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The 18650 - 13s10p Project - 48V x 34Ah

15922 Views 95 Replies 8 Participants Last post by  PStechPaul
The 18650 - 14s10p Project - 48V x 34Ah

This is to chronicle a project to construct a 1.6 kWh module from 18650 cells. The configuration is 13s10p (edit: upgraded to 14s10p) using 130 (140) "Grade A" Panasonic NCR18650B (edit: unprotected) li-ion cells. These cells are nominally rated at 3.4 Ah and guaranteed to 3.25 Ah.

The first order of 100 cells (at US$3.28 each - $272/kWh) was received from Shanghai today and I have started testing individual cells, and plan to test all cells. Shipment was by air and arrived in 4 days. Shipment adds another $1 to the price, and with transaction costs amounts to $1.24 a cell. I was told it is not possible to ship via sea, and must be by air.

It has been observed that cells purchased from some suppliers in China have contained some fake mislabeled re-cycled cells or fake low-quality low-capacity Chinese cells. I plan to test for capacity, weight, impedance, and thermal behaviour during charging.

Photo of shipment - each cell has an individual white paper box with a safe handling warning, and a pair of these boxes are inside a green box with the same warning. These boxes were not from the original manufacturer (Panasonic / Sanyo). Each cell had a sticker that covered the manufacturer's label that says NCR18650B without giving the capacity. The sticker says "18650 3400 mAh 3.7V". There are also lot numbers on the cell's wrapping and probably on the steel casing, which may give a clue to the origins of the cell.


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fwiw, I did find this:

which implies for lack of cell current monitoring, it is better to match levels by capacity and impedance for longevity, in simulation. The more cells in parallel and the less you push them to their current and capacity limits, the less critical.

And they state that the capacity changes and impedance changes aren't uniform even if they all start out the same.

soo, you pays your money, you takes your chances :)

6. Conclusions and future work
The primary results from the experimental and simulation work presented highlights that cells with different impedances and capacities connected in parallel do not behave in a uniform manner and can experience significantly different currents. The distribution of cell current is shown to be a complex function of impedance, including the high frequency aspects typically ignored for single cell models, and the difference in SOC between cells. As a conventional BMS design does not monitor current within parallel units, some cells may be taken above their intended operating current, or be aged more quickly due to increased charge throughput and ohmic heat generation – shortening the lifespan of the overall battery pack.
but... (SOH=State of Health)
This implies that they should age slower than the other cells, and so it is expected that gradually the SOHs of the cells within the parallel unit will converge.
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Here are the cells in the 13 x 10 spacer, gradually getting filled up, as they are tested. I will need to order some more cells in order to fill this one up. (eBay search "18650 spacer"). They come in sizes 3x5, 4x5, 4x6, 1x3, 1x2, 1x1, and snap together. I wish they made them larger as a multiple of popular configs.

To connect the cells, nickle strips of 8mm wide are strung and attached with a spot welder to the negative terminal. (eBay search "battery nickel strip")

I am thinking of building a spot welder per the design of Ian Hooper at What a great guy. But I wonder if I could rent one for this job?

What about the individual cell fuse wire for the positive terminal? Are those welded or soldered? What AWG? I am designing for peak 30 seconds 1.5C or 5A per cell. I think 10A for the fuse is the right size?

What about those 1mm flat tabbing wire used in soldering solar cells. Has anyone used these? I will test them, and I will guess they blow at 10A or a bit under?


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Being in parallel, the stronger cells will supply more current proportionally while the weaker supplies less, and thus the weaker cell cannot get over-discharged. Their voltage will always remain equal. If the current is cut, and the five cells separated, the five in parallel will have same voltage. One weak cell cannot disproportionally reduce the voltage when in parallel.
But your BMS is not monitoring current, so how will you know when the weak cell has been depleted? How will you not over-charge the weak cell? Do you have test data or experience that indicates your statements hold true?

karter said:
"Cells in a parallel group cannot be individually monitored for voltage or SOC, or protected against over discharge, so need to be well matched."
But your BMS is not monitoring current, so how will you know when the weak cell has been depleted? How will you not over-charge the weak cell? Do you have test data or experience that indicates your statements hold true?
The weak cell is in parallel with a large number of strong cells. How can it get depleted without also depleting the strong cells?

