Do cells on final connection before controller work harder ?

[wb9k]

Center of pack cell fails are usually the result of higher operating temps in that portion of the pack. The fix there is to keep operating temps even across the pack. For the end-cell issue, a slight modification in balance strategy could mitigate the damage, though the better fix is to get interconnect resistance values down as low as possible. My experience suggests that is easier said than done however, and balance strategy modification is probably also necessary to keep the effects to a minimum in a real system.

Glad to see I'm not the only one who thinks this is an interesting topic. It's been holding my fascination for some time now as this is a behavior that has surfaced again and again in a variety of contexts I've seen. Exciting times!

 

[Punx0r]

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.

This describes where I'm at as well.

Wb9k, if you could post a photo of your diagram that would be most interesting, and I hope, illuminating.

I suspect this may be one of those problems that doesn't respond well to thought experiments and I'm over-thinking it.

 

[arkmundi]

As has been well expressed by many others, we're in a total mind f#&)^kk, mind blowing age. I suggest we think less before we explode like those proverbial pouch cells on over-charge. :lol: :lol: :lol:

 

[wb9k]

OK, the quick sketch I've put togther is attached. To keep things simple, let's first discuss a 1P pack--no parallel cells to complicate things.

The drawing models each cell as having two "barrier regions" across which current flows from one electrode to the next. First is the "separator/electrolyte region" where Peukert losses happen. It is obvious that this is contained within each individual cell. The second region is less intuitively obvious, the "interconnect region" where a mechanichally connected conductor ties two different electrodes together. Just as with losses across the actual cell, losses in this region are also local to the impedance node. In other words, LOCAL HEATING MEANS LOCAL LOSSES. If the whole pack were bearing the losses equally, the whole pack temp would rise roughly equally. But it doesn't. High Peukert loss cells heat to higher temps than cells around them because they must dissipate more of their available power to make the current being demanded by the load available to use. This is well established and very easily observed with a thermal camera. Even with equal capacity to other cells in the pack, the high IR cells will discharge faster than the others.

Similarly, high impedance connections between cells don't heat the pack evenly, and this should be our first clue that the corresponding losses are also almost entirely local. Only the cell next to the high impedance is driving losses across this junction because as far as the electrons are concerned, the interconnect is an intrinsic part of the cell (or cells) it is connected to--it's just another local cell impedance. The minor difference here can be seen in packs with multiple cells in parallel. If only one cell has high IR, only that cell imposes losses for the group. In the case of a high impedance connection, individual cells only are affected if the bad connection is a parallel tie to the group. If the bad connection is the main connection between two cell groups, then the whole cell group is affected directly.

Categorically proving this is not difficult to imagine but relatively tedious to execute. It will be a while before this makes the top of my list, especially since the more I think about it the more obvious it all seems. I have also wondered if the connections from the pack to the load have a similar effect on end cells in a pack. Luke's observation that beefing those connections up ameliorates the battering of a pack's end cells seems to suggest that they do. This means we need to be very careful about considering parasitic losses in the pack itself and its connections to the actual load. The losses imposed can be significant, and the effects tend to compound over time as the deeper discharge and higher heat take their toll on the affected cell(s) and the situation becomes worse and worse over time.

I hope that helps clarify my thoughts here. Shoot me full of holes if you can!

 

[Punx0r]

Thanks, I know get that the resistance of the exacerbates peukert losses.

I think my understanding of the peukert effect is lacking, though. My understanding is the electrochemical reaction in the cell shifts electrons from cathode to anode against the cell IR.

V = IR suggests that pulling X amps from a cell will produce a voltage drop (sag) proportional to the current and IR. That voltage drop would result in energy wasted as heat in the cell, with a resulting drop in watt-hours available from the cell. However, peukert effect seems to result in a reduction in voltage, Wh *and* Ah.

Can I take it as true that a high-IR cell, of equal capacity to its low-IR, series-connected neighbour will be become discharged sooner? I.e. it won't just get hotter or exhibit a lower voltage?

