You are all destroying your LiFePO4 cells! - DIY Electric Car Forums

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#1
11-07-2011, 02:48 PM
 bgeery Member Join Date: Oct 2011 Location: Yucca Valley, CA Posts: 73
You are all destroying your LiFePO4 cells!

OK, so I'm exaggerating a little bit. But you are going to get fewer life cycles than specified by the manufacture if used under that standard conditions.

A fully 100% charged battery has a resting voltage of 3.33 volts per cell at 25 Celsius.

The only way to insure we never overcharge the battery is to charge at 3.33 volts until it tapers down to 0.00 amps. That's a fully 100% fully charged cell. For those that want to charge to less than 100%, just discontinue that charge before current drops to 0.00 amps.

The problem with this is that it takes longer to charge this way; so the manufacture gives us a formula of at 25 Celsius, apply .5C, until the cell reaches 3.6 volts, then disconnect the charge. This is 100% charged.

I see allot of people charging at 3.5-3.8 volts per cell until 0 amps. This is overcharging the cells to some degree. Yes, they may accept some additional charge above 100%, but it's past the manufacture's definition of the 100% charged point.

Either charge at 3.33 volts until it tapers down to 0.00 amps, or go with the manufacture's formula of .5C charge rate to 3.6 volts and disconnect. Either one is the only known safe and accurate charging procedures to reach 100%.

How's that for a first post?

PS: Other charging Formulas could be developed by the manufacture, if they choose to do so. We could also, if we knew the exact conditions they use to determine and define 100% state of charge.

Last edited by bgeery; 11-07-2011 at 03:08 PM.

#2
11-07-2011, 03:59 PM
 frodus Senior Member Join Date: Apr 2008 Location: Portland, OR Posts: 3,778
Re: You are all destroying your LiFePO4 cells!

..... except you won't cause any current to flow if the battery is at 3.33V and the charger/power supply is at 3.33V. There has to be a voltage difference between the two.... period.

The larger that differential between supply and load (battery) the larger the amps you'll draw from the supply. Thats how you cause current to flow. Most chargers we use are CC/CV and they current limit to XXAmps, and voltage limit to YYVolts.

As long as you don't A) go over the manufacturer max charge current B) Manufacturer max temperature or C) manufacturer max voltage, you're going to be within the limits of the cell. That's how the manufacturer actually rates cell life-cycle, using those specs.

We're not overcharging them at all. It's actually the only way you can charge a battery. You have to pull the battery voltage above it's nominal voltage and cause current to flow into it.
__________________
Travis Gintz
Electro Motive Force, LLC.
#3
11-07-2011, 04:33 PM
 bgeery Member Join Date: Oct 2011 Location: Yucca Valley, CA Posts: 73
Re: You are all destroying your LiFePO4 cells!

Quote:
 Originally Posted by frodus ..... except you won't cause any current to flow if the battery is at 3.33V and the charger/power supply is at 3.33V. There has to be a voltage difference between the two.... period. The larger that differential between supply and load (battery) the larger the amps you'll draw from the supply. Thats how you cause current to flow. Most chargers we use are CC/CV and they current limit to XXAmps, and voltage limit to YYVolts.
Agreed. But if the cell is at 3.33 volts resting voltage, it's already at 100% state of charge, and any further charging is an overcharge. You don't want further current flow.

A battery under 100% SoC will rest at something less than 3.33 volts until it's fully charged. The charge will be slower and slower as it approaches 3.33 volts, but the current will flow, until the cell reaches the true 3.33 resting voltage equilibrium with the charger and is at 100% SoC, and current flow will cease.

Using higher than resting voltage on the cell does nothing but increase the rate you can charge; avoiding the long trickle charge that would occur as the cell approached 3.3 volts and less and less current flowed. That's a valid practical compromise, but it doesn't get a 100% charged cell. It gets a 100.2% or 99.6% charged cell.

Quote:
 Originally Posted by frodus As long as you don't A) go over the manufacturer max charge current B) Manufacturer max temperature or C) manufacturer max voltage, you're going to be within the limits of the cell. That's how the manufacturer actually rates cell life-cycle, using those specs. We're not overcharging them at all. It's actually the only way you can charge a battery. You have to pull the battery voltage above it's nominal voltage and cause current to flow into it.
The manufacture gives us exatly *one* charge curve that they rate the battery at, and if you are not charging in exactly the same conditions, you can't tell your state of charge with true accuracy. You can come close enough for practical use, but you still don't really know exactly.

