I noticed that when charging and discharging a lithium polymer battery, the incoming/outcoming energy I calculated is very near to the specifications of the manufacturer.

How is that possible if any battery has a "real-life" efficiency? In practice, my test data say that my battery is nearly 97% efficient (-3% of the designed capacity or measurements error). I suppose I should give the battery more energy in order to achieve a capacity of 3600 mAh.

The battery has a capacity of 3600 mAh and I charge it with 1 A, which is actually \$\approx\$0.72 A for 5 hours (0.72C). The battery heats only about 1-2 °C. The ambient temperature is 25°C and stable. Battery resistance is < 70 \$\text{m}\Omega\$.

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    \$\begingroup\$ Your 0.27C value implies that you charged your battery for 1 hour. Is this true? How did you determine that your battery was completely empty when you started charging (hint - this is a very unlikely state for a LiPo ...)? \$\endgroup\$ – brhans Jul 24 '18 at 15:02
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    \$\begingroup\$ Consider losses rise with age rapidly near end of life. \$I^2ESR*t=\$ losses in watt-hours vs capacity in watt hours for t in hours. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 24 '18 at 15:03
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    \$\begingroup\$ "How is that possible if any battery has a "real-life" efficiency?" - What kind of question is this? \$\endgroup\$ – Ale..chenski Jul 24 '18 at 15:54
  • \$\begingroup\$ @brhans you're perfectly right. I corrected the error. \$\endgroup\$ – Nic1337 Jul 25 '18 at 8:24

Efficiency is a complicated idea to apply to a battery of any chemistry.

The losses in a cell are going to be the impedance of the cell during charging and discharging.

If you have cells of a known resistance, an known steady currents for charge and discharge, you can calculate the losses by using P=I^2R.

The problem is that these losses are dependant on the current. Current varies significantly between applications. As an example, I have designed things which charge at 0.5A or 2.2A. And they output anything from 1A to 10A (or more). Charging at half an amp is nice an efficient, but very slow. Higher charging currents lead to more thermal losses.

You've then got the complexity of how high currents and how they effect the cell chemistry itself. In a very top level and basic view, the higher current used to charge a pack, the lower the usable capacity tends to be.

Then there are lots of questions about how to rate the capacity and how it's valued etc. I'd recommend you do some serious research into battery chemistry if you are going to use cell capacity as your yard-stick.

TL:DR - losses from cells are pretty much only P=I^2R losses, don't get confused about capacity when talking about efficiency.

  • \$\begingroup\$ I think this answer solved the question. Please vote the question so I can check as solved. \$\endgroup\$ – Nic1337 Jul 25 '18 at 10:34

Your treatment of "real-life" battery efficiency is flawed. The number "97% of rated capacity" is just a measure of current capacity of your battery under your particular charge-discharge conditions. This number may differ from the nameplate of the battery, since the manufacturer might have different charge-discharge-termination conditions than your experimental setup, and should be taken as a ballpark estimate. Or it could indicate a degradation in SOH - State Of Health of this particular battery if your test conditions are exactly as used by manufacturer. Or it could indicate nothing if your environment was colder than the test conditions of manufacturer.

The efficiency of a battery (aka Coloumbic efficiency) is defined as a difference between "charge in" and "discharge out", or, as you said, the difference between incoming/outcoming energy. The loss of energy comes from dissipation over internal (parasitic) resistance (See Tony's comment above), plus some battery irreversible aging (degradation of solid-state chemistry), and self-discharge.

With the charge current of 1 A you should be able to charge your 3600 mAh battery in approximately 4 hours, which should equate to approximately 4.2 V * 1 A * 4hr = 16 W-hr. For more accurate number you should integrate the actual product {V(t) * I(t)} over time until the charger stops. This is your input energy W(in).

The output should be measured in the same way, as an integral{V(t) * I(t)}dt during discharge. If you discharge your battery at a similar rate, say, 1A, the rough estimation would be 3.6 V * 1 A * [measured time] = W(out). The ratio W(out)/W(in) would be the battery "real-life" efficiency.

In my own research on aging of some phone batteries the "real-life" efficiency was about 88%, W(in) = 10.4 Wh, and W(out) = 9.2 Wh at 9.88 Wh nominal{1}. BTW, I have never seen this parameter specified by any cell manufacturer. Interestingly that calculated losses over DC-estimated ESR (~140 mOhms) are only about 60% of the total measured losses, which means that the single-ESR model is not entirely accurate.

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{1} Test conditions: charge at 2.048 A with 4.35V, cut-off at 256 mA, discharge at ~800 mA to 2.9 V.


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