Do cells on final connection before controller work harder ?

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Jun 25, 2013
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HI ALL,

Ive noticed on my friends bike (18s 4p 20ah Lipo pack) that the group of lipos in the last connection before controller where lowest voltages by 0.2v then series connection 2, 0.1v higher and series connection 3, 0.1v higher again. So for example in order, Controller then group 3 cells 3.7v, group 2 cells 3.8v then group 1 cells 3.9v...

Does this mean that group 3 cells experience more draw as they are first inline?

The packs are built buy doing the parallel groups first and then joining them in series. Would it be better to join in series first and then join them in parallel?

Any help much appreciated :)
adam
 
No. Current is the same through all parts of a series circuit. Where you will find possible imbalance is in parallel connected cells, where the parallel links are suboptimal and differing resistances means the load is shared unequally between different cells (or series strings). However, this will only result in a temporary imbalance under load, as the cells in a parallel group will balance themselves when the load is removed, so measuring resting voltages is unlikely to reveal anything amiss.

In short, I think your observations are just coincidence, or possibly an artefact of the charging process.
 
It depends on how the pack is built. For current to flow, it needs a potential difference, both +pos and -neg terminals. The current will be the same across that line. But cell interconnection can be serial-parallel or parallel-serial or serial & parallel combined. The more interconnection, the more lines that current can cross. What happens as a pack ages is that it develops internal plating at the cathode & anode, though less for the latter. This increases internal resistance. And that IR will mean that current flow will differentiate along the path of least resistance. Hence cells will discharge at different rates, within the bounds of electrical-chemical conductivity. Since there are two terminals, this may mean those cells may be forced to always conduct and you'll see voltage differences. Again, depending on how the pack is built. And more because of different rates of IR increases than because of its position, though the two are related. To complicate matters, battery packs often have a BMS that is intended to keep cells in balance, no matter.
 
Take a rubber band and cut it. Now pull on the ends until it breaks. Where will it break? Can you predict in advance? There is a modus of elasticity at work, a physical phenomena - it will break when stretched hard enough, and at the point where the bands elasticity is weakest because of physical degradation. Yes, if you pull electrons hard enough (beyond their C-rate), it will damage the cell, often exhibiting as puffing or spotting in pouch cells.

But what is the basic question? Is it: What is the best cell to buy? Or is it: How can extend the life of my battery? Or is it: Do I need a BMS on my pack? I take the KISS approach - 1> build a pack from the best (highest C-rate, most cycles) cells, which are undisputedly A123, at least for the moment, 2> eliminate the BMS by keeping the pack charged at 80%<SOC<20%, and 3> having way more amps than needed (need max draw of 20 amps and have I estimate closer to 100 amps potential in my pack).

Back to the rubber band analogy. My pull is limited by how strong I am. If I now double, triple, x-multiple the rubber bands, at some point I can not break it because its beyond my physical ability to do so. That's how I make my battery packs. And no more than is needed, because battery cells are expensive, especially the A123 brand.
 
Yes it makes sense for the current to be running equal throughout the circuit however yes i understand that I.R will slow down certain parts of it. Thats a great way of thing about it using the elastic band analogy.

Ill test and compare the parallel groups I.R

We have mostly been charging using a 900W charger so about 12 amps. So well within the c rating however I feel a slower charger is needed.

What does spotting look like in pouch cells? Ive experienced a fair few puffed cells but I carnt imagine what you mean with spotted cells?

Many thanks guy that sure does help me understand more so far.
 
adrock8519 said:
Yes it makes sense for the current to be running equal throughout the circuit however yes i understand that I.R will slow down certain parts of it. Thats a great way of thing about it using the elastic band analogy.