For example, how many people are charging at 3.5-3.8 volts until the charge tapers to 0.0 amps? I see top balance supporters mentioning this all the time here. Where does that extra power go between the 3.8 volts and the 100% charged resting voltage of 3.33 volts? It's not a whole lot of energy, but it is an overcharge of the battery. My guess is that it's oxidizing some of the plates and diminishing some minor amount of cell capacity.

#4
11-07-2011, 04:44 PM
 DavidDymaxion Senior Member Join Date: Dec 2008 Posts: 1,377
Re: You are all destroying your LiFePO4 cells!

Great first post. I throw this out as a discussion point, and not as hard science. Jack mentioned this in his videos. (All hail Jackton! All hail Jackton!)

The voltage for a cell is Vcell = V0 + I*R

So suppose you have a cell with 5 milliOhms of resistance, you want to charge at 10 Amps.

Vcell = 3.33V + 10 Amps * 0.005 Ohms = 3.38V

Suppose you are regenning at 100A (or have a super powerful charger):

Vcell = 3.33V + 100 Amps * 0.005 Ohms = 3.83V

Anyway, it's a thought that if the cells are cool enough and the currents are low enough, perhaps this isn't hurting them at all?
Quote:
 Originally Posted by bgeery OK, so I'm exaggerating a little bit. But you are going to get fewer life cycles than specified by the manufacture if used under that standard conditions. A fully 100% charged battery has a resting voltage of 3.33 volts per cell at 25 Celsius. The only way to insure we never overcharge the battery is to charge at 3.33 volts until it tapers down to 0.00 amps. That's a fully 100% fully charged cell. For those that want to charge to less than 100%, just discontinue that charge before current drops to 0.00 amps. The problem with this is that it takes longer to charge this way; so the manufacture gives us a formula of at 25 Celsius, apply .5C, until the cell reaches 3.6 volts, then disconnect the charge. This is 100% charged. I see allot of people charging at 3.5-3.8 volts per cell until 0 amps. This is overcharging the cells to some degree. Yes, they may accept some additional charge above 100%, but it's past the manufacture's definition of the 100% charged point. Either charge at 3.33 volts until it tapers down to 0.00 amps, or go with the manufacture's formula of .5C charge rate to 3.6 volts and disconnect. Either one is the only known safe and accurate charging procedures to reach 100%. How's that for a first post? PS: Other charging formulas could be developed by the manufacture, if they choose to do so. We could also, if we knew the exact conditions they use to determine and define 100% state of charge.
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#5
11-07-2011, 05:21 PM
 piotrsko Senior Member Join Date: Dec 2007 Location: RENO eNVy Posts: 1,346
Re: You are all destroying your LiFePO4 cells!

my \$.02:
It is a CHEMICAL reaction. the current is what is forcing the chemicals to physically revert / change to some sort of different state. When the charge current goes to zero, then theoretically all the chemicals have been changed, all the electron holes in the collector plates are empty again. Not exactly true, but close enough for this argument. As you overcharge you may even apply a slight coating made from the collector plates to the materials in the electrolytic thus altering the characteristics, possibly forever. In one of the battery books I have read, as little as 1 percent chemistry change destroys the cell. heat does this process also.
ymmv
#6
11-07-2011, 05:23 PM
 frodus Senior Member Join Date: Apr 2008 Location: Portland, OR Posts: 3,778
Re: You are all destroying your LiFePO4 cells!

Stop for a minute and really try to stop stating these things as facts....

Quote:
 Originally Posted by bgeery Agreed. But if the cell is at 3.33 volts resting voltage, it's already at 100% state of charge, and any further charging is an overcharge. You don't want further current flow.
As soon as you start charging it, the voltage will rapidly increase, and within a very short time it will match the voltage of the charger and the current will drop to zero. You're not overcharging it. This happens almost immediately.

Quote:
 A battery under 100% SoC will rest at something less than 3.33 volts until it's fully charged. The charge will be slower and slower as it approaches 3.33 volts, but the current will flow, until the cell reaches the true 3.33 resting voltage equilibrium with the charger and is at 100% SoC, and current flow will cease.
Right.... but hardly any current will initially flow into the cell if you try charging a lifepo4 cell at 3.33V and when it does get to 3.33V, no current will flow. The system would then be in equilibrium. The cell wouldn't take long to get to 3.33V, even if it's at 2.7V. This is exactly how I recondition low cells. You HAVE to pull the voltage above nominal. This is how current flows.

