Batteries Charging Routine

[Woosh]

That can give very unpredictable results. The battery voltage will almost certainly start to fall unless the charge current is very small. There is a difference between (1)setting the charger to 4.1 volt/cell and (2)interrupting the 4.2v charger at 4.1v/cell. In the former case, the charger raises the voltage under limited constant current and then exposes the cells to 4.1 volt/cell (constant voltage) until the charge current tapers down. This saturates the cells to 4.1 v.
In the latter case, the cells wont even be at 4.1 volts EMF due to battery internal resistance and they wont be exposed to the constant voltage and certainly wont be saturated at 4.1 v.
I have tried to explain this by drawing in green lines in the drawing below. In this rather extreme example to illustrate the problem of a quick charging setup, interrupting the 4.2 charger at 4.1 volts only gives 60% charge. (The severity is high because the charge rate is high, = 1C)
If a 4.1 cell charger was used in this case instead of interrupting the 4.2 charger, the voltage slope (red line)would start to plateau at 4.1v instead of 4.2 and the plateau constant voltage/floating and saturation would take place without interruption at this 4.1 voltage.
I hope this explains. (Its the only diagram I could quickly find)
View attachment 52921

I understand your point.

That can give very unpredictable results. The battery voltage will almost certainly start to fall unless the charge current is very small. There is a difference between (1)setting the charger to 4.1 volt/cell and (2)interrupting the 4.2v charger at 4.1v/cell. In the former case, the charger raises the voltage under limited constant current and then exposes the cells to 4.1 volt/cell (constant voltage) until the charge current tapers down. This saturates the cells to 4.1 v.
In the latter case, the cells wont even be at 4.1 volts EMF due to battery internal resistance and they wont be exposed to the constant voltage and certainly wont be saturated at 4.1 v.
I have tried to explain this by drawing in green lines in the drawing below. In this rather extreme example to illustrate the problem of a quick charging setup, interrupting the 4.2 charger at 4.1 volts only gives 60% charge. (The severity is high because the charge rate is high, = 1C)
If a 4.1 cell charger was used in this case instead of interrupting the 4.2 charger, the voltage slope (red line)would start to plateau at 4.1v instead of 4.2 and the plateau constant voltage/floating and saturation would take place without interruption at this 4.1 voltage.
I hope this explains. (Its the only diagram I could quickly find)
View attachment 52921

I understand your explanantion. The reason for that is fast charging creates a film of passivation on the surface of the electrodes causing the resistance to charging to go up, therefore the real capacity is less than what the apparent charging voltage would suggest. When the charging is near complete, the charging current reduces, dissipating the passivating layer and the capacity is what you would expect. However, Stuart's numbers are quite a bit more than I expected. I expected up to 10%, Saneagle measured it at 3.1%. Stuart found it at 24%.

 

[StuartsProjects]

When my battery, which had been charged to 42V, was discharged, it went from 42V to 41V almost instantly, but that says nothing about what the capacity would be when the charge is stopped at 41V.

 

[saneagle]

With the battery I was testing, 24% of the total charge in occurs from 41V to 42V.

That's impossible. You made a mistake somewhere.

Look at what you wrote in post #102. How can you lose 24% of your charge instantly without setting fire to your house?

 

[Woosh]

I think it's possible but what Stuart found if repeateable may be attributable to the small capacity 5ah of his pack. You don't see it on your pack because there is very little passivation on a large pack. Passivation is non linear and increases rapidly with changing current.

 

[Nealh]

24% of the time taken to charge from 41v - 42v isnt't due to capacity but the slower time that blancing takes , so the balancing % time is disproportionate to the total charge time.
The charge goes from 2a etc,etc , to a lowly paltry 50 - 100ma for balancing.

 

[StuartsProjects]

24% of the time taken to charge from 41v - 42v isnt't due to capacity but the slower time that blancing takes , so the balancing % time is disproportionate to the total charge time.

The 24% was the amount of Ahr, not time.

