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Let's consider a coaxial cable which supplies a dipole antenna. My reference for all my statements is here.

enter image description here

  1. My reference says:

Note that the current along a transmission line should be of equal magnitude on the inner and outer conductors, as is typically the case. Observe what happens when the coax is connected to the dipole. The current on the center conductor (the red/pink center core of the coax, labeled IA) has no where else to go, so must flow along the dipole arm that is connected to it. However, the current that travels along the inner side of the outer conductor (IB) has two options: it can travel down the dipole antenna, or down the reverse (outer) side of the outer conductor of the coaxial cable (labeled IC in Figure 1).

Well, I don't understand why the current in the outer conductor should be divided in IB ("direct") and IC ("Reverse"). I see it in this way: the coaxial cable is supplied by a voltage or current source, with its terminals connected one to the shield and one to the inner conductor. The current in the inner conductor is equal to the total current in the shield. The last one, because of skin effect, flows mainly on the shield surface, so it goes partially in the inner side and partially in the outer side of the shield. Both current have the same direction, as the current that flows any conductor surface at RF.

I'd say that IA is equal to the total current flowing in the shield surface. And a dipole with arms connected to the whole shield conductor and to the inner conductor will receive equal currents. Which is the mistake in my analysis?

  1. As I see, the unbalanced nature of the coaxial cable is due to its shield having both an inner and an outer surface. So, it's a parasitic effect of the coaxial cable, and I'd say that it's absent for instance in simple 2-wires transmission line. Is it correct?

  2. Again, the problem is the parasitic current IC in a coaxial cable. Well, what does it has to do with this usual representation of a balun as a converter from a grounded signal to a floating signal?

enter image description here

Even in a 2-wires transmission line, which I'd say it's balanced, we consider one conductor as GND, like in the following picture (indicated as "0").

enter image description here

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    \$\begingroup\$ Generally, a balanced circuit will have two paths that are identical but operating 180 degrees out of phase, and both of these are referenced to a common ground. The key is the antiphase or differential use of the circuit. The main benefit is reduced sensitivity to noise from external sources. This is because the external aggressor will influence both paths equally, applying common mode noise, which the circuit will largely ignore. So really, you can think of the term "balanced" having special meaning in electronics, beyond just the matching currents of a wire and it's return. \$\endgroup\$
    – Troutdog
    Feb 19, 2021 at 20:46
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    \$\begingroup\$ Kinka, might this question appear a little easier to figure out a decent answer to if you forgot the dipole part of the question and concentrated on an load that was not the right characteristic impedance. I only scanned your question and answers so I may have missed a crucial point. \$\endgroup\$
    – Andy aka
    Feb 20, 2021 at 23:07
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    \$\begingroup\$ Maybe some help \$\endgroup\$
    – Andy aka
    Feb 20, 2021 at 23:08

3 Answers 3

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The signal from the transmitter comes along the insulation layer between the shield and the center wire. Conductive metal has some induced current - it's needed to make the wave able to propagate in the insulation layer. The induced current exists on the inner surface of the shield and the outer surface of the center conductor.

When the wave meets the end of the line it meets actually a highly non-symmetric antenna. One piece of it is the red rod and the other piece of it is the grey rod and the outer surface of the shield. Quite far from anything balanced. But it can still do its job as an antenna. I have used portable HF radios with dipole antennas at frequencies 3...20 MHz. When there was coaxial feed with no balun the radio communication still happened, but the antenna cable became a part of the antenna and the whole radio did the same. If one touched loosely the metal of the radio or morse key when transmitting the skin started to smoke at the touching point. Harmful, but not serious due the low power (about 10W max total transmitter output). I guess the balun was not included because the radio did its job without one and every extra part increased system complexity and unreliability.

But you asked why the coaxial cable is unbalanced, not the why antenna construction is unbalanced. That's because its conductors are not equal. It's shield can collect substantially bigger noise voltage from the external electric fields than the center conductor which is surrounded by metal - nearly complete Faraday cage except the ends have holes.

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  • \$\begingroup\$ Thank you for your answer. Can you explain me why doesn't the input current signal from the transmitter goes in both surfaces (inner and outer) of the shield? I imagine that If I connect a voltage or current source to a conductor, the input current goes across all of it \$\endgroup\$
    – Kinka-Byo
    Feb 19, 2021 at 21:06
  • \$\begingroup\$ The current flows on the whole cross section of the shield If it 's DC or has so low frequency that the skin depth ( due the skin effect) is equal or more than the thickness of the shield. At 10MHz the skin depth is about 0.02 millimeters. The shield generally is much thicker. The perfect proof is possible only with differential equations. Connecting an AC source between the shield and center wire generates TEM waveform and it exists between the conductors, not outside the cable. The fields of the TEM waveform induce substantial currents only at the skin depth. I must skip the proof. \$\endgroup\$
    – user287001
    Feb 20, 2021 at 9:23
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The current in the inner conductor is equal to the total current in the shield.

