Paralleling uneven voltage cells?

[rg12]

If I have two identical cells, one at 3.75V and the other at 4.10V and I spot weld a nickel strip to the positive and negative of both to parallel them, how fast will it take the two cells to level eachother?
Will it be a super fast almost making the nickel red kind of current share or a steady slow speed?

 

[hemo]

The transfer will be instant between the two and very very hot, I don't know the math involved but someone else more knowledgable will.

 

[goatman]

if im playing with batteries like that ill use a single strand of copper wire, 3 magnets and use my finger to hold the wire to the cell and if the wire gets hot you will naturally pull the wire off the cell. a single strand will act as a fuse wire. a 25r will melt the wire at about 0.5v difference if I remember right.

 

[amberwolf]

the answer is:

don't parallel cells of different voltages




either charge up the low ones,

or

drain the high ones

until they are as close to equal as you can get them.



to determine the current flow you'd see from any particular cell to any other, you need to know the difference in their voltages (high cell voltage minus low cell voltage) and the dcir of each cell (which changes with state of charge, so it will change as the current flows beteen them, changing the rate of current flow at the same time that the flow changes because the voltage difference changes). and also the resistance of the connections.

anyway, with the voltage difference and the dcirs, you add the two dcirs and the connection resistances, and then divide the voltage by that, to get the amps. (ohms law)


pulling some numbers out of thin air, that may have nothing to do with your actual parts:

let's say you have 1.5v difference, and oh, say, 1mohm total dcir and connection resistances. that's 1.5v / 0.001 = 1500a initial current. since that will cause huge voltage sag in the source cell, that will drop just about instantly to some much lower level, but still very high--possibly hgiher than the cells are designed to handle.

if the current lasts long enough to heat them internally enough, they could start burning. it's not very likely, but...the possiblity exists.

even if they don't, they could be permanently damaged in ways you can't know, and therefore cannot predict how they will behave under any particular situation, so they could become a fire risk--you simply can't know.


now that we've seen what could potentially happen to the cells...you can then calculate what will happen in the conections themselves.

if, let's say, the nickel itself is 0.1mohm, between the cells, total. 0.05mohm on each end. if you had that initial 1500a, you can figure the voltage drop across each nickel strip is 1500 * 0.00005 = 0.075v.

watts is v * a, so 1500a * 0.075v = 112w

112w is a heck of a lot of heat for a very small space like the nickel between two 18650 cells. thankfully the current can't stay that high for more than an instant due to voltage sag in the source cell. if it's a drop of 100x then you're down to a bit more than a watt over that small area, which is probably not enough to heat it very much, but if it only drops 10x then that's still 10w, and soldering irons start at not much above that, and fairly quickly get hot enough to begin melting solder.

i doubt there's enough energy in the difference between the two cells to do that (but if you had two banks of already paralleled cells that wer at sufficinetly different voltages, and had sufficient ah among the high cells vs the low ones, there could be time enough at the high currents to make enough heat in the nickel to do something.




since i don't know what your numbers are, i can't say if any of this would happen in your case. you'd have to determine tha once you have the necessary numbers.

 

[serious_sam]

Besides the safety issues, it's bad for the cells anyway.

If you have to parallel unequal voltage cells, use a resistor to get them within a reasonable potential difference before permanently connecting them.

 

[john61ct]

Also depends, same chemistry or not?

e.g. LFP with NMC both fully charged 3.35V vs 4.1x not at all a good idea

Lead not as fast less of an issue.

Too many variables, basically unless you already know the answer don't mess with it.

Boom bad!

 

[serious_sam]

Also depends, same chemistry or not?

Read the first sentence of the first post...

 

[dogman dan]

Short version. Don't do it.

Damage to the cell itself is the real reason, although it could also get the strip hot enough to cause it to become a resistor.

If you can't charge them to closer to the same voltage, then for sure you don't want that piece of dog shit cell in your string.

 

[spinningmagnets]

If you are building a pack from individual cells, there are cheap data logging charge/discharge devices that are handy to use. Maybe $35? Like the Opus or Littokala.

They are popular for building DIY powerwalls from near-free cells that are only half-used. Cycle a cell twice to verify its actual capacity, then charge to a voltage that's the same for all cells in the group before assembly.

That being said, you can equalize several cells by hooking them up together, but each cell must have a resistor. As far as the resistance value of the appropriate resistor...if you have plenty of time on your hands, err on the side of too much resistance. That way, it will be a very slow and cool equalization.

If you want to experiment, please do so in a way where a runaway cell can burn with no damage to your home or garage. If a cell begins to go into a runaway condition, it goes from warm to flames in just a few seconds.

