[ElectriCar]

I've not seen this addressed and I've often wondered about it so I'll throw it out there. Does a lithium battery pack really need a BMS or do just the larger modules EV's use with the plastic housing?

Apparently the Tesla doesn't have a BMS in the sense of monitoring each battery as most home built EV's do. I've wondered for a while whether a BMS is really needed if you have it set up properly.

The Tesla uses 6800 of the Panasonic 18650 battery at 2.9ah. The pack is comprised of 11 modules, each with it's own microprocessor sending voltage and temp reading to the cars main processor. Each module apparently has about 618 cells. Below is link to a summary of the battery packs configuration and it's properties.

The batteries, unlike typical EV batteries use internal pressure relief mechanism and a passive over current protection mechanism built in and is contained in a steel canister to help dissipate heat which also extends the cell life. Also the pack is assembled with two fuses, one on each terminal which disconnects the cell in the event of a short somewhere external to the cell.

Seems there is a lot of fault protection measures not used in typical plastic housed models.

http://www.google.com/url?sa=t&sour...OnBBi0YUQ&sig2=J4FdtkbBiVAHHoyYSp2JLA&cad=rja

 

[JRP3]

Tesla doesn't monitor each cell in a parallel string but monitors each string as a single cell, I think. Conversions that do use a BMS monitor each of the large prismatic cells because they are in series. My feeling is that it's better to get cells as close as possible in capacity and resistance and not bother with a BMS at all.

 

[jockepocke]

Well, no, you don't really need the BMS, if you make sure to take other precautions! What they will probably do in the tesla is closely matching batteries so they all have the same capacities, and either top/bottom balance them all before putting them in the final series/parallell configuration, and make sure not to discharge more than the weakest cell can handle and/or stop charging when the first battery is full.

There was a thread that later turned into a discussion about this balancing:

http://www.diyelectriccar.com/forums/showthread.php/fla-trouble-bad-cells-whyi-52255.html

Started about FLA, but info some pages later are about Li, and wether BMS is necessary!

 

[steven4601]

I have an idea : put your batteries in series by a very low-impedance connection. Add parallel strings. and make 'high'-impedance connections between the parallel batterries. That way individual differences should balance out by putting a high number of cells (20 or more). During high discharge the small differences may cause a small amount of energie being burned off by the 10 ohm resistors, but generally this loss is very little compared to battery losses. Putting your BMS on one side or in the center of the string should provide a way to balance the pack.

string of 3S 2P

[neg]---------[|+------|]+------|]+------[pos]
        |        |        |              |  
        |       10E      10E             |
        |        |        |              |
[neg]---------[|+------|]+------|]+------[pos]

<

 

[dtbaker]

Does a lithium battery pack really need a BMS or do just the larger modules EV's use with the plastic housing?

Apparently the Tesla doesn't have a BMS in the sense of monitoring each battery as most home built EV's do. I've wondered for a while whether a BMS is really needed if you have it set up properly.

Engineering-wise it would be great to monitor EVERY cell and be able to interact with smart charger ,controller, and warning lights....
but economically, it just is hard to justify the cost of cell-level BMS.

Lots of large format prismatic cell users decide to install some type of BMS because there are only 30 to 50 cells (not thousands). The small cell users sometimes opt for string level BMS to spot difference between strings indicating problems with cells somewhere in the string.

Whether or not you NEED a BMS becomes pure personal preference and ability to deal with the risk. In a DYI ev, nobody expects 100% idiot-proof fabrication, assembly, or driving habits, and so have to make decisions and weigh the costs individually. If the pack cells are well-balanced at whichever end you consider the most likely for you to end up at (top or bottom), it is somewhat unknown if or how much a pack will ever go out of balance under normal conditions...

Lots of people, including me, are taking a 'wait and see' path... I have balanced my pack as best I can and will probably install 24ga wires to bring a wiring loom back to spot easy for me to check with a multi-meter. Planning to check manually 'periodically' at the end of my charge cycle (because I will rarely go below 50% DOD) to make sure I don't have a few rogue cells at the top, and see how it goes.

 

[crostiburger]

I have an idea : put your batteries in series by a very low-impedance connection. Add parallel strings. and make 'high'-impedance connections between the parallel batterries. That way individual differences should balance out by putting a high number of cells (20 or more). During high discharge the small differences may cause a small amount of energie being burned off by the 10 ohm resistors, but generally this loss is very little compared to battery losses. Putting your BMS on one side or in the center of the string should provide a way to balance the pack.

string of 3S 2P

[neg]---------[|+------|]+------|]+------[pos]
        |        |        |              |  
        |       10E      10E             |
        |        |        |              |
[neg]---------[|+------|]+------|]+------[pos]

<

I think would work fine..., and it's simply brilliant.

 

[Huub3]

I have an idea : put your batteries in series by a very low-impedance connection. Add parallel strings. and make 'high'-impedance connections between the parallel batterries. That way individual differences should balance out by putting a high number of cells (20 or more). During high discharge the small differences may cause a small amount of energie being burned off by the 10 ohm resistors, but generally this loss is very little compared to battery losses. Putting your BMS on one side or in the center of the string should provide a way to balance the pack.

string of 3S 2P

[neg]---------[|+------|]+------|]+------[pos]
        |        |        |              |  
        |       10E      10E             |
        |        |        |              |
[neg]---------[|+------|]+------|]+------[pos]

<

Steven,

I was wondering what the added value would be of the high-impedance paths. Why not choose these also low-impedance, so the cells always even out? Basically this is how parallized packs are build up.

