N o a h ?

A long digression on dimensioning metal strips for cell-to-cell connections

Ampac­ity is a kind of bull­shit met­ric that expresses how much cur­rent a con­duc­tor can “safely” carry, i.e., with­out get­ting too “hot.” You can find ampac­ity charts of all kinds sprin­kled around with vary­ing degrees of cred­i­bil­ity. Most com­monly it shows up relat­ing to the gauge of alu­minum or cop­per wires (because that’s what the NEC pub­lishes), but it’s applic­a­ble to the metal strips you use to con­nect bat­tery cell ter­mi­nals, too.

At this point in my life, I don’t feel com­fort­able using the charts some­one posted on a forum with­out check­ing the math first. Let’s give it a try.

Accu­rately deter­min­ing ampac­ity plunges us into the under­world of fluid dynam­ics because the heat trans­ferred to the envi­ron­ment from the metal is a func­tion of the air­flow around its sur­face (i.e., con­vec­tion). Since we wrap the bat­tery in PVC heat shrink tub­ing, we also need to con­sider con­duc­tion through it.

Fourier’s law of ther­mal con­duc­tion is $$\dot{Q}=\kappa \frac{A}{L}∆T\mathrm{\text{,}}$$ where $\kappa$ is the ther­mal con­duc­tiv­ity of the mate­r­ial, $A$ is the cross-sec­tional sur­face area, $L$ is the dis­tance between the ends of the mate­r­ial, and $∆T$ is the change in tem­per­a­ture between the ends.

For con­vec­tion, by New­ton’s law of cool­ing: $$\dot{Q}=hA∆T\mathrm{\text{,}}$$ where $h$ is the heat trans­fer coef­fi­cient of the fluid (in our case, air).

We can frame these equa­tions in terms of an over­all tem­per­a­ture change to find the total heat trans­fer: $$\begin{array}{cc}\hfill ∆T& =∆T_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}+∆T_{\mathrm{a}\mathrm{i}\mathrm{r}}\hfill \\ \hfill & =\frac{\dot{Q}}{A}\frac{L_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}{{\kappa }_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}+\frac{\dot{Q}}{A}\frac{1}{h}\hfill \\ \hfill & =\frac{\dot{Q}}{A}(\frac{L_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}{{\kappa }_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}+\frac{1}{h})\hfill \\ \hfill \dot{Q}& ={(\frac{L_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}{{\kappa }_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}+\frac{1}{h})}^{-1}A∆T\mathrm{\text{.}}\hfill \end{array}$$

Transfer of energy from Ohmic heating

The power dis­si­pated by the con­duc­tor itself will be the result of Ohmic heat­ing: $${\dot{Q}}_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}=I^{2}R\mathrm{\text{.}}$$ Pouil­let’s law is $$R=\rho \frac{L}{S},$$ where $\rho$ is the elec­tri­cal resis­tiv­ity of the mate­r­ial, $S$ is the cross-sec­tional area per­pen­dic­u­lar to the cur­rent flow, and $L$ is the dis­tance the cur­rent is car­ried. By sub­sti­tu­tion: $${\dot{Q}}_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}=I^2\rho \frac{L_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}}{S}\mathrm{\text{.}}$$

Let $A=L_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}w$ and $S=zw$, where $w$ is the width of the metal strip and $z$ is its thick­ness. We want to trans­fer the heat gen­er­ated by the metal strip through the insu­la­tion into free air. Solv­ing for $S$: $$\begin{array}{cc}\hfill I^2\rho \frac{L_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}}{S}& ={(\frac{L_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}{{\kappa }_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}+\frac{1}{h})}^{-1}{L}_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}w∆T\hfill \\ \hfill \frac{1}{S}& ={(\frac{L_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}{{\kappa }_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}+\frac{1}{h})}^{-1}\frac{L_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}w∆T}{I^2\rho L_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}}\hfill \\ \hfill S& =(\frac{L_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}{{\kappa }_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}+\frac{1}{h})\frac{I^2\rho }{w∆T}\mathrm{\text{.}}\hfill \end{array}$$ As I’ve been told by numer­ous cred­i­ble peo­ple but needed proof to believe myself, $L_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}$ is irrel­e­vant in this cal­cu­la­tion. Inter­est­ingly, though, $w$ is not! We can try to max­i­mize it later as that will give us the best behav­ior with the insu­la­tion.

Can we put 50 A over it?

I’m com­fort­able say­ing I won’t be using my RC mod­els at ambi­ent tem­per­a­tures above 30 °C and I’d like to keep the bat­tery under 50 °C, so that gives us a $∆T$ of 20 K. We want $I$ to be 50 A.

The thick­ness of PVC heat shrink tub­ing unsur­pris­ingly depends on how much you shrink it (more shrink­age puts more mate­r­ial into the walls). An ideal thick­ness $L_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}$ is about 0.25 mm. PVC has a ther­mal con­duc­tiv­ity ${\kappa }_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}$ between 0.12 W⁄m·K and 0.17 W⁄m·K at room tem­per­a­ture. Com­mer­cially avail­able pure nickel has an elec­tri­cal resis­tiv­ity $\rho$ around 7.0×10^-8 Ω·m at room tem­per­a­ture. The heat trans­fer coef­fi­cient $h$ for still air is, at a min­i­mum, about 5 W⁄m²·K.

Using the worst-case val­ues, when we put every­thing into our for­mula: $$\begin{array}{cc}\hfill z{w}^2& <(\frac{L_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}{{\kappa }_{\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{u}\mathrm{l}}}+\frac{1}{h})\frac{I^2\rho }{∆T}\hfill \\ \hfill & <(\frac{0.25\mathrm{m}\mathrm{m}}{0.12\mathrm{W}\mathrm{/}\mathrm{m}\mathrm{\cdot }\mathrm{K}}+\frac{1}{5\mathrm{W}\mathrm{/}{\mathrm{m}}^{\mathrm{2}}\mathrm{\cdot }\mathrm{K}})\times \frac{{\left(50\mathrm{A}\right)}^2\times (7.0\times {10}^{-8}\mathrm{\Omega }\mathrm{\cdot }\mathrm{m})}{20\mathrm{K}}\hfill \\ \hfill & ⪅1763\mathrm{m}{\mathrm{m}}^{\mathrm{3}}\mathrm{\text{.}}\hfill \end{array}$$ As a prac­ti­cal mat­ter, the max­i­mum thick­ness of nickel we can rea­son­ably spot weld is 0.3 mm. Using that value for $z$, we find $$w⪅\sqrt{\frac{1763\mathrm{m}{\mathrm{m}}^{\mathrm{3}}}{0.3\mathrm{m}\mathrm{m}}}⪅77\mathrm{m}\mathrm{m}\mathrm{\text{.}}$$ Since our cells are only 26 mm across and 65 mm long, this con­fig­u­ra­tion doesn’t work, even if we fold the strip over the side (reserv­ing at least half of the side for a strip on the other ter­mi­nal if needed).

An addi­tional obser­va­tion: the PVC insu­la­tion con­tributes vir­tu­ally noth­ing to the over­all cal­cu­la­tion. The heat shrink tub­ing has an R-value of $$\frac{0.25\mathrm{m}\mathrm{m}}{0.12\mathrm{W}\mathrm{/}\mathrm{m}\mathrm{\cdot }\mathrm{K}}\approx 0.002{\mathrm{m}}^{\mathrm{2}}\mathrm{\cdot }\mathrm{K}\mathrm{/}\mathrm{W}\mathrm{\text{,}}$$ whereas free air’s R-value is $$\frac{1}{5\mathrm{W}\mathrm{/}{\mathrm{m}}^{\mathrm{2}}\mathrm{\cdot }\mathrm{K}}=0.2{\mathrm{m}}^{\mathrm{2}}\mathrm{\cdot }\mathrm{K}\mathrm{/}\mathrm{W}\mathrm{\text{.}}$$ We may as well just exclude the insu­la­tion from our cal­cu­la­tions from now on. It will make our lives a lit­tle eas­ier.

Cop­per has an elec­tri­cal resis­tiv­ity of 1.7×10^-8 Ω·m. Will we have more luck with it? $$\begin{array}{cc}\hfill z{w}^2& <\frac{1}{5\mathrm{W}\mathrm{/}{\mathrm{m}}^{\mathrm{2}}\mathrm{\cdot }\mathrm{K}}\times \frac{{\left(50\mathrm{A}\right)}^2\times (1.7\times {10}^{-8}\mathrm{\Omega }\mathrm{\cdot }\mathrm{m})}{20\mathrm{K}}\hfill \\ \hfill & ⪅425\mathrm{m}{\mathrm{m}}^{\mathrm{3}}\mathrm{\text{.}}\hfill \end{array}$$ The max­i­mum thick­ness of cop­per we can weld is 0.1 mm due to its excel­lent con­duc­tiv­ity. Unfor­tu­nately, $$w⪅\sqrt{\frac{425\mathrm{m}{\mathrm{m}}^{\mathrm{3}}}{0.1\mathrm{m}\mathrm{m}}}⪅65\mathrm{m}\mathrm{m}\mathrm{\text{,}}$$ which still won’t work for our cells.