This is my stumbling block: Either the chemical reaction is transporting electrons, but they are not making it to the anode due to the IR, or the efficiency of the reaction is being impeded - chemical energy that ought to be shifting electrons is being wasted exothermically instead? Or something else?

What I'm trying to get my head around is that we have say, 100A from cell A entering cell B and 100A will leave cell B no matter what it's IR, but the voltage will drop (and heat produced). It's as though a high IR cell is expending more energy passing on that 100A than a low IR one, but how?

Apologies for the lack of understanding I'm no doubt showing here, but I'd really like to get a handle on this and learn something worthwhile...

 

[PaulD]

OK, the quick sketch I've put togther is attached. To keep things simple, let's first discuss a 1P pack--no parallel cells to complicate things.

Interesting idea, but I can't really see why the end cells would see more abuse besides additional heating from a high impedance connection to a busbar. The current is the same in every cell in a 1P battery. Coulombic efficiency of lithium ion batteries is nearly 100%. Ah in = Ah out. The additional resistance of a busbar connection wouldn't affect the adjacent cell (except for the heating).. the voltage between the electrodes of the cell only would depend on temperature, current and SOC.
Oh and AFAIK, Peukerts equation really only applies to lead-acid AFAIK. Unless you are just using "Peukert effect" interchangeably with "cell resistive losses".. maybe I'm misunderstanding your theory.

 

[wb9k]

Can I take it as true that a high-IR cell, of equal capacity to its low-IR, series-connected neighbour will be become discharged sooner? I.e. it won't just get hotter or exhibit a lower voltage?

Yes. The voltage of the high IR cell sags because it is being loaded more heavily than the others by its own internal resistance. This results in additional POWER lost (IxR) from the finite amount available from the cell.

This is my stumbling block: Either the chemical reaction is transporting electrons, but they are not making it to the anode due to the IR, or the efficiency of the reaction is being impeded - chemical energy that ought to be shifting electrons is being wasted exothermically instead?

Yes again, exactly--electro-chemical energy is being exothermally wasted.

What I'm trying to get my head around is that we have say, 100A from cell A entering cell B and 100A will leave cell B no matter what it's IR, but the voltage will drop (and heat produced). It's as though a high IR cell is expending more energy passing on that 100A than a low IR one, but how?

Picture a load of two resistors in series, one is 1 Ohm, the other is 10 Ohms. 1 Amp of current is flowing through the series, but the power dissipated by the 10 Ohm resistor is ten times greater than that dissipated by the 1 Ohms resistor. Similarly, if you have two cells in series driving an Amp of current, and one cell has an IR of 1 mOhm, and the other has an IR of 10 mOhms, the 10 mOhms cells will waste ten times more power on Peukert loss than the 1 mOhm cell.

 

[wb9k]

OK, the quick sketch I've put togther is attached. To keep things simple, let's first discuss a 1P pack--no parallel cells to complicate things.

Interesting idea, but I can't really see why the end cells would see more abuse besides additional heating from a high impedance connection to a busbar. The current is the same in every cell in a 1P battery. Coulombic efficiency of lithium ion batteries is nearly 100%. Ah in = Ah out. The additional resistance of a busbar connection wouldn't affect the adjacent cell (except for the heating).. the voltage between the electrodes of the cell only would depend on temperature, current and SOC.
Oh and AFAIK, Peukerts equation really only applies to lead-acid AFAIK. Unless you are just using "Peukert effect" interchangeably with "cell resistive losses".. maybe I'm misunderstanding your theory.