One might think so, but it is not necessarily the case.

it goes partially in the inner side and partially in the outer side of the shield. Both current have the same direction, as the current that flows any conductor surface at RF.

Again, they current on the outside of the shield does not necessarily flow in the same direction as the current on the inside of the shield.

I'd say that IA is equal to the total current flowing in the shield surface.

Again, one might think so, but it is not necessarily true.

a dipole with arms connected to the whole shield conductor and to the inner conductor will receive equal currents.

Not necessarily

it's absent for instance in simple 2-wires transmission line. Is it correct?

A ladder line, or for lower currents, a twisted pair, is much, much, less likely to be unbalanced.

Which is the mistake in my analysis?

Your mistake lies in assuming the currents are balanced. And, my guess is that you believe the currents must be balanced because you are unfamiliar with displacement current. My guess is that you are reasoning that if the currents in the inner and outer conductor are equal when they leave their source, they must be equal everywhere down the coax. If they become unequal, so you might reason, where does the the "missing" current go? To understand the answer to that, consider where the current "goes" when it is in the antenna? There is no current at the far ends of the dipole, but there is current entering? The answer is that there are voltage changes, and these voltage changes give rise to what are termed displacement currents.

Well, what does it has to do with this usual representation of a balun as a converter from a grounded signal to a floating signal?

A balun converts from an un -balanced to a bal -anced configuration, not from "grounded" to "floating".

Even in a 2-wires transmission line, which I'd say it's balanced, we consider one conductor as GND, like in the following picture (indicated as "0").

One might consider one conductor "GND", however, if there is AC flowing in the transmission line, then there are (electric) potential differences between various points on that "GND" wire. Given these potential differences between points, they cannot all be at 0V.

Addendum: If the currents through a coaxial cable are balanced, then the cable will generate no external magnetic field. A current transformer surrounding such a coax will show no signal. If, however, a current is induced in the (secondary of the) current transformer, we have practical proof that the currents are not equal. I thought it important to mention this, as it gives us a way to move out of the realm of theory, and into the realm of testable hypotheses. The experiment has in fact be performed with the results showing net current (algebraic sum not zero) through coaxial cables feeding antennae.

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  • \$\begingroup\$ Thank for your answer. Precisely, where does the displacement current you are talking about flow? About the reference conductor in a transmission line, in this previous question we had concluded that it's a common practice to imagine one conductor of a transmission line as a single GND node (electronics.stackexchange.com/questions/540327/…) \$\endgroup\$
    – Kinka-Byo
    Feb 19, 2021 at 21:59
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    \$\begingroup\$ Displacement current is the "current" that "flows" through a dielectric when a capacitor is being charged. It "flows" from a point which is experiencing a rising voltage, to a point that is experiencing a falling (or rising negative) voltage. It is not a flow of actual charges, although there are charges flowing to create the rising (or falling) voltage. // It may be common practice to consider one wire in a transmission line as "ground". However, if the transmission line were curved around so that the "far" end is near the "near" end, one can test that they are at different potentials. \$\endgroup\$ Feb 19, 2021 at 22:13
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The current in the inner conductor is equal to the total current in the shield.

If it's correctly terminated it will be equal and opposite.

The last one, because of skin effect, flows mainly on the shield surface, so it goes partially in the inner side and partially in the outer side of the shield.

No, because of the magnetic effects of the inner conductor a perfectly terminated coax cable will have all the shield current flowing on the inside of the shield.

If there's a current unbalance then the excess or deficit will flow on the outside of the shield because of the skin effect.

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  • \$\begingroup\$ I might be wrong, but "skin effect" specifically describes current flow trough the outer layers of a conductor, not outside of it. That is, unless when you say "outside of the shield" you actually mean "in the outward-facing surface layer of the shield", as opposed to the inward(dielectric)-facing surface of the shield. \$\endgroup\$
    – Maple
    Feb 19, 2021 at 22:44
  • \$\begingroup\$ Just a caution. A "correctly terminated" coax attached to a dipole antenna requires a balun. On the other hand, impedance mismatch between characteristic impedance and termination does not cause unbalance. See for example The myth of SWR causing coax radiation \$\endgroup\$ Feb 19, 2021 at 23:38
  • \$\begingroup\$ Termination has got nothing to do with this. \$\endgroup\$
    – MAM
    Feb 24, 2021 at 11:13
  • \$\begingroup\$ requiring termination I allows me to ignore several edge cases. \$\endgroup\$
    – Jasen
    Feb 24, 2021 at 21:23

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