I have lots of used Li-Ion 18650s. I want to test their capacity.

There are however 2 important parameters that I cannot find anywhere which is amperage and cut off voltage. Some people use 500mA some people use 1000mA. Some discharge to 3.0V some do that up to 2.8V.

Depending on which parameters you choose you will get different capacity. Since capacity is defined for cells it's calculation must adhere to some standards for these parameters.

To wrap it up - what is the amperage and ending voltage for capacity measuring for single 18650 cell?

  • 2
    \$\begingroup\$ A look at the specific cell's datasheet might answer your question about cut-off voltage. \$\endgroup\$ – Electric_90 Feb 8 at 12:30
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    \$\begingroup\$ There is no reason to believe that the capacity calculation "must adhere to some standards". Read the datasheet. \$\endgroup\$ – Elliot Alderson Feb 8 at 12:41
  • \$\begingroup\$ Take care when combining different cell types and or cells with unknown soc. Compensations currents between the cells can be very high. Also the internal resistance of the cells should be the same when combining them. \$\endgroup\$ – A.R.C. Feb 11 at 9:54

If there's a significant difference in measured capacity between 500mA and 1A discharge currents, you should stick to lower value, i.e. 500mA (or even less). Unless you're interested in capacity under specific load, you should aim to measure the maximum value, and cells typically have higher capacity at smaller discharge current.

The cut-off voltage cannot be reasonably estimated and should be taken from the datasheet (idem for the full-charge voltage). If the additional capacity between 3.0V and 2.8V is insignificant to you, you can stop at 3.0V for all cells (or even 3.2V, which already correspond to 95-98% discharge; IMO 5% precision is plenty for the purpose of sorting used cells). What you should avoid is discharging cells with 3.0V cut-off down to 2.8V.


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