I dont recommend equalizing cells like this, but since some builders are going to do this, I just want to emphasize taking precautions.

 

[999zip999]

I would have a clay pot the one without the hold in the bottom thing it . Just throwin that clay pot if they start to smoke. Always have a backup plan. I have taken the cover off used tool packs before just to see them start to smoke.

 

[docware]

If I have two identical cells, one at 3.75V and the other at 4.10V and I spot weld a nickel strip to the positive and negative of both to parallel them, how fast will it take the two cells to level eachother?

First of all question; why anybody would exercise such crazy thing ? Moreover, how would you spot weld sparking second nickel strip ?

In the real case it depends mainly on type of the cell. Current depends on the voltage and resistance according to Ohm´s law. DCIR resistance of the cell is main constituent, however you have to include also resistance of the spot welds and resistance of the nickel strip to the total sum. Resistance depends on nickel strip dimensions, number and quality of the spot welding. We can calculate roughly 1 – 2,5 miliohm /strip, that is cca 2 – 5 miliohm total.

Here is course of the equalizing current for two Samsung 25 R cells sitting in the holder and connected trough measuring shunt 10 miliohm. DCIR of 25R is about 20 miliohm at 25 °C. In my case additive resistance of the wires and transient resistance of all contacts is about 52 miliohm, total resistance with measuring shunt is cca 102 miliohm. Cell A voltage 3,751 V, cell B voltage 4,101 V, resulting initial current 3,486 A. At time 3 600 s is current 20 mA, at 4 800 s is current 8 mA.




With nickel strips, total resistance about 45 miliohm (for Samsung 25R) and the same voltage difference 0,35 V would be initial current cca 7,8 A.
Discharging Samsung 25R Li-ion cell with initial 7,8 A is not so big issue as charging of the second cell with such high initial current.

Comments on proper voltage equalizing from other members are comprehensive enough hopefully. You want to have all cells equalized within few milivolts (or rather bellow 1 milivolt) before final spot welding, also because of the proper balancing of all paralel groups from the very beginning.

 

[eMark]

Having a difficult time understanding (24S*7P* 17.5Ah 25R battery) the reason rg12 even considered spot-welding two cells of a 7P-group together when there's a difference of 350mv (3.75v vs 4.10v) between two of the cells of a 7 P-group. Is it a typo? Maybe, he meant 3.95v vs 4.10v which is still a difference of 150mv.

You want to have all cells equalized within few milivolts (or rather bellow 1 milivolt) before final spot welding, also because of the proper balancing of all paralel groups from the very beginning.

Docware may have meant "bellow 10 milivolt" (e.g. 3.765v vs 3.755v) instead of "bellow 1 milivolt" (e.g. 3.756v vs 3.755v) difference between the 7 cells in each of the seven P-groups. Even less than 10mv may not be possible if salvaged 25R cells ... depending on condition (differences in cell storage capacity). Has rg12 checked to see if all 168 cells are capable of being balanced to within 10mv of one another (e.g. 3.750v - 3.760v) before spot-welding? From his post one could conclude that he's not using brand new 25R cells with some at nominal voltage and some higher.

What do you guys consider the allowable millivolt difference between the 168 cells in a 24S7P 25R battery (new cells vs salvaged cells) that a DIYer can safely allow when spot-welding busbar connections on each of the 7P-group cells (e.g. 10mv, 20mv, 40mv, 80mv)?

 

[spinningmagnets]

Al that matters when considering the parallel connection of unequal-voltage cells is the cells in each parallel group, compared to each other in that same parallel group.

By that I mean that, if one P-group is at 3.0V, and another P-group is at 4.0V, they can be attached in series to produce a 7.0V sub-pack with no harm. Eventually, the BMS will slowly bring both P-groups up to 4.2V, to result in a sub-pack of 8.4V

In a 7P pack as described, you only need to be concerned with each group of seven cells being close in voltage to each other.

Of course, I definitely recommend that all cells should be brought to as close a voltage as possible before pack assembly.

 

[rg12]

Having a difficult time understanding (24S*7P* 17.5Ah 25R battery) the reason rg12 even considered spot-welding two cells of a 7P-group together when there's a difference of 350mv (3.75v vs 4.10v) between two of the cells of a 7 P-group. Is it a typo? Maybe, he meant 3.95v vs 4.10v which is still a difference of 150mv.

You want to have all cells equalized within few milivolts (or rather bellow 1 milivolt) before final spot welding, also because of the proper balancing of all paralel groups from the very beginning.

