Recovering cells from an undervoltage condition, a wiki

I also would have guessed less. Like I said, even at 0.1 C, it takes 10s of seconds or maybe a single minute to go from around 2.0 V to 2.5 V in my experience. One minute at 0.1 C would be 0.17% of the capacity. Maybe reverse sag tricked me, or it took way longer than I thought, but it seemed near instant to me.

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I’m going off my memory of some testing last year but perhaps I’m in error. I’ll test it later today or tomorrow.

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I’ve done work with AI/machine learning for a long time. I just don’t think LLMs are worth a shit. Garbage in, garbage out is the first thing anyone doing data science should learn, but for whatever reason, the current LLM hype seems to think that if you pour a mount of garbage in, you’ll somehow get a genius out.

All I see is a fancy autocomplete that acts like a confidently incorrect random person on the internet a lot. Also, I work in physics, and while I haven’t messed with the newest models, when the recent hype started, talking to AI about physics always ended up in effectively technobable that was about the quality of talking to a random conspiracy theory crackpot at a bar. If I’m going to talk to something like that, I at least need it to be in a bar so I have the necessary amount of alcohol to tolerate it.

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Okay, no more multi-tasking for me…with a clearer mind now I am definitely wrong and I apologize for the confusion. Yes, you two are right, the capacity gain will be tiny. I have edited my post.

I’ll continue to do the test anyway…my penance I guess.

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Okay, penance done…

Reliance RS40 drained from 2.498V to 1.000V at 100mA delivered 20mAh. Umm…yea…I was verrrrrry wrong.

Now to bring the cell to zero volts a couple of times, short it for a couple of hours, and then tear it apart. :smiling_imp:

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reviving from 1v is described in the section related self-discharge with this data sheet but with other data sheets it’s written elsewhere.

In those other datasheets, is the spec for end-of-discharge and/or BMS shutoff above 1V? If so then the discharge must stop there and the only way 1V can be reached is via self-discharge or quiescent current draw from the BMS.

So I’m confused regarding your statement.