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I acquired 6 used li-ion cells 1 year ago. Since I wanted to know for certain how well the cells would hold a charge after 1 year, I conducted an experiment. I recently tested the cells. Below is some data related to my testing:

Charger info:
Voc = 5.25 (Open circuit Voltage--no cell in charger)
Imax = .68A (max charge current=680mA--no cell in charger)

Cell Info:

Manuf:  Panasonic
   Id:  CGR18650CE
    V:  3.6 (nominal)
  mAh:  2150 (typical capacity when fully charged)
+Chrg:  CC=1.43A (max), 4.2V (max)     // "+Chrg"=Charging
-Chrg:  CC=2.04A (max), cut-off @ 3V   // "-Chrg"=Discharging

Notes:
- All cells were charged at a CC=.68A to a CV=4.2V. Charging was cutoff when the charge current reached .01A@4.2V
- The resting V of all cells after 24 hrs was at least 4.05V.

Here are the cell measurements after 1 year of resting:

Cell V___ A___ VA(W)  Est mAh (i.e. VA/3.6*1000)
1    4.16 3.29 13.67  3802
2    4.01 3.23 12.95  3598
3    3.95 3.18 12.56  3489
4    3.85 3.09 11.90  3305
5    3.95 3.17 12.52  3478
6    3.77 2.95 11.12  3089

The multimeter was verified to be operating properly using a bench power supply.

We can see from the data that some cells held their charge better than other cells. My question is, how can the cells have such a high capacity when the datasheet indicates their capacity is 2150mAh? I know the cells are at least 5 years old. I also realize that my estimated mAh capacity is not based on actual discharge tests, but is there some other way of estimating the capacity of such cells using a multimeter & a bit of math?

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  • \$\begingroup\$ Was your voltage reading off-load? What load did you have on to measure the current? \$\endgroup\$ – HandyHowie Oct 7 '15 at 19:47
  • \$\begingroup\$ Yes, the Voltage reading was terminal voltage read with just a Voltmeter. \$\endgroup\$ – zeffur Oct 7 '15 at 20:09
  • \$\begingroup\$ You don't explain very well how you estimated the cell capacity. Obviously, the estimates are totally wrong, which is not surprising because you have not taken any data which would allow an estimate. Connect the cells to some a constant current load, or if you don't have one, a lamp or resistor and discharge them and record voltage and current at as many intervals as you can. You can stop the discharge at, say, 3V. You should aim for a current of around 500 mA or so, or if you are in a hurry, just do it at 1A (approximately). \$\endgroup\$ – mkeith Oct 7 '15 at 20:17
  • \$\begingroup\$ The load for the test was provided by the Ammeter. \$\endgroup\$ – zeffur Oct 7 '15 at 20:26
  • \$\begingroup\$ Did you short the cells with the multimeter in current measurement setting ??? \$\endgroup\$ – Bimpelrekkie Oct 7 '15 at 20:33
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I don't get the point where you get from the cell voltage to the capacity. What you need to do is to discharge the cell with a constant current and measure the time until 2.7V is reached. This is done best with a electrical load or 4 quadrant source.

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  • \$\begingroup\$ Volts x Amps = Watts. Wh/nominal V = Ah (I noted that in the "Est mAh (i.e. VA/3.6*1000)" column heading for the data that I provided. I know it is necessary to perform a proper discharge test to get an accurate capacity for an energy cell--that's not the point of the question that I asked--which is "...is there some other way of estimating the capacity of such cells using a multimeter & a bit of math?" \$\endgroup\$ – zeffur Oct 7 '15 at 20:21
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    \$\begingroup\$ The answer is no. The momentary cell voltage does not tell you much. On a worn cell the "full" voltage is also 4.1V but it will drop very fast once you draw current. On a good cell the voltage will remain at 3.8V for a long time and then decline sharply as the battery becomes empty. \$\endgroup\$ – optronik Oct 7 '15 at 21:24
  • \$\begingroup\$ I agree that V is not the only important factor to consider when trying to determine the stored energy in a cell. By looking at my data, you will see that I also measured the output of each cell using the internal resistance of the Ammeter. From that I calculated/estimated Wh. From that figure & the nominal V of the cell, I estimated available capacity. \$\endgroup\$ – zeffur Oct 7 '15 at 22:11
  • \$\begingroup\$ Now, I know that number is not correct, however, it should be somewhat useful as it provides a relative difference from cell to cell (i.e. stronger cells have higher Wh & est. mAh & weaker less of each)... Now the question is--can my method be correlated to some approximate energy capacity of a cell? Finding a reasonably accurate method to determine energy capacity of a used cell with V and A measurements & a bit of math. Or, if someone else has another method, I'd certainly like to consider it. \$\endgroup\$ – zeffur Oct 7 '15 at 22:11
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    \$\begingroup\$ Since the cell voltage is not linear with the capacity I see no feasible way to do a proper estimation. \$\endgroup\$ – optronik Oct 7 '15 at 22:36
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No, you cannot estimate battery capacity quickly and accurately with a multimeter. Measuring battery capacity is complex, and the most accurate methods require a full discharge cycle (and even these aren't perfect). Otherwise, you need techniques like coulomb counting during charge and discharge, or look at the battery's response in the frequency domain to AC pulses. There are specialized ICs built for these purposes.

