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Consider a coaxial cable with shield not connected to anything. My textbook seems to suggest that there will be leakage current between the "signal" and the "shield".

I have two questions:

1) When the shield isn't connected to anything, how can any current be there and where will the current flow to?

2) How does connecting the shield to common mode voltage reduce both leakage current and capacitance?

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    \$\begingroup\$ A wire inside a metal shield is a capacitor! \$\endgroup\$
    – Hearth
    Commented Mar 21, 2020 at 18:18
  • \$\begingroup\$ Ok I see how two metals at distance form a capacitor. Pardon if I'm dumb but how is it possible to charge/discharge the capacitor when one terminal is not connected to anything? \$\endgroup\$
    – across
    Commented Mar 21, 2020 at 18:20
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    \$\begingroup\$ Whereabouts in the quoted text does it imply the shield is unconnected when you say this Consider a coaxial cable with shield not connected to anything. My textbook seems to suggest that there will be leakage current between the "signal" and the "shield". \$\endgroup\$
    – Andy aka
    Commented Mar 21, 2020 at 18:24
  • \$\begingroup\$ @Andyaka Oh Looks I just assumed that.because the author was proposing to connect the shield to common mode potential. Do you mean they must be "grounding" the shield before? If so, I guess it makes sense. The leakage current was flowing to the ground through the capacitor. I still don't get how putting common mode potential on shield reduces the capacitance though... \$\endgroup\$
    – across
    Commented Mar 21, 2020 at 18:29
  • \$\begingroup\$ Leakage current might reduce because the voltage difference between signal and shield was reduced. But how can capacitance be changed with voltage? Isn't capacitance purely a geometric property? \$\endgroup\$
    – across
    Commented Mar 21, 2020 at 18:34

1 Answer 1

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1) When the shield isn't connected to anything, how can any current be there and where will the current flow to?

The shield is ordinarily connected to ground or some other low-impedance path to ground. If it wasn't, it wouldn't be that useful in stopping noise. With such a connection, the (grounded) shield forms a capacitor between itself and the leads inside, biased at the common-mode voltage.

Because no insulator is perfect, there is a small leakage current as well, modeled as a resistance to ground. The greater the voltage difference between shield and leads, the greater the leakage.

2) How does connecting the shield to common mode voltage reduce both leakage current and capacitance?

Driving the shield to the common-mode voltage reduces the DC bias between the shield and the signal lines. With a smaller (or practically no) bias, there is no DC leakage. It also cancels out the common-mode AC voltage so it doesn't influence the signal pair either.

Some relevant info:

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  • \$\begingroup\$ Ah pulling the shield to common mode potential reduces the voltage between "signal" and "shield". Consequently, less leakage current XD You're so awesome! Thank you so much:) \$\endgroup\$
    – across
    Commented Mar 21, 2020 at 18:56
  • \$\begingroup\$ If I may ask one quick q: when you/textbook say "grounding the shield" it is the circuit ground and not the earth right? \$\endgroup\$
    – across
    Commented Mar 21, 2020 at 18:57
  • \$\begingroup\$ Yes, the system (circuit) ground. \$\endgroup\$
    – hacktastical
    Commented Mar 21, 2020 at 20:02

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