Ill test and compare the parallel groups I.R

We have mostly been charging using a 900W charger so about 12 amps. So well within the c rating however I feel a slower charger is needed.
Its why I recommend that everyone get the datasheet for their cells, either from who you bought it from or the manufacturer. Li-ion cells will often have different C-rates on charge and discharge. Yes, you'll not want to exceed the recommendation for your cells - don't guess - know.
What does spotting look like in pouch cells? Ive experienced a fair few puffed cells but I carnt imagine what you mean with spotted cells?
I posted a picture of what happened to me at: http://endless-sphere.com/forums/viewtopic.php?f=14&t=42871&start=125#p864429
Many thanks guy that sure does help me understand more so far.
Best!
 
each pouch sees the same current. does not matter where they are in the series.

pouches swell up when the voltage drops down to the LVC level while discharging current through it.

for the ping packs, he uses 5Ah pouches and solders them together in parallel to make a higher capacity cell. i have found that the swollen pouch can be any of the pouches in that group.

my educated guess about why one pouch will swell versus the others connected in parallel relates to the internal resistance each pouch has of it's own.

when one pouch develops resistance to the charging current then that charging current flows around that pouch and as the other pouches in parallel charge up the voltage climbs and so that cell or channel can reach the balancing voltage along with other channels even though it has stored less charge and that channel is lower in capacity than the other channels.

on discharge the pouch with less charge stored can be exhausted of current carrying ions before the other pouches in parallel so that pouch becomes over discharged while the other pouches in parallel can continue to carry current but it now is heating up from lack of ions to carry the current so it begins to puff up.

this is why the ping packs can have a cell in the middle of the pack that goes bad before the others.

the pouches on the end are the ones that have the most physical damage from the swelling of all the pouches with age. since it is on the end it is like the outside of the balloon and it is deformed and stretched the most so that the middle of the pouch protrudes far out and the top and bottoms and corners of the pouch are restrained by the shrink wrap that holds the pack together.

as the end pouch gets pushed farther and farther out it pulls on the tabs at the top and can sometimes tear the tab off in worst cases and usually the pouch will begin leaking electrolyte at the seam where the tab comes out.

his newer pouches and packs show much less damage like this but the old packs almost always died this way, especially those with 4 pouches in parallel, 48V 20Ah.

for a long time i have recommended people take their ping pack out of the shrink wrap and rebuild the pack by cutting it apart in the middle and inverting it so the former end pouches are facing each other in the middle with the rounded outer ends pushed up face to face and the former inner pouches now on the ends and then add some hardboard end plates to each end, then compressing the pack enuff to push the middle rounded ends tight against each other so that as the pack continues to swell with age the swelling forces the pouches to become more flat again in the middle and the forces are allowed to develop slowly over time as the pack ages.

this is the only way to get the maximum life cycles out of the ping packs.
 
dnmun said:
each pouch sees the same current. does not matter where they are in the series {emphasis added}}.
pouches swell up when the voltage drops down to the LVC level while discharging current through it.....
... for a long time i have recommended people take their ping pack out of the shrink wrap and rebuild the pack by cutting it apart in the middle and inverting it so the former end pouches are facing each other in the middle with the rounded outer ends pushed up face to face and the former inner pouches now on the ends and then add some hardboard end plates to each end, then compressing the pack enuff to push the middle rounded ends tight against each other so that as the pack continues to swell with age the swelling forces the pouches to become more flat again in the middle and the forces are allowed to develop slowly over time as the pack ages.
this is the only way to get the maximum life cycles out of the ping packs.
Thanks for that dnmun - learned from that post. So in this context, the question is really: How can we extend the life of our packs? That's a good question. In part, its why I decided to build my own battery packs. With careful forethought, planning and building, a pack can be made so it can be disassembled. Why do that? To rotate the cells just as you suggested, but also to replace cells. I believe that certain cells will die before others, and by replacing those cells, the life of the pack can be extended.

I put my A123 AMP20 12S1P pack on the shelf after making my A123 26650 12S8P pack. Because I didn't see the whole compression problem that wb9k elucidated. I've got years & years in my second pack. But my first AMP20 pack too, when rebuilt. I'll do that in good time, rebuilding it to get the compression thing right. And so have a battery pack to use when pack two - the 26650 - grows cranky. Then with that pack out of service, do the diagnostics and maintenance on it.

