There has been a lot of discussion on how to figure out the state of charge on Li-Ion cells by measuring their resting voltage.

I picked up some information on high current draws that gives the following values:

4.2V – 100%

4.1V – 87%

4.0V – 75%

3.9V – 55%

3.8V – 30%

3.5V – 0%

Please note that resting voltage means the cell has stabilized at room temperature and the voltage has also stabilized.

I decided to check a brand new 18650 cell at a defined current draw. This cell is a Pila 600P rated at 2200 mAh. The test current was 2 amps with a low voltage cut off of 2.8 volts.

At a 2 amp current draw, this is what I observed:

4.20 volts – 100%

3.97 volts – 80%

3.85 volts – 60%

3.77 volts – 40%

3.72 volts – 20%

3.58 volts – 0%

This cell tested at 2000 mAh capacity at 2 amps. I ran 2 amps for 400 mAh, then stopped the test to let the cell and voltage stabilize. I then continued to do this 5 times to come up with the values listed.

Luna brought up a question involving the state of health of the cell. I got to thinking and realized that I have a Sony 18650 that has been abused by over charging, over discharging, and has been cycled to near death. I charged it up and ran a similar test on it.

On major difference was that I discharged at 1 amp.

Here are the results for a "damaged" cell, that should be thrown away, at a 1 amp current draw down to 2.8 volt cut off:

4.20 volts - 100%

3.95 volts - 80%

3.85 volts - 60%

3.68 volts - 40%

3.45 volts - 20%

3.31 volts - 0%

It is interesting to note that at this lower discharge rate, there was less voltage bounce back during the last half of the discharge.

Tom

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