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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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The cells arrive with 3.51V charge or about 25% SoC - a bit low.

The first cell tested provided a whopping 3,462 mAh of capacity (272 Wh/kg, relaxed 4.20V - unrelaxed 2.86V at -0.25A discharge current). This is rather surprising as I have never seen any cell, including new brand name laptop cells, that could exceed its marked capacity, even at discharge rates as low as 0.1A. The manufacturer's own capacity rating is 3350 mAh (typical), 4.20V - 2.50V at -0.65A discharge.

I would now need to test the same cell at 0.5A to get a better idea of its charge capacity.

At 45.8 grams each, the cells are about 2.7 gr lighter than spec, which puts them at 3.67 kg/kWh, probably the lightest storage available.

A second cell results in 3,466 mAh under the same conditions.

 

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Hi kennyhobby - the cells are unprotected, at least according to the order, and the price reflects that. The labeling is standard for Panasonic. The standard label neither mentions the manufacturer or the capacity. That is why the distributor affixes the capacity sticker.

The 3200 mAh "Rated capacity" on the data sheet is confusing. In another spec, it says "Nominal Capacity (typical) 3350".

The spec weight is 'max 48.5'. I suppose different lots may have some weight variation, or that overtime, they manage to reduce weight.

 

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Hi Karter2 - thanks for the links to the 18650 testing site.

Yes, that is exactly the cell I have received. Here is an image.

The label at the bottom of the cell wrapper says "E6 7130".

 

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Can you carefully measure the overall length (without shorting the ends)?

They all measure 65mm plus a tad over. About 65.2mm.

There are now two batches. E6-7130 and E6-6X21.

I have measured the capacity of about 10 cells (once each). If measured at 0.25A, they are above 3400 mAh with one at over 3500 mAh. If measure at 0.5A, they are above 3350 mAh.

 

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Karter2, do you know why China or the shipping companies require cells to be labelled for capacity, when they know a lot of cells exported are fakes and have unbelievably wrong capacities? Is there a promotion subsidy for high capacity exports?

 

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A quarter of the cells have been tested, and all have passed. The variation between cells is slight and the physical appearances are identical.

I am using 3 different discharge testers and at varying discharge currents. So it has been difficult to arrive at a testing standard. After some compensation I get the following at 0.25A discharge:

Minimum capacity 3,300 mAh
Maximum capacity 3,500 mAh
Average capacity 3,444 mAh

Testing is from 4.17V relaxed to 2.8V unrelaxed.

At 0.5A discharge, I get about 4% less capacity: 3,310 mAh on average.

In order to verify these cells, I don't think it is necessary to capacity test more than 50% of the cells.

 

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How does one test for cell DC impedance?

When the cell discharges, the voltage drops within 5 seconds (unrelaxed state), and then when the current is cut, the voltage returns back to almost its original level (relaxed state).

So I believe in addition to drop due to impedance, the electrochemical process also results in drop of voltage.

How does one separate the two? Just measuring the drop in voltage on a 1A discharge will be the combination of the two effects, and not just due to the DC impedance.

Any ideas how to measure just that?

 

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Thanks to all for the suggestions.

Hmmm ... as dcb says, I am not sure if it is a good idea to put all weak cells in the same group. For simplicity, assume discharge not more than 1C, and the configuration is 10s5p. Assume 45 of the cells are 3000 mAh to cutoff and 5 of the cells are 2500 mAh to cutoff (the weak cells).

If the 5 weaks are grouped together, then cutoff for the group happens at 2500 mAh per cell (i.e. when 5*2.5A has been drained from the weak group). This means the other 9 groups each has 5 * 0.5 = 2.5Ah of unused capacity. So total dead capacity = 2.5 * 9/10 = 2.25Ah.

On the other hand, distribute the 5 weak cells, one per group. So cutoff will happen at (4*3000+2500)/5 = 2900 mAh. There will be 5 groups each with 100 mAh of dead capacity per cell. So total dead capacity = 5 * 0.1 * 5/10 = 0.25Ah. Which is much less.

When a weak cell is grouped in parallel with four strong cells, I don't think it is the case that cutoff will happen at 5 * 2500 mAh. Rather, cutoff will happen at 4*3000 + 2500. 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. In series, it will disproportionally reduce the voltage, due to the "first past the post" cutoff.

 

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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 ZEVA.com.au. 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?

http://www.ebay.com/itm/10-ft-Flat-...d=182350096443&_trksid=p2047675.c100005.m1851

 

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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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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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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.

 

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Not worth making anything when a proven unit like that costs under $100.
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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.

 

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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.

 

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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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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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Thank you PStechPaul, Karter2, dcb. I will look into the Arduino data acquisition. Plan to build the 100 kWh BMS using an Arduino someday. Thank you for the valuable information.

(Trivia: Did you know that on the Stalinesque forum Stack Exchange - Electrical Engineering, you are not allowed to say "thank you" in the comment? Avoid that place like wildfire.)

 

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What devices and functions would you recommend for this project?

Pack is 48V, 34Ah, 1.6 kWh, rated at continuous 1.5C (50A), Peak 2.2C (75A), charge 0.5C (16.7A).

The 13s10p pack will need the following at the minimum. A 800W charger, a balancer-protection, fuse, i/o plugs, ammeter, voltmeter, 4x thermistors.

Anything else?
Should the charger be installed internal to the enclosure? (power supply will be external)
What size fuse? 100A?
What kind of output plug? What kind of input plug?
Shall the voltmeter also measure individual cell groups voltage? If so, it needs to switch between 13 groups. What is the best way to do this? Use analog CMOS switch?
Charge timer?
On/Off switch or contactor?