Docware may have meant "bellow 10 milivolt" (e.g. 3.765v vs 3.755v) instead of "bellow 1 milivolt" (e.g. 3.756v vs 3.755v) difference between the 7 cells in each of the seven P-groups. Even less than 10mv may not be possible if salvaged 25R cells ... depending on condition (differences in cell storage capacity). Has rg12 checked to see if all 168 cells are capable of being balanced to within 10mv of one another (e.g. 3.750v - 3.760v) before spot-welding? From his post one could conclude that he's not using brand new 25R cells with some at nominal voltage and some higher.

What do you guys consider the allowable millivolt difference between the 168 cells in a 24S7P 25R battery (new cells vs salvaged cells) that a DIYer can safely allow when spot-welding busbar connections on each of the 7P-group cells (e.g. 10mv, 20mv, 40mv, 80mv)?

Who said anything about the pack that is in my signature?

First of all thanks everyone for the detailed responses

As I said in the initial message that I was only wandering and not willing to do this and even mentioned the fact that it might just get the nickel strip to being red hot, so am not a noob thinking it's a good idea as the difference is too high.

I usually never mess with used cells so I receive all matched brand new cells that are exactly the same voltage and IR.

 

[999zip999]

I like for the final connection is a magnet with a piece of 24 gauge or 20 gauge wire little four strand wire in the hit him with the magnet and see what happens see how long it takes a burn that wire. Yes connecting outside in a safe place not on the kitchen table

 

[fechter]

What do you guys consider the allowable millivolt difference between the 168 cells in a 24S7P 25R battery (new cells vs salvaged cells) that a DIYer can safely allow when spot-welding busbar connections on each of the 7P-group cells (e.g. 10mv, 20mv, 40mv, 80mv)?

I'd get them well under 10mV. You can temporarily short them together with some skinny wire and tape and let them equalize overnight and they will all be good for welding. The thing to avoid is destructively high currents when you connect them. docware's graph really tells the story. If your starting differential is too high when you put the skinny wire across it, the wire will burn like a fuse and hopefully not damage the cell too bad or start a fire. You can strip a length of stranded wire and untwist it to get single strands.

 

[eMark]

Who said anything about the pack that is in my signature?

Thought your battery listed at the bottom of your post was the future DIY build you were referencing (7P) ... "24S 17.5Ah 25R, Greentime Controller, QS203 5KW"

In a 7P pack as described, you only need to be concerned with each group of seven cells being close in voltage to each other.

Of course, I definitely recommend that all cells should be brought to as close a voltage as possible before pack assembly.

Being that the OP mentioned two cells having a difference of 350mV (3.75V v 4.10V) which we all agree is way too much with the OP possibly thinking the two cells might balance each other without any harm to either.

As docware indicated DIYers should strive for 1mv when using new brand name 18650 cells (which is possible) ... 2-3mv is close enough with useable salvaged cells having similar capacity and amperage rating assuming they can maintain that for several hours. What if voltage strays to 6-9mV after 24 hrs? Would you still consider them useable being they're still under 10mV ?

 

[spinningmagnets]

This is just my opinion. If I charge a bunch of cells to 4.10V, and the next day (as I am about to assemble them), if I find one cell has self-discharged to 4.00V, I would set it aside. Any cell that self-discharges will drain the entire P-group over time (when stored for winter).

I'm not saying that a cell that has drifted a little bit is worthless. I have a lot of work flashlights that use a single 18650. But if I have plenty of cells that don't drift at all, I wouldn't use a cell that drifts for any job.

If you want to equalize seven cells just before assembly into P-groups, and they are all verified to be of identical capacity and they all hold a steady voltage, you can put all seven on a DIY equalization board where every cell has a resistor to limit current. Start with a resistor that has a high value. If you want to speed up the process, slowly swap-in resistors with slightly lower values until the largest-disparity equalization results in cells that are slightly warm. Personally, I am very patient...and I would equalize all cells very slowly.

After a pack is fully assembled, lets just say one P-group is at 4.15V and another P-group is at 4.05V. Then, the BMS will go through a few cycles, and all groups will end up at 4.20V

Of course, what I do is within what I am willing to risk. I could be wrong.

 

[docware]

The cells after roughly 80 minutes equalizing were within 1 milivolt, after resting another 60 minutes were within 1,5 milivolt. It´s obvious that with proper procedure, taking enough time and starting first with resistor, it´s no problem to have all good cells within 1 milivolt. Just take enough time. Discard self-discharging cells during the process. 1 Ohm resistor with proper power rating for laid case can cover all you need. Use DMM with suitable resolution and repeatability.

 

[methods]

This comes up over and over.
docware has the data

I regularly connect single 18650 cells in parallel that are off by even a volt.
You have to consider the potential and all of the inline resistances including connection points, internal resistance, and wires.

The next data to see is life-cycle data on two test cases
1) Lifetime test on 10 cells, all the same
2) Lifetime test on 10 other cells, but subject those to a tiny charge over-current 10 times each

What is the impact on lifecycle?