I figure I may not need a third pack, that I'll be able to keep maintaining these two packs in that way, buying new cells when needed. At some time I'll have replaced all the cells. But the remainder of my investment into the build is preserved, especially the battery harnessing, box and racking. I hope, however, that A123 will place on the market a factory assembled battery pack engineered specifically for the LEV market, or that Harley, Yamaha, Honda or such will have contracted for an A123 pack that can be used. Or, a better battery emerges like the Lithium-Titanate.
 
It has been reiterated several times in this thread that current is equal in all series elements in a pack. While this is true, there are other dimensions to consider--the situation is not nearly so simple as "alll current is the same, therefore every cell is dissipating the same power". In fact, this is NOT true.

Adrock, what you are seeing is not a coincidence. It is the result of the interaction of a number of factors in your pack design. To say exactly what's gone on would take a lot of experience and a full knowledge of the history of the pack's intimate behaviors and environmental conditions. However, there are some relatively common phenomena that I think are being underappreciated here.

I have worked with hundreds of new and used modules and stared at data from thousands more. I noticed a long time ago that cells at the ends of the modules seem to take the most abuse and generally degrade the fastest, all else being equal. There can be multiple reasons for this. Boundary conditions in typical measurement circuits can contribute, but I'm convinced that the bigger influence is the additional impedance of bus bars that are mechanically connected between modules wired in series. The welded connections within a module tend to have much lower impedance than the impedances of the bus bars--think hundreds of microOhms vs. dozens of milliOhms. This additional impedance affects only the cell adjacent to it--one adjacent cell during charge, and the other adjacent cell during discharge. It looks exactly like Puekert loss in the context of the pack--that is, it is a parasitic load that is borne ONLY by the adjacent cell. This means in practice that those cells will discharge the most deeply, will suffer the most extreme electrical conditions (I'm ignoring heat effects for the moment), and thus degrade the most quickly. I have recently discovered I don't like "split packs" for hobbyists for this reason--they foster imbalance unless you are very careful to insure that the impedance of the connection between EVERY series cell group is very near equal.

This makes sense if you think about how a series battery pack is "connected". None of the electrodes in a pack are touching, and the connections between them effectively become a part of the two electrodes it connects. That pair of electrodes is separated from the next electrode pair by a seperator and electrolyte. Just as a problem in the separator or electrolyte causes extra loss across only that cell, so does a high impedance connection between two electrode sets. It's just another means to exactly the same result.

It took me a long time to wrap my brain around all of this, and I came to this mostly on my own. I'm sure I'm not the first to see this, but I have yet to hear it articulated by anyone else.
 
wb9k said:
It has been reiterated several times in this thread that current is equal in all series elements in a pack. While this is true, there are other dimensions to consider.... I noticed a long time ago that cells at the ends of the modules seem to take the most abuse and generally degrade the fastest, all else being equal....
OK, but that was not my experience. Of my original 21 AMP20 cells, I've lost 8, have 12S in the working pack, and 1 last spare to go. Yea, the Victpower grey market cells - let's not bring that up again. Of those 8 cells, 7 were in the middle, cells 5, 6 & 7 in the series. 1 was at the end. So yea, I replace a cell and then had to replace the same cell, positionally again. Not a fluke. I've continued to believe that without adequate compression, cells would fail again. Note too that its around the seams that you see electrolyte leaking.
 
Spotting which leads to leakage is a completely different failure mechanism. That is caused by an isolation fault between the anode and the pouch. The phenomenon I'm describing will happen with either prismatic or cyclindrical cells. It is not related to compression. Compression and isolation problems will kill a cell way faster than what I'm talking about here. But you can see the exaggerated imbalance relatively quickly in a poorly designed split pack. It's not a severe problem, but if you are taking voltage logs during charge and discharge, this kind of thing has a way of rising to the surface and getting your attention after a while. The cells that work the hardest are most often next to higher impedance connections.
 
