less stress on the cells.
(and more costs for those added cells :wink: here copper is relatively cheap)
 
carboncanyon said:
... Would it then be better to build each pack 14s16p and wire them in series to make a 28s16p battery?
I have made one pack consisting of two 10s5p and one 1s5p wired all in series to 21s5p.
During driving I can see a voltage drop on the 36mm² copper wires (6x6mm² about 10cm short) used for the series connecting between the two 10s5p sections and the one 1s5p.
The BMS measures a 40mV voltage drop at cell 11 and 21 while drawing 165A, only from the used wire connection .
If I would have had the space I would have built one 21s3p and one 21s2p and wired them in paralell, then all batteries have the same connection style and resistance.
During driving this is no big deal. Only problem is during charging, because if I charge with 0,5C (50Ah) the bms measures a 12mV voltage difference at cell 11 and 21 and the BMS tries to balance this difference, which is only wire resistance based.

If I ever have to take the battery out, I will upgrade the wires to two 70mm².
 
Jk-B2A24S 200A continous.
Neither the ANT nor the JK do shut off
charging for measuring, I checked both, and the chargers would go crazy.
My Junsi and my ISDT rc-chargers do shut off the current before measuring.

It could be possible to tell the BMS that there is a wire resistance between a few cells, but I have not tried it yet.
 
carboncanyon said:
If I split the pack in two, can I use two BMS's?

If you mean in parallel, sure. Then it is just like paralleling any two smaller capacity batteries. The only complication of this is that if either half shuts off for any reason, the other half will be supplying the entire system load, so both the BMS and the cells would need to be capable of this.



If you mean in series...you'd need BMSs that can handle the full series voltage of *both packs*, because when one turns off (for HVC, LVC, or any other condition), the voltage then is applied across the open circuit, which is the FETs of the BMS that turned off.



If so, does that complicate charging?

If in parallel, no. You just use your regular charger and each pack will turn on and off as needed just as if it was one big one, and the charger will fill them up.

If in series, with separate BMSs, you'd have separate charge ports for each (even if that's the same as the discharge port), and you'd need separate chargers for each of those, that supplies just the voltage needed for each half of the pack. (half the full voltage of the system, if it's equally split). The chargers must also be isolated, which means that none of the AC input wires connects to any of the DC output wires. Many, but not all, chargers are done this way. Some current-limiting PSUs that can be used as chargers, like the Meanwell LED PSUs I use, HLG-600H-54A, are isolated. You can test your charger, with power disconnected at both ends, using a multimeter set to ohms or continuity, and test each input wire to each output wire, and vice-versa; you should get no continuity, or an extremely high ohms reading (hundreds of kohms or higher), if it's isolated. Note taht if the casing of the charger is grounded to the DC side (either positive or negative), you cannot let the chargers touch each other. It's fine if the casing is grounded to the AC ground pin (just as long as the DC side is not).
 
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