Technical Battery Discussion








Thanks Timo.





The roadster batteries are 2200mah,3.7V and 44g.











Does anyone know how much litium the battery contains?



Yes, I miscalculated: 1,6 billion!

i see lots of talk about the specs but nothing on cycles, does anybody know how well these things will do in the long run?

the amount of lithium in the battery pack can be estimated by math. Lithium only exists in the LiCoO2 cathode and the LIPF6 electrolyte. The electrolyte is typically 1 mol/L so there is 1 mol of Li in 1L of electrolyte.

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...

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