I mean a "soft short" (high impedance) between the anode and the aluminum pouch material. This can happen if the interior poly layer of the aluminum pouch is breached. When the pouch is held at anode potential, Li in the electrolyte alloys onto the aluminum, causing it to become brittle and crack. The black spotting around the cell is the hallmark symptom of this failure mode. Eventually, such cells will leak electrolyte, which can lead to wider isolation issues at the pack level. This was common in the grey market cells because much of that material had been recalled and scrapped by A123 for having this very issue.
 
wb9k said:
I mean a "soft short" (high impedance) between the anode and the aluminum pouch material. This can happen if the interior poly layer of the aluminum pouch is breached. When the pouch is held at anode potential, Li in the electrolyte alloys onto the aluminum, causing it to become brittle and crack. The black spotting around the cell is the hallmark symptom of this failure mode. Eventually, such cells will leak electrolyte, which can lead to wider isolation issues at the pack level. This was common in the grey market cells because much of that material had been recalled and scrapped by A123 for having this very issue.
Ah, thanks for that assessment. So its more that they were the victpower cells than a lack of compression or anything else. OK, I got a good 20 months of operation on that pack and I'm good to go for another 20 months, since by now all the bad cells have been replaced (hopefully). I'm in the process of thinking about my second eBike build and will most likely go back to using the AMP20 prismatics. This time, new from http://www.buya123batteries.com/, and a better design to hold the pack in compression.
 
wb9k,

I'm struggling to understand this.

I started writing this post thinking you were wrong, but in trying to describe how, I've cast doubt on my own doubt...

Comparing a high resistance series like to a high IR cell is an interesting analogy, but I have my suspicions over its validity.

As I understand it, a high IR cell will provide the same current as the other cells in the pack (as it must to obey Kirchoff's current law), but the cell will also suffer losses due to self-heating, which must involve the consumption of some current. I first thought this current must be taken from the total circulating through the pack, the same as adding a resistive load to the output terminals, but then what's the explanation for the high IR cell being depleted before the others?

Perhaps a battery cannot simply be considered as an electrical junction. I like to think of it (and generators) as electron pumps.

Is the high IR cell sourcing more current than it's neighbours due to the effects of its electrodes? If so, it's tempting to extrapolate that to a high-IR interconnect which, as you say, simply connects two electrodes in adjacent cells. However, my stumbling block there is that attaching a resistive load to the output terminals is the same - a resistance connecting two adjacent electrodes and a load doesn't unbalance a pack. To my mind, neither would splitting the series string in half, placing the load between the pack halves and remaking the circuit.

Perhaps this is one of those things that suddenly becomes clear with the correct diagram...
 
The current is generated by the Lithium Ion migration from anode to cathode. But high IR is indicative of lithium plating. So there is in fact less capacity in the cell. The chemistry of the electrolyte is degraded and the resistance of the cathode is increased. I believe you may be right that some current may show up as heat, so completely wasted as well. It'd be good to find some academic battery scientists to chime in. These guys are often more than delighted to talk to the public & media. So we have a far more enlightened point of view informing our understanding.
 
The current in the string is even.

The heating in the string is never even.

The mechanism that creates the heat isn't in additional current consumption (that remains dead-even in a series string), but in additional voltage drop due to increased resistances.


This effect of the first and last cell in the string getting hammered harder than the others is something I observed with many pack designs, from pouch cells to 26650/18650 cans.

I can tell you what stops it for pouch cells (and may work for can's as well, but I've not tried it). Make the connection points to the ends of the cell strings from what seems like very 'excessive' thickness copper, and get the connection resistance as low as mechanically possible. Once you do this, you find the ends of the strings stop drifting down in relation to the other cells. A good ballpark would be calculating what size bus cross section you think you need, then multiply by perhaps 5x for your end taps on the cell string.

Similarly on this topic, don't solder cell tabs, you definitely are damaging them. Don't weld them in a process that uses multiple types of metal either. Any process that heats the cell's tabs has already failed IMHO.


arkmundi said:
But high IR is indicative of lithium plating.

