Lithium Ion 18650 cell capacity

[Harald Kapp]

Therefore, should I perform the same discharge test on the cells down to a cut off voltage of 2.5V?

Right

3200mAh = 3.2A
3.2A x 0.2C = 0.64A

Mathematically incorrect, but you get the gist . 0.64 A is correct.

 

[BobK]

I think your question has been answered. You discharged to 3V, they discharged to 2.5V when rating the capacity. This would account for the difference.

To answer your question though, you need a constant current sink to draw 620mA.
Here is a simple circuit:



The supply labeled 2V must be a stable reference voltage independent of the battery, since the battery voltage will be changing. You should get 620mA with something close to 1.2V on the base of Q1.

Bob

 

[Ronnie_Space]

Right

Mathematically incorrect, but you get the gist . 0.64 A is correct.

Thank you.

 

[Ronnie_Space]

I think your question has been answered. You discharged to 3V, they discharged to 2.5V when rating the capacity. This would account for the difference.

To answer your question though, you need a constant current sink to draw 620mA.
Here is a simple circuit:

View attachment 34752

The supply labeled 2V must be a stable reference voltage independent of the battery, since the battery voltage will be changing. You should get 620mA with something close to 1.2V on the base of Q1.

Bob

Thank you for your time and help. I would like to try this. Can you recommend a transistor for Q1? ( I would not know where to start).

I am pretty sure I have an adjustable DV power supply I could feed 2V from.

 

[Audioguru]

2.5V is too low to discharge a cobalt-Lithium-ion cell. It makes their capacity rating higher and you will be replacing the over-discharged cell soon.

 

[Arouse1973]

To get close to the capacity claimed within the cells data sheet you have to discharge and charge to the spec they say. If that is what you have put above then doing this should get you close.

You will need a low voltage constant current sink to allow you to do the discharge test. Using just a resistor will introduce errors because as the cell voltage reduces the discharge current also reduces so it's difficult to do charge calculations and that's charge as in Coulombs, not anything to do with charging

Thanks
Adam

 

[(*steve*)]

If they want to calculate capacity by effectively "red lining" the battery on both the charge and discharge phase, my recommendation is that you don't want to repeat their tests.

Decide on your own charge and discharge limits, test with these, and live with the smaller capacity and longer life of the cells.

In some applications the charge limit is 3.92 volts. Serious capacity loss, but equally serious longevity.

 

[Ronnie_Space]

Thanks for all the replies.

I was really looking to quantify if the cells are indeed what they say they are (3200mAh), but it sounds like quite a specific charging/discharging process to do so.

However, I guess in a real world scenario my previous test proves realistically what I am going to get out of the powerbank... which is 16500mAh, meaning the powerbank has a conversion rate of 85% (19200mAh), which is what the vendor has claimed.

Ronnie

 

[BobK]

No, that is not what the conversion rate would mean. The 85% is for the conversion from the battery voltage to the regulated output.

Bob

 

[Ronnie_Space]

No, that is not what the conversion rate would mean. The 85% is for the conversion from the battery voltage to the regulated output.

Bob

In my initial test I fully depleted the powerbank and charged using a USB multimeter to measure capacity, repeating four times I am measured an average 16523mAh. To me this suggested that whilst the pack claims to be 19200mAh capacity, realistically I will get 85% of that stated capacity from it.

 

[marcobouillon]

It is important to carry out discharge tests on the cells to assess their performance, their autonomy and their real capacity. According to the LG INR18650 MH1 3200mAh battery specification you provided, the standard discharge should be done at a constant current of 0.2C to 2.5V.

To discharge the cell at a current of 0.2 C, you need to divide the nominal capacity of the cell (3200 mAh) by 5. Then you can use a suitable resistor to achieve the desired discharge current. In this case, the discharge current should be 640 mA (0.2 C * 3200 mAh = 640 mA). You can use an 11.5 ohm resistor to get this discharge current (V=IR; V=3.67V - 2.5V=1.17V; I=V/R; I=1.17V / 11.5 ohms = 101.7 mA; 640 mA / 101.7 mA = 6.29; so two 7 ohm resistors in parallel would supply approximately 6.29 ohms, giving a discharge current of approximately 640 my).

It is important to note that fully discharging a battery can reduce its life, so it is recommended not to discharge the battery below 3V to prolong its life. You can also monitor the battery voltage while discharging to ensure it does not drop below the recommended cut-off voltage.

 

[(*steve*)]

A constant current sink would be better. Also you want (in your example) about 5.8 ohms. Two 11.5 ohm resistors (not a standard value) would be close. 4 22 ohm resistors in parallel would be close enough. Two 7 ohm resistors (again, not a standard value) would give you 3.5 ohms, and a nominal discharge rate of just over 1A.