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Assuming an antenna system where a transmitter feeds an ideal coaxial cable which is then connected to an ideal ladder line without a balun:

It is often said that common-mode currents appear in the transition between a coaxial cable and a ladder line simply because the coaxial cable is unbalanced while the ladder line is balanced. My question is related to whether this mechanism is related to how the load/antenna looks like. Let us say that the two wires of the ladder line end up in a load without reference to ground (probably impossible to achieve with a real antenna). This can for example be achieved with a simple ideal resistor (dummy load) in the end of the line (and hanging the ladder line in completely free air). Let us also for the sake of the example assume that the characteristic impedance of the ladder line is 50 ohm such as in the coaxial cable - in order to avoid standing differential waves. Could there in this case be common mode currents appearing on the coaxial shield?

Then what if the characteristic impedance of the ladder line is not the same as the coaxial cable and we therefore have (differential) reflections at the transition - does it make a difference for the common mode?

And thirdly, what if the characteristic impedances of the ladder line and the coax are the same, but the resistive load at the end of the cable does not match the characteristic impedance of the cables so that we get (differential) reflections at the load - does it make a difference for the common mode?

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  • $\begingroup$ Note that there is a difference between ideal coax and ideal dummy loads, and coax and resistors in the real world. There can be leakage between the inside and outside of the shield braid, etc. You can experiment with measuring this difference by holding a small radio receiver up next to the coax, the "balanced" line, and the dummy load, vs. measuring in the far field. And people have accomplished DX QSOs from their "dummy" loads. $\endgroup$
    – hotpaw2
    Nov 4 '21 at 16:14
  • $\begingroup$ Good point, I should have written "ideal coax" and "ideal dummy loads". :) $\endgroup$
    – rubund
    Nov 4 '21 at 20:32
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Yes it radiates. How much depends on the lengths of the wires.

The point at which you connect the coax to the two-wire line is a bit like a dipole feedpoint, with about half the voltage applied to it. This is why:

  1. The voltages on the lines look like this at the junction. (Remember when you zoom in to a region much smaller than a wavelength, there's no RF magic; all the usual rules of circuits apply as normal):
    enter image description here

  2. Coax is "shielded" so the outside world sees the voltage on the (outside of) the outer conductor. On a two wire line, the fields are mostly constrained to an area about double the spacing of the lines. Because the spacing is << wavelength, from far away the two wire line looks like a single wire with the average voltage on it. Zooming out:
    enter image description here

  3. From far away now, you have a wire with a half voltage source in the centre. One side is the outer of the coax, the other is the two-wire line, just the common mode current. How much this radiates depends on the length of the parts of the wire. The case that radiates the most would be a (physical) quarter-wave of coax and a quarter-wave of two-wire line. From far away now:
    enter image description here
    So you have a dipole antenna formed by the whole coax on one side and the pair of wires on the other. The dipole currents are the common mode currents.

I'm ignoring some more subtle effects. For example the dipole radiation resistance forms a load which is connected at the feedpoint, which might interact with the voltages on the lines. So the 0.5 V might change when this is considered.
Also your transmitter is likely to be grounded, so the antenna will work differently.
Finally, if the load and/or line impedances are mismatched it's possible there would be some impact on the effective voltage applied at the junction. Remember in the zoomed-in picture, there's no impedance (on the two wire line) that would prevent it from being about half the voltage on the coax. I think if the two wire line results in a short circuit there (if it's a quarter wave open or half wave short), then the voltage may be zero at that point.

This is a great question, and directly applicable to hams as we often do connect coax to two-wire line in a G5RV / ZS6BKW type design. A common-mode choke helps prevent the radiation effectively disconnecting one side of the voltage source that appears at the junction. It's easiest implemented on the coax side with a coil or ferrite.
At microwave frequencies the solution is a Marchand Balun which enforces the balance on the two wire line by embedding the coax in a symmetrical structure.