Comparison to empirical evidence

The cross-sec­tional area of the cop­per strip, 6.5 mm², lines up with what I would expect from pub­lished ampac­ity charts for a sim­i­larly spec­i­fied cop­per wire. So I don’t think the math is wrong. But this devi­ates so much from the rec­om­men­da­tions given by peo­ple who build tons of these bat­tery packs that I sus­pect I’m miss­ing some­thing.

Could the cells them­selves act as a heat sink to the strips? I think it’s very pos­si­ble. Li⁺ cells are ther­mally anisotropic, with good con­duc­tiv­ity in the axial plane (i.e., on a cylin­dri­cal cell, from ter­mi­nal to ter­mi­nal) and very poor con­duc­tiv­ity in the radial plane. Roughly, we may have some­thing that looks more like this:

Mod­el­ing the cells and other con­nected strips as rec­tan­gu­lar prisms with the same cross-sec­tional area $A$ as the strip we’re solv­ing for: $$\begin{array}{cc}\hfill {\dot{Q}}_{\mathrm{u}\mathrm{p}}& =hA∆T\hfill \\ \hfill {\dot{Q}}_{\mathrm{d}\mathrm{o}\mathrm{w}\mathrm{n}}& ={(\frac{L_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}{{\kappa }_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}+\frac{z}{{\kappa }_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}}+\frac{1}{h})}^{-1}A∆T\hfill \\ \hfill {\dot{Q}}_{\mathrm{t}\mathrm{o}\mathrm{t}}& ={\dot{Q}}_{\mathrm{u}\mathrm{p}}+{\dot{Q}}_{\mathrm{d}\mathrm{o}\mathrm{w}\mathrm{n}}\hfill \\ \hfill & =hA∆T+{(\frac{L_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}{{\kappa }_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}+\frac{z}{{\kappa }_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}}+\frac{1}{h})}^{-1}A∆T\hfill \\ \hfill & =[h+{(\frac{L_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}{{\kappa }_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}+\frac{z}{{\kappa }_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}}+\frac{1}{h})}^{-1}]A∆T\hfill \\ \hfill I^2\rho \frac{L_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}}{S}& =[h+{(\frac{L_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}{{\kappa }_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}+\frac{z}{{\kappa }_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}}+\frac{1}{h})}^{-1}]L_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}w∆T\hfill \\ \hfill \frac{1}{S}& =[h+{(\frac{L_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}{{\kappa }_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}+\frac{z}{{\kappa }_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}}+\frac{1}{h})}^{-1}]\frac{L_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}w∆T}{I^2\rho L_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}}\hfill \\ \hfill S& ={[h+{(\frac{L_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}{{\kappa }_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}+\frac{z}{{\kappa }_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}}+\frac{1}{h})}^{-1}]}^{-1}\frac{I^2\rho }{w∆T}\mathrm{\text{.}}\hfill \end{array}$$

The ther­mal con­duc­tiv­ity of nickel ${\kappa }_{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}}$ is about 91 W⁄m·K at room tem­per­a­ture and I found some stud­ies that put ${\kappa }_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ for LFP around 11 W⁄m·K for the axial plane. These should pro­duce such low resis­tances that, once again, I’m com­fort­able ignor­ing them and sim­pli­fy­ing our equa­tion sig­nif­i­cantly: $$\begin{array}{cc}\hfill S& ={[h+{\left(\frac{1}{h}\right)}^{-1}]}^{-1}\frac{I^2\rho }{w∆T}\hfill \\ \hfill & =\frac{1}{2h}\frac{I^2\rho }{w∆T}\mathrm{\text{.}}\hfill \end{array}$$