Peukert losses happen in all kinds of cells, LA is just rather extreme and a sharp contrast to Li ion. Peukert losses are one form of "cell resistive losses", the interconnect losses are another. Different mechanisms, but essentially the same result. If you look at the last example above, you'll see why what seem like even tiny numbers can make a difference in this context. We're not used to seeing any material difference between 1 mOhm and 10 mOhms in any circuit or connection, but with batteries we need to pay special attention. If you take a pack with perfect cells and replace one cell in the middle with a cell with double or more the IR (this can be the difference between 0.5 mOhms and 1 mOhm, which sounds trivial to most ears), the pack will not have the same usable capacity as it did with the cells all equal. I know this for a fact from various experiences in the plant at A123. The high IR cell will hit LVC earlier than the others, stranding what should be usable energy in the remaining cells. Higher IR always means less usable capacity with all else being equal. Always.

I have the advantage of having seen this in action from many different angles in all kinds of contexts from brand new production parts, to stuff I've worked in my garage, to modules being returned from the field after years of hard use (where the cumulative data logging of the whole modules' life tells the story in stark detail.) I didn't just cook this up in my head one day, it arose from a need to explain a behavior that is very clearly real and shows itself in virtually every large battery pack I've ever seen and most of the small ones too unless the effect is swamped by some other larger problem.

 

[John in CR]

wb9k,

I hope you're pulling everyone's leg with this 1st cell bearing the greatest load and deepest discharge stuff. That can't be the case or in packs not regularly balanced we would see a staggered out of balance condition ALWAYS develop due to cumulative resistive losses being different for each cell in a series, but it doesn't happen. The simple answer is that those end cells get the most damage. If the buss bar is higher resistance then it's likely transmitting extra heat to the first cell. With packs where we commonly see this first cell issue, as Dnmun mentioned the packaging itself distorts the shape of the of end cells much more than others. Ping packs with their duct tape and cardboard or RC Lipo packs compressed by heat shrink are good examples. In other packs it easy to imagine damage from stacking, bumping, or just a bit of grit getting on the face of end cells after series connections are made but before compression is applied added.

 

[arkmundi]

I hope you're pulling everyone's leg with this 1st cell bearing the greatest load ....

I'm in no position, nor are most of us. It'd be great, for instance if the MIT nanotech laboratory had an address where used up A123 cells could be sent for analysis. In other words, some battery scientists with the fancy equipment, like the latest generation electron microscopes. Take a cell, freeze it and thin slice it into layers like they do living tissue, so as to digitally image it and reconstruct the whole cell as a massive 3D image. Do that with enough cells, and with the scientific mind, analyze the data set, you've got a worthy publication. Advance the art of making batteries.

I don't think any of us ES'ers are in the position to know what's going on really. But being the curious folk we are, can speculate. There is now a noted phenomena that there are battery pack cell degradation rate differences that are dependent on the cell's position in the pack, serially, maybe also in parallel. What has not been suggested or discussed is the role that a BMS may play in all this. It can not be ruled out as a possible factor.

My take away from the discussion, which I very much appreciate, is that during the course of monitoring my battery pack health, I may want to make note of cell IR drift, and has been suggested, possibly rotate the cells. If I have a 12S 26650 {as I do}, then break the string in half and rebuild with the outside cells in. It'd take a little bit of doing, but would be manageable. See if I can extend the life of my pack, when I'm at that point. Because I can, because I'm curious.

First, I need to rebuild my 12S string monitor so its easier to use, easier to apply a balance charge when indicated.

 

[wb9k]

wb9k,

I hope you're pulling everyone's leg with this 1st cell bearing the greatest load and deepest discharge stuff. That can't be the case or in packs not regularly balanced we would see a staggered out of balance condition ALWAYS develop due to cumulative resistive losses being different for each cell in a series, but it doesn't happen. The simple answer is that those end cells get the most damage. If the buss bar is higher resistance then it's likely transmitting extra heat to the first cell. With packs where we commonly see this first cell issue, as Dnmun mentioned the packaging itself distorts the shape of the of end cells much more than others. Ping packs with their duct tape and cardboard or RC Lipo packs compressed by heat shrink are good examples. In other packs it easy to imagine damage from stacking, bumping, or just a bit of grit getting on the face of end cells after series connections are made but before compression is applied added.

