Current Carrying Capability of Nickel Strips - Help Understanding

[YoshiMoshi]

I'm simply trying to understand how to select nickel strip for my battery pack. Note that I'm just simply trying to understand the math here only and the concepts behind selecting nickel strips, and nothing more. I'm not suggesting anyone do anything unsafe.

I'm building a 5S3P Li-Ion pack from 21700 cells. I have found various charts and tables depicting nickel stirp dimensions and their acceptable currents. But I'm not sure where these values come from. An example of one is below.

I'm trying to minimize the thickness of my nickel strips, by evaluating how wide I can make my strips. Typical cheap spot welders have difficulty spot welding strips thicker than 0.15 mm. The largest cross sectional area on this chart is 12 mm wide and 0.15 mm thick, with optimal current carrying capacity of 17 A (from that table).

My BMS has a continuous discharge current of 40 A, and a maximum instantaneous current of 80 A. Do I select nickel strip for 40 A instead of 80 A? I would assume so. So lets assume 14 mm thick and 50 mm long (3 series connections of

Bulk resistivity of pure nickel is 6.99 x 10^(-8) ohm meters
Strip is 14 mm by 0.15 mm by 50 mm, current stated at 40 A
Resistance is length / (width * thickness) * bulk resistivity
( ( 50 mm ) / (14 mm * 0.15 mm) )( 6.99 * 10^(-8) ohm m)( 10^3 mm/m) ~ 0.001664 ohm
Power is I squred R or (40 A)^2 * ( 0.001664 ohm ) ~ 2.663 W
Voltage across strip is I * R or ( 40 A )( 0.001664 mohm )= 0.0666 V

So pased on a loss of 2.663 W from the nickel strip is 4 mm by 0.15 mm by 50 mm considered optimal to carry 40 A? I'm not exactly sure how much power loss is considered "optimal" "acceptable" or "poor".

Additionally, I was wondering how much current should I consider for the balance leads coming off the BMS? Meaning B1, B2, B3 and B4 below.



Thank you for any help in understanding this!

 

[docw009]

Which chart to use is a good question. At 9.9A for 8 x .15 nickel, the above chart is 2X the Matador chart's 4.2A recommendation, I buy nickel from Shonan on amazon, and they say that width is safe for 7A, putting them in the middle, I would use the seller's advice, which suggests that14 x .15 nickel would be about 14A per strip. .

It's a lot less than your 40A calculation based on power loss, but I would go with the conservative approach, Better to have plenty of headroom,

 

[YoshiMoshi]

Which chart to use is a good question. At 9.9A for 8 x .15 nickel, the above chart is 2X the Matador chart's 4.2A recommendation, I buy nickel from Shonan on amazon, and they say that width is safe for 7A, putting them in the middle, I would use the seller's advice, which suggests that14 x .15 nickel would be about 14A per strip. .

It's a lot less than your 40A calculation based on power loss, but I would go with the conservative approach, Better to have plenty of headroom,

Thanks for the reply! It does seem odd that different charts have different values. I've been trying to understand Matador's chart. It seems like he hasn't been online for a long time. I really wish I could ask him about it!

Here's the research I've done so far. I feel like I'm so close to understanding this topic, but I still have some open questions about this topic.






My main questions at this poitn are:
1) Matador's chart does not appear to take into consdieration the length of the condcutor. Why? Was this done on purpose or was this overlooked?
2) What are the 0.003344818 and 0.0094278187 values in Matador's excel sheet for calculating ampacity? Where do they come from? Matador uses the same equation regardless of the material.
3) The thermal resistance equation in electrician2 website http://www.electrician2.com/articles/ampacity.htm that he references looks nothing like Matador's equation. Why?
4) What is the thermal resistance of the conductor to ambient air? What temepratures of the conductor are considered acceptable?

From the Electrician2 website. In theory I should be able to calculate ampacity via the equation below, and make my own table.

T_C = Temeprature of the conductor
T_A = Ambient temeprature
p = resistivity of the material your using
L = length of the conductor
W = Width of conductor
T = Thickness of conductor
R Bar = Thermal resistance from conductor to ambient

The only problem is I don't know what to use for T_C or R Bar.

Please let me know if I have made any errors, or if something is not clear and that maybe I made a mistake! Really looking for some guidance in understanding how ampacity is calculated! The charts are great, but each chart is slightly different, and some widths are not included within them.

 

[YoshiMoshi]

Update.
So I've made some more progress.
Equation 1 from this source: http://www.electrician2.com/articles/ampacity.htm

The term "thermal resistance" is used in this source. Trying to search for "thermal resistance of air" does not lead to much. The lack of parantheeses for units of measurent of different online source add confusion as well.

I beleive the proper term for R_CA is actually "thermal resisitivty". The term "thermal resistance" is often misused to discuss "thermal conductivity"
Per this soruce: https://thermtest.com/thermal-resou...-materials-and-gases#:~:text=Air: 0.024 W/m•K
The thermal conductivity of air is 0.024 W/(mK). The thermal resistivity and thermal conductivity are inverse relations. So we can get the thermal resistivity of air as 125/3 (mK)/W

The R value in Equation 1 is also the resistance per unit of length, and not just purely resistance. So we disregard the length.


So in my case with a width of 5 mm and a thickness of 0.15 mm


Does this look correct? Now all I have to do is select the desired temeprature of the conductor and temerpature of ambient air in Kelvin? What values should I select for these values?

 

[Hummina Shadeeba]

It’s easy to just lay another stack of nickel on top and better to not try to do the minimum. I’d get some WIDE strips like the eskate people use and then you won’t have to stack

 

[Chalo]

So much for those "40 amp" 21700 cells, yeah? Maybe they should start making them with threaded studs like they should have from day one.

 

[Cowardlyduck]

Much of the above is irrelevant as there are many more factors than just the nickel size that contribute to amp carrying capabilities. Things such as cell end cleanliness when welding, quality/number of welds, position of welds, ambient temps, pack insulation, etc.
So given many of the above factors are nearly impossible to always have consistent, I would just say overbuild the pack and be done with it.

As an example, I am currently building a 20S10P pack with double stacked rows for the parallel cells. I'm using Samsung 50S cells capable of '25A' (real world 10-15A) and the draw will typically be under 50A with potential peaks up to 65-75A.
I will be using 0.2x8mm nickel and I typically just say it has a max amp draw of 10A to keep it simple. If using 0.15 I say 7A.
If layering nickel I typically assume a loss of capability of about 2A per layer.

Since I will have 5 series 'links' per parallel group I simply intend to have 2 x 0.2 x 8mm layers of nickel. This gives me a conservative value of 18A per link or 90A total. This could theoretically peak to over 150A without issue. The point is, I've added a healthy margin, so I'm not worried about it!

In the past for different pack layouts where there are less series links or higher amp draw required I've simply added more nickel layers (sometimes up to 4 layers), or switched to a copper/nickel sandwich. Both work well!

Hope that helps!

Cheers

 

[YoshiMoshi]

Much of the above is irrelevant as there are many more factors than just the nickel size that contribute to amp carrying capabilities. Things such as cell end cleanliness when welding, quality/number of welds, position of welds, ambient temps, pack insulation, etc.
So given many of the above factors are nearly impossible to always have consistent, I would just say overbuild the pack and be done with it.

As an example, I am currently building a 20S10P pack with double stacked rows for the parallel cells. I'm using Samsung 50S cells capable of '25A' (real world 10-15A) and the draw will typically be under 50A with potential peaks up to 65-75A.
I will be using 0.2x8mm nickel and I typically just say it has a max amp draw of 10A to keep it simple. If using 0.15 I say 7A.
If layering nickel I typically assume a loss of capability of about 2A per layer.

Since I will have 5 series 'links' per parallel group I simply intend to have 2 x 0.2 x 8mm layers of nickel. This gives me a conservative value of 18A per link or 90A total. This could theoretically peak to over 150A without issue. The point is, I've added a healthy margin, so I'm not worried about it!

In the past for different pack layouts where there are less series links or higher amp draw required I've simply added more nickel layers (sometimes up to 4 layers), or switched to a copper/nickel sandwich. Both work well!

Hope that helps!

Cheers

Thanks, that's what I figured. Way to many variables are required to calculate it. Requires knowledge of both electrical and mechanical engineering. Doing a finite element analysis simulation in COMSOL, like best option. Something I haven't done in about 14 years. So I guess I'll just reference charts. Do you know of any good charts that I can use for nickel strips and the current carrying capacities? It seems like each chart has different values.

Thanks for the help!

 

[Cowardlyduck]

No chart can predict the future. I would just make a middle road assumption based on all of them combined. That's why I just go with a rough/rounded figure of 10A for 0.2x8mm that I often use. I also have some 0.15x7mm and I assume 7A for that.
Keep it conservative and you won't run into any trouble.

Cheers

 

[MaxBicycleMax]

Balance leads carry bugger all and can be tiny.
0.5mm2 20awg thereabouts

 

[docw009]

By the way, I think pulling 40A continuous out of a 5S-3P 21700 isn't good for battery life, Overdesign it.

 

[YoshiMoshi]

With regards to Matador's ampacity table. I finally broke the formula. What he did was took the power stream data American Wire Gauge Chart and AWG Electrical Current Load Limits table with ampacities, wire sizes, skin depth frequencies and wire breaking strength for copper wire and assumed that the relation was the same for nickel stip.


Then he plotted 1/ampacity on the y axis, and the reisstance on the x axis.
He then found a line of best fit of y = 0.0033448180x + 0.0094278187
Where x is the resistance of the conductor and y is one over the ampacity,

Therefore
Ampacity = 1/(Resistance * 0.0033448180 + 0.0094278187)

 

[leffex]

There is a mathematical route or a simple one.

What brand and model of 21700 cell are you going to use, YoshiMoshi?
If you use the weakest 21700 they can output 6A each or 12A for 30 seconds or 24A for a shorter duration. What happends is that the more amps you steal the more heat is generated per efficiency of the cell goes down the more you push it. A cell rated 45A would have very low losses at those amps and will not make the pack heat up. As the pack is heated to the limit for a cell type 70-90 degrees celsius they are not able to sustain amps without overheating or making other problems for you.

CE products have a 60 degree celsius maximum outside temperature with rollback or security like temperature auto fuse in place to care for that. (I think ) (water boiler for example )

As the length of the nickel is short it will have little effect on temperature. If temperature per definition is increased it will flow back to the cells which will take it up with a loss for that resistance going to heat. My question here is: is it really worth it to put 8 sheets of nickel for a time expense of the build or more for the pack to really get that full efficiency from the cells or could you just get higher performance or amp rated by switching to another cell instead?

If the battery will be used with a tool with a professional user the performance of your build is very important. Is this a pack for yourself to use lightly and rarely then the performance of the pack parameters are less important.

Balance leads doesn't normally carry current. The series connection does. Parallel connection generally does not need to be thick either for the same reason.