At the relavent voltages the amount of charge in was;

41V, 3.56Ahr
41.5V,3.81Ahr
41.8V,4.32Ahr > constant Volt charge begins.
42V, 4.64Ahr > charge stops.

 

[Woosh]

24% of the time taken to charge from 41v - 42v isnt't due to capacity but the slower time that blancing takes , so the balancing % time is disproportionate to the total charge time.
The charge goes from 2a etc,etc , to a lowly paltry 50 - 100ma for balancing.

he discharged the battery using the same load resistor. 135 minutes from initial 42V to 30V but 106 minutes from 41V, so 29 minutes difference or 21.5% capacity.
What he demonstrated is the problem with fast charging. If you stop before the constant voltage charging stage, you can't charge at more than 80%

 

[saneagle]

The 24% was the amount of Ahr, not time.

At the relavent voltages the amount of charge in was;

41V, 3.56Ahr
41.5V,3.81Ahr
41.8V,4.32Ahr > constant Volt charge begins.
42V, 4.64Ahr > charge stops.

You can't draw any meaningful conclusions from those measurements. During that time, the charge current was decreasing. The charger holds the voltage higher than the battery voltage. That was very noticeable on my 20A battery charging at 0.1C. On yours charging at 0.4C, the efeect would be much more. As the charge reaches 100%, the charge current reduces right down so the charger no longer holds the voltage up and the voltage you measure is closer to the actual battery voltage.

In other words, when you measured 41v, the actual battery voltage was much lower, something like 39v. Next time, while the battery is on charge in the middle of its range, measure the voltage, then disconnect the charger and watch the voltage drop to the actual battery voltage. That's the explanation for your anomalous results. QED.


Additional info: During the charge cycle on my battery at 95 watts, I noticed that when the battery reached 53.9v, the charging reduced to 65w and went down as follows:
53.7v 95W
53.9v 65w
54.0v 52w
54.1v 21w

I stopped recording at that point, but it went down to 6w, then stayed there for a bit before switching off at around 54.2v, which was too low, so I changed to another charger. The main point is the gradual decline in charge current. I can't say that any other chargers do that. You have to test yours to see. Regardless of that, there's a massive difference in charge current between the CC and CV stages. The effect of either holding up the battery voltage during CC and not holding it at CV is the same.

 

[Sturmey]

I am inclined to wonder if charging a battery can be compared to cooking a chicken. The chicken can be crispy and brown on the outside but frozen on the inside. The chicken needs time for the temperature to get to the center.
The cells also need a bit of constant voltage/floating/saturation time after the initial high constant current for the voltage to equalize on the plates. Putting them on a charger and pulling them off before they have floated is like pulling a chicken out of oven before its cooked. They will look charged like the chicken looks cooked. (Just something I am thinking about while I am waiting for my dinner to be cooked)

 

[StuartsProjects]

The chicken can be crispy and brown on the outside but frozen on the inside. The chicken needs time for the temperature to get to the center.

If by that you mean that cooking a chicken at a lower temperature (charge current) for longer, then yes its likey the chicken stands a better chance of being fully cooked (fully charged).

The problem I have with that analysis is that the longer you cook a chicken, in general, the smaller it gets.

OTOH, Lidl use to do a German sausage that expanded as you cooked it, so maybe not all meats are the same .

 

[Sturmey]

If by that you mean that cooking a chicken at a lower temperature (charge current) for longer, then yes its likey the chicken stands a better chance of being fully cooked (fully charged).

The problem I have with that analysis is that the longer you cook a chicken, in general, the smaller it gets.

OTOH, Lidl use to do a German sausage that expanded as you cooked it, so maybe not all meats are the same .

