Checking Bats

Got some new 18650 and wanted to check internal res. Got a dig. multi. now what?

Checking internal resistance is hard.

Hopefully one of the folks who know this stuff will chime in.

sixty545 - hope you see this!

not a hard thing, but this is a DC measurment, not AC as factories usually give it

first you have to measure the unloaded cell voltage

after that you have to load it with a resistor, and paralell it, you have to measure the loaded cell voltage,

usually, I use a 3-5ohm resistor (a 5W because of the heating up)


and here is the formula

r -> internal resistance

R -> load resistance

Uo -> unloaded cell voltage

Uk -> loaded cell voltage

in this case: ((4.14V-4.05V)/4.05V) * 3.5ohm= 0.078ohm

+1 thing, the cell tempreture should be at room temperature ~ 20-25°C

I love this thread already. :)

Viffer, do you have a source you can quote for that formula? I'd love to see an explanation for that calculation. :)

Thanks!

Viktor

There are a lot of formulas. You have to use Kirchoff's and Ohm's laws simple.

here are two solutions below:

Uo: unloaded battery voltage

UL: loaded battery voltage

RL: load resistor

Ri: internal resistance

I: current

I use the first formula, because it is much easier to memorize.

situation: the battery be under load, e.g. in a tailcap reading (flashlight measurements) for current (amperage).

Question: what type of voltage is it when i quickly interrupt the circuit, quickly take out the battery and quickly measure the battery's voltage?

this way, for example i was able to measure 0.8460V (i call it "remaining offline voltage") in an Eneloop AA cell which recovered to a steady 1.1680V after a few hours.

Apparently the cell's voltage under load at the point when i interrupted the circuit was *not* 1.1680V but i am measuring a sag of at least (1.1680V - 0.8460V =) 0.322V.

Assuming i would be infinitely quick with the measuring of the battery's voltage (see my Question above), would then the measured battery's voltage equal the cell's voltage under load (limes, in the limit)?

( Or is there any (other) way to measure the cell's voltage under load DIRECTLY? )

This isn't a good example, because an eneloop at 1.1680V almost or fully discharged. And that's the reason why you measured 0.8460V when you took the battery out.

If you charge the cell up to 20-30% at least, the cell voltage will jump back very quickly to the unloaded cell voltage.

Oh, lovely! I love it! :D

Thanks!

Viktor

I use the internal resistance feature in the menue of my icharger 106b+ and it gives me a read out good enough to compare to my other batteries.

yeah, this is the easier way, but more expensive

Earlier I found some equipments on ebay, which can measure internal resistance, but they are too expensive for me:

http://www.ebay.com/itm/20R-Internal-Battery-Resistance-Impedance-Meter-Tester-/220873511070?pt=UK_BOI_Electrical_Test_Measurement_Equipment_ET&hash=item336d16509e

http://www.ebay.com/itm/Mini-Internal-Resistance-Battery-Tester-KT-407-/140592412844?pt=LH_DefaultDomain_0&hash=item20bbf5f8ac

Maybe putting that money to a decent charger that does IR or use the methods shown on this thread to work out th IR of the batteries

Earlier I thought that it would be good to buy a 106b+, but I realized I don't need this, because I have a lot of nimh chargers including an ipc-1l, a lot of liion chargers etc, so why?

And beside that, I should buy a good power supply for the 106b+, and finally I should pay $100 for these, then I could throw away my other chargers.

If you buy the cheapest IR meter off ebay if you dont want to do the math for IR then you will have another item for your batteries, its great having 3 or 4 things to do one job, but the math is the cheapest option as a icharger 106B+ plus a power supply is around $130 usd with internation shipping not every ones cup of tea

Does this process only apply to li ion? if so why?

It can be applied to all types of batteries.