2
$\begingroup$

Can anyone tell me, inside a transmitter, what stops RF current from flowing from the transmitter along the outside of the coax and causing similar problems as when a half wave dipole is connected to coax without a balun ?

Edit : added " from the transmitter", i am asking about current flowing from the transmitter along the outside of the coax towards the antenna.

$\endgroup$

4 Answers 4

5
$\begingroup$

What you need to realize is that coaxial cable is a waveguide, and the wave is coupled in on the inside, not the outside.

Taking a small step back: you know how a dipole antenna works, roughly (assume it's directly soldered on to the output of a differential amp): The amplifier causes harmonic voltage oscillations at the "beginning" of the two branches. Because "just a moment ago", the voltage at the neighboring points was still a bit different, that causes a harmonically changing current to flow in the dipole branches.

Current that flows along the length of the conductors causes a magnetic field, circling around the conductor. So, that changes to, harmonically, with the same frequency. Around the conductor. So, this has an effect on the environment of the dipole, not the dipole "metal sticks" themselves.

Changing magnetic field causes an electrical field, with a phase shift relative to the change of the magnetic field. That in turn causes a changing electric field, again in the environment: What we have is energy shifting from the magnetic to the electric field, and vice versa, alternatingly, and even more relevant to us: reaching further in space! We've caused an oscillating electromagnetic field, and what we got is a wave that radiates. Awesome; the physics for radio now stands! (Fast forward: billions of people watching cat videos on their phones. Who would have thought that differential equations involving electric and magnetic fields would have had that effect?)

So, now we do that, but we do it inside the coax. Same math/physics still works, but there's boundary coditions. These are:

  1. The center conductor is made of something very well-conducting. That means there cannot be an electrical field point "radially" inside the center conductor – electrical fields within a perfect conductor don't exist (if they existed, they would instantly be cancelled by the current they cause to flow). What can exist are electrical fields parallel to the surface, both in "forward" direction along the length of the coax, and in circular directions "around" the conductor
  2. The outer conductor (the "shield", which is a bad name, but commonly used; it's not shielding the waveguide, it's half of the waveguide) is also made of something very well-conducting. So, again, no electric field "radially", but sure longitudinally (i.e., along the length) and on the inner surface "rotatingly".

This has a couple interesting consequences. Namely, these waves that propagate, they have electric fields, but these are not "free"; they need to fulfill an equation that says "hey, at these points, your radial component needs to be zero, no discussion". That's fine – there's still "solutions" that allow for these waves to exist and propagate in the space between the inner and outer conductor.

So, back to your question: Why no current on the outside?

Well, what we've ruled out so far is that the electrical field from the inside leaks to the outside – that's the second condition there. But what about the displacement currents that might flow on the outer conductor? (Remember, electrical field along the surface is allowed to exist!)
Wouldn't they also work, just as in a dipole, to make an antenna out of the outer conductor? I just explained how current flowing in a conductor makes a dipole an antenna!

The truth is that the current flows on the (inner) surface of the outer conductor, because that's the place "in touch" with the electrical fields parallel to the surface. Now, while that does induce magnetic fields, these cannot extend "into" the conductor, to reach the other, outer side. The reason is simple: you put a changing magnetic field perpendicularly through a conductor, it inducts a current that is opposed and cancels the field.

Hence, these fields, both the electrical and the magnetic field, in the components that matter to the outside (electrical: radially, magnetic: parallel to the surface) can't exist at any depth into the outer conductor. We call that depth skin depth.

That is, as long as the conductor is perfect, that skin depth is arbitrarily thin, and the thinnest foil can effectively separate the inside from the outside. (And that is exactly why you can see, when you cut the center conductor of a coax cable, that its surface is plated with copper, silver or some other high-conductivity material, and the core is often made from aluminium or steel for cost and/or mechanical strength reasons: Deeper than a couple micrometers, the conductivity doesn't matter anymore, all the juicy currents happen within a skin depth from the surface.)