Stuart, look at the graph below and see what you think of it. The Samsung is under charge at 1 Amp. At approx 237 minutes, it has reached a voltage of 4.1 volts (red line) and has reached a capacity of 4 Ah (blue line). If cell is allowed to go to full charge (315 minutes),it will have reached a capacity of 4.8 Ah. So this works out at about 83.33% capacity or a 16.66% capacity loss when charger is interrupted at 4.1v. However you may have to make a small offset to take account of his charging from 2.8 volts but it doesn't seem much. If you are using a 2 Amp charger, I would expect that the capacity loss would be higher. Anyhow, I hope this helps and that I have interpreted it correctly and have not led to any confusion. The cell is a very good cell. The graph is from

Test of Samsung INR21700-50E 5000mAh (Cyan)

 

[Woosh]

I like Sturmey chicken analogy.
And the conclusion of this thread is?
You should wait until the LED turns green before switching off your charger. Your battery needs time to spread the charge inside its electrodes. You can undercharge your battery up to 20% if you use a fast charger and unplug it too early.

 

[StuartsProjects]

So this works out at about 83.33% capacity or a 16.66% capacity loss when charger is interrupted at 4.1v.

So the lost capacity, 16.66%, is in line with my discharge tests and is considerably more than some suggestions that the capacity loss is only a few percent.

 

[Woosh]

So the lost capacity, 16.66%, is in line with my discharge tests and is considerably more than some suggestions that the capacity loss is only a few percent.

yes, your battery received a lot less charge than sangeale and me expected, it's because you charged it at 0.4C and interrupted the charging before the constant voltage stage, Saneagle and me use much larger capacity battery, charge the battery at 0.1C and waited for the charger to complete the CV stage. That's why we did not relate to your numbers. In fact, if you make a graph of 2 plots of the voltage / remaining capacity for your two charging methods, you will see that the 16% discrepancies are spread along the plots, wider at the high voltage side, showing the lack of electrode saturation when interrupting the charger at 41V and skipping the CV stage.
I have seen this problem reported by customers before but couldn't remember. The remedy was to leave the charger on for 24 hours.

everybody's numbers are compatible with all the explanations I have seen here. If you want to charge only to 41V, you need that smart charger saneagle linked to in the other thread, you would have lost only 3.1%, same as Saneagle's big battery.

 

[saneagle]

So the lost capacity, 16.66%, is in line with my discharge tests and is considerably more than some suggestions that the capacity loss is only a few percent.

It's because you didn't actually charge to 4.1v. What you saw on the meter was not the battery voltage. Switch the charger off and watch the voltage immediately drop.

 

[Woosh]

He discharged the battery from 41V. It's because the CV phase was skipped, the battery has less charge than we expected. So yes, he did see 41v and his battery has 24% less than if it were charged normally to 41.7V .
Sturmey 's graph in post #111 made the point very well. The capacity of the charge is lagging behind the battery voltage in the CC phase until the CV phase when it slowly catches up.
Similar to what you see in hysteresis.

 

[Sturmey]

Just to mention something that Padja is suggesting in his ES testing.
Cells come in different qualities when it comes to cycle life.
Some are top quality ,some are medium quality and some are poor or of unknown quality e.g unbranded cells.
He seems to be suggesting that good quality cells such as the (Samsung 50e) because they are good can be fully charged and dont benefit that much from reducing there depth of discharge. He suggests only very minimal reduction of 5% max depth of Discharge/3 - 41.5 v fully floated if you want to get a very high cycle life in this case.
His suggestion is that it is the medium to poor quality cells that get the most benefit from fairly drastic reduction in their voltages/depth of discharge e.g 3.4 to 4.1 v.
This seems to make sense to me. Its the poorer quality batteries that wear/come under pressure the most when subject to 4.2 v. But the good quality cells seem to hold up well.
There is separate issues with high discharge cell which has its own tread.

 

[saneagle]

He discharged the battery from 41V. It's because the CV phase was skipped, the battery has less charge than we expected. So yes, he did see 41v and his battery has 24% less than if it were charged normally to 41.7V .
Sturmey 's graph in post #111 made the point very well. The capacity of the charge is lagging behind the battery voltage in the CC phase until the CV phase when it slowly catches up.
Similar to what you see in hysteresis.

Yes, he saw 41v, but that was the charger voltage, not the battery voltage. If he switched off the charger, he'd see a battery voltage of about 39v. You've all been making it more complicated than it is.