A pouch filled with gas would be a better indicator of lithium plating. High Ri can occur due to a structure being mechanically broken by the creation of metallic lithium crystals in cases of extreme charge rates at low temperatures. This is not related to this mechanism, nor do cells with lithium plating on the anode have high Ri, they have fantastic Ri while it's plated, they suck after you've discharged it and you're left with a crumpled anode structure.

arkmundi said:
So there is in fact less capacity in the cell. The chemistry of the electrolyte is degraded and the resistance of the cathode is increased. I believe you may be right that some current may show up as heat, so completely wasted as well.

Are you saying the capacity is reduced because of the mistaken belief metallic lithium formed in the cell? I can assure you, unless they are being charged in very cold conditions at high rates, you've never had any metallic lithium in your cells, and be glad about that. Increased temperatures can and do age the cells more rapidly, and this does cause components of the electrolyte to decompose more rapidly and cause surface reactions with active materials that make them inactive.

arkmundi said:
It'd be good to find some academic battery scientists to chime in. These guys are often more than delighted to talk to the public & media. So we have a far more enlightened point of view informing our understanding.

The solution is to have extremely low resistance cell interconnects on the ends of the cell strings, combined with even pressures on the surface of the cells. In a pouch cell, if your pressure against the face of the cell isn't even, the areas of highest pressure have the shortest ionic path, and hence carry the bulk of the current demand in local concentrated points that fail prematurely from being over-stressed.

To do a battery right:

1. Extremely low, and EQUAL cell interconnect resistances, including at the ends of the string (for at least 6" or so before dropping to a smaller thickness current conductor).

2. No processes that involve heating the cells, especially on the tabs. This is a delicate vacuum sealed hot-melt glue joint sealing the tab/pouch interface, and you can compromise it and/or destroy the cell simply be applying a bit of heat. You may have no awareness you've compromised the cell for months, maybe a year even before it starts puffing or worse.

3. You must provide even and flat mechanical support for cells, and ONLY ever clamp through the Z-axis. Never squish the sides of a pouch, never even let them carry the weight of the cell on it's side.

4. Thoroughly cleaned and de-burred surfaces contacting pouches ONLY! If you scrape through the extremely thin polyethylene layer on the outside of the foil (which you can do simply by setting it down on a table that has 1 tiny spec of a sharp sand particle on it), your cell will fail (can take months, but will start a corrosion spot and fail there).

5. Do NOT permit the foil pouches from cells at different potentials to contact each other. In theory all the pouches are galvanically isolated. In practice, the inside layer often has defects and pin-holes, which will eventually manifest as spotting and inner aluminum foil layer of the pouch de-laminating and failing. Some internal pin-holes will last many years before failing if they never see voltage stress. Introduce voltage-stress on that pin-hole (from touching other pouch foils at other potentials) and the same failure that would have taken years can take a week.

6. Don't flex or bend your cells. Period. I am aware they are soft, and tempting to pick-up and handle. You are breaking them when you do, set them on a clip-board or something to move them, with them always laying dead-nuts flat.

7. Lastly, you must protect the cells from all moisture intrusion, even water vapors (as they condense when the temp differential permits). This last step can be quite a challenge for DIY pack builders, and as a result many DIY packs fail from humidity related corrosion.


If you follow those 7 steps, you're still not guaranteed a long-term successful pack, but at least you aren't guaranteeing a pack failure by design.
 
Its a complex process....
http://epg.eng.ox.ac.uk/content/degradation-lithium-ion-batteries
Degradation_mechanisms_cChristoph_Birkl_small.png
 
Interesting discussion.

I've observed the end cells behaving differently in several packs but never fully figured out why. I'm not sure I understand why resistance between cells would be different than say, taking all the resistance and lumping into one spot. It's still in series, and the resistance and current will determine how hot it gets. As long as it's not a significant amount of heat, it shouldn't matter? I'm not saying it's wrong, I just don't quite understand why it effects the cells.

Another thing I've observed is with many BMS topologies, the end cells definitely charge to a different level due to differences in the voltage drops in the balancing wires. If all the shunts are on, the current in the balance wires between cells will be balanced out and nearly zero, where the end cell wires have to take the full balancing current which throws the voltage off a little.
 
