9
\$\begingroup\$

Recently I posted an answer mentioning the very classic "lithium batteries like partial discharges, so design your system for limited depth of discharge". But then I wondered: with partial discharges, the number of charge/discharge cycles increases as well for the same energy delivered, so the gain in available cycles lifetime will decrease. For example, the battery of a cellphone discharged at 50% in the morning, recharged, discharged at 50% in the afternoon and recharged overnight requires twice as many cycles as a cellphone discharged at 100% and recharged once a day to last as long. I thought it would be interesting to look into that.

I went ahead and as usual, I submit my findings for any SE user's approval and welcome anyone to add to it.

I should point out this only covers regularly used batteries, not those sitting on a shelf for periods greater than a few days. Even so, they do tend to age independently on the cycles but I do not have data on that - perhaps experts could shed some light on that.

\$\endgroup\$
  • 1
    \$\begingroup\$ if you don't let them discharge at all they will last forever \$\endgroup\$ – Jasen Jan 5 '15 at 0:13
  • \$\begingroup\$ Nice touch of sarcasm. You're right I forgot to point out there is still an expiry date for batteries regardless of the cycles count. I'll add that tomorrow. \$\endgroup\$ – Mister Mystère Jan 5 '15 at 0:25
  • 1
    \$\begingroup\$ At BU he says that after 1 year, at 100% charge, you have 65% (40°C) or 94% (0°C) of your capacity remaining. If stored only 40% charged, the figures are 85% and 98% instead. Pretty scary shelf life figures! \$\endgroup\$ – tomnexus Jan 5 '15 at 2:28
11
\$\begingroup\$

My quick look into it:

The lifetime of lithium batteries decreases with the depth of discharge, looking like the following (this curve is for lead-acid batteries, but Lithium is stated as following a similar curve): enter image description here

(source)

If the 100% DoD value is taken as a reference, one can plot what I call the "isoenergy" curve (I gave it a 2sec thought) which is basically how many cycles are required from the battery to deliver the same amount of energy as 100% discharges over its entire lifetime: $$isoenergy(DoD)=\frac{100\%DoDlifetimecycles}{DoD}$$ For example, 50% DoD require twice as many cycles as 100% DoD, 25% four times etc.

The results with this particular example:

enter image description here

Conclusion, it still holds that the depth of discharge should be minimised as much as possible.

\$\endgroup\$
  • \$\begingroup\$ is there a curve for percentage of lifetime wear on the cell per depth of discharge? \$\endgroup\$ – Jasen Jan 5 '15 at 0:14
  • \$\begingroup\$ They must exist, but all of those I've come across are VS temperature for fixed DoD. \$\endgroup\$ – Mister Mystère Jan 5 '15 at 10:01
  • 1
    \$\begingroup\$ Note also the "calendar" effect. LiIon have a finite life once they start to be used (or before?) and will just "wear out" wity age even if they have little or no use. There seems to be about zero comment on LiFePO4 calendar lifetimes but they seem to be relatively free of this. | My understanding is that main LiIon degradation mode in cycling is mechanical. Li depletion changes the physical volume and the battery works itself to death. Whereas LiFePO4 have the permanent olivine framework even when Li is not there. \$\endgroup\$ – Russell McMahon Jan 5 '15 at 13:23
5
\$\begingroup\$

I agree that you get a gain in whole of life capacity with decreasing DOD - from memory the figures that I have seen suggest a greater gain with decreasing DOD in the say 10%-80% DOD range - but I won't guarantee my recollections to be correct.

However, there are several other factors that may be more important and/or useful.
If you are in a position to tolerate reduced capacity discharges and/or multiple recharges per day, better gains may be obtained by limiting the top end of charging.
LiIon cells are usually charged in a CC/CV mode with CC usually at C/1 rate and with Vmax (typically 4.2 V/cell) being reached at around 70%-80% of total capacity, with the balance being input in CV mode at reducing current (set by the battery chemistry). Charge termination occurs at some selected Imax x k with (0.05 <= k < 1)
K = 1 corresponds to terminating charging at the CC/CV transition. It is well recognised that smaller values of k give somewhat increased total energy capacities but disproportionately decreased life cycle. k is quite often set at 0.25 or even 0.5, aggressive charging may set k to 0.1 or even 0.05.
Your curves suggest that even at a usually unacceptably low DOD of 10% total lifetime energy stored in less than 50% more than with 100% DOD. I have not got time at present to locate references but I am (essentially :-) ) sure that gains of better than 50% are gained by use of k=1 (no CV cycle) and this has the bonus of very fast charging (under 1 hour)(eg 48 minutes at C/1 from fully empty if CC/CV transition occurred at 80% energy level). Discharge to 100% DID is also "not helpful" and setting some minimum DOD with this sort of scheme is also useful. Some thing like 20% to 30% remaining capacity and 80% max capacity still returns 50% to 60% of overall capacity, leaves a 20% to 30% emergency buffer when needed and is liable to be superior to simple bottom end DOD control.

Another aspect that provides increased cycle life and overall whole of life energy storage increases is setting Vmax to less than the usual 4.3V/cell at 25C. Publoshed results suggest that even 0.05V decrease (to 4.15V) gives useful gains, 4.1V more so and 4.0V very much more. These reduced levels are accompanied by significant decreases in stored capacity per cycle.

This useful Battery University page discusses various LiIon life extension methods.
Table 4 suggests a 4 x increase in cycle life by decreasing Vmax to 4.0V from 4.2V with an only 20% decease in energy capacity per cycle - a gain or 3+ x the usual capacity.

The tables below are copied from the above page.

enter image description here

Utilising some mix of Vmax reduction, max DOD restriction and minimisation of current reduction in CV mode looks likely to produce very major whole of life capacity gains. For any given acceptable reduction in capacity some optimum mix could be established. Sounds like a PhD :-).

Also see:

BU - Lithium based batteries - why they are better

BU - Charging LiIon


Better still - use LiFePO4 / LifeYPO4 :-)

\$\endgroup\$
  • \$\begingroup\$ Thanks a lot for this addition. Are the first numbers (before the link to BU) sourced from BU as well or from your experience? \$\endgroup\$ – Mister Mystère Jan 5 '15 at 10:03
  • \$\begingroup\$ @MisterMystère - Who can say :-).... rereads ... Mostly out of my head BUT information has come from all over over some while - and I've been playing with batteries "seriously " for about 7 years. (Initially mainly NiCd (briefly) and then NimH. LiIon and LiFePO4 more recently. I don't know if they cover every point there but where they do I don't think we'd disagree. BU is a significant source of input but not the majority source by any means. I seldom find that they say things that I disagree with totally - we may occasionally see things from different perspectives.I'm happy to .... \$\endgroup\$ – Russell McMahon Jan 5 '15 at 13:17
  • \$\begingroup\$ .... discuss at more length / justify (after a fashion) any thing I claim which seems may be wrong. Occasionally it will be wrong bt usually it's just sumarised received knowledge or matching experience which I hope I manage to recallcorrectly :-). \$\endgroup\$ – Russell McMahon Jan 5 '15 at 13:19
1
\$\begingroup\$

One problem with these sorts of analysis is the question of what constitutes a "dead" battery. Most uses will involve a maximum permissible loss in capacity which is different depending on the use. EVs generally are very dependent on range so very little loss of capacity is acceptable. Home storage will continue to provide significant savings even if there is a large loss in capacity and this is why it is suggested that EV batteries could be reused as home storage units after being removed from a vehicle.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.