I don't usually get into technical discussions on this blog but I thought it might be worth while to continue the topic started on Jack Rickard's EVTV blog. Simply put Jack has stated that pushing the voltage of a LiFePO4 cell above 4.3 Volts, the breakdown voltage of the electrolyte, will not harm a cell if the cell is not full, and that in fact there is no such thing as electrolyte breakdown voltage. This of course flies in the face of just about everything I've read on the topic, and even contradicts what battery researcher Jay Whitacre states in his LiFePO4 battery lecture video, a video that Jack actually provided a link to.
Go to the 50 minute mark in Jay's video and listen as he talks about electrolyte solvent breakdown above 4.3 volts, equivalent to water electrolysis at 1.3 volts. If a cell voltage is held above the breakdown voltage of the electrolyte the components of the electrolyte begin to separate out. This can happen even if the cell is not yet full. In normal use this should not be an issue, but it is something to be aware of.
For further technical reading about electrolyte breakdown voltage there is this rather comprehensive review by Kang Xu: Nonaqueous Liquid Electrolytes for Lithium-Based Rechargeable Batteries
The pertinent information starts in section 5.2 page 4325
Try CV charging from the outset and you will find it takes forever to up that cell from say, 2.6 to 3.6 volt.
ReplyDeleteOne volt at the cells internal resistance. How long? Oh hum, I'll leave it to you. :)
That is why the initial voltage is meaningless. They simply regulate the current 'cos the volts are adjustable. I=V/R
Andy, your comment still misses the entire point and I never said anything about charging a cell at constant voltage. Jack claims that voltage does not matter, that there is no such thing as electrolyte breakdown voltage, and that putting in enough current to drive cell voltage to 8 volts during charging has no negative effect as long as the cell is not full. This is simply not correct as all the evidence I have provided has shown. If a cell is held above the breakdown voltage of the electrolyte, somewhere between 4.3 and 4.5 for most lithium cells, components of the electrolyte, specifically the electrolyte solvent, will begin to breakdown.
ReplyDeleteYou might want to ask Jack why my last post, quoting the Kang Xu document Jack linked to, specifically discussing electrolyte solvent breakdown, has mysteriously disappeared from his blog. Apparently Jack is more interested in appearing to be right than actually dealing with facts.
I see we have hundreds of comments here on your bizarre vendetta. I have asked you to leave our blog, and now I have enforced that. I don't know WHAT you are on about JP, but I'm not having it any longer.
ReplyDeleteAnd a visit here indicates just how popular a topic it really is anyway. You took something out of context that you didn't understand at the time, and tried to twist it into a "hang Jack" vendetta using materials you don't unerstant. But in my "opinion" you've now found the perfect place to entertain yourself with that HERE with the thousands who are interested in such nonsense.
Jack RIckard
An ANONYMOUS Blog? You have to be kidding....
ReplyDeleteJack Rickard
The anonymity of a blog has nothing at all to do with the information therein. I don't write this for a popularity contest either. I guess yours is since you've deleted two of my recent comments in a weak attempt to appear to know what you are talking about on this topic. I have no vendetta other than trying to correct inaccurate information, which for some bizarre reason you feel compelled to repeat. As I mentioned on your blog, before you deleted the comment, at one time I too thought voltage didn't matter if the cell was not full. When someone brought up the topic of electrolyte breakdown voltage I realized the error after doing some investigation.
ReplyDeleteRegardless, I notice you continue to ignore the heart of the matter or even attempt to discount the references I've made. Bottom line is that electrolyte breakdown voltage is a real and well known phenomenon, backed up by industry experts, including the two you provided, and that holding a cell above the electrolyte breakdown voltage will cause the electrolyte to deteriorate. You insist on promoting the idea that elevated voltages don't matter, and if people actually believe what you are claiming they very well may damage their cells. I fail to see how that is in any way helpful to your viewers. You constantly criticize others for promoting false information, and, ironically, censorship. You're now guilty of both.
Andyj said...
ReplyDelete" Try CV charging from the outset and you will find it takes forever to up that cell from say, 2.6 to 3.6 volt.
One volt at the cells internal resistance. How long? Oh hum, I'll leave it to you. :)"
As it turns out Andy, not very long. I do that every charge cycle only I don't go over 3.465vpc. I can even charge at 55A into a 200Ah battery and it doesn't take long at all to fill the pack. The last part of the charge, from ~68V to 69.3V for my 20 cell pack takes less than a half hour, the rest of the time my chargers are cranking out at max output. There is also less than 1% of the charge left to do at this point too.
About the 106 page article it does talk about the breakdown of electrolyte at higher voltages. What isn't clear, however, is whether this is at a fully charged state or at any SOC. I'm going to assume that this is at any SOC. What might be the case, however, is that the terminal voltage may be at something like 8V but that the voltage at the electrolyte might be well below the breakdown voltage.
For all practical purposes it may not matter anyway. The most I can pump into my pack with my two chargers is 55A any way. This is only 0.275C so the terminal voltage doesn't get very high.
