# Is charging an empty Li-Ion cell to 50% more or less efficient than charging it from 50% to 100%?

Many devices that are powered by a Li-Ion battery can be recharged over 5V DC supplied by a USB connection.

Is the 5V voltage internally converted by the charging logic to something lower, and is that process energy efficient (e.g. by using a DC-DC converter), or is some of the power dissipated as heat (e.g. by simply using a resistor to reduce the voltage)?

If so, is there a difference in efficiency when the battery is empty vs. when it is almost fully charged?

My practical reason for asking is that I am using a portable USB battery pack to charge my devices and I want to use the energy stored in it as efficiently as possible. It would be useful to know whether charging from 50-100% is more or less efficient than charging from 0-50%.

• Most would use a form of buck converter and they tend to be more efficient at higher loads, but I'm not sure Li-Ion battery chemistry comes into play so interesting question. – PeterJ Mar 7 '14 at 10:28

Obviously this depends on the device itself; but there are quite a lot of Li charging ICs that just dissipate the difference in a MOSFET.

However the graph isn't linear at either end. Take a look at the graphs on http://www.ibt-power.com/Battery_packs/Li_Ion/Lithium_ion_tech.html

The greatest efficiency is in the constant current region, once you get past the initial charge from flat. So in practice you should keep Li batteries between 20% and 80% charge both when charging and discharging, if you want to maximise efficiency and cell life.

Note that the software reporting "state of charge" may lie to you about where "full" and "empty" are - they may already be the 80-20 points.

• I'm wondering if the OP might really be concerned with power loss rather than efficiency if he's trying to save a few joules in his charging device. Even though efficiency may be high during the constant current phase, losses might be significantly lower in real terms once current has fallen off below 100mA? – Andy aka Mar 7 '14 at 11:59
• I'd think higher current would mean higher losses in the battery's internal resistance. I'd also guess that when the battery is most discharged, these losses are the highest, since R is bigger (though for Li-ion, it's quite small even when at its maximum). – Phil Frost Mar 7 '14 at 12:10
• Surely the important thing is the efficiency with which current through the battery is converted to "state of charge", i.e. that which can be extracted from the battery later? I'll admit that that's not exactly what the graph shows. – pjc50 Mar 7 '14 at 13:34

My practical reason for asking is that I am using a portable USB battery pack to charge my devices and I want to use the energy stored in it as efficiently as possible. It would be useful to know whether charging from 50-100% is more or less efficient than charging from 0-50%.

While charging the battery at say 900mA (I'm using the graph in pjc50's answer for reference), the efficiency may be 90% (irrespective what the battery's terminal voltage is). The power delivered to the battery will vary from: -

$3.70V\times 0.9A = 3.33W$ to a value of $4.2V\times 0.9A = 3.78W$ and this means the losses (real energy lost as heat) are about 0.36W.

When near-fully charged and the current reduces to (say) 100mA, the power delivered is only about 0.42 watts and it is likely that the chip that controls the charge to the battery can go into quite a low-power idle mode and may consume only 20mW.

The point I'm trying to make is that losses (rather than efficiency) are the important factor when trying to conserve the limited energy from a portable battery pack. Clearly I've used some hand-waving to emphasize the point I'm trying to make but if, when fully charged, the battery charge circuit shuts down into fairly low power consumption, does it really matter?