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My current understanding: The line in the center of the coax and the inner side of the shield are the current paths, as in the signal/field propagates between these two, (i can't therefore imagine it matters if the outer shield connection is grounded or whatever since I would assume its sort of a faraday cage assuming the skin depth is not significant). Anyway for this I would strongly guess the currents must be the same and opposing in the shield and center piece so no fields escape, image below shows my understanding...

Coax field propagation

Now when attaching a dipole, if theoretically it was perfect, I would assume the voltage is just 180 out of phase of the other pole; now connecting one pole to the shield -in my eyes- doesn't cause any problems since its just equally opposed by the center fed pole in a way it can pass in the coax. I've looked online and despite many different answers, one said its the difference in the currents that end up flowing on the outer side of the shield, I guess this is only caused by imperfect dipole design or where there is a difference in a multitude of things that mean the two are not equally opposite, now if this is true I see a choke balun can be used to create a high impedance on the outer shield...

My question is if the above is true where does that reflected current on the shield now go? I can't see it been shoved in the inner shield because again there would be different current in the center and shield.

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My question is if the above is true where does that reflected current on the shield now go?

The outer surface of the outer conductor acts like an antenna. You will notice that in an antenna, for example a dipole, the current is not the same everywhere. This seemingly violates Kirchhoff's Current Law, which might be reworded to state that the current into a volume is the same the current out of that volume.

This conundrum was solved by Maxwell, who introduced the concept of displacement current. The current we are first made aware of when we learn electronics is conduction current -- i.e. the movement of charges. Maxwell realized that this current was not the whole story, and there was something else, that behaved like conduction current (for example it causes magnetic fields) but was not conduction current. He called this displacement current (although he used the same name for a slightly different concept earlier). The displacement current density at a point is equal to the rate of change in the electric field at that point. (May seem odd, but it solves what are otherwise problems).

In a (for example, dipole) antenna, charges flow down a conductor, but because the conductor does not form a complete circuit, the charges accumulate as they approach the end of the conductor. This accumulation of charges causes a change in electric field. The conduction current decreases from the center of a dipole antenna to the end, but at the same time, the displacement current increases. When the two types of current are added together, Kirchhoff's Current Law (KCL) is preserved. The algebraic sum of all the current (both kinds) flowing into a fixed volume is 0.

The same thing happens in the antenna formed by the outer surface of the outer conductor of a coaxial cable that connects a transmitter to a dipole without a balun. At the transmitter end of the coax, the conduction current on the inner conductor is equal but opposite to the conduction current on the outer conductor. However, at the connection to the antenna, the conduction current in the outer conductor has a different magnitude from conduction current in the inner conductor. Where did that current go?

The answer is that electrons accumulate and then rarify on the surface of the outer conductor. This constitutes displacement current. The conduction current on the outer surface of outer conductor "stalls" and becomes displacement current, exactly as the conduction current in a dipole antenna "stalls" and becomes displacement current.

Knowing of the existence of displacement current is essential to an understanding of antennas in a way that does not conflict with other known laws of electricity.

I guess this [difference in currents] is only caused by imperfect dipole design or where there is a difference in a multitude of things that mean the two are not equally opposite

No. The problem of currents on the outer surface of the outer conductor of a coaxial cable is not (necessarily) the result of imperfections in the balance of the dipole antenna. It is the result of connecting an unbalanced transmission line (the coaxial cable) to the (otherwise) balanced load.

The outer surface of the outer conductor of the coax acts like an independent conductor. It is as if one hung a wire physically parallel to the coax, but electrically parallel to one of the arms of the antenna.

schematic

simulate this circuit – Schematic created using CircuitLab

In the diagram, since it is free to do so, the current splits between the right-hand arm of the antenna, and the outer surface of the outer conductor of the coax. Instead of a simple dipole, you essentially have an antenna with three arms, connected in an asymmetric fashion.

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  • \$\begingroup\$ I feel i need to research a fair bit more to fully understand why in my head but seems to have summed it up thankyou. I noticed this is a response to my question, does this mean the comment of my understanding above it is correct, i used some of the top answer to this question to help... ham.stackexchange.com/questions/538/… \$\endgroup\$ Jan 19 at 13:49
  • \$\begingroup\$ I have added to my answer. The addition may help. \$\endgroup\$ Jan 19 at 14:24
  • \$\begingroup\$ Thanks, so this means the quoted question was answered wrong? i think its this part of your answer "However, at the connection to the antenna, the conduction current in the outer conductor has a different magnitude from conduction current in the inner conductor" , in specifics 'why' would the outer conductor have a different magnitude? I see youve also explained to me that it splits into the inner of the sheild and outer, but what exactly determines this extra current? \$\endgroup\$ Jan 19 at 14:35
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    \$\begingroup\$ @Georgekirby suppose at a certain moment, current is flowing "up" the inner surface of outer conductor. It reaches the antenna. Some of it flows into the right arm, and some of it flows down the outer surface of the outer conductor. When that happens, the net current (inner surface minus outer surface) of the outer conductor does not match the current on the inner conductor. \$\endgroup\$ Jan 19 at 19:26
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    \$\begingroup\$ Conduction current (not displacement current) flows from the transmitter through the inner surfaces of the coax. It gets to the antenna. Some conduction current from the coax inner surface of outer conductor then flows back down the outer surface of the coax. In the process of flowing back down, it stalls creating a buildup of charges, creating displacement current, and radiation. This is not a site for tutorials, so this is my last comment on this question. \$\endgroup\$ Jan 19 at 23:20

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