KeepPower IMR 18650 3.6V/3.7V 1600mAh High Drain Li-ion Rechargeable Battery

Min Capacity: 1500mAh

Typical Capacity: 1600mAh

4.2 volts max voltage and 3.6-3.7 volts nominal voltage

Button top: Yes

Dimension: (D)18.50mm*(H)66.50mm

Weight: Max. 48.0g

This is a high current cell with fairly low capacity like all high current cells.

This cell is not really interesting for application with 1A or 5A current draw, even at 10A it is not that interesting, but above 10A this cell will still deliver current.
And as can be seen the ouput between the two cells has a very good match, even at 30A.

This cell can be safely drained in about 3 minutes. That is safe for the cell, it might not be safe for the load or connections (30A is a lot of current).

Conclusion

This is a very good cell, but only for special applications, i.e. where the current is in the two digits range.

Except high drain tool, is there any flashlights which can take full advantage of 10A+?
Only my DRY Shocker needs over 5 amps.
Thanks once again for the test/review.

How hot do all these high power 18650 cells get at 10+ amp?
(I see you post max. temp rise for some nimh cells,but no such data for liions)
For example if this cell has 40mohm resistance,at 30Amp disipated power is: P=I^2*R=900*0.04=36Watts!?
I know they last only few minutes,but this is still huge amount of heat.

I just went to your web page that has all your reviews and such. It been a long time since I've been there and it has grown impressively. Thank you HKJ for all your efforts. Anyone wanting to purchase flashlight related stuff should check out your site before purchasing.

Your calculations are a bit off (it is I*V*R, not I^2*R). Power at 30A is ~100W, not 900W, and internal resistance is 0.05 ohm. 100*0.05 = 5W. That’s not too much to cope with for just 3 minutes.

Nope, power loss in a resistor (or cell IR) is I*I*R -> 30*30*0.05 -> 45 watts. 50 milliohms IR is huge for a 30 amp cell… should be under 10 milliohms.

I do not have a temperature sensor on that test station (It requires a bench multimeter).

I doubt you can calculate power that way, the internal resistance is not a regular resistor, but a combination of chemical processes and resistances in the construction. The last part will generate heat according to ohms low, but I doubt the chemical processes follows ohms law.

The internal resistance is calculated from the slope on the "Protection test", the lowest value I got from the 10 runs was 0.045 ohm.

Could you tell me how exactly do you calculate resistance from protection test?(I think I know why resistance numbers aren’t accurate with this method) .dU/dI method (at around 50% SOC ) directly from discharge graphs should give more realistic results.For example,this cell has dU=50mV for dI=2A (between 3-5A,or 5-7A,basically same result) which gives internal resistance of 25mOhm,which sounds more realistic(but still little to high,how do you measure voltage,4 wire?)Of course,internal resistance isn’t constant during discharge,it’s a function of voltage,but if I remember correctly,above method is close to method that it’s used by most cell manufacturers(AC 1kHz).

It is explained in the linked article. AC impedance is always considerable lower than DC resistance, i.e. do not expect the same numbers as published by manufacturers. I wrote a bit about it here: http://lygte-info.dk/info/Internal%20impedance%20UK.html .

I do use four terminal, but has a (very) few mOhm in the connection.