No, I'm not pulling your leg, and your post tells me you haven't read carefully enough. This is not a pouch cell phenomenon, it affects cylindricals too. Brand new packs, old packs, they all do it, but you won't be able to see it in a pack with defects like the ones you list here. The pack has to be well built and free of other defects (even minor ones) for the end-cell behavior to be dominant enough to actually see. Otherwise, the other build problems will cause issues elsewhere that swamp the effect.

The more this discussion goes on, the more convinced I am that I am correct. It seems incontrovertible proof is going to be needed to persuade some folks.

 

[Punx0r]

wb9k,

Thanks for your continued input. That both you and Luke have described this issue gives me confidence that you are right.

Picture a load of two resistors in series, one is 1 Ohm, the other is 10 Ohms. 1 Amp of current is flowing through the series, but the power dissipated by the 10 Ohm resistor is ten times greater than that dissipated by the 1 Ohms resistor. Similarly, if you have two cells in series driving an Amp of current, and one cell has an IR of 1 mOhm, and the other has an IR of 10 mOhms, the 10 mOhms cells will waste ten times more power on Peukert loss than the 1 mOhm cell.

This is the only point I need a bit more clarification on. The example with resistors is clear, but these are passive devices, whereas cells are, to my mind at least, not. With the resistors the extra power dissipated in the 10ohm one is lost from the entire circuit. With cells we are saying that the extra power loss is borne by a single cell.

What about an example? We have a series string of equal IR cells. Let's say it's delivering 100V and 10A. If we add a 1 ohm IR to one of the cells (or its interconnect), what is the voltage and current now?

I'm trying to distinguish the difference between that scenario and adding 1 ohm to the load and using ohms law as we normally would.

 

[wb9k]

You're on the right track. Resistors are purely passive, so I see the disconnect you are having.

Batteries are complex and dynamic...much more so than appears at first blush. In this context, we have to understand that cells exhibit behaviors of both passive and active devices. They are active in that they can source power, but there is also passive behavioral element that puts a load on the active behavior we seek. And yes, you use Ohm's law to calculate it. To find load impedance (IR) of a cell, load it at 50% SOC at a given C-rate for a several seconds. Measure the voltage drop and use Ohm's law to calculate the cell's internal resistance. (I like to also take a charge measurement--the numbers are close, but not the same). Now you can simply use Ohm's law to calculate how much power will be wasted by this loading effect. The amount depends on the current being drawn and on the cell IR. High current applications will be more profoundly affected--if the pack is being asked to regularly deliver, say, 2C or more. There--now you can calculate Peukert losses for any cell.

Consider now the similar effect of poor interconnections between cell groups. Outliers here tend to be more extreme, especially in hobbyist packs. I think it no stretch to say that many packs out there waste more power on their interconnects than they do on Peukert losses.

 

[dnmun]

did you attempt any experiments to evaluate this line of reasoning? seems like it would be difficult to demonstrate because of the length of time involved creating this effect. maybe putting it into SPICE would push out something to analyze.

 

[wb9k]

did you attempt any experiments to evaluate this line of reasoning? seems like it would be difficult to demonstrate because of the length of time involved creating this effect. maybe putting it into SPICE would push out something to analyze.

I've got an experiment in mind. Shouldn't be too difficult, but finding the time to run it is harder at the moment. I was talking with our lead test guy yesterday (very sharp), who agrees that I am correct and told me he can provide reams of data to back it up. I'm finding this is an idea that is out there in various flavors, but large segments of the engineering community among us seem to resist (no pun intended) this line of thought.

It would be interesting to see if SPICE would yield realistic results. How good are the battery pack models it uses? That's really the question. If it reflects actual reality (demonstrated in an actual test), then somebody at SPICE is clearly already aware of this. Either way, the non-virtual test will be needed to categorically prove what reality actually is.