Alas, nothing is perfect, and certainly the outer conductor of your coax is not a superconductor. That means some field does leak to the outside, and your coax is a leaky waveguide. But: We've got that under control, relatively well. (Surprisingly well, actually)

So, for all practical purposes, a waveguide operated within the recommended frequencies and powers will be more or less leak-free. When you start doing higher frequencies with higher powers, your waveguides need to get more expensive: the quality and thickness of your outer and inner conductors matter more, and these currents that flow on the surfaces really start, even in a well copper-coated-then-silver-plated conductor, to see a bit of ohmic losses, which heat up the material, and you add another dimension of problems to your waveguide. Great. (Also, the dielectric, the nonconducting material between inner and outer conductor, is a lossy thing, gets hot, and/or stops being as nice and linear. That's why high-power coaxes are often mostly air-filled instead. Small mechanical problem: without some rigid material between inner and outer conductor, how do you keep the inner conductor exactly in the center? That's when you stop doing coax and just do hollow waveguides instead.)

$\endgroup$
7
  • $\begingroup$ So what you are basically saying is that so long as the transmitter is enclosed in a sealed metal box, then no current can get out and flow on the outside of the coax ? If i have a transmitter with a metal chassis but a plastic case, then it can ? $\endgroup$
    – Andrew
    Apr 12, 2023 at 12:26
  • 2
    $\begingroup$ no. the box around the transmitter has nothing to do with it. As long as the transmitter transmits into the cable, the fields stay on the inside, and so do the currents. $\endgroup$ Apr 12, 2023 at 12:47
  • 1
    $\begingroup$ because the electrical voltage that the transmitter introduces is between the center conductor of the coax and the outer conductor, the electrical field exists between these two, and thus the wave and the propagating power is inside the coax. Remember, there's an electromagnetic wave propagating in the space between the inner and outer conductor. The signal is not transported as a current that runs through the conductors from the transmitter to the antenna, but in the field! $\endgroup$ Apr 12, 2023 at 14:12
  • 1
    $\begingroup$ So, whether you keep the outer conductor at a fixed potential and only wiggle the voltage at the feedpoint of the inner conductor, or whether you pull them both in opposite directions, that doesn't matter to the wave inside: what matter is the difference between the potential of the inner and the outer conductor, because that defines the E-field. So, in an unbalanced feeding, you keep the outer conductor fixed relative to "ground" (whatever that means!), but you don't have to do that. $\endgroup$ Apr 12, 2023 at 14:15
  • 1
    $\begingroup$ @user10489 I'm sorry, you're wrong there – coaxial cable is a coaxial waveguide. You're right in that the number of modes that typically propagate in coax is quite limited – usually, in coax line, you use frequencies low enough for only a single TEM mode to propagate (and that mode has no cutoff frequency, so it can always exist, regardless of the frequency). If you happen to have that handy, Pozar's Microwave Engineering, 3rd ed, Section 2.2, subsection "The Telegrapher Equations Derived from Field Analysis of a Coaxial Line";the time dependency or wave number $k$ have no effect in there! $\endgroup$ Apr 13, 2023 at 5:45
3
$\begingroup$

Nothing inherently "stops" current from flowing on the outside of the coax shield, and it certainly will flow on the outside of the shield, through the chassis of your radio, and to anything else connected to the chassis, if a potential is developed (for instance by RF in the outside environment). We can counteract that tendency by increasing the impedance of the outside of the shield (e.g. using chokes) and by providing that current a low-impedance path to somewhere other than the shack (e.g. by grounding the coax where it enters the building).

But I don't think that's the question you're trying to ask. I think the question you're trying to ask is "when a transmitter transmits, why doesn't that cause a current on the outside of the coax?". Because the transmitter uses the coax as a circuit. Current that leaves the transmitter on the center pin returns via the shield, so it's inducing a differential-mode current which remains confined to the center and the inside of the shield. Only common-mode current can travel on the outside of the shield, so in order to put current there, the transmitter would have to work the center and the shield of the coax both against something else... and transmitters are simply built not to do that. The practice of connecting the coax shield to the transmitter's chassis (which is hopefully grounded) makes it difficult to do that even by accident.