The manufacturer specifies that the charge voltage is to be no more than 3.6-4.0Volts (depending on your cell). As long as you're at, or below that, you'll never overcharge the cell.

Quote:
 Using higher than resting voltage on the cell does nothing but increase the rate you can charge; avoiding the long trickle charge that would occur as the cell approached 3.3 volts and less and less current flowed. That's a valid practical compromise, but it doesn't get a 100% charged cell. It gets a 100.2% or 99.6% charged cell.
The cell is 100% charged when current stops flowing, but the cell has to be pulled higher than it's nominal voltage, even if it's just a tad. Higher voltages just cause it to charge faster, but if you're only at nominal voltage, you're not going to charge.

Quote:
 The manufacture gives us exatly *one* charge curve that they rate the battery at, and if you are not charging in exactly the same conditions, you can't tell your state of charge with true accuracy. You can come close enough for practical use, but you still don't really know exactly.
Have you even looked at a charge curve from a manufacturer? It shows voltage and all the one's I've seen are above way 3.3V.
http://liionbms.com/pdf/thundersky/TS-LFP100.pdf
http://liionbms.com/pdf/psi/PC40138F1.pdf
http://liionbms.com/pdf/shandong/200ah.pdfhttp://liionbms.com/pdf/goldpeak/GP18EVLF.pdf
http://liionbms.com/pdf/huanyu/HYP-3.2V-100Ah.pdf

That charge curve is at a given C-rate. The lifecycle is at that c-rate. As long as the cell never goes above the manufacturer's recommended charge voltage, you're going to charge to 100%. Most of the charge curves I've seen are at 1C. True, you could charge at 3.5V, which many do, but 3.5V > 3.33V, so charge flows.

Quote:
 For example, how many people are charging at 3.5-3.8 volts until the charge tapers to 0.0 amps? I see top balance supporters mentioning this all the time here. Where does that extra power go between the 3.8 volts and the 100% charged resting voltage of 3.33 volts? It's not a whole lot of energy, but it is an overcharge of the battery. My guess is that it's oxidizing some of the plates and diminishing some minor amount of cell capacity.
Thats how I charge, every time. My charger is set to a certain max voltage, and a certain max current. Set it and forget it. It turns off when current goes to 0.

It's not power we're looking at... it's energy. What you aren't looking at is that Energy = Wh = the area under the curve.

Lets say you have a 10Ah 3.2V nominal cell. Situation 1: Charge voltage is set to 3.5V. Charge current is 1C, or 10A. Situation 2: Charge voltage is set to 3.7V. Charge current is the same.

The slope of the charge curve is steeper with situation 2, so it gets to 3.7V faster. Then it holds that voltage until current drops to 0. The area under the curve will be XXWh.

The slow of the charge curve is less with situation 1. Not only does it take longer to rise to the 3.5V charge level, it takes longer for the current to drop to zero.... so it takes longer to charge. The area under the curves is essentially the same (some is blown off due to IR of the cell).
__________________
Travis Gintz
Electro Motive Force, LLC.
#7
11-07-2011, 05:24 PM
 bgeery Member Join Date: Oct 2011 Location: Yucca Valley, CA Posts: 73
Re: You are all destroying your LiFePO4 cells!

Quote:
 Originally Posted by DavidDymaxion Great first post. I throw this out as a discussion point, and not as hard science. Jack mentioned this in his videos. (All hail Jackton! All hail Jackton!) The voltage for a cell is Vcell = V0 + I*R So suppose you have a cell with 5 milliOhms of resistance, you want to charge at 10 Amps. Vcell = 3.33V + 10 Amps * 0.005 Ohms = 3.38V Suppose you are regenning at 100A (or have a super powerful charger): Vcell = 3.33V + 100 Amps * 0.005 Ohms = 3.83V
Anyway, it's a thought that if the cells are cool enough and the currents are low enough, perhaps this isn't hurting them at all?[/QUOTE]

Very interesting. So that might be the "formula" for coming up with a recipe for charging in excess of 3.3V and insuring exactly 100% SoC is reached but not exceeded.
In your example then, if I wanted to charge at 100 amps, I'd apply 100 amps, and charge would be complete to moment the cell reached 3.83 volts. Cease charging and you have a 100% SoC cell. Of course, in a string of cells that all have different resistances, I'm not sure how that would interact.

