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Consider a lithium ion battery that is continuously cycled at different depths of discharge (say, 20% vs. 40% vs. 80%) and at a constant c-rate:

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It is well known that the larger depth of discharge will correspond to a faster rate of degradation for the battery. However, I wonder what the relationship is between the depth of discharge and the energy efficiency (kWh discharged divided by kWh consumed in charging). Is this expected to depend on depth of discharge? If so, are there any good references characterizing this relationship?

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    \$\begingroup\$ "it is well known..." any source for this? \$\endgroup\$
    – Solar Mike
    Mar 15, 2021 at 18:10
  • \$\begingroup\$ @SolarMike sure -- I added a reference to a paper that models degradation based on DoD; there are many others in this direction. These papers tend to model degradation as growing more than linearly in DoD, which implies that a single cycle of DoD x has more than twice the degradation impact than two cycles with DoD x/2. However, if the statement bothers you I can remove it; I'm interested in the impact on energy efficiency, not on degradation. \$\endgroup\$
    – josliber
    Mar 15, 2021 at 18:26
  • \$\begingroup\$ So, those sources, do they consider your question at all or do they publish papers in that direction? \$\endgroup\$
    – Solar Mike
    Mar 15, 2021 at 18:28
  • \$\begingroup\$ @SolarMike I haven't found a paper that addresses my question, hence the question here to see if any of the experts could point me in the right direction! \$\endgroup\$
    – josliber
    Mar 15, 2021 at 18:29

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Based on ohms law, if you were drawing constant power from the battery, then yes the efficiency would go down. If the load was purely resistive, then efficiency would actually improve. The reason for this is that your battery has an ESR which acts as a resistor. The more current that is drawn from the battery, the greater the losses across the internal resistance of the battery. If you draw 5A from a battery with an ESR of 6mOhm, the losses of that are 150mW. If you draw 1A from the same battery the losses from the battery are 6mW.

When the voltage of the battery starts to drop, your current will increase proportionally in the case where you're drawing a constant power such as an inverter connected to the battery. If the load is purely a resistor, as the voltage of the battery drops, the current will also drop proportionally.

Depending on what battery chemistry you use, will also play a big role the efficiency. If you have a more linear decrease in voltage with the DOD, then the losses in the battery will increase linearly. If you have a chemistry like LIFEPO4 where the voltage of the cell consistently stays at 3.3-3.2V though the majority of the discharge cycle, the losses will not really increase much until you reach the tail end of the discharge cycle where the voltage begins dropping off which I'm pretty sure is around 70%-80% DOD based on tests that I've done.

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  • \$\begingroup\$ Thanks -- this is informative. Are you aware of any benchmarking that is done of the relationship between DoD and energy efficiency, e.g. on battery fact sheets or in publications? \$\endgroup\$
    – josliber
    Mar 15, 2021 at 20:24
  • \$\begingroup\$ @josliber I recommend batteryuniversity.com. They have a lot of reliable information on there. Here's a link I found on a quick search regarding battery ESR with relation to SOC. Notice different battery chemistries can act completely different with the ESR in relation to SOC. batteryuniversity.com/learn/archive/… \$\endgroup\$ Mar 15, 2021 at 21:04

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