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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.
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:
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.)
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.
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.)
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.