My main issue is with those holding cells at 3.5-3.8 volts and forcing more and more current into the cell, until reaching 0.0 charging current.

Quote:
 Anyway, it's a thought that if the cells are cool enough and the currents are low enough, perhaps this isn't hurting them at all?
In theory, if the current is flowing into the cell after 100% SoC, it has to be going somewhere. In a lead acid battery, the forced energy electrolyze the water in the electrolyte creating Hydrogen and oxygen, and I believe also can warp the plates. Not a big deal to add water. Maybe in LiFePo4 it's just converted to heat, but I don't think so. Jack would know.
#8
11-07-2011, 05:36 PM
 EVfun Senior Member Join Date: Mar 2010 Location: Seattle, WA Posts: 2,206
Re: You are all destroying your LiFePO4 cells!

Quote:
 For example, how many people are charging at 3.5-3.8 volts until the charge tapers to 0.0 amps? I see top balance supporters mentioning this all the time here.
I have heard of this being used by a few people to top balance a pack in parallel but I have never heard of this being a regular charging routine. It is a drastic way of sharply finding a full charge, similar to those who bottom balance by getting the resting voltage down to 2.5 to 2.8 volts to find a matched empty.

I did my top balance by charging to 3.65 volts until the current tapered down to about 0.5 amps (60 amp hour cells.) It was mostly the time that it took my to get them all the same. That might cause some wear, but it is a single cycle. My charging routine is 12 amps until the voltage reaches 3.50 volts, then hold 3.5 volts for 40 minutes (around 2.5 amps ending current.) The next morning the cells consistently read 3.34 volts each and after 24 hours they consistently read 3.33 volts each.

Here is the Thunder Sky charging documentation, from "Thunder Sky LiFeYPO4 Power Battery Performance Test Instructions":

Quote:
 5.2.4 Charging At 20°C±5°C temperature, the cell is discharged at a current of C3 till voltage of the cell reach 2.8V, and then start to perform constant current charge at a current of C3 till voltage of the cell reach 4.0V. 5.2.4.1 Low temperature charging At -18°C±5°C temperature, the cell is discharged at a current of C3 till voltage of the cell reach 2.2V, and then start to perform constant current charge at a current of C3 till voltage of the cell reach 4.2V.
#9
11-07-2011, 06:45 PM
 dougingraham Senior Member Join Date: Jul 2011 Location: Rapid City, SD USA Posts: 2,037
Re: You are all destroying your LiFePO4 cells!

Quote:
 Originally Posted by bgeery A fully 100% charged battery has a resting voltage of 3.33 volts per cell at 25 Celsius. The only way to insure we never overcharge the battery is to charge at 3.33 volts until it tapers down to 0.00 amps. That's a fully 100% fully charged cell. For those that want to charge to less than 100%, just discontinue that charge before current drops to 0.00 amps. I see allot of people charging at 3.5-3.8 volts per cell until 0 amps. This is overcharging the cells to some degree. Yes, they may accept some additional charge above 100%, but it's past the manufacture's definition of the 100% charged point. Either charge at 3.33 volts until it tapers down to 0.00 amps, or go with the manufacture's formula of .5C charge rate to 3.6 volts and disconnect. Either one is the only known safe and accurate charging procedures to reach 100%.
Where did you get the 3.33 volt number? I have not seen that on any of the manufacturers data sheets. The only reference I have found to a float voltage is 3.45V found on the A123 systems data sheets that comes with the 26650 cells. And since those are plain Jane LiFePo4 it seems like a very reasonable number that could be used on other brands as well. I have floated the A123 cells at 3.45V for several days at that voltage with no ill effects.
__________________
Doug Ingraham
Rapid City, SD
1985 Mazda RX-7 GSL (9800+ EV miles)
Saving the planet one EV mile at a time!
#10
11-07-2011, 06:47 PM
 bgeery Member Join Date: Oct 2011 Location: Yucca Valley, CA Posts: 73
Re: You are all destroying your LiFePO4 cells!

@ frodus:
I find it hard to write a reply without nested quotes. sigh.

As soon as you start charging it, the voltage will rapidly increase, and within a very short time it will match the voltage of the charger and the current will drop to zero. You're not overcharging it. This happens almost immediately.