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  • $\begingroup$ Interesting explanation. It somehow feels so intuitively explained that it is hard to believe it can be true. So the left part of the 0.5V sinusoidal voltage source is not connected to the shield of the coax, how do you then explain that the coax shield is part of this dipole? $\endgroup$
    – rubund
    Nov 4 '21 at 20:15
  • $\begingroup$ No, perhaps my diagram isn't clear. From far away, a piece of coax just looks like a piece of wire. It's shown in black. I'll tweak it. $\endgroup$
    – tomnexus
    Nov 4 '21 at 23:20
  • $\begingroup$ My point is that only the 0V side of the ladder line is connected to the shield of the coax. How can the 0.5V source then generate currents (common mode) in the coax shield? $\endgroup$
    – rubund
    Nov 5 '21 at 8:21
  • $\begingroup$ So this answer helped me to find this article: learnemc.com/introduction-to-imbalance-difference-modeling which seems to explain things really well! :) $\endgroup$
    – rubund
    Nov 6 '21 at 22:40
  • $\begingroup$ @ruben well look at that diagram under point 2 ;-) I hadn't seen this but it's a good write up. EMC problems do yield to first-principles thinking, at least at first. $\endgroup$
    – tomnexus
    Nov 7 '21 at 5:30
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The thing about the coax that I didn't know for the longest time is that the inside of the shield and the outside of the shield are essentially 2 separate wires, that are only connected at the ends (or anywhere the shield opens up). This is because of the skin effect at the frequency and thickness of the shield. When you connect the coax to ladder line, you are also connecting that 3rd wire (or 5th wire however you want to look at it). Some of the current will go into that and it throws the whole thing off and yes it will cause common mode currents on the coax because of this. I could be wrong but I think also common mode currents on the ladder line which isn't balanced now.

the impedance miss-matches create reflections of course and those will superimpose on whatever signal you have going on. I would think of these 2 separate things that get mixed together at the end, so the balancing problem and then the reflection problem. If you only wanted to look at the reflection by itself and then look at how the unbalanced line effects it, then I don't know the answer :D

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Mike Waters
    Nov 12 '21 at 14:16
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    $\begingroup$ @MikeWaters just to let you know that I flagged this, also I'm very disappointed in this site and I'm not going to participate anymore. $\endgroup$
    – pgibbons
    Nov 12 '21 at 18:00
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The characteristic of coax that causes current to split up between the load or antenna and the outside of the braid is the fact that the braid has two surfaces, inner and outer.

Ladder line has two wires with no inner or outer, the conductors only have one surface each and so at the ends there is no where for the current to split up to, as with coax.

If a balanced transmission line such as ladder line is connected to a resistive load which has the same impedance, then all the energy traveling along the transmission line is absorbed by the load and then in this case dissipated as heat. If the load was a lossless balanced antenna of the same impedance, then also all of the energy would be absorbed but in that case radiated as EMF waves by the antenna.

So the answer to your question is no, there will be no common mode current on either wire of ladder line when connected to a dummy load.

If the impedance of the load is a different impedance to the ladder line, a portion of the incident energy traveling towards the load is reflected back down the transmission line.

If you were to connect coax which is unbalanced to ladder line which is balanced, regardless of what the impedance of each is or whether or not they are matched, at the junction, because the shield of the coax has two surfaces, the current on the inside of the coax will split up between the ladder line and outside of the coax, in the ratio of the impedance to ground of that side of the ladder line to the outside of the coax.

If the two lines are not the same impedance, current will also be reflected back down the coax on the inside of the shield and the outside of the center conductor. The magnitude of this reflected current is determined by the ratio of the differential impedance across the two wires of the two transmission lines, and not by the impedance of the wires at the junction to ground.

In a balanced system, at every point, current in both wires are of equal amplitude but opposite in polarity, and so the magnetic fields cancel and there is no net radiation, however this fact doesn't stop current splitting up at the ends if there is more than one path for the current to flow.

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  • $\begingroup$ How do you really define the "impedance to ground of that side of the ladder line"? I am thinking since the load is a resistance and not an antenna, the impedance should be extremely high towards the ground on the ladder line. And if the path with the smallest impedance would take most of the current, then nearly nothing would go towards the load and everything return via the coax shield - (especially if the ladder line is very short) :s $\endgroup$
    – rubund
    Nov 4 '21 at 20:30
  • $\begingroup$ Whomever downvoted my answer, I'd like to know why. $\endgroup$
    – pgibbons
    Nov 5 '21 at 18:24
  • $\begingroup$ It wasn't me, and i agree, down votes should be accompanied with a reason so we can all learn something ! My answers get down voted all the time with no explanation. $\endgroup$
    – Andrew
    Nov 5 '21 at 22:23
  • $\begingroup$ @Andrew Do you have an answer to the question in my previous comment? Thank you. :) $\endgroup$
    – rubund
    Nov 6 '21 at 16:09
  • $\begingroup$ @rebund i woud like to point out that my answer is reflected in the article you mention learnemc.com/introduction-to-imbalance-difference-modeling, one point being that if you connect coax to ladder line, at the junction, current splits up between the ladder line and outside of the coax because coax is not balanced. This is exactly what the article says, but i never get points for most of my answers, i must always be wrong. $\endgroup$
    – Andrew
    Nov 8 '21 at 3:31

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