Does this allow our 0.3 mm-thick nickel strip to work? $$\begin{array}{cc}\hfill w^2& <\frac{1}{2\times 5\mathrm{W}\mathrm{/}{\mathrm{m}}^{\mathrm{2}}\mathrm{\cdot }\mathrm{K}}\times \frac{{\left(50\mathrm{A}\right)}^2\times (7.0\times {10}^{-8}\mathrm{\Omega }\mathrm{\cdot }\mathrm{m})}{0.3\mathrm{m}\mathrm{m}\times 20\mathrm{K}}\hfill \\ \hfill & ⪅2920\mathrm{m}{\mathrm{m}}^{\mathrm{2}}\hfill \\ \hfill w& ⪅\sqrt{2920\mathrm{m}{\mathrm{m}}^{\mathrm{2}}}⪅54\mathrm{m}\mathrm{m}\mathrm{\text{.}}\hfill \end{array}$$ It seems more rea­son­able. I sup­pose the dif­fer­ence between this approx­i­ma­tion and what elec­tric skate­board­ers actu­ally observe is a mat­ter of taste or how con­ser­v­a­tive we want to be.

Stated another way, if we use the NEC’s 30 K $∆T$ and work back­ward from one of the val­ues pro­vided in a semi-anony­mous ampac­ity chart, for exam­ple that you can pass 56.67 A con­ti­nously through a 0.2 mm by 30 mm nickel strip “opti­mally,” we can deter­mine what they’ve assumed the over­all heat trans­fer coef­fi­cient to be: $$\begin{array}{cc}\hfill 0.2\mathrm{m}\mathrm{m}\times 30\mathrm{m}\mathrm{m}& =\frac{1}{h_{\mathrm{t}\mathrm{o}\mathrm{t}}}\times \frac{{\left(56.67\mathrm{A}\right)}^2\times (7.0\times {10}^{-8}\mathrm{\Omega }\mathrm{\cdot }\mathrm{m})}{30\mathrm{m}\mathrm{m}\times 30\mathrm{K}}\hfill \\ \hfill h_{\mathrm{t}\mathrm{o}\mathrm{t}}& \approx 42\mathrm{W}\mathrm{/}{\mathrm{m}}^{\mathrm{2}}\mathrm{\cdot }\mathrm{K}\mathrm{\text{.}}\hfill \end{array}$$ For each side, again assum­ing air is the pri­mary insu­la­tor, this would give a value for $h$ around 21 W⁄m²·K, which is on the high side for nat­ural con­vec­tion but not totally out of the ques­tion. In real­ity, it prob­a­bly means the heat spreads out over more sur­faces than we’re con­sid­er­ing in our very sim­ple model.

Complications and other heat sources

Ear­lier, I said we can fold the strip around the edge of the cell if its width exceeds the cell diam­e­ter. For me, that raises more ques­tions about both Ohmic heat­ing and ther­mal dis­si­pa­tion. The elec­tric field is no longer cen­tered in the strip and some of the ther­mal energy will be directed side­ways.

For now, I’m putting this in a cat­e­gory of poten­tial con­cerns (no pun intended) that cre­ate sub­tle dif­fer­ences in behav­ior but prob­a­bly don’t mate­ri­ally change the spec­i­fi­ca­tions. Depend­ing on what I observe as I build bat­ter­ies, that could change.

Another obvi­ous prob­lem is the resis­tive and elec­tro­chem­i­cal heat trans­ferred from the cells them­selves, which is way greater than the heat gen­er­ated by some thin metal strips. Even though the sur­face area of the cells is larger than the strips, at a suf­fi­cient cur­rent draw, the heat becomes sig­nif­i­cant. For exam­ple, the Lithium Werks ANR26650M1B has an inter­nal resis­tance of 10 mΩ, so at 50 A it gen­er­ates 25 W. I briefly tried to explain away this heat with nat­ural con­vec­tion but came up quite short: the tem­per­a­ture dif­fer­ence is some­thing like 900 K, i.e., effec­tively unbounded because the cells will destroy them­selves long before that. Are bat­ter­ies doomed to retain heat beyond their safe oper­at­ing range at “high” cur­rents with­out enor­mous heat sinks or fans?

Conclusion

The accept­able cross-sec­tional sur­face area of a rec­tan­gu­lar strip with an elec­tric cur­rent pass­ing through it is $$z{w}^2=\frac{I^2\rho }{h∆T}\mathrm{\text{,}}$$ and the hard part, of course, is fig­ur­ing out what $h$ should be.