When current is drained from a group of weak and strong cells, it is not supplied equally from the cells. The strong cells supply more than the weak, hence the voltage of the weak cell does not drop faster, because it is being replenished by the strong cells. If the weak cell's voltage is not dropping, how can it be depleted?

Weak cell cannot be overcharged because it will raise the voltage of the group and the strong cells will absorb the current, and not the weak cell. All cells will have same voltage. So how can the weak cell be overcharged? Also known as Kirchhoff's Law.

Are you saying that if I have a 1000 mAh cell in parallel with a 3,000 mAh cell and I charge the group with 2,500 mAh, that the weak cell will be overcharged because it gets 1,250 mAh? Impossible. But I don't think you are saying this. Please provide example with numbers to clarify this matter. :)
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the thing not being discussed is that internal resistance is more important than capacity when it comes to determining how the current will be shared among cells, and the two are not directly related. They both are subject to change over the life of the cell, in less than predictable fashion too. But it seems to affect the lifespan of the healthier cells, if the pack is pushed hard and there are few in parallel, and if the pack is pushed beyond 80/20, regularly.

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And internal resistance can be unrelated to cell capacity. So if cells of same capacity are put in parallel, under dynamic condition with a load, they can go out of whack if their internal resistances are very different.

However note that at steady state, when drain is 0A, the parallel cells will quickly balance out, no matter the disparity in internal resistance.

As long as all cells have the same full capacity voltage of 4.2 and empty capacity voltage of 3.0, then parallel cells will balance out at steady state if the internal resistance are different. And no cell, weak or strong, can get over-discharged or over-charged. Of course at 10C, it is a different matter, and internal resistances, thermal behaviour, electrochemical differences, and connection and junction resistance come into play.
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Perhaps a good way to measure internal resistance would be to measure the voltage difference for current of 0.5C and 1C. That would eliminate the effect of the unloaded (relaxed) condition.
Karter2, major - I am not so sure that delta measurement will remove the electrochemical relaxation effect. But agree that is the better way to measure.

Assume a 3,500 mAh cell at 4.0V relaxed. I drain 1A and within a few seconds it drops to 3.8V. The fact that this drop is not immediate shows that there is another process in play, besides the internal resistance. Call it the ECRE effect (electrochemical relaxation effect).

So now I drain another 0.2A for a total of 1.2A, and I get a ΔV. Why would ΔV be composed purely of DCIR and no ECRE? If both DCIR and ECRE were in effect from 0A to 1A, why would ECRE cease to be in effect from 1A to 1.2A, when 1A was arbitrarily chosen?

When the manufacturer does a 1 kHz AC test, I believe the ECRE is cancelled out. But not in a DC test. I would think I need a 1 kHz high power (1A) signal generator with 3.6V bias.

Or is there an easier method?

As it seems ECRE takes a little while to impact the ΔV, I will try to take the measurement immediately. But I don't have a waveform analyzer with storage.

On the other hand, one can argue that for practical purposes, "apparent DCIR" = DCIR + ECRE. And for fake/real testing, it is just fine to measure the apparent DCIR and not the theoretical DCIR?

However when calculating thermal generation, I suppose one needs the theoretical rather than apparent DCIR?
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fwiw, grouping all the high resistance cells together creates other problems, i.e. heat concentration.

I think "reasonable" safety factors on state of charge/discharge and max current are in order, depending how much variance you see in your cells, regardless if you make each level the same capacity/resistance or make each cell within a level the same capacity/resistance.

My gut is that if they are all within %10 that distributed is better. And charge to %80 and discharge to %20 and limit to 3c. But that is a swag, and your use case may differ.

plus a sensitive bms should tell you when something has changed at a level.
My gut is that if they are all within %10 that distributed is better.
but then again, in this simple example the lowest resistance "cell" sees 37 amps and the highest resistance sees 30.3 amps. Even though the distributed pack has slightly less resistance (9.933 ohms instead of 10), just being +-10% resistance can mean a large swing in the current demands per "cell" (~%20, no surprise there)

Group by resistance at least, a quick test.