For more info on estimating capacity and state-of-charge, see here: https://batteryuniversity.com/learn/article/how_to_measure_capacity

and for info on how one IC manufacturer does it, see here: https://www.maximintegrated.com/en/design/partners-and-technology/design-technology/modelgauge-battery-fuel-gauge-technology.html

As to why your back-of-the-envelope method didn't work, to measure power from the battery you need to multiply voltage provided and current being drawn at the same time. The open-circuit voltages you see will immediately drop when you apply a load, and in general, for most of the discharge cycle, the voltage will be about 3.6V, so multiplying current draw by the voltage you see now will be an overestimate. And multiplying that by a burst of short-circuit current won't tell you anything meaningful about capacity.

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To measure cell capacity accurately you should charge a cell to a specified voltage, for example 4.1 or 4.2 Volts. Then DISCHARGE the cell with a constant current, for example 1 A, until the unloaded voltage (when no current flows) has reached 3.6 V (for example).

It is also not completely clear to me if you assumed that the charging current is constant. Because it should not be ! A proper Li-Ion charger will charge with a lower current when the cell's voltage is quite low and when it is nearly full. Only when the cell is around 40 % - 80 % full the charge current should be at it's maximum value.

You measured the charging of the cells which is not a 100% efficient process, some of the energy will be converterted into heat. This can explain why the measured capacity appears higher. The cell probably did not absorb all the mAhrs.

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  • \$\begingroup\$ Does your response answer the question "...is there some other way of estimating the capacity of such cells using a multimeter & a bit of math?" I charged the cells 1 year ago & allowed them to rest for 1 year. The measurements are recent & they were obtained using a multimeter (Voltmeter & Ammeter). \$\endgroup\$ – zeffur Oct 7 '15 at 20:19
  • \$\begingroup\$ What I explained is rougly the procedure that a battery manufacturer would use to measure the capacity of a cell. You followed a different procedure so the result cannot be compared. No you cannot repeat this procedure using a multimeter, you need a programmable load that can draw a constant current. You could use a resistive load (like 4 ohms) and every minute write down the voltage until the cell is 3.6 V then you would need to integrate the currents over time. That is complicated so that is why constant current discharging is used. \$\endgroup\$ – Bimpelrekkie Oct 7 '15 at 20:30
  • \$\begingroup\$ I agree with you & I was already aware of how a manuf. conducts such testing using a test setup that does exactly as you described. My question is more along the lines of how a person with a multimeter using the Voltmeter & Ammeter might be able to get a reasonable approximation of the capacity of a cell after they have charged it. The Ammeter uses a constant resistance. I use my V & A measurements & a bit of math to derive the other data in the table. As you can see & probably calculate, the max est mAh rating is 3802 which is ~1.77 times greater than the rating provided by the manuf. \$\endgroup\$ – zeffur Oct 7 '15 at 20:36
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    \$\begingroup\$ @zelfur - yes, your question was about finding an accurate way of measuring Lion capacity with just a multimeter and the answer must be no. Also, in your measurments, you measured watts, not watt-hours if I understand your experiment correctly. If that is the case, your estimate if capacity has no meaning at all. Also, when measuring the amps, was the voltage being measured simultaneously, or was the voltage measured first and then amps? Because when you draw 2-3 amps from an 18650, the voltage will drop significantly while providing the amps, and that is the voltage you want to measure. \$\endgroup\$ – Filek Oct 11 '15 at 6:24
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    \$\begingroup\$ I wish there was a way to use the multimeter only, but the more I think about it, the more I realize that the readings tell us very little about capacity. I just don't see a way around the old fashioned discharge or partial discharge to estimate capacity. \$\endgroup\$ – Filek Oct 11 '15 at 6:30

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