Lithium-ion battery Internal resistance myth

Question

I am making a battery tester, for lithium ion batteries in particular.

I want to measure the internal resistance, but after testing few cells, I am skeptical of my results. 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? Since new and old cells all have similar result, is it safe to say that internal resistance is irrelevant to the health of a li-ion cell?

Answer 1

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.

Answer 2

It depends on what you mean by internal resistance. The "effective" resistance can be measured from the DC current and DC voltage drop. However, as pointed out by mKeith, the DC voltage drop, \$dV_{DC}\$ has two parts:

\$dV_{DC} = V_{drop} + I*R\$

\$V_{drop}\$ is a voltage drop caused by drawing the current I, but it is different than the ohmic resistance R. If you want to know how much power is wasted as heat, that number is \$I^2 *R\$. mKeith points out 2 different ways of measuring R. One is using an AC voltage. Many battery spec sheets report R at 1 kHz. However, this value of R is typically less than the DC value of R, due to the finite capacitance of the cell. The second method recommended by mKeith is to measure the instantaneous voltage drop when a DC current is turned on. This method will give the DC value of R, as long as the measurement is not truly "instantaneous". You want to wait long enough, say 0.1 s, for the battery capacitance to fully charge, but not so long that the voltage drop appears. These effects are discussed in detail in the reference provided by jbord39 (http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3247723/). However, a careful reading of the paper shows that contrary to the assertion of jbord39, the authors specifically state that just measuring the DC voltage drop is not sufficient for obtaining R. Note that Figure 3 has a good illustration of the voltage drop. A load is applied at 10 seconds and removed at 30 seconds. However, the battery voltage at 31 seconds is less than the voltage at 9 seconds, even though the current is 0 in both cases. This is partly due to the voltage drop as the state of charge of the battery decreases, but also due to the voltage drop caused by a change in the chemical state of the battery. The voltage slowly increases with time from 31 to 65 seconds as the battery slowly returns to its original chemical state.

Answer 3

the correct way to measure the resistance is to start with a small preload constant load eg 100 -200 ma and then add the rest 800 mA to reach 1 amp.
so now u have the change in voltage for an 800mA load. now calculate dV/dI= internal resistance.