4
\$\begingroup\$

As a rule of thumb, charging should be terminated when the charging current drops to 0.1C at constant voltage phase of lithium battery charging. Membranes would have over stressed otherwise, which means "damage" to the battery.

What if we set a lower CV? For example, what if we set 4.1V as CV phase for a Lithium Ion battery and keep it under charge indefinitely? Would it still cause any damage?

\$\endgroup\$
  • \$\begingroup\$ Surely all this info is easily googled? \$\endgroup\$ – Andy aka Feb 23 at 16:52
  • \$\begingroup\$ @Andyaka Can you find such an information on Google? Do not hand over a link, just answer if you can or not. \$\endgroup\$ – ceremcem Feb 23 at 17:16
  • 2
    \$\begingroup\$ @Andyaka there's plenty of opinion if you google, but neither BatteryUniversity nor the battery manufacturers will answer this question definitively. I personally don't see how floating at 4.1 can be any worse than charging to 4.2, then recovering down to 4.1 and recharging to 4.2, rinse and repeat, but that's just me. \$\endgroup\$ – Neil_UK Feb 23 at 17:19
  • 1
    \$\begingroup\$ 'Optimum' is rarely the right word to use unqualified. The whole phrase is 'optimum for a purpose'. Keeping the battery at 50% SOC is different to keeping it at 4.1v, or the thick end of 100% SOC, which is what I'm sure the OP means. Cycling between 4.1 and 4.2 which is what most commercial chargers do is also a long way from 50% SOC. \$\endgroup\$ – Neil_UK Feb 23 at 18:00
  • 1
    \$\begingroup\$ See any MPPT IC datasheet, (BQ24650 for example), they terminate charging at 0.1C and "automatically restarts the charge cycle if the battery voltage falls below an internal threshold". I'm not aiming maximizing the lifespan of the Lithium battery. I want to know the limits to avoid a damage when a dedicated charger is not available. \$\endgroup\$ – ceremcem Feb 23 at 19:34
6
\$\begingroup\$

if we set 4.1V as CV phase for a Lithium Ion battery and keep it under charge indefinitely? Would it still cause any damage?

A Lithium Ion battery has limited lifespan at any voltage, but it degrades faster at higher voltage and temperature. So the question is, what amount of degradation do we call 'damage'?

In the graph below (from Aging of Lithium-Ion Batteries in Electric Vehicles by Peter Keil) we see that these particular cells (Panasonic NCR18650PD) show a distinct reduction in capacity after being stored at charge levels above 60%, but little difference between 88% (4.1 V) and 100% (4.2 V) charge at 25 °C. As storage temperature increases the degradation becomes more affected by state of charge, but variation at the high end is only significant well above normal ambient temperature.

enter image description here

Even though the voltage when continuously 'floated' would be a bit higher, this data suggests that lowering the voltage may not have much effect above ~3.8 V (60% charge) - at least for this type of cell. However it might be different for other cell chemistries.

Then you must balance the 'damage' against the loss of functionality at lower voltage. For longest life you should float below 3.8 V, but this provides less than 60% capacity. If having to top up before running on battery power is not acceptable then a slightly shorter lifespan is probably a small price to pay for not needing a battery with twice the rated capacity.

On the other hand, floating at 4.1 V more closely matches the long-term resting voltage of a fully charged cell. This may reduce the chances of catastrophic failure due to high voltage stress, while not greatly affecting the available capacity.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ The exact related part to the question is "This may reduce the chances of catastrophic failure due to high voltage stress" part, however this needs citation. \$\endgroup\$ – ceremcem Feb 24 at 3:20
  • \$\begingroup\$ In theory, the foolproof voltage that can't make any harm to the battery can be measured in the following way: At the termination current (0.1C) at standard CV (4.2V), voltage of the power supply would be slowly decreased to the point that the current drops to zero or a little bit less. This voltage must be the absolute maximum floating voltage for a lithium (ion) battery. This is just a theory though. \$\endgroup\$ – ceremcem Feb 24 at 3:51
  • \$\begingroup\$ At 4.2V a Li-ion battery is supposed to be (just) under the voltage that causes damage. But what if has a weak separator, insufficiently oversized anode, or slightly off chemistry? These are considered defects and therefore not an effect of charging to max voltage, but real batteries are not always perfect. I don't have a cite (which is why I said 'may') but just finding a good study of calendar aging has hard enough. It seems nobody is interesting in testing the 'common wisdom' that floating at 4.2V is bad... \$\endgroup\$ – Bruce Abbott Feb 24 at 10:22
  • \$\begingroup\$ The main thrust of my answer was actually to debunk the idea that floating at 4.2V is a very bad idea. But I only found one study of one type of cell. Meanwhile manufacturers are telling us to avoid leaving their batteries fully charged, eg. my Nissan Leaf manual recommends only charging to 80%. Do they know something we don't? Or are they just being over-cautious? I wish more studies were available that could provide definitive answers. \$\endgroup\$ – Bruce Abbott Feb 24 at 10:35
0
\$\begingroup\$

I conducted a simple experiment with a used 18650 Li-ion battery cell. The cell has no capacity label on it, which I guess it was originally between 1500mAh and 3000mAh.

1st Charge

I started charging by setting the charge voltage to 4.2V, limiting the charge current to 0.5A with a lab power supply and observed the terminal voltage.

When the terminal voltage reached to 4.19V, I pushed the current limit switch to full forward and saw that the "nominal" charge current (the constant charge rate) is 0.7A.

If the nominal charge rate were 0.5C (usually it is), then 0.5C = 0.7A, which means that the battery is probably rated with 1400mAh. So probably the cut-off current should be around 0.1C=140mA.

enter image description here

1st Discharge

Then I fully discharged the battery under constant current of 0.5A. It took 1.42 hours to reach down to 3.0V, which means the battery is effectively 710mAh.

enter image description here

enter image description here

It's obvious that the battery represents a very unfamiliar discharge graph from a new one.

2nd Charge

After fully discharged, I charged the battery again under 4.2V by limiting the current to 0.5A.

Charging Power

The experimental "SOC vs. Charging Power" table is as follows:

      | Charging curr.| Charging Voltage | Charging Voltage
SOC   | @4.2V         | @0mA (1st charge)| @0mA (2nd charge)
----- | ------------- | ---------------- | -----------------
~65%  | 700mA         | N/A              | N/A 
x1    | 350mA         | 4.08V            | 4.09V
x2    | 290mA         | 4.11V            | 4.11V
x3    | 200mA         | 4.15V            | 4.14V
x4    | 150mA         | 4.15V            | 4.15V
100%  | 100mA         | (missed, sorry)  | 4.16V
x6    | 0mA (for 10 hours)

I obtained the corresponding "zero current voltages" by setting the current limit to zero and reading the output voltage of power supply.

Conclusion

The safe voltage that the battery could be floated indefinitely can be determined by setting the current to zero at the rated charging termination current and reading the terminal voltage at that point.

For example, for the battery we tested here, the safe voltage that the battery can be floated indefinitely is 4.16V if we select the termination current as 100mA.

Regarding to Bruce Abbott's comments and my experiment, it is rational to assume that floating the battery at 4.2V indefinitely is also acceptable and should give no harm.

This experiment should be repeated with a brand new Li-ion battery for more accurate results.

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.