So I am an engineer and really want to understand the battery. It is my understanding that there are 6831 NCR18650 Panasonic batteries in the "Battery pack". 11 moduals in series each with 9 "bricks" in series and each brick with 69 18650's in parallel. So 69 * 9 * 11 = 6,831. Are you with me so far?

Now the hard part. On Panasonic's website i get specs for the NCR18650 showing a nominal voltage and capacity of 3.6 VDC and 2.9 AH, respectively. Lets do the voltage first. 9 bricks X 3.6 volts X 11 moduals = 356.4 volts. But the Tesla specs say the battery is 375 volts. Backing into the nominal voltage needed to get 375 volts each battery has to have a nominal of 3.78 volts. If someone knows the answer to this puzzle I would greatly appreciate an explanation. Thanks in advance.

Now lets do kWH. 69 cells X 2.9 AH = 200.1 AH/brick X 9 bricks X 11 moduals X 3.78 volts /1000 = 74.9 kWH. So how do I match this up to the 56 kWH claimed by Tesla? At 75% the number is 56kWH. Does that mean that only 75% of the batteries nominal capacity is "usable"? Is this because there is about a 25% loss between input energy (from the HPC) and actual usable stored and and delivered energy from the battery to the inverter/motor?

Finally, the new 18650 battery is supposed to be 4 AH instead of 2.9 AH in 2012. The 2.9 AH battery weighs 44 g while the new 4 AH battery weighs 54 g. 6831 * 44 g = 300.564 kg or 663 lbf. Since the entire battery weighs 990 lbf, the battery enclosure, coolant and electronics must weigh 990 - 663 = 327 lbf. Therefore if the new batteries weigh 6831 * 54 g = 368.874 kg or 814 lbf, then a new battery pack for my Roadster with the new batteries should weigh 814 + 327 = 1,141 lbf. For the extra 151 lbf of weight, one should get an increased nominal range of 245 * 4.0/2.9 = 338 miles minus a little for hauling around the extra weight. Do I have it right on all counts?

Now the hard part. On Panasonic's website i get specs for the NCR18650 showing a nominal voltage and capacity of 3.6 VDC and 2.9 AH, respectively. Lets do the voltage first. 9 bricks X 3.6 volts X 11 moduals = 356.4 volts. But the Tesla specs say the battery is 375 volts. Backing into the nominal voltage needed to get 375 volts each battery has to have a nominal of 3.78 volts. If someone knows the answer to this puzzle I would greatly appreciate an explanation. Thanks in advance.

Now lets do kWH. 69 cells X 2.9 AH = 200.1 AH/brick X 9 bricks X 11 moduals X 3.78 volts /1000 = 74.9 kWH. So how do I match this up to the 56 kWH claimed by Tesla? At 75% the number is 56kWH. Does that mean that only 75% of the batteries nominal capacity is "usable"? Is this because there is about a 25% loss between input energy (from the HPC) and actual usable stored and and delivered energy from the battery to the inverter/motor?

Finally, the new 18650 battery is supposed to be 4 AH instead of 2.9 AH in 2012. The 2.9 AH battery weighs 44 g while the new 4 AH battery weighs 54 g. 6831 * 44 g = 300.564 kg or 663 lbf. Since the entire battery weighs 990 lbf, the battery enclosure, coolant and electronics must weigh 990 - 663 = 327 lbf. Therefore if the new batteries weigh 6831 * 54 g = 368.874 kg or 814 lbf, then a new battery pack for my Roadster with the new batteries should weigh 814 + 327 = 1,141 lbf. For the extra 151 lbf of weight, one should get an increased nominal range of 245 * 4.0/2.9 = 338 miles minus a little for hauling around the extra weight. Do I have it right on all counts?

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## Comments

Also that 4Ah battery weights only 46g not 54g. That is unless you have some better and newer information than I do. Those that go to Tesla are also modified somehow, which probably does something to the weight, not sure what though.

