# Li-ion cell charging efficiency

7 views

### steve church

Aug 8, 2005, 1:16:24 PM8/8/05
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I've just built a 2 cell charger using a BQ2057 i.c. I'm monitoring the
charge characteristics very closely for the first few times. I'm charging a
800mAh battery with a current of 400mA (at the constant current part of the
curve). How long should I expect it to remain at this current? Presumably if
the battery does not get warm it must be converting all the enery going in,
to chemical enery. Is this correct?

### budgie

Aug 8, 2005, 8:46:20 PM8/8/05
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On Mon, 8 Aug 2005 18:16:24 +0100, "steve church" <st...@rivend.freeserve.co.uk>
wrote:

>I've just built a 2 cell charger using a BQ2057 i.c. I'm monitoring the
>charge characteristics very closely for the first few times. I'm charging a
>800mAh battery with a current of 400mA (at the constant current part of the
>curve). How long should I expect it to remain at this current?

Very close to one hour.

>Presumably if
>the battery does not get warm it must be converting all the enery going in,
>to chemical enery. Is this correct?

The short answer is yes. The slightly longer answer involves knowledge of
whether the chemical process the charger is driving is endothermic or
exothermic.

And in relation to your subject line, all out testing on 18650 cells gave a
calculated charging ("coulomb") efficiency of 98-99%.

### Evgenij Barsukov

Aug 9, 2005, 11:49:22 AM8/9/05
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Note that this is coulomb and NOT energy efficiency. Energy efficiency
includes losses due to internal battery resistance, which are I^2*R
as well as entropic heat losses.

If we neglect the second for simplicity, it can be seen that I^2*R
strongly depends on current. Considering R of about 0.1Ohm for 800mAh
cell, the heat generation over the constant current period of charge at
400mA (which will be about 2 hrs) is I^2*R*2 = 0.032 Wh

Total energy of this battery is 0.8Ah * 3.6V = 2.88 Wh so energy loss at
C/2 charge is only 1.1%.
At 1C rate (800mA) it would be I^2*R*2 = 0.064 Wh e.g 2.2%.
Still not much, but still not _all_ the energy is converted as originaly
implied by the question.

Regards,
Yevgen