I am making a battery tester, lithium ion battery for particular. and I want to measure the internal resistance, but after testing few cells, I am skeptical for my results, since most of them, new or old are around 500-800 mOhm, totally not close to 150 mOhm range as it should be.
The photo: For this instance, the open voltage is 4.18V and load voltage is 3.41V with 1525 mA discharging current; 4.18V-3.41V/1.525A= 0.504 ohm;
Is this the correct way to measure internal resistance? and since new and old cells all have similar result, is it safe to say that internal resistance is irrelevant to health of a li-ion cell?
Best Answer
No, the method is not correct at all (if you wish to duplicate manufacturer's specifications). There are two ways. Apply a 1 kHz sine wave current to the battery. Measure V and I at 1kHZ to calculate R. The current should be small, but large enough to get a useable reading on your oscilloscope. Use AC coupling so you can see mV readings. Use a sense resistor on the low side of the battery to measure current.
The second way is to apply a low-duty cycle step current load. Measure the instantaneous voltage drop at the moment the step load turns on (you need an oscilloscope to do this). The load could be, say, C/10 or C/5. I usually do this with a function generator, a resistor, and a FET. The function generator turns the FET on. The resistor is attached so when the FET turns on, it becomes a load current to the battery. The resistance is calculated using V/I = R, where V is the voltage change, and I is the step load current value.
If you measure the load voltage and no-load voltage with a slow instrument such as a battery analyzer, you will be measuring load droop caused by chemical rate effects at the anode and cathode. I guess this number could be meaningful, but it is not how the manufacturer measures internal resistance.