 

[7circle]

Have a pondering thought on the thermal dynamics manifesting in the end cells and when these are the ends of the series string too, and the relating non-uniform heat distribution throughout these end cells.

Luke-forphysics mentioned this.
The ends are the surface area that the heat is transfered through and thus must be the coolest.

Batteries have an optimum temperture operating point giving a better energy output due to resultant elevated voltage of the reactions.

But what if the wiring to the terminals may cause heat build up around the output terminal.

What causes Ri (internal resistance, or Vout = Vcell(T,C) - Vloss(I,T,C ) The Vloss is a function of Ampere rate, the Temperture and the State of charge (SOC).

So just saying IR causes the cell to work harder is missing the detail of what's going on.
I assume when "IR" is being mentioned it is as Internal Resistance, not I x Ri.

The power loss I^2.Ri (Watts) is being released as thermal energy. If it can't be dissipated the point sources will get hotter.
A hot spot may not be noticed when the end cell has has a large exposed surface area allowing the surface to feel similar to the rest of the pack, because it is releasing the heat. But their is still a hot spot which may be above thresholds allowing electrolyte chemical side reactions adding to Ri in the electrodes.

I don't think I'm adding much to the discussion. Just clarifying it.

Main point it what causes cell degradation?

If within a cell one area is hotter say 10 degrees different to another then the Voltages produced by the Li+ ions migrating to reactions at the electrodes will be different.
The terminal could be cooler if the wiring is large and allows good heatsinking away from the battery cell.
If the area is cooler and thus having a lower cell voltage, then even more Li+ reactions will be occuring at this point as the lower Voltage is less energy to exchange n the chemical reaction.
This could allow degradation of the electrode materials due the higher concentration of deposition of reactants..

Then there is the effect of mechanical loading on the end cells. The ends have to deal with stack weight of momentum transfer under vibration. The closer to the middle the forces (or pressure) are less extreme or say at the average.

Even the wires themselves could be shacking around if not restrained, transferring stresses into the battery tabs.

Moving cells around could just cause more problems if the tabs are being resoldered, causing more heat stress.

But I would like to mention that Ri is not related to Peurket effect.
Peurket effect is a reduction of C (Ah) due to higher I (ampere (C/s))
Ri doesn't change the Coulombic efficiency. (Ah-out / Ah-In)
It will effect Energy Efficiency (Wh-Out / Wh-In)
But Coulombic efficiency changes are key to detecting cell degradation it seems after reading and watching some of the views of scientists working on these cells.

At higher currents I guess there will be probabilities of electrons being transferred through the cell as a loss.
That's the funny perrogative of the uncertainty of tiny electrons but Li+ ions are a bit more predicable.
... Now I'm getting really out of bounds. I'lll exit stage left.... to the hiddden monoteny.

 

[Punx0r]

I think it was me that started with the IR thing. Yes, Internal Resistance. Probably should have stated is as Ri, IR just seem to be the prevailing shorthand when discussing batteries on this forum.

One final question: If high interconnect resistance loads the previous cell, what of the effect on the last cell in the string on the resistance of the cabling to and from the load? And the load itself?

1) Is this not the same effect as an interconnect, but much greater?
2) Surely only one of the two end cells would suffer, due to the unidirectional flow of the current?

 

[wb9k]

I think it was me that started with the IR thing. Yes, Internal Resistance. Probably should have stated is as Ri, IR just seem to be the prevailing shorthand when discussing batteries on this forum.

One final question: If high interconnect resistance loads the previous cell, what of the effect on the last cell in the string on the resistance of the cabling to and from the load? And the load itself?

1) Is this not the same effect as an interconnect, but much greater?
2) Surely only one of the two end cells would suffer, due to the unidirectional flow of the current?

1) No, because the other end of the load is tied to the opposite end of the pack. The actual load itself does load the whole pack equally.
2) I'm still working out the details of exactly how this manifests at the ends of the pack and the directional effects of current as well. I think I've got it figured out, but I'm going to work this all out on paper more neatly before I go further with it here.