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What about the individual cell fuse wire for the positive terminal? Are those welded or soldered? What AWG? I am designing for peak 30 seconds 1.5C or 5A per cell. I think 10A for the fuse is the right size?
Tesla use an Ultrasonic welder i believe for their ((aluminium) fuse wires.
There has been much debate as to what rating fuse wire to use, but i believe The Tesla wires blow between 15-20 amps.
Remember, for them, its main function is to protect the cells/pack from a full internal cell short.

Spot welder...this may be of interest..
It looks like the source of the current is the LiPoly.

I was thinking of using A123 nano-LFP 26650 cells, as I have a whole bunch of them, and Hooper's circuit. So replacing the capacitors with the LFPs. Each cell can deliver 120A pulse.
Not worth making anything when a proven unit like that costs under $100.
Not worth making anything when a proven unit like that costs under $100.
Point taken.

I looked up eBay and the cheapest I could find for a spot welder was $200.

If the one in the video is $100, I will check it out. It looked a bit underpowered, and I don't think LiPoly can deliver 700As like Hooper's does.
Its not necessarily the current needed as much as the control, pulse time etc.
Read the full thread, ....and this one also if you are serious about welding those tabs..
Do I have any other option but welding - I may build 8 modules?

BTW, The power supply will be a plug-in and used for more than one module. But should the charger be installed inside the module enclosure? This way the connection is better, and less floating things to carry. But it increases the module size by about 3 cm in length.
There are many ways to assemble and connect 18650s into packs, ranging from mutliple spring clip cell holders, soldering, welding, various "dry contact" assemblies, even using those tiny neo magnets to clamp wires on ( yes, it works !). There are also various commercial pack assembly kits, and contract build services....the choice depends on many variables, not least of which are budget and available skills.
This "no weld" system is well respected for its integrity..
However, the majority of DIY pack builds are generally spot welded or some still risk soldering.
You could also consider silver-bearing conductive epoxy or other adhesive, but it's rather expensive: ($33 for 0.09 oz) ($122 for 0.5 oz) ($49 for 0.2 oz pen) ($14 for 2.5 gram 0.09 oz syringe)

There are less expensive alternatives with nickel, carbon, or perhaps copper, but look at the volume resistivity. It may not require very much if the conductor and the cell are smooth and flat. You could fasten the connecting strip with a tiny drop of conductive adhesive and then reinforce it with a larger glob of plain epoxy over the assembly.
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50 of 100 cells have been tested. They all pass with flying colors. Some of the cells that were charged and discharged with the same testers register within +/- 1 millivolt of one another. All physical appearances are identical and same as genuine Panasonic NCR18650B.

Test is from 4.18 settled to 2.8 unsettled at 0.5A

Minimum 3,300 mAh
Maximum 3,500 mAh
Average 3,440 mAh

If more than 3 or 4 fake cells were included in the batch, one or more would have been detected. There is no reason why the distributor would want to throw in a few fake cells. If the remaining cells include fakes, they will show up in weak groups as the pack is discharged.

I will also be doing a nail test and an impact test with video!

I think I have found a method to measure the internal resistance. I will test DCIR with different base currents, and different SoC levels, and see what we get, and how it compares to the ΔV/ΔI formula.
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If you are using a data logging type charger/discharger, you can simply compare discharge curves at different loads and even directly plot DCIR against dod/capacity.
Are you logging data for future comparason ?
I wish I had a data logger. Can you refer me to a nice one suitable for this task?

What future comparison? You mean as the cell ages? It could be for that too. But at the moment my interest is to detect fakes.

One problem is that when current is increased or decreased, it takes a while for the voltage to adjust to the new condition. So the impedence will depend on when the measurement is taken.

To see how wrong the formula (V0 - V1)/(I1 - I0) is, note that V0, I0, and I1 are constants. But as the cell is discharging, V1 decreases in voltage and is time dependent. But the DCIR does not linearly decrease with cell voltage, it is pretty constant. So the formula cannot be right.
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