JHZR2
Staff member
Hi,
Trying to figure out an electrical question pertaining to batteries.
One might say that a battery's maximum "power", or discharge current in reality is based upon what provides a practical value of the equation:
Vt=Voc-IR where Vt is the terminal voltage, Voc is the terminal open circuit voltage, and IR is current times internal resistance.
So for, say, a 30mOhm internal resistance in a 1600mAh Li-ion 3.7V cell, one would notionally be able to discharge at up to 23A, since 23*.03=0.69, and one would not want to discharge a Li-ion cell past 3.0V.
Yet at the same time, manufacturers generally only rate their cells at up to 3C or 10C in some power designs, which corresponds to 6-16A for a typical small Li-ion cell. Is that purely being conservative, or am I missing something on the load side?
As I understand it, maximum power is obtained when load and source impedances match. I suppose then you could say that:
P = V^2 / (2R) where the R is the internal resistance of the battery, but matched identically to the load.
Then, one would say that for that cell: Vt=Voc-2IR, right?
So then for the max 0.7V drop, max current would be 11A roughly.
Does this make sense from a basic electrical perspective, and am I missing something in terms of how a max rating for a small cell is created?
Hope this is clear enough in explanation and my thought process.
Thanks!
Trying to figure out an electrical question pertaining to batteries.
One might say that a battery's maximum "power", or discharge current in reality is based upon what provides a practical value of the equation:
Vt=Voc-IR where Vt is the terminal voltage, Voc is the terminal open circuit voltage, and IR is current times internal resistance.
So for, say, a 30mOhm internal resistance in a 1600mAh Li-ion 3.7V cell, one would notionally be able to discharge at up to 23A, since 23*.03=0.69, and one would not want to discharge a Li-ion cell past 3.0V.
Yet at the same time, manufacturers generally only rate their cells at up to 3C or 10C in some power designs, which corresponds to 6-16A for a typical small Li-ion cell. Is that purely being conservative, or am I missing something on the load side?
As I understand it, maximum power is obtained when load and source impedances match. I suppose then you could say that:
P = V^2 / (2R) where the R is the internal resistance of the battery, but matched identically to the load.
Then, one would say that for that cell: Vt=Voc-2IR, right?
So then for the max 0.7V drop, max current would be 11A roughly.
Does this make sense from a basic electrical perspective, and am I missing something in terms of how a max rating for a small cell is created?
Hope this is clear enough in explanation and my thought process.
Thanks!