1
\$\begingroup\$

I have a Li-Ion battery cell, with a given datasheet providing the following discharge curve:

enter image description here

I thought to estimate its DC impedance (resistance) by, at a given capacity, making the ratio between the delta between open voltage vs voltage at given load, and its current. So, considering for example the discharge rate of 1C:

\$ R = \frac{V_{noLoad} - V_{1C}}{I_{1C}} \$;

But considering that from the datasheet I have only the 0.2C as the most light load, I thought to use 0.2C voltages instead of the no load voltage (open voltage):

\$ R = \frac{V_{0.2C} - V_{1C}}{I_{1C} - I_{0.2C}} \$.

I'd like to know if I am completely off-road, and maybe suggestions on the topic. Moreover, analyzing some points at different state of charge and different loads, seems that higher the current, the higher the resistance, as shown also from the curves. But also that with a higher current, the average resistance is lower, despite the increased drop.

It may be not so precise, but it could be useful since I would like to find just an upper and lower bound on the resistance values.

\$\endgroup\$
  • \$\begingroup\$ Just to clarify something, shouldn't the X-axis be Discharge Time in h instead of Discharge Capacity in mAH? \$\endgroup\$ – Macit May 4 '17 at 15:30
  • 1
    \$\begingroup\$ @Majid_L Actually, it's right the way it is. This method lets you more easily compare the discharge characteristics, because it intrinsically scales all of the curves based on how fast you're discharging it. You can easily see, for example, that there's less total energy output by discharging at 2C as opposed to 1C or 0.5C, because the area under the curve is the energy extracted from the battery. \$\endgroup\$ – Hearth May 4 '17 at 15:32
1
\$\begingroup\$

Generally to determine the resistance the cell in question is given a known load, and the voltage across the battery is measured. Comparing the loaded to the unloaded readings the cell's resistance can be calculated.

Generally these tests are done with a DC load but in cases where you need more in depth information AC load tests can be done in a similar manner to determine the overall cell impedance.

This battery university article has some more information on the subject: http://batteryuniversity.com/learn/article/how_to_measure_internal_resistance

That site has tons of useful knowledge on batteries in general, definitely worth going through.

\$\endgroup\$
  • \$\begingroup\$ Indeed. But the resistance is not so linear as they said in the article. Ideally, very ideally, I'd like to make an estimation from the datasheet, where they already have made the various measurements and characterizations. \$\endgroup\$ – thexeno May 4 '17 at 21:33
  • \$\begingroup\$ Unless there's some information containing the loaded cell voltage at some known load, there isn't really a way to guess. You can try looking up similar cells to get a possible ballpark. Realistically if you want accurate information however you'll need to do the testing yourself. \$\endgroup\$ – alphasierra May 4 '17 at 23:14
  • \$\begingroup\$ I think that the known load is what shown on the graph. You have some different loads (i.e. some different currents drain from the battery at a given voltage) \$\endgroup\$ – thexeno May 5 '17 at 8:22
1
\$\begingroup\$

Batteries like Caps has ESR which can be measured at DC or some f and not related directly to Capacity from vendor to vendor.

It may be measured by incremental V/I with 1/C load.

ESR rises sharply with <5% SoC and slowly with aging until one cell is dead in a string. 10 mOhm is a nom. value for 18650 cells. Some are better and worse by >50%. Smaller mA capacity LiPo's may be 30 mOhm nom.

Your mileage will vary.

A car battery tested with a cranking amp. CA of 800A is always done at7.5V with 12.5 full,charge thus ESR is 5V/800A =0.63mOhm roughly.

\$\endgroup\$
  • \$\begingroup\$ So, from what I have understood, basically my initial guess-math on impedance (from datasheet plots) was right, i.e. measuring the drop form a given load, considering the difference of the drop between two different loads. Correct? \$\endgroup\$ – thexeno May 10 '17 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.