Hey! So I'm trying to find an internal resistance of a cell battery (Li-Ion)
So, basically what they have given me: A battery open circuit voltage connected in series with a resistor (R0) and then, with two RC-paralel circuit (R1//C1; R2//C2)
How should I define the internal resistance? Is there a formula for it or should I just define R0 as the internal resistance of the cell?
Hello,
literally, the term of battery internal resistance implies the sum of all the resistive elements in the battery equivalent circuit, however, R0 series resistance adds the most to the battery total resistance so using the term of internal resistance for R0 is more or less true in a general case.
Regards,
Bence
Hey Bence Csomos ! Thank you for your quick answer!
I forgot to mention, the circuit is an equivalent thevenin circuit, so as you may know, it is an series R0 connected to two RC parallel branchs.
But you told me that the R0 series resistance adds the most to the battery. What happens if the value of R2 (RC-branch 2, in a Li-Ion) is higher than the value of R0? Should I keep my calculation of heat dissipated energy with R0 (R*I²) or should I add R1+R2+R0?
Thank you!
The calculation methodology depends on your time-scale used. In a nutshell:
If the battery is at equilibrium state (its OCV is stable and capacitors are fully discharged) and you start to discharge it, the first thing you can observe is an instantanous voltage-drop due to R0 (serial resistance). This is because the capacitors behave as shorts and they short the resistors in RC branches. As you see, in this case, the effect and so the values of R1-R2 etc. are negligible. Then, after some milliseconds the capacitors start to being charged and due to their increasing impedance, some of the currents start flowing through the R1-R2 etc. Eventually, the current reaches a steady-state value where all the resistances add up and the heat generated are related to the resultant resistance. Please note that now we are talking about Ohmic losses only, however, there are thermodynamic losses as well which are not considered in the given equivalent model.
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