It may be the end cells stay cooler rather than hotter via being close to the ends with a greater area exposed to conduct and radiate thermal energy.
 
liveforphysics said:
The current in the string is even.

The heating in the string is never even.....

Excellent post, all great advice for prismatic cells and you've nailed it regarding connections between series cell groups. All the same is where it's at, and the lower the better.

I've come up with a way to diagram this that I think will make this easier for some to understand. I just put it down on the white board in the lab for the first time today. My colleague from the lab next door seemed to think it made perfect sense. We've talked about this phenomenon before, but it never seemed to click for him until today with the drawing. I'll try to sketch something out I can post in the next couple days. Brutally busy this week between work and tweaking my new toy :twisted: .
 
arkmundi said:
The current is generated by the Lithium Ion migration from anode to cathode. But high IR is indicative of lithium plating. So there is in fact less capacity in the cell. The chemistry of the electrolyte is degraded and the resistance of the cathode is increased. I believe you may be right that some current may show up as heat, so completely wasted as well. It'd be good to find some academic battery scientists to chime in. These guys are often more than delighted to talk to the public & media. So we have a far more enlightened point of view informing our understanding.


There are a lot of possible factors and mechanisms that can lead to cell degradation and/or failure. Li plating (which perhaps happens most often in LFP by overcharging) is one of them, but there are many others and usually it's more than one that's at play in real life scenarios. At the same time, there are lots of possible problem scenarios that involve no Li plating at all. It is important to realize that as cells degrade, they tend to lose capacity and their IR (internal resistance) goes up. However, I have seen absolutely no predictable correlation between the two. High IR is no guarantee that capacity is low and vice versa.

High IR causes loss of usable power not by forcing a cell to deliver more current, it just makes the cell work harder to keep up with everyone else, and the lost power is expressed as heat. This is Peukert loss, and it is a cell-level phenomenon. When one cell in a group of parallel cells has elevated IR ALL of the loss from that happens in the individual cell. The other cells in parallel with the lagger prop the lagger up through the discharge cycle, so in that way the whole cell group is affected, but that elevated IR loss really just comes from the one cell. During charging, the lagger charges the slowest because charge energy is also sacrificed as heat. How that ends depends on your management system, but most systems would stop charging at some point, and the cell group with the lagger would settle at a lower voltage than the others because the other cells end up trickle charging the lagger for quite a while before stasis is reached.

This can be tough to get the brain around for a lot of guys....Kirchoff's law has been mentioned in this thread. It applies of course, but the traditional model that describes the law is not really complete for the battery discussion. The usual approach assumes a theoretically perfect power supply and discusses only how current flows in the load. With battery packs, we have so see the power source itself and all the interconnects as part of the load. The traditional model pretends parasitic loads don't exist. Everything we are talking about in this thread amounts to a parasitic load that originates in the "power supply" portion of the Kirchoff explanation as usually diagrammed. When we're plugged into the wall we don't care about those losses, no matter how real they are. But in EV's, losses in heat translate directly into lost range and at this moment in time we need every Watt we can get out of our batteries.

Hopefully that will help clarify some corners of the discussion.
 
wb9k said:
.... Hopefully that will help clarify some corners of the discussion.
Yea, thanks for that post. Some of us are perverse enough to enjoy discussions of this sort. Current era batteries are fascinating in their complexity. Best! And pass that on to the A123 team!

From the Battery Space 20AH Prismatic LiFePO4 tests thread:
RLT said:
The packs I've built with cylindrical cells, it usually seems that the cells that go bad are in the center, where you have to take the whole dang thing apart to replace one or two bad cells.... I haven't had to mess with pouch cell packs yet, and with any luck, never will
Confirming my own experience in a different format and cell. Have no clue as to why the different positional first-to-fail cells arise, but it is a curious phenomena. In any case, for a practical guide in extending pack longevity, there seems to be an argument for rotations. End becomes middle and middle becomes end, so does not matter which.
 
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