Yes, I said this is a slower method of charging, but is guaranteed to find and stop at exactly 100% SoC. The cell voltage only rises in response to the charge current. Taper the current and the cell voltage will drop again and current flows, this happens until the cell is actually at 100SoC and does not fall.

The manufacturer specifies that the charge voltage is to be no more than 3.6-4.0Volts (depending on your cell). As long as you're at, or below that, you'll never overcharge the cell.

The cell is 100% charged when current stops flowing, but the cell has to be pulled higher than it's nominal voltage, even if it's just a tad. Higher voltages just cause it to charge faster, but if you're only at nominal voltage, you're not going to charge.

Yes, but they don't say hold 4.0 volts all the way to 0.0 amps. And a discharged cell's *static* voltage is not 3.3 volts, thus will accept current regulated at 3.3 volts, at a slower and slower rate, until the cells' *static* voltage reaches 3.3 volts. This is 100 SoC.

Lets make it clear. I'm not saying charging this way is practical for everyday practice, as it may or may not take too long to reach 100% SoC. But, I am saying it's the only way to know you are actually *at* a perfect 100% SoC.

Have you even looked at a charge curve from a manufacturer? It shows voltage and all the one's I've seen are above way 3.3V.
http://liionbms.com/pdf/thundersky/TS-LFP100.pdf
http://liionbms.com/pdf/psi/PC40138F1.pdf
http://liionbms.com/pdf/shandong/200ah.pdfhttp://liionbms.com/pdf/goldpeak/GP18EVLF.pdf
http://liionbms.com/pdf/huanyu/HYP-3.2V-100Ah.pdf

That charge curve is at a given C-rate. The lifecycle is at that c-rate. As long as the cell never goes above the manufacturer's recommended charge voltage, you're going to charge to 100%. Most of the charge curves I've seen are at 1C. True, you could charge at 3.5V, which many do, but 3.5V > 3.33V, so charge flows.

no, you charge a >3.33 so the charge flows faster. But in charging in excess of 3.33 volts, you are forcing the electrons into the cell, instead of alloing them to flow in naturally. That means it's up to you to discontinue the charge a some time befor you have tapered down to 0.0 amps.

Looking at your very first link for the LFP 100 clearly shows that the cell reaches 100% SoC before you taper the current down to zero. It looks to me the current should taper to 4 amps or so and stop. The battery is 100% charged. Continuing to 0.0 amps at *any* voltage in excess of 3.3 is an overcharge of the cell. How much does this damage the cells each cycle? That's a good question that I don't have an answer for.

Thats how I charge, every time. My charger is set to a certain max voltage, and a certain max current. Set it and forget it. It turns off when current goes to 0.

If that certain voltage is >3.3 volts, then you are slightly overcharging the cell each cycle.

It's not power we're looking at... it's energy. What you aren't looking at is that Energy = Wh = the area under the curve.

Lets say you have a 10Ah 3.2V nominal cell. Situation 1: Charge voltage is set to 3.5V. Charge current is 1C, or 10A. Situation 2: Charge voltage is set to 3.7V. Charge current is the same.

The slope of the charge curve is steeper with situation 2, so it gets to 3.7V faster. Then it holds that voltage until current drops to 0. The area under the curve will be XXWh.

The slow of the charge curve is less with situation 1. Not only does it take longer to rise to the 3.5V charge level, it takes longer for the current to drop to zero.... so it takes longer to charge. The area under the curves is essentially the same (some is blown off due to IR of the cell).

I get what you are saying, and you have a point about the resistance of the cell requiring something greater than 3.33 to allow a charge. So lets call it 3.331 volts then-- or whatever small amount is need to overcome the couple of milliohms of resistance in these cells.

I get the point that charging at higher voltages get the charge dome quicker. The point is, these fast charges (and that's what they are as far as the cell is concerned) need to terminate at some point before the current tapers to 0.0 amps. The higher above 3.3 volts you are charging at, the higher point in the current curve you need to terminate the charge. And in any case, with these fast charges, you never will reach 100% charge, and will always overshoot or undershoot by some percentage. The better you picked your termination current point, the closer you will be to 100%, but it will never be exactly right.

One alternative might be to fast charge to say 99% and then slow charge at 3.331 volts to 0.0 Amps. That will get you to 100% SoC much quicker, and insure you don't overshoot and overcharge.

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