$\endgroup$
1
$\begingroup$

This question has shifted slightly since this answer, so I'm adding to this answer.

Current on the "outside" of the coax is properly called common mode current, because it has a polarity in common with the current on the center conductor rather than opposite (or differential current).

This doesn't occur normally or on correctly functioning coax, and it is not sourced from the transmitter. The transmitter puts a balanced signal into the coax, and frequently has a balun on the inside of the radio at the coax connector to make sure of this. The current on the coax shield from the transmitter is opposite and balanced with the current on the center conductor, so these two currents are attracted to each other, causing them to flow on the inside of the shield and outside of the center conductor.

Common mode current can be induced when a portion of the antenna couples with the coax, generally due to geometric arrangement of the coax and the antenna. This current travels on the outside of the coax because it is the same polarity as the current on the center conductor, and thus is repelled from it.

The easiest way to fix common mode current is a balun or unun. You can get inline versions of those. Or you can make an air balun by making a coax coil of 6-10 turns and taping it together (or use velcro or zipties or whatever).

But you said without a balun so...

Sometimes, if you are super careful, you can prevent common mode current by getting the geometry of the coax vs. antenna perfectly symmetrical. But this is unreliable (as it is hard to get perfect), although it helps an existing balun work better if you get it close.

Depending on antenna type, you can also try a gamma or delta match, which can partially replace a balun by forcing the antenna to be more symmetrical.

And some antennas like the bazooka or double bazooka have a balun built into the design of the antenna, so they don't need another one...

If the geometry between the antenna and the coax causes the antenna to couple to the coax, the antenna can carry current opposite from (half of) the antenna that is induced by the antenna's field. This current is not balanced with the current on the center conductor of the coax, so instead of canceling, it will cause the coax to radiate, basically becoming part of the antenna. This radiation will travel the length of the coax (attenuating as it radiates) until it is blocked, possibly by something like a balun. If you have no balun, it can travel all the way back to the transmitter, where it can then flow into the outside of the case of the radio, possibly causing RFI, RF burns, and other issues. (It stays on the outside of the radio's case both due to skin effect and due to the balun inside the radio blocking it at the coax connector.)

$\endgroup$
4
  • 1
    $\begingroup$ I believe that your answer is about the connection between the antenna and the feed line, but the question is about the connection between the transmitter and the feed line. $\endgroup$
    – Kevin Reid AG6YO
    Apr 12, 2023 at 2:49
  • $\begingroup$ @KevinReidAG6YO, thank you, yes you are correct. $\endgroup$
    – Andrew
    Apr 12, 2023 at 4:32
  • 1
    $\begingroup$ I don't distinguish the difference between the two connections. You can put the balun at either location and get nearly the same effect. The only difference is how much of the coax radiates and thus is part of the antenna. Really, the coax, balun, and antenna work together as part of the antenna system. From that point of view, no matter where you put the balun, that is where the coax connects to the antenna, because the antenna includes the coax between the balun and the rest of the antenna. If you put it (only) at the transmitter, all the coax is antenna. $\endgroup$
    – user10489
    Apr 12, 2023 at 11:34
  • 1
    $\begingroup$ Let me add that if you have an unbalanced system such that the antenna is interacting with the coax shield, and have no balun at all, not only the coax is part of the antenna, but also the case of your radio.... which is why there are currents there and the risk of RF burns. $\endgroup$
    – user10489
    Apr 12, 2023 at 11:56
1
$\begingroup$

I'm not sure this answer is correct because it's coming from intuition only. The inside of the shield and the outside can be thought of as 2 wires, plus the center conductor. The shield current is given both options, inside or outside. But only the inside is closely coupled to the center conductor and the outside is "far away" so the current doesn't like it. In a circuit board's ground plane, the AC return current wants to go beneath the other conductor to minimize the loop area, even though it is given the choice to go anywhere. I think it's the same reason.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .