Hi all,
I was performing EIS on both a commercial Li-ion cell and a coin cell which was fabricated with the harvested electrodes from the commercial cell.
The coin cell's impedance values are quite higher compared to the commercial cell. What can be the reasons?
I guess the LiPF6 electrolyte used when making the coin cell can be one reason. But I was told this can be due to smaller electrode areas of the coin cell. But isn't it the current per unit area that matters? which is guess is similar in both cases.
Hoping for a generalized explanation of why a coin cell can have higher impedances compared to its own mother-commercial cell.
Many thanks in advance.
Without data, I can only say my guess: many parameters can effect: electrolyte (concentration, grade...), separator, the pressure on the electrode or sometimes disassembly the commercial cell process can effect that electrode.
Many thanks, Artur and Giang.
I wonder whether the geometric size factor can cause a large variation between the two EIS?
Dear Buddhi Wimarshana ,
in order to find some more help for your interesting, but special, questions, we need an additional standard information, about:
1) What is the (type or geometry of this) commercial Li-ion cell ?
2) What is your commercial EIS-setup, as well as say, if it's specs 'allow', such a very low-Z (commercial Li-ion) cell, (at high freq.) to be, reliably, measured ?
3) What is your coin (half ?) cell ?
4) During the measurements of the two Z-spectra, their (WE) SOC is 'the same' (or similar) ?
5) Also, could you, please, show/send the two (area normalized) Z-spectra in the same graph.
The SOC of the cells are really important when you are doing EIS measurements. Did you disassemble the cell at very high or very low SOC levels? This could easily change the SOC when you reassemble your coin cells and change the EIS spectra. To avoid minimum changes it is a good practice to disassemble your cells at 50 % SOC. Hope this helps!
Dear Buddhi,
in principle, the higher the contact area (i.e. the size of the cross-section), the higher the thickness of the electrode, and the higher the specific active area of the electrode (that can be approximated by (3*solid fraction/average particle radius)), the smaller the resulting impedance, measured in Ohm. These three parameters multiplied together determine the actual surface area of your electrode.
The figure is a simulation of the impedance spectra of a commercial battery while varying the contact area. You see how there is an increase in the magnitude of the impedance when the area decreases. It is like the specific impedance of your battery, measured in Ohm*m2 and that depends on the electrochemical performances of your specific cell, that should not change if you use the same materials of the mother cell, gets distributed over a smaller area, and hence you see a higher resulting impedance in Ohm.
Of course, also the other parameters mentioned in the other comments can play a role.