The 2.1 AH clears up the kWH. I found the 54g on a website somewhere. 48g is a lot better news. Looks like all of them are the same physical dimensions (2.1, 2.9, 3.1 and 4.0). Do they all have the same discharge ability? Do you know what the xC rate is? Wouldn't the 3.1's give 47% more range (3.1/2.1 = 1.47)? In any event, it is exciting to think that much more range and or lower weight for the same range is just over the horizon. Can't wait.

Any clarification on the nominal voltage? Is the 2.1 AH battery nominal 3.78 volts? My model airplane LiPo's are all 3.7 v.

rsdio,

If you look at the calculations again the batteries themselves made up 663# of the 990# weight so I assumed the balance of the weight was due to the enclosure, cooling system and electronics (about 327# worth).With the 48 gram weight for the 4 AH I get 1050# total pack weight (663*48/44 = 723 + 327 = 1050#

Model S 300 mile pack batteries are supposedly the 3100mah, 3.6v and weigh 44.5g.

I'd say it's a safe bet that the batteries will be imporved with respec to power/weight ratios, but far more important will be improved costs.Frankly, right now, with 300 miles of range, and a 45 minute recharge, they are good enough to be fully competitive with gas powered jobs.

even though you have 6831 cells, only the capacity (Q) of 69 of them (those arranged in parallel) is used for discharge at the nominal voltage of 375V (given by the arrangement in series of the remainder 99). E = Q x V so E = (69 x 2.2)A x 375V = 56.9 kWh. the capacity coming from the cells arranged in series is only there for "balancing". If they were "empty" then the capacity of the 69 arranged in parallel would have to be averaged over the entire number of cells (6831).

as you can see an increase in capacity is most wanted as it allows for both an increase in energy (driving range) and power (acceleration, etc) EVEN if the individual cell operates at a much lower voltage. For example, if you had double the capacity in a single cell you would only need 35 of them in parallel for the same energy (driving range) and 195 in series which would now give a much higher nominal voltage. V = 195 x 3.5 = 683V for a 3.5V cell or 585 for a 3.0V cell, etc.

so lets hope there is interest in post lithium ion batteries such that the chemistries of such batteries are ironed out in due time .

Based on the total capacity of the unit cell (18650) of ~3000 mAh and the reversible specific capacity of LiCoO2 of 140 mAh/g, we can estimate 21.5g of reversible LiCoO2 in one cell. however, the total capacity of LiCoO2 is ~280 mAhg, but only ~50% of is reversible, so we have to multiply 21.5 x 2 = 43g LiCoO2 to obtain total amount in the unit cell.

now the molecular weight of LiCoO2 is 97.87 g/mol and that of Li is 6.941 so Li weighs 7.1% of total active material in cathode. Then 43g x 0.071 = 3.05g total Li in the cathode of one 18650 cell with ~3 Ah capacity.

I dont know how much electrolyte one 18650 cell contains but it must be as much as the free volume (~50%) of its celgard separator which must be the same geometric area as the cathode. The geometric area of the cathode in an 18650 cell is 265.65 cm2. So if i assume ~280 cm2 geometric area for the celgard separator with 0.1 mm thickness and ~50% porosity, i get ~ 280 cm2 x 0.01 cm = 2.8 cm3 / 2 = 1.4 cm3 or 1.4 ml of electrolyte.

at a a typical concentration of 1M LiPF6, a volume of 1.4 ml corresponds to 0.269 g LiPF6 (mw of LiPF6 is 191.905 g/mol). there is 3.6% Li in LiPF6 by weight so 0.269 g x 0.036 = 0.0097 g Li.

to add everything up we have 0.0097g + 3.05g = 3.06g Li in one 18650 cell. Model S has 7000 cells so a total of 21.4 kg Li per battery pack or just about 47 lbs of lithium.

i hope i made only a few mistakes in my math.

if anyone can pitch in how much that costs...