Hello everyone,
I would like to estimate my cell's utilizable volume of solid materials, in this particular case, the LCO and carbon content in my 18650 cell. I do the following, don't I?
Let's say I know these data for a fresh cell at 25°C:
- Nominal capacity (Q0 = 2,6Ah = 9360 As)
- Molar mass of Li (MLi = 7 g/mole)
- Max. concentration of Li in cathode (c_cathode_max = 56000 mole/m3)
- Max, concentration of Li in anode (c_anode_max = 31000 mole/m3)
I can calculate the nLi amount of Li in the cell using Faraday Law:
n [mole] = Q [As] / F [C/mole]
Then, I can calculate the total volume of solid material in both of the electrodes using:
Vs_cathode = nLi/c_cathode_max
Vs_anode = nLi/c_anode_max
Looking the equations above, I have to decide what electrode has the total nominal capacity of the cell. Or in other words, what electrode I should calculate the whole amount of charge for.
I counted on the anode since this electrode contains all the Li-ion when the cell is fully charged. However, as I know the more the charges can receive the cathode coming from the anode, the more the efficiency is getting better.
How do you calculate Vs in this case? What is the convention for it?
Regards,
Bence
Hey Bence,
You should rather look at the utilizable volumetric capacities of the anode and cathode at low C-rates. For LiCoO2 this is about 707 mAh/cm3 (based on the LiCoO2 utilizable gravimetric capacity of 140 mAh/g and crystal density of 5.05 g/cm3) and for the graphite it is about 784 mAh/cm3 (utilizable gravimetric capacity of 347 mAh/g and crystal density of 2.26 g/cm3). Typically, the total capacity of the anode is chosen slightly higher than the capacity of the cathode, to avoid localized Li-plating on the anode at the end of battery discharge. The chosen ratio of the anode-to-cathode total capacity is usually referred to as the N/P ratio, which is typically set as 1.1-1.2.
Thus, you can assume that the nominal battery capacity is determined by the total capacity of the cathode, which is the product of its volumetric capacity and volume of the active cathode component. In the case you mentioned, the total capacity equals 2600 mAh, thus the volume of active LiCoO2 equals 2600 mAh / 707mAh/cm3 = 3.68 cm3 and that of graphite with e.g. 1.2 N/P ratio equals ~ 2600 mAh × 1.2 / 784 mAh/cm3 = 3.98 cm3. This means that active materials occupy ~45% of the total battery volume (~17 cm3), which is a value also reported in the literature.
If you’re interested in structure of cylindrical 18650 and 21700 batteries, here are two reads which may be helpful:
1. M. Lain, J. Brandon, E. Kendrick, Design strategies for high power vs. High energy lithium ion cells, Batteries, 5 (2019)
2. J. B. Quinn, T. Waldmann, K. Richter, M. Kasper, M. Wohlfahrt-Mehrens, Energy Density of Cylindrical Li-Ion Cells: A Comparison of Commercial 18650 to the 21700 Cells, J. Electrochem. Soc., 165 (2018), A3284-A3291.
Hope it helps!
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