Let's say there are two identical Lithium-Ion batteries. (is it any different for Ni-MH batteries?)

  1. One gets charged at 0.1C until the EMF voltage (which is capacity-dependent) reaches 4.3V.
  2. The other battery gets charged at 1C until the terminal voltage reaches 4.3V. Then, charging stops.

Which battery has worn out more? Battery 1 or 2?

I know that battery 2 will not be charged fully. I am not intending to. What I intended to find out is whether charging my mobile phone quickly to ~70% extends the lifespan of the battery significantly compared to charging slowly to 100% or nearly 100%.

Does the same also apply to NiMH? (at their respective voltage)

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    \$\begingroup\$ Your "test" is confusing as you use two charging currents but also two "end of charge" methods. In 1 you stop charging and measure the voltage but in 2 you keep charging while continuously measuring the voltage. Both batteries might not be charged to the same level, in 2 you will charge less as the measured voltage will be somewhat higher due to the battery's internal resistance. So in my opinion this is not a fair test making the question unanswerable. In a fair comparison you should only change one thing between tests. \$\endgroup\$ – Bimpelrekkie Mar 6 '19 at 20:47
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    \$\begingroup\$ "Worn out more"? How do you measure that? \$\endgroup\$ – Dave Tweed Mar 6 '19 at 21:02
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    \$\begingroup\$ EMF means "voltage". So "EMF voltage" means "voltage voltage". What do you mean when you say "EMF voltage"? \$\endgroup\$ – TimWescott Mar 6 '19 at 21:36
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    \$\begingroup\$ Indeed such a question would get downvoted, also because it is comparing apples to pears. Known facts are that charging to 70% instead of to 100% will increase battery lifetime. Also slower charging increases battery lifetime. To find what the effect will be of a combination of the two where one is exchanged against the other, requires complex testing. There are many factors involved like how you use your phone that influence how much lifetime is affected. Simple solution: buy a phone where the battery can be (easily) replaced. \$\endgroup\$ – Bimpelrekkie Mar 7 '19 at 13:48
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    \$\begingroup\$ "... where one is exchanged against the other, requires complex testing ..." and probably varies from one battery design to another. \$\endgroup\$ – TimWescott Mar 7 '19 at 16:01

Batteries are nonlinear chemical cells that age from temperature rise, excess voltage and insufficient voltage. LiIon are fairly linear on the VI curve from 3.8 to 3.4V but also have a secondary large capacitance with a higher ESR that requires "soak" time this is done during CV as the current declines and cutoff is usually defined at 10% CC rate.

Studies ( research Battery University) have shown that the total lifetime Ah capacity can be doubled by reducing the capacity 10 to 20% by reducing the CV terminal voltage down to 4.0V The ageing is somewhat exponential above this and starts by increasing self-leakage then ends up as self-heating with more voltage going from 4.1 to 4.3, so most chargers use 4.20+/- x0 mV. but reducing CV to 4.0V will multiply your lifespan in terms of cycles before capacity drops 50%. Even raising the threshold for Depth of Discharge to 50% can increase lifespan. It just means shorter use time and more batteries needed but ends up extending the charge cycles more than 4x.

I don't have the charts I computed at one time and I am just doing this from memory.

YOur mileage may vary with battery chemistry, and degree (<< 1%) of mismatch in series connected cell capacity as the weakest cell ages the fastest and becomes 1st to drop below 3V accelerating Dendrite growth which can short the cells internally.

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