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I have been trying to reconcile the fact that current flows on the inside of a coax shield with the common practice of grounding the shield. These ideas seem contradictory at the moment. For a 2-wire transmission line it seems intuitive that both wires have a varying voltage wave along them, creating an electric field between the wires and also a potential difference along the wires, due to the peaks and troughs of the wave. Hence, current flows backwards and forwards in each wire.

For coax the argument is the same, until we connect the shield to ground. Now it seems that I have locked the shield potential, so now it seems that all the voltage variation is on the centre conductor, while the shield remains at a fixed potential. If the potential is fixed, how can there be current flow on the inside of the shield?

I know there is a high impedance between the inside and outside of the shield, but I can't see how that solves anything. Clearly I am misunderstanding the electromagnetic effects in some way. Appreciate any advice.

This diagram tries to illustrate the issue: enter image description here

At a given moment, the shield has a sinusoidal current distribution, which to me suggests a potential difference along the shield. Yet we have the shield grounded?

Thanks, Ian

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  • \$\begingroup\$ The current flows in the inner layer of the shield. \$\endgroup\$
    – Marko Buršič
    Mar 19, 2017 at 9:09
  • \$\begingroup\$ Thanks, but if the coax is several wavelengths long then current is flowing both forwards and backwards at various points on the shield. Surely that requires a potential difference between those points. How can there be a potential difference if the shield is tied to ground? \$\endgroup\$
    – Ian_B
    Mar 19, 2017 at 9:26
  • \$\begingroup\$ Unless you're using superconductors, everything has a non-zero resistance. Current indeed implies a voltage drop. (And no ground is ideal.) \$\endgroup\$
    – CL.
    Mar 19, 2017 at 11:41
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    \$\begingroup\$ You are in the classic pitfall of thinking ground = 0V everywhere; it does not, and is destinctly noticeable at high frequencies. \$\endgroup\$
    – Peter Smith
    Mar 19, 2017 at 11:57
  • \$\begingroup\$ @Peter Smith, ok, but how should I be thinking about the ground? \$\endgroup\$
    – Ian_B
    Mar 19, 2017 at 19:49

3 Answers 3

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I think the varying voltage actually travels along the dielectric between the conductors.

As for current, current can flow in a loop regardless of voltage at a given point (e.g. for one loop with a 5V source and a 5 ohm load, the current will be 1 amp both before and after the load).

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  • \$\begingroup\$ Thanks for your reply. I understand your comment about current for simple DC circuits, but it doesn't seem so simple for transmission lines that are several wavelengths long. In that case, the AC current is moving both backwards and forwards, depending where you are on the transmission line. This is my difficulty. If you have current moving in two directions on a coax shield, then surely there is a potential difference between those points. Yet, this seems at odds with connecting the shield to ground. \$\endgroup\$
    – Ian_B
    Mar 19, 2017 at 9:19
  • \$\begingroup\$ "the varying voltage actually travels along the dielectric between the conductors." Slight correction, if I may. It's the power that travels inside the dielectric. The current travels in the outer surface of the center conductor and returns in the inner surface of the shield. The voltage itself doesn't "travel", thought the waveform of the voltage may appear to travel. EDIT: Reference \$\endgroup\$
    – Davide Andrea
    Sep 23, 2021 at 1:25
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That is a very interesting observation, Ian. The waveform in the drawing represents the measurement from a reference at the middle of signal and ground.

However the shape is, whichever side gets grounded, current has to return to the source through the paired wire or any other provided path (leak). For a transmission line, distributed-element model may help understanding.

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schematic

simulate this circuit – Schematic created using CircuitLab

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