How Does Current Affect Li-Ion Battery Efficiency?

[Timeforheroes0]

Hi,

I am trying to figure out how Li-Ion batteries operate in regards to efficiency. I understand they have rated voltage and current etc. and to operate them at these parameters if possible. However, I'm wondering is it simply the current that has effect on their efficiency. By Efficiency I mean the amount of energy leaving the battery that get transferred to the appliance divided by the amount of energy leaving the battery.
Is the energy dissipated in the battery solely due to heat dissipation?
If it's operated below rated voltage what effect does this have?
I also understand that Li-Ion operate better at higher temperatures, but does this not just mean that they operate at a higher voltage at these temperatures but aren't actually more "efficient"?

Thanks for any help, trying to get my head around this!

 

[mfb]

Is the energy dissipated in the battery solely due to heat dissipation?

Heat is where the energy mainly ends up, yes. Maybe some unwanted chemical reactions in addition, especially if it gets hot.

If it's operated below rated voltage what effect does this have?

This is the same as operating it above the rated current.

I also understand that Li-Ion operate better at higher temperatures, but does this not just mean that they operate at a higher voltage at these temperatures but aren't actually more "efficient"?

I would expect more efficiency, but not a higher voltage at zero load.

 

[Timeforheroes0]

Heat is where the energy mainly ends up, yes. Maybe some unwanted chemical reactions in addition, especially if it gets hot.

This is the same as operating it above the rated current.

I would expect more efficiency, but not a higher voltage at zero load.

Ok, I see. Why is voltage below rated voltage the same as operating it above the rated current?
Is there anyway of quantifying battery losses beyond P=I^{2}*R?

Thanks!

 

[mfb]

You can approximate the battery as a perfect power supply with an additional resistor in series. A larger current will lead to a larger voltage drop at this (virtual) resistor, so the output voltage goes down.

Battery loss is then given by I²R = I(U-U₀) with the zero load voltage U₀.