I noticed this when I did my charging test at a much lower C-rate. I had a meter on it to show the battery voltage. As soon as I switched the charger on, it jumped 2V without time to add any charge. You can see from that last data I gave you that the charger ramps down the current when it approaches the top of the charge. The lower the current, the less the difference becomes. Stuart can easily demonstrate that by switching off his charger during charging to see the difference in voltage in the middle of the charge. It's going to be quite a lot charging at 0.4C.

 

[Woosh]

He did see an immediate drop but only 0.3V-0.6V from 42V to 41.3V and from 41V to 40.4V. Stuart is experienced in electronics, I have no reason to think that he did not measure voltages correctly. He uses a single load resistor to mimic his bike ride, one full battery gives him about 2 hours 15 minutes from 42V and 1 hour 46 minutes from 41V, hence 29 fewer minutes or 29/135=21.4%, much more than you found (3.1%) or I expected (up to 10%).
The explanation is given by Sturmey's roast chicken analogy, the Lithium ions (or equivalent heat) are not reaching the part where they do with constant voltage charging, as illustrated in his post #111.
Think about how the difference in voltage between the two electrodes is created in the first place. You'll see that the capacity can be different depending on the way the battery is charged. Flecc made the same point with his car. That's why I reminded you the hysteresis.

Rest volts 41.7V	Rest volts 40.4V
41V, 0.01Ahr,0min
40V, 0.384A,10min	40V,0Ahr,0min
39V,1.16Ahr,31min	39V, 0.295Ahr,8min
38V,1.63Ahr,44min	38V,0.748Ahr,20min
37V,2.25Ahr, 62min	37V,1.26Ahr,35min
36V,2.72Ahr,75min	36V,1.75Ahr,50min
35V,3.23Ahr,91min	35V,2.25Ahr,64min
34V,3.7Ahr,105min	34V,2.72Ahr,79min
33V,3.97Ahr,114min	33V,2.97Ahr,87min
32V,4.2Ahr,122min	32V,3.18Ahr,94min
31V,4.41Ahr,129min	31V,3.37Ahr,100min
30V,4.59Ahr,135min	30V,3.53Ahr,106min

 

[saneagle]

He did see an immediate drop but only 0.3V-0.6V from 42V to 41.3V and from 41V to 40.4V. Stuart is experienced in electronics, I have no reason to think that he did not measure voltages correctly. He uses a single load resistor to mimic his bike ride, one full battery gives him about 2 hours 15 minutes from 42V and 1 hour 46 minutes from 41V, hence 29 fewer minutes or 29/135=21.4%, much more than you found (3.1%) or I expected (up to 10%).
The explanation is given by Sturmey's roast chicken analogy, the Lithium ions (or equivalent heat) are not reaching the part where they do with constant voltage charging, as illustrated in his post #111.
Think about how the difference in voltage between the two electrodes is created in the first place. You'll see that the capacity can be different depending on the way the battery is charged. Flecc made the same point with his car. That's why I reminded you the hysteresis.

Rest volts 41.7V	Rest volts 40.4V
41V, 0.01Ahr,0min
40V, 0.384A,10min	40V,0Ahr,0min
39V,1.16Ahr,31min	39V, 0.295Ahr,8min
38V,1.63Ahr,44min	38V,0.748Ahr,20min
37V,2.25Ahr, 62min	37V,1.26Ahr,35min
36V,2.72Ahr,75min	36V,1.75Ahr,50min
35V,3.23Ahr,91min	35V,2.25Ahr,64min
34V,3.7Ahr,105min	34V,2.72Ahr,79min
33V,3.97Ahr,114min	33V,2.97Ahr,87min
32V,4.2Ahr,122min	32V,3.18Ahr,94min
31V,4.41Ahr,129min	31V,3.37Ahr,100min
30V,4.59Ahr,135min	30V,3.53Ahr,106min

Go to any of your half-charged batteries, measure the voltage, then plug in the chargere and watch it jump up, then remove the charger and watch it go down to where it was.

The one thing you're missing is that it's not possible to have 23% of the charge between 41v and 42v. It's 3%. His data is false because his method is false.