I measured the capacitance between to 100m cables to 10nF with a calibrated LCR meter. Both of the cables are inside a bigger one:

enter image description here

The black part is the isolation. Brown part is the copper itself.

I also tried to calculate the capacitance with this below formula but I get a value of 1.5nF. The epsilon_r of the cable isolation is 2 so total electric permittivity is 2 * 8.8542*10^(-12). I used 100m for l. a = r = 0.7mm.

Is my formula wrong? What am I not considering here?

enter image description here

  • \$\begingroup\$ It's just a formula with no definition of a or d and you never mentioned what electric permittivity value your real cable has and most of the colour inside the drawn cable is brown = copper. \$\endgroup\$ – Andy aka Nov 14 '20 at 17:52
  • \$\begingroup\$ d = 30mm, a = r = 0.7mm as you can see in the image. The electric permittivity is 2 * 8.8542*10^(-12). The brown part is the copper as I mentioned \$\endgroup\$ – electrococuk Nov 14 '20 at 17:54
  • \$\begingroup\$ It's all brown inside the big circle except for small black circles. Please link to the online calculator. \$\endgroup\$ – Andy aka Nov 14 '20 at 17:55
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    \$\begingroup\$ Then d becomes very much smaller than 30 mm. And if the big copper is earthed than you are using the wrong calculator tool. \$\endgroup\$ – Andy aka Nov 14 '20 at 18:06
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    \$\begingroup\$ The problem becomes that of a coaxial cable and, there is no effective capacitance between the two internal conductors because "earth" gets in the way. This means that you are trying to find a value that is meaningless. Because you are trying to find a value that is meaningless, you haven't got any information in your question that gives a meaning to what ultimately you are trying to achieve. Unanswerable as it stands. \$\endgroup\$ – Andy aka Nov 14 '20 at 18:23

Is my formula wrong? What am I not considering here?

You are not recognizing that the solid copper volume that surrounds the two insulated conductors shortens the distance dramatically. Then if that solid copper is earthed: -

Yes the big copper is earthened. Which tool do I have to use? Or which formula?

And if the big copper is earthed than you are using the wrong calculator tool. So, with it being earthed the capacitance between the cores is largely immaterial and of no consequence.

You can certainly measure capacitance but what does that tell you that is useful to know. Now the two cores should be treated as two separate coax cables and if you did that and asked someone what the capacitance is between the two coax cable inner conductors/cores then you would just get blank stares because it is meaningless without further context.

But when I use a time domain reflectometer I can still measure the cable length.

TDR relies on an anomaly at the end of the cable i.e it has an open circuit. TDR does not use a capacitance measurement; it emits a very thin pulse and measures the time for it to hit the end of the cable and cause a reflection that is then received after it travels back up the cable. Totally different thing entirely.

  • \$\begingroup\$ Ok thank your for your insights. I thought for the TDR to measure this there needs to be some sort of capacitance between the two wires. Because when I look at the characteristic impedance formula there is a jwC in it. And I know that you can't use a TDR on a single cable which has no return path. \$\endgroup\$ – electrococuk Nov 14 '20 at 19:05
  • \$\begingroup\$ The return path for your cable is very complicated in this scenario. \$\endgroup\$ – Andy aka Nov 14 '20 at 19:21
  • \$\begingroup\$ Ok thanks again. I guess as long the TDR works I don't need to know why exactly and how the return path looks like. But do you know how the return path could be determined? I guess I need to use some sort of EM simulation software. \$\endgroup\$ – electrococuk Nov 14 '20 at 19:38
  • \$\begingroup\$ The main return path assuming you terminate both cores in their identical matching impedance is via the main copper earth. But, given that you may not drive both coes antiphase you may get different effects. Not knowing precisely what you are trying to achieve on this winding road means I can't give better advice and realistically I haven't got all day to think about this and neither have you possibly. So, state unambiguously what you are trying to achieve and I'll try and make my next comment more useful. \$\endgroup\$ – Andy aka Nov 14 '20 at 19:44
  • \$\begingroup\$ I amt trying to build some sort of primitive tdr and just want to understand this phenomenon better. I didn't terminate the ends of the cores. It was open circuit \$\endgroup\$ – electrococuk Nov 14 '20 at 19:49

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