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I would like to verify my understanding of the following scenario with regards to why coax is 'unbalanced' and how a choke interacts with this unbalance:

  • Consider a half-wave resonant dipole, if one leg develops a +1V referenced to earth, the other leg will be at -1V, not 0V, referenced to earth. Additionally, if the two drawn capacitors are equal in capacitance, both legs have an equal impedance to earth. This is what balanced means.
  • Pretend I solder the blue leg of the dipole to the shielding of some coax cable. The black vertical line emphasizes the thickness of the shielding.
  • Coax: the shield sits at earth, while the core voltage fluctuates above and below earth voltage. It is unbalanced because as drawn, the core is shielded, and its path to earth is only through the red leg and its capacitor. Meanwhile, the shield has infinite capacitors from the outside of the shield to earth all in parallel. This presents a different impedance to earth for the shield vs the core.
  • If the red antenna leg is at +1V, the core is at +1V, the blue antenna leg will be at -1V while the shield is at 0V. Thus, a current Icm flows across the outside of the coax shield, to the blue antenna leg back to earth (coupling etc). This is the "common mode" current, and it is in the same direction as the core current.
  • I now put a ferrite ring around the coax cable.
  • I1 and I2 are differential but not balanced (equal and opposite voltages with different impedances to earth). Icm and the core current are common mode and their flux's add. They are impeded by the choke. Since I2 is equal and opposite I1, its magnitude is also reduced.

Diagram

Two questions:

  1. Is that an accurate intuition as to why coax is considered unbalanced in addition to the effects of adding a choke around the coax? I am self teaching and would like to know which resources I can rely on.

  2. Adding a ferrite ring choke decreased all currents; since V=IR, does that mean that the voltages on the ends of the dipole would decrease as well? Can a dipole antenna stop receiving if a large enough choke was placed around its coax cable?

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    \$\begingroup\$ The ferrite as drawn here would not discriminate between I3 and I4. Both are attenuated. Coax is unbalanced because for a differential signal, the shield has lower Impedance to common ground than the inner conductor. If you do not consider the external common ground, then all two-conductor cables are balanced \$\endgroup\$
    – tobalt
    Nov 20, 2023 at 5:37
  • \$\begingroup\$ Does this answer your question? Why is a coaxial cable unbalanced? \$\endgroup\$
    – tobalt
    Nov 20, 2023 at 5:42
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    \$\begingroup\$ Yes that is what I mean, but I would hope someone with a broader view of the topic could write a clear answer so +1 \$\endgroup\$
    – tobalt
    Nov 20, 2023 at 6:23
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    \$\begingroup\$ This drawing is widely used on the Internet but I think it's wrong. It poses a problem in the explanation, because if you move the FEEDER away from the radiating dipole, you'll see the current in the shield decrease. I1 doesn't split into I3 and I4. In your antenna you have a differential current I1 and a common-mode current which is induced by the antenna's radiation in both the shield as well as in the center conductor. \$\endgroup\$
    – Vincent
    Nov 20, 2023 at 7:54
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    \$\begingroup\$ "if one leg develops a +1V referenced to earth, the other leg will be at -1V, not 0V". That is not a given. It is only true if both legs are driven with balanced current. \$\endgroup\$ Nov 20, 2023 at 17:06

2 Answers 2

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This is my view of the problem

enter image description here

According to the electromagnetic theory of guided TEM waves: the current flowing at the surface of the central conductor and that at the inner surface of the shielding must be equal but flowing in opposite direction. A coaxial cable is an current balun circuit. But that is true inside a coax, not outside.

The blue current is the differential current of the RF generator. Ideally, it is balanced.

In reality, the red current has found a new path outside the coax because of the junction with the antenna's right arm (the inner and outer surfaces of the shield are both connected to each other). According to Kirchoff's law, the red current must return to the RF generator because the red current is part of the blue current.

The green current is a common-mode current induced by the antenna's radiation on the feed line. This current couples with the environment, creating a loop with parasitic capacitances.

Green current disturbs the antenna by unbalancing its arms. Red current reduces system performance. Both contribute to the shield's radiance.