Ah'll be Bach....you be Mozart. :lol:

 

[adrock8519]

Thank you guys for all of your inputs. I really learn allot from this thread.

I think the cause of the packs voltage being different between the series connections where due to my friend taking more than the rated ah from the pack. There for, cell groups hitting the cliff earlier than other groups along the string due to IR in the interconnections? I have suggested to him to charge when he has used the rated AH of the pack rather than using the voltage as an indication. Also heat from the controller also in the frame bag will be effecting cells.

Plus if you guys have a look at my loom I simply use 6x parallel leads and using 4x 5AH packs I have 2 connectors spare to add more capacity if desired. In your options do you think leaving these two unused connections is okay or does raise the impedance to unaccountable levels. Maybe I should measure the IR of the parallel sections with the parallel leads attached. What would you class as an acceptable difference of IR between the packs?

thanks again every one especially wb9k

 

[dnmun]

or we can all go Chopin at the mall! or maybe Chopin at the bit.

if your friend does not balance his pack then there is no way to make informed statements about what is happening to his pack on discharge. internal resistance has a big effect during charging also. if the cell has high resistance to charging then the charging current goes to the other cells and leaves the cell with high internal resistance undercharged unless there is a BMS balancing network capable of pushing charge into that cell at the end of charging during the balancing phase.

so you cannot make assumptions based on what you think he has done with his packs.

 

[adrock8519]

He does balance the packs with these. However I dont feel they are sufficient enough to balance such big capacity packs. http://electriflyer.co.uk/smart-guard-2-lithium-battery-checker-balancer/

Yeah I never though about the Ir whist charging.

 

[johnrobholmes]

I have noticed this with pouch cells over the years and also wondered why. After reading this thread I have two theories.



End cells have the least compression. This in itself can lead to early cell retirement.


End cells take the brunt of voltage ripple. I know from experience that measured ripple is different depending on the measuring point. Just like trying to scope a gate on the wrong side of the resistor, as we go interior to the battery I would assume pack impedance drowns out remnants of switching that the power caps don't snub. I've never tried to measure this though, so my assumption of how ripple voltage is dissipated may be incorrect. It does support the idea that lowest impedance connections from battery to controller are best for end cell life.

 

[Russell]

Just adding my limited anecdotal evidence to the mix:

The end cell on my LiFePO4 battery died early because the thin tab failed prematurely due to the physical stress of having the main positive lead soldered to it.

The end cells on both of my LiPo 12S (2x6S) packs bloated first but by then the rest of the cells were getting weak too.

-R

 

[docnjoj]

So compression of the pouches like dnmun and WB9K stated (plus support of the cables at their connection points could really lead to longer lasting packs. I am seriously considering (2) 16000ma Multistar packs for Christmas. I will leave them alone for about 4 weeks just using and balance charging and then when I know they balance and work I will open them and put them back together with pegboard sides and filament tape. They may last a while and not ignite. I have my old grill out on the porch ready for charging them. Does this sound reasonable?
otherDoc

 

[wb9k]

I have noticed this with pouch cells over the years and also wondered why. After reading this thread I have two theories.



End cells have the least compression. This in itself can lead to early cell retirement.


End cells take the brunt of voltage ripple. I know from experience that measured ripple is different depending on the measuring point. Just like trying to scope a gate on the wrong side of the resistor, as we go interior to the battery I would assume pack impedance drowns out remnants of switching that the power caps don't snub. I've never tried to measure this though, so my assumption of how ripple voltage is dissipated may be incorrect. It does support the idea that lowest impedance connections from battery to controller are best for end cell life.

Uneven compression can be an issue, but this usually fixes itself as the pack settles after initial assembly. We let modules sit for 72 hours prior to any in-factory balancing for this very reason. After a couple days, compression throughout the stack equalizes and the effects imposed by the initial unevenness disappear.