Li-ion charge termination sensing method

[mudman cj]

This question is aimed at Silverfox, but anybody should feel free to weigh in on it.

I have seen recommendations from Silverfox to charge Li-ion cells until the charging current is down to between C/10 and C/20. This is supposedly the safest way to charge Li-ion cells. That said, there are other methods in use, and some of these are even known hazards.

My question is if there are issues with sensing the proper charge termination point by voltage. I presume such a charger would apply a voltage in excess of the desired charged cell open circuit voltage, so something like 4.25V for a one cell charger, and end charge at near 4.20V.

The reason I suspect an issue is because I recall that Li-ion cells are ready to be replaced when they are down to 80% of original capacity when fully charged. As we are all aware, you can roughly determine the capacity of a cell by its open circuit voltage, but does this extend to the open circuit voltage corresponding to the proper charge termination point? In other words, do older cells gradually charge to lower and lower voltages as they age?

If so, then do aged cells with reduced capacity only charge to an open circuit voltage corresponding to that level of charge when using a current-sensed charge termination method? And would they be damaged in some way like dendrite formation within an electrode were they to be pushed beyond that C/20 charging point as with a 4.20V termination point?

 

[SilverFox]

Hello Mudman Cj,

I am not sure I completely understand your question, but let me see if I can shed some light on charging Li-Ion cells.

Li-Ion cells are supposed to be charged with a Constant Current/Constant Voltage algorithm. In the CV stage of the charge, the charge is supposed to be terminated when the current drops below a certain rate. The ending charge rate varies from 0.1C to 0.03C. An ideal charging rate is 0.7C, but most cells can be charged at 1.0C.

Let's put some numbers to this...

Let's start with a 2000 mAh cell. The maximum charge voltage is set to 4.20 volts and the maximum charge current can be set to somewhere between 1400 - 2000 mA. The cut off for the charge should occur when the charge current drops below around 100 mA (0.05C).

You hook the cell up to the charger, and the charger begins charging at your maximum charge rate, which let's call 2000 mA for this example. This is the CC portion of the charge. The cell will be charged at 2000 mA until the voltage of the cell reaches your maximum voltage of 4.20 volts.

Once the cell reaches 4.20 volts, the current drops off. This is the CV portion of the charge. The voltage is held at 4.20 volts and the current gradually drops off. When the current drops below 100 mA, the charger should terminate the charge.

Here is a graph that shows the CC/CV charge.

As Li-Ion cells age, their internal resistance increases. The effect of this increase in internal resistance is that when the voltage is held at 4.20 volts, and then shut off, the voltage of the cell will be reduced.

Now, I believe part of your question may have been that once you see your cell ending up with lower than 4.0 volts upon the completion of a normal charge, what would happen if you increased the charge voltage to 4.25 volts.

Charging to a voltage higher than 4.20 volts will accelerate the degradation of the cell. A well used cell has already degraded to the point that it should be retired, so the added degradation of charging to a higher voltage will accelerate the degradation that much faster.

You may be able to get a few cycles out of the cell, but the degradation will increase rapidly and the increase in internal resistance will cause the cell to heat up during charging. This process eventually ends up with the cell overheating and rapidly venting with flame.

Tom

 

[mudman cj]

As Li-Ion cells age, their internal resistance increases. The effect of this increase in internal resistance is that when the voltage is held at 4.20 volts, and then shut off, the voltage of the cell will be reduced.

Charging to a voltage higher than 4.20 volts will accelerate the degradation of the cell. A well used cell has already degraded to the point that it should be retired, so the added degradation of charging to a higher voltage will accelerate the degradation that much faster.

Tom

Hi Tom,

Thanks for your thorough answer. That provides a better background to my question for the benefit of others. I think we are on the same page here, but I want to be sure.

What is the magnitude of the difference in open circuit voltage measured across a cell that is near retirement compared with a new cell (which would give 4.20V when fully charged)? Will they charge to only 4.10V [Edit: Whoops, that should have been more like 4.00V @ 80%] open circuit when they have about 80% remaining capacity if charged with 4.20V applied until current is 0.05C?

Wouldn't the higher internal resistance of such a cell cause it to be charged at <0.05C with a 4.20 applied voltage if charging was not terminated until a voltage of 4.20V was measured across the cell during charging?

 

[SilverFox]

Hello Mudman Cj,

Using the proper Li-Ion charging algorithm, a cell is ready to be retired when it drops from 4.2 volts at the end of the charge, to 4.0 volts in around 15 minutes after the end of the charge. A cell at a resting voltage of 4.0 volts is at roughly 80% of its total capacity.

I think you need to take another review of the CC/CV graph. The CC portion of the charge continues until the 4.20 voltage is reached. As the cell reaches 4.20 volts, the charge changes to the CV phase where the voltage is held constant at 4.20 volts while the current drops off.

The charge is not terminated when the cell reaches 4.20 volts, only the CC portion of the charge.

Once again to throw some numbers out, at a 0.7C charge rate, the CC portion of the charge will take around half of the charge time. The remaining time is spent in the CV portion of the charge. When you bump the rate up to 1C, the CC portion is reduced and the CV portion increases. This is why higher charge rates don't really reduce the charging time all that much. If you charge at 2C or 3C, you spend very little time in the CC portion of the charge and a lot of time in the CV portion. This is also why a 1C charge rate takes around 90 minutes to charge a cell.

Going back to the original example with the 2000 mAh cell, we have the charger charging at 2000 mA until the cell voltage reaches 4.20 volts. Then the charger adjusts the current down to maintain the cell at 4.20 volts until the current drops below 100 mA. Then the charger should shut off. When the charger shuts off, the cell voltage will drop from 4.20 volts to its resting voltage, based on the cells internal resistance.

A new cell may come off the charger with a resting voltage of 4.195 or 4.19 volts, and a cell ready for retirement will settle in at 4.0 volts.

Tom

 

[mudman cj]

Thanks Tom. That sag in the open circuit cell voltage over 15 minutes was part of the answer I was looking for.

My questions about charge termination sensed by cell voltage came from a charger that is claimed to use this method in its description. Perhaps that description just came about due to a misunderstanding of the way that charger actually works; because it is clear to me now that were a charger to terminate charging upon reaching the CV phase, it would not result in a fully charged cell.