One solution is to use a ferrite to attenuate currents on the shield's outer surface.

enter image description here

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  • \$\begingroup\$ Nice diagram! I take it that the blue currents cancel each other out so net flux/signal attenuation occurs, while the red and green currents see the full inductance of the choke? The comment by @tobalt stating the center conductor current would decrease once the choke is applied confused me a bit, but I can find multiple sources published by ARRL stating similar information to your diagram \$\endgroup\$ Nov 29, 2023 at 3:44
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    \$\begingroup\$ Yes, a toroidal ferrite core is a "magnetic flux" absorber. A differential current is not affected because a current I + (-I) flows through its inner surface. The magnetic flux is cancelled out. The toroidal ferrite core has nothing to absorb. Unlike the common-mode current, the latter cancels out. In reality, the toroidal ferrite core isn't perfect: it's not efficient at all frequencies, it doesn't have infinite impedance, and it has a leakage inductance that somewhat attenuates the differential current. All these factors mean that it affects the HF generator signal somewhat. \$\endgroup\$
    – Vincent
    Nov 29, 2023 at 9:06
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Here are some equivalent circuits you may find illustrative:

enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

Consider for sake of argument, the values of impedances don't matter, just that they are there, and what the AC steady state current paths are.

Note first, we can model the antenna as some triangle of impedances (between lines, and to ground); or equivalently a tee (see: delta-wye network equivalent). We can further convert this to a pure pair of impedances Zdiff and Zcm, using an ideal 1:1 CT transformer in parallel with Zdiff. Note that no current flows in Zcm if the two ends of the transformer move perfectly equal and opposite, and no current flows in Zdiff if they move perfectly together (equal).

Then let us add the feedline choke. We have some impedance (the choke) dropping between the defined (and as-given) ground point, the unbalanced source (by definition: it arrives on coax, and has one side grounded), and the antenna CM impedance.

Note that the CM parallel equivalent Zcmc1 || Zcmc2 appears in the CM equivalent circuit, which if XFMR3 is ideal, either one will suffice, but both possibilities are shown anyway.

If we want to reduce feedline currents by say 20dB, we need Zcmc to be a good 20dB larger than Zcm. If it's small and resistive (the antenna in CM looks like a 1/4 wave monopole cleaved down the middle and splayed in half), then not too much is needed (mid to upper 100s ohms); on the other hand, if the extended structure including feedline, tower, etc. looks more like a 1/2 wave or 1 wave (or whatever other multiple), it might need to be a whole lot more than that (10kΩ+?).

On the upside, if Zcm is large by nature, that implies good CMRR by nature, which is an interesting property. Again, I don't know what practical values are here, how it varies with height above GND, feeder and tower lengths, etc.

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  • \$\begingroup\$ I am trying to digest the basic model: Could the node near "E" in "equivalent" be mapped to the shield-dipole node in my diagram such that a signal from transmitter towards antenna sees Zdiff as the left capacitor in my diagram, and Zcm1 as outer shield coupling to nearby objects as a "relative earth", and Zcm2 as the impedance from shielding to true earth? \$\endgroup\$ Nov 21, 2023 at 14:41
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    \$\begingroup\$ Sorry, I should add the reference diagram to be clear -- does that help? \$\endgroup\$ Nov 21, 2023 at 20:36
  • \$\begingroup\$ It all makes sense now! Thank you for clarifying with the diagram. With regards to question #2 in the original post, it seems that a CMC would reduce the current in the core conductor; could a very large CMC stop a signal from a fixed voltage source from reaching the antenna? \$\endgroup\$ Nov 21, 2023 at 20:51
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    \$\begingroup\$ If XFMR3 is ideal (no leakage inductance), then V(Zdiff2) = V(V3). For other values, you'll have to do the AC steady-state analysis and figure it out. Looking at it in black-box terms (add CMC --> core wire current decreased) doesn't seem useful to me here; flipping it around seems more meaningful (reduce Zcm from infinity towards some given (finite) value --> core wire current changed; not necessarily increased). Then it becomes obvious that Zcmc and Zcm work together, independently of the V3/Zdiff loop (aside from taking Vcm = V(V3)/2 as shown, where Vcm is the center-tap node). \$\endgroup\$ Nov 21, 2023 at 21:28

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