I can say that no appreciable energy is transferred cell to cell on high internal resistance cells as you might buy for a dollar on Amazon.

But - of course
If you get ultra low internal resistance cells (super high C-Rate)
or
You get a bunch of cells in parallel (so paralleling their internal resistances down)
Then yea - more current for a given potential across the conductor

...

I dont worry about it AT ALL from a safety, fire, or heat standpoint - hooking two cells together in parallel.
Not until I see conclusive data to the contrary.

Give me a pile of 3.7V cells and a pile of 4.0V cells
I will hook them all up in parallel

Not saying I am right, just saying what I do.
Looking forward to LifeTime Test Data that is definitive (I know it is out there)

-methods

EDIT


Please confirm that is in Seconds and not milliseconds

If that is the case, and that data is true and correct, then significant power transfer is occurring. My testing did not show this, not against $1 cells or even $10 Vape cells. Perhaps I need to repeat my test, if that test data is correct.

I will do it right now (hold)

 

[methods]

Quick Math

Current in excess of 1A, Area under the curve, greater than 600 seconds

600 seconds = 10 minutes
10 minutes = 1/6 of an hour
1Ah * 1/6 = 1/6th Ah minimum (much more) transferred

Ok, the data is not insane

(Grabbing two cells from my bin)
3500mAh 20A rated

Cell 1 = 4.072V as read with a calibrated FLUKE 87V
Cell 2 = 4.075V
(Damn! they are balanced, no test)

Will discharge one cell and return to prove posted data is correct.

-methods

 

[methods]

Repeating the Scientific Test, because that is what Science does right?
Eliminates circular arguments by agreeing upon something that we can both confirm.



No amount of prose can change a mans mind...
No amount of 3rd party test data can change a mans mind...
But a test?
Run with your own, known test equipment?
Using your own proven knowledge of math?

... Will have to agree with whatever the results are.

-methods

 

[methods]

Test Results Part 1

Is Ohms Law Correct?



Current thru one resistor = 568mA
Current thru two resistor of near equal value in parallel = 1.051A

V = I * R

1.051A / 2 = 525.5mA
1000mA = 1A
10% = 52.5mA
568mA + 52.5mA = 620.5mA
568mA + 52.5mA =515.5mA
515.5mA < 568mA < 620.5mA = TRUE

So the rule of thumb for paralleling resistors is true, to within 10%, which is Engineering standard.
(Wasting time while apx 3.7V * 1A = 3.7W burn off...)

3.7V * 3.5Ah = ~13Wh in the battery
...

Now we are almost ready for our test.
Sorry for multiple posts - but - this must be important if we are talking about it again.

This should be SQUASHED by 5 posters at once, all posting legitimate test data - RIGHT?

-methods

 

[methods]

TEST RESULT #2

Resting voltage immediately after test = 3.968V
Temperature of resistors >60C (burns my finger)

Proof of power transferred out of battery

Resting voltage after a few seconds = 3.977V
3.977 > 3.968
Proof that battery voltage is still recovering (so delay)

Resting voltage after a few seconds = 3.981V
Still rising (this is what I would program the automated tester to do... wait until voltage stabilizes

Resting voltage after a few seconds = 3.982V
Curve (Rate of change or dv/DT - change in voltage over time) has settled.

Continuing test

Cell 1 = 4.075V
Cell 2 = 3.984V
4.075-3.984 = 91mV

91mV is far in excess of 10mV or 20mV

Continuing test

TEST PROCEDURE
1) Parallel the negative of each cell
2) Connect the positive terminals thru Fluke 87V, in high current mode, in high resolution mode


96mA Observed IN MY TEST SETUP

TEST SETUP
Spring loaded battery carriers (pictured above)

TEST RESULTS
"You can absolutely parallel single high rate cells in parallel to balance them in a configuration like this"

FURTHER TESTING
"Determine what the current is if you hard-short them (Doing that now)

-methods

 

[methods]

FINAL TEST

TEST SETUP
Batteries pressed together, negative to negative, with high force
One FLUKE lead on the Positive of each
Fluke set to high current mode

Reality Check
Shorting out a cell like this results in a reading of 26A

TEST RESULTS
360mA observed

Thoughts
"The test setup is EVERYTHING on something like this. You would have to use heavy gauge wire, or large sheets as Fechter suggested, and a low value shunt... to even see high currents"

SO

The question becomes: "How do people parallel cells together for normalizing their voltage"?

My test data shows that it does not matter, so long as there is some smaller gauge wire connecting them.

...

Who wants to contradict my findings with ACTUAL TEST DATA?

-methods