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I'm trying to figure out the maximum voltage of some cables in a vacuum system I've inherited, and I'm confused by the DC voltage ratings given by this supplier as they step from the 3 kV to 10 kV rated cables. In each case the difference between the conductor diameter and outer diameter is 3 mils, which I assume to be made up of the kapton insulation. So the thickness of the kapton insulation is half this, and according to this kapton datasheet the dielectric strength is about 7 kV/mil at 1.5 mil thickness. I assume that the worst-case geometry looks something like this, where the cable touches a grounded plane:

geometry of insulated cable and grounded plane

and I would think that you could approximate (looking close in so that the curvature is irrelevant, which if anything will overestimate the field and underestimate the max voltage) the peak field strength as \$E_{breakdown} = V_{max}/d\$, where \$d\$ is the insulation thickness, so that the peak voltage is just \$V_{max} = E_{breakdown} \times d\$. For a 1.5 mil layer of kapton insulation, this would then give \$V_{max}\ = 7 \space\mathrm{kV/mil}\times 1.5 \space\mathrm{mil} = 10.5 \space\mathrm{kV}\$, which would tell me that any of these cables should be able to withstand 10.5 kV. I could understand a safety factor of two, but if this calculation is right then the 10 kV cable has almost no safety factor. The only difference that I can see between the cables is the gauge of the conductor, and I don't see how that would affect the breakdown characteristics.

There is the following note on the store page:

The electrical and thermal ratings specified are of safe operating limits determined by various factors including material properties, mechanical design, and the intended operating environment. All electrical ratings are based on operation with one side in dry atmosphere and the other side in vacuum of less than 1 x 10^-4 Torr.

But that still leaves the question of why voltage rating would increase with thicker conductors but constant insulation thickness.

For bonus points: why does this other supplier advertise a cable with a 5 mil kapton layer as only rated to 2 kV?

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    \$\begingroup\$ Could this be related to the concentration of electric field lines due to geometry? As a result. a point will attract lightning more than a large round or flat surface will. I imagine the same is true with a very thing conductor vs a very thick one. \$\endgroup\$
    – DKNguyen
    Sep 28, 2021 at 2:57
  • \$\begingroup\$ @DKNguyen yeah I thought of that but got stuck trying to prove that it was actually the case (pretty obvious if you keep the charge density constant and reduce radius, not as clear if it's the voltage you keep constant). Gonna write something up now \$\endgroup\$
    – llama
    Sep 28, 2021 at 20:05
  • \$\begingroup\$ @DKNguyen hmm I worked through it and from what I can tell, at these wire gauges, it only changes things by a few percent (unless I got something wrong somewhere) physics.stackexchange.com/questions/668699/… \$\endgroup\$
    – llama
    Sep 28, 2021 at 22:10

1 Answer 1

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First of all, the difference in diameter is not 3 mils, some cables have 1 or 2 mils difference as shown here in red.

enter image description here

Second, the type of Kapton used by the two manufacturers is different as indicated by the different temperature ratings. Compare the values highlighted in blue in both pictures.

enter image description here

Third accuglass's datasheet states that they use Kapton F and you use a datasheet of Kapton FPC.

Probably each manufacturer uses a different type of Kapton as the "special sauce" in their products. Manufacturing processes possibly also affect the material properties of Kapton, so only the cable manufacturer knows the dielectric strength of their insulation material.

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  • \$\begingroup\$ Ken Grimes - Hi, Thanks for providing an answer. Please see this site rule from the help center which explains that when you include any content from elsewhere into a post here (e.g. photo, image or text), you must add a proper reference, which includes a link back to the original web page. Therefore please edit your answer to include the source web page link for each image which you copied into this answer. Thanks. (After you have done that, please reply in a comment so that I can remove this comment.) \$\endgroup\$
    – SamGibson
    Jan 2 at 16:44
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    \$\begingroup\$ Also the diameter depends upon the cable and the voltage. The only 10kV is 40 mil conductor and 43 mil outer, which make OP math correct. \$\endgroup\$ Jan 2 at 16:59
  • \$\begingroup\$ @StainlessSteelRat Yeah I specifically referred to the 3-10 kV rated ones, I'm pretty sure I'm capable of subtracting one number from another \$\endgroup\$
    – llama
    Jan 4 at 14:52

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