I'm not sure how it would be possible for the terminal voltages to be significantly higher than electrolyte voltages. Any thoughts on how this would occur? I have discussed electrolyte breakdown voltage with a battery researcher, he posts as MRTTF on seekingalpha.com, and I did ask him specifically about higher voltages independent of SOC. He said it's the same negative effect and did not mention any difference between terminal and electrolyte voltages.
ReplyDeleteThe only thing I could think of was some voltage drop between the terminals and the plates and then possibly a voltage drop across the graphite crystals and the surface layer. I have no idea how much the drop could be, however. Given the low resistance of these cells it can't be very much.
ReplyDeleteIt seems that if Li plating occurs at low temperatures while charging that it could also happen at 25°C if the charge rate is high enough. This is assuming that part of the reason plating occurs at low temperature is because of the time it takes for the Li-ions to intercalate. If other ions are "stacked up" behind it the potential for them to plate out increases. Maybe charging too fast could have a similar effect. This is of course speculation on limited knowledge of what is really going on.
The paper did say that "the stability of an electrolyte can also be quantified by the range in volts between its oxidative and reductive decomposition limits." (p. 4304) Clearly there is a voltage potential above which the electrolyte will begin to break down.
Actually chaps. When I said CV charging. That is CV only charging. You are stating CC charging has no function other than to stop your charger from emitting smoke and dying an early death?
ReplyDeleteInteresting.
I'm not sure what you are getting at. CV charging just means you don't allow the cells to go above a set voltage. For current to flow that voltage has to be higher than the cell's resting voltage and to get it near full it needs to be higher than the cell's finishing voltage at a specific C rate.
ReplyDeleteCC = Constant current. This is so you don't fry the charger or overheat the cells. It's normally set to AH or simply time.
ReplyDeleteThen normally follows with CV charging; constant voltage... When the charger and pack are close to parity on charge, very little current crosses. It only tops up the last bit.
If its CV charging all the way and is quick enough, who cares? :)
On the video you link to, CV charging was not bothered with. He demonstrated constant current until enough charge was reached regardless of voltage. Thats fine so long as the discharge and charge curves are repeated in parity.
You can't do CV charging the whole time, voltage needs to rise during charging above resting voltage. You can't charge a cell resting at 3.1V using a constant voltage of 3.1V. I think all you are talking about is having a set finish voltage above which the cell is not allowed to go, which is basically CC charging, not CV. In fact all I use most of the time is CC charging because I'm not trying to completely fill the cells. That's because we know that cells last longer if not fully charged all the time. Typically I use 20 amps until the average pack voltage is around 3.41-3.42 and just stop, no CV phase at all. This gets me around 85% SOC, depending on pack temperature.
ReplyDeleteHi JP,
ReplyDeleteEhm. I've seen that number of 4.3V before. And I still believe that is just another way of saying: Don't exceed 1C. I'm not sure, that's why I'm dropping it here: If a 100Ah cell has an IR of aprox 8 mOhm, and a max open circuit voltage of 3.5V. A 100 amps should rise that voltage with 0.8V, giving 4.3V. If the IR is a constant during charging, Amps and Voltage are just synonyms.
I'm not quite getting your point. You can charge above 1C and not push voltage above 4.3V. During a demonstration Jack cranked up the current enough to push a cell to 8 volts and claimed it was just fine to do so. This flies in the face of the physical constraints of the electrolyte in the cell which becomes unstable above 4.3V or so.
ReplyDeleteMe neither.
ReplyDeleteI never understood, and still don't understand the IR concept of a Li-ion battery. In general people talk about internal resistance. But in spec sheets and scientific papers they call it impendance. I first thought, OK, that's becaus it's not really a resistance, but a chemical proces that results in a voltage rise and drop that looks like resistance. But impendance isn't a word for 'things that look like resistance' it's specifically for alternating current. It's the resistance only alternating current experiences, and not direct current. And than I got lost. What's so special about AC and batteries? Why do they care? And is there no resistance in case of DC? I never found out.
But I they really have a constant resistance, regardless of AC or DC, then there is just no difference in saying:
Don't exceed 4.3V while charging because you will end up killing them.
Or don't exceed 1C (or whatever comes close to 4.3V with U=I*R) because you will end up killing them.
This last sentence is often stated in specsheets. And the recommendation is mostly even lower: 0.3C. But I think that's not because your 4.3V limit. I think batteries are just like humans: They age faster under heavy stress. Treat them with care and respect most of the time, and they will give a lot if you really need it for a long time.
Sure, higher C rates, even if they don't push voltage above 4.3V, can still cause local internal heating which can also degrade the cell.
ReplyDeleteYour basic tenet is 100% correct, in the context of batteries or anywhere else. If you exceed the breakdown (read electrolysis) voltage of the background electrolyte or the medium (the two are often NOT the same*), electrolytic breakdown will occur. This does not mean that the desired process (eg Li deposition) will stop but the other process will lead to irreversible damage, the exact extent of this will depend on the circumstances.
ReplyDelete*The single largest consumer of electricity in the chemical industry is the electrolysis of near-saturated brine. It is the sodium chloride that is preferentially electrolyzed, not the water. This will not be the case if you electrolyzed sodium sulfate.