RE: The effect of a BMS when measuring LIBs

Gough Lui — Tue, 22 Sep 2020 00:00:48 GMT

I think those results are what I would have expected based on the fact most BMSes would be using a MOSFET (often an N-channel type) to act as a switch and it's usually going to have some level of resistance (Rds-on). This will very much depend on what MOSFETs are chosen, whether there are multiple in parallel, etc. Higher current BMSes may well have a much lower influence on the resistance measurement, for example the ones which are used for power-tools, portable vacuum cleaners, e-scooters/e-bikes, etc. However, as many of them achieve the high current capacity with more MOSFETs, this means larger circuit boards and more chips which might add more capacitance to the system - this is probably going to create more of a deviation in the reactance graph. What would probably defeat this test entirely is if the BMS had a capacitor on the output in some way, in which case then you might be measuring the characteristics of the capacitor more than the cell itself.

As for the influence of wiring, you're seeing perhaps close to 18mohm of difference. At a guess, you may have about 40cm total of (I'm assuming) 24AWG copper wire in the shot, that would be about 3mohm. Where did the rest of the 15mohm go? Perhaps in the spot-welded tabs on each cell (the length of the tab and the contact spot) and solder joints at a guess. Perhaps you can eliminate some variables if you clipped the probes onto the tabs of the cells to see whether you can resolve the resistance contribution of the tabs and spot-welds, as the baseline seems to be "direct" to the terminal of the cells.

- Gough

RE: The effect of a BMS when measuring LIBs

jw0752 — Mon, 21 Sep 2020 03:59:43 GMT

Hi Kai,

Thank you for the great information on the effects of the BMS on the Lithium batteries. Would you mind explaining to me the reason for testing batteries at different frequencies and what that means. Is it that a load is applied and removed? Is this done as a square wave or is it done sinusoidally?

Thanks John