Regards,


Huub

 

[steven4601]

It prevents cells from burning up due to minor cell differences.

Edit:

What the high impedance path does : It prevents a bettery battery falling in for the other during heavy (dis)charge. (Cheap) Batteries do not match. Allowing them not to match improves the current share between the cells through the length of the entire string. I can add a simple simulation to explain it further it if required.

 

[dtbaker]

i would be very nervous having that many potential toasters in my car...

 

[steven4601]

I am not saying you have to do what I posted.
Just I would not hardwire the cells in parallel and hope for the best. It will work, but if you want to make sure all cells get an equal share and yet allow a BMS to manage all parallel cells this might be a cheap solution.

If you calculate it through, the energy dissipated in the resistor is minute compared to the loss in the battery.
Lets say the batteries are at an average 0.35V out (bad case of total 5mili ohm difference) during discharge of 70Amp per string. (210A total) That is 35mA through the resistor, 0.035^2 * 10 = 12.25mW. If the battery itself is 7mR, 70^2 * 7mR = 34 watts in that battery heating it up! If you'd hardwire the three batteries together, two 10m Ohm a third 5m Ohm cell. All of the 210A will pass through them, just the 5m Ohm one will have to handle 105A, and the two 10m Ohm's 52.5A each.! 105A^2 * 0.005 = 55 watts, that cell will get hotter quicker, drain quicker, and peters out the first. Guess which are next ? Yes the lethargic 10m Ohm ones.

Also worth notice is that failure mode of a (thick)film is quite subtle compared to an overstressed Lithium battery.

Hope it makes sense.

 

[Huub3]

Steven,

hmmm, whether it makes sense I am still not sure about, but you certainly got me thinking. I was missing a part of this in my consideration. Would be good to do a simulation indeed.

I am a bit disputing large voltage delta's over neighbouring cells. When charging or resting, the voltages will allign, also over a 10 ohm R. Furthermore for lithium, small voltage differences would mean already large charging states. So we should avoid that through adding these resistors, we would have one or more cells go into deep discharge (or overcharge) without it being visible from the outside.

Could you do such a simulation (in a simple model), that might be very enlightening indeed.

Regards, and thanks for explaining this further!


Huub

 

[steven4601]

Tomorrow hopefully Ill have time to make a simple simulation demo.


How would it not be visible to see a cell reversing if the 10 ohm resistor would be made up from a network of three 4.7 Ohm resistors in star formation?

Also, adding resistors does not enhance the quality of the measurement as in voltage per cell, In fact, it worsens the voltage per cell measurement!

 

[Huub3]

Steven,

Tomorrow hopefully Ill have time to make a simple simulation demo.

Thanks, great! will be my first visit tomorrow, as I feel I am missing a point somewhere.

Also, adding resistors does not enhance the quality of the measurement as in voltage per cell, In fact, it worsens the voltage per cell measurement!

I think I was trying to convey the same, that the resistors do not help monitoring individual cell voltages. On the other hand, your proposal was aimed at reducing the need for a BMS, so that might be fine.

One point that I would like to consider, is the case of a fail open cell. In that case the additional resistors might cause the cells in the same string, and just "before" and "behind" the bad cell to take less of the total load, meaning that not only the neighbouring cells (same "layer") of the dead cell get additional load, but also cells at a larger distance. THis effect reduces with the amount of parallelized cells.

Tomorrow I might try to visualize my point on this.

Regards,


Huub

 

[steven4601]

Cheap BLDC hall sensors (like in sensored spindle drives in cd/dvd recorders) could be used to detect a dramatic open cell failure for each string. Would not even add more than a few dollar/euro's in total for the parallel strings. Could also use accurate current transducers / LEM sensors if you want to measure string current in tens of miliamp accuracy.

Meh, so many possibilities, so little time.

 

[steven4601]

Here two simulations. I did not add the graph as its is DC and does not make sense to include it here. I did add the currents in the diagrams.

An impedance imbalance of 5 and 10mili-ohm has been added to show what happends.

The low impedance simulation shows that battery V6 is being loaded with 101A compared to the 51A for the other two next to it.

The power dissipated in R4 and R3 is less than 5mW in the simulation with 10 Ohm parallel resistance.


//steven

 

[Huub3]

Steven,

indeed, somehow I misinterpreted this concept differently, and missed the point completely.

I was not aware of the importance of matching internal resistances of parallel packs, but your graph shows it clearly. And even when matched at the beginning, a divergence over lifetime could lead to some instable behaviour.

I am unsure whether higher temps lead to higher resistances or lower (I hope higher, so PTC), and whether deterioration due to temperature loads leads to higher or lower resistances.

Looking a that graph, I am a bit bothered by the fact that one cell more or less influences a whole sting. In the picture all cells of the middle string have now 9 A more than the cells of the left and right string.

What if in the next "layer", the unbalance would be in the left or right string, then this difference would be smaller, isn't it?

I am having troubles grasping the deeper concept after this idea, that's the reason I ask these questions, that are most likely simple to see for you.

Thanks a lot for doing this analysis,


Huub

 

[steven4601]

You right. If you grab a bucket with 0's and a bucket with roughly the same amount of 1's and you throw these together in a larger bucket. mix them thoroughly. Now you grab a hand full of these mixed numbers.

How many 0's you have if you counted 1000 1's ? probably also 1000 0's. Its chance but its likely ;-)

In other words / on topic:
The longer the strings, the higher the probability the total string impedances will match.





Edit: Lithium based battery impedance initially drops as temperature rises. After it heats up it starts rising again IIRC followed by a thermal runaway.