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Does anyone know how a current balun, when connected between coax and a series center fed half wave dipole, prevents RF current from splitting up at the join between coax braid and balun and traveling down the outside of the coax ?

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I have added another picture to further clarify what i am asking - when transmitting, how does the current balun in the drawing below prevent Ishield-outer ? Noting that the current arrows indicate instantaneous direction only, because this is the current of an RF AC waveform flowing through the wires of the coax that i am talking about.

enter image description here

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    $\begingroup$ There are some good answers with drawings, for the first few linked questions below - eg. here and here . $\endgroup$
    – tomnexus
    Apr 13 at 14:29
  • $\begingroup$ @tomnexus Thanks tomnexus, those links are on topic and very helpful. $\endgroup$
    – Andrew
    Apr 13 at 23:27
  • $\begingroup$ Question, as per user10489's answer, re this comment "A balun or unun prevents (or reduces) common mode current by providing a low impedance path for differential current and a high impedance path for common mode current." As thus described, When the current balun is connected between coax and antenna, does the high impedance that the balan presents to CM current inside the coax also stop current slitting up at the join of coax to balun at the antenna end ? And if so exactly HOW does it do this ? Everyone already knows that is what a current balun does, but how does it do it ? $\endgroup$
    – Andrew
    Apr 13 at 23:35
  • $\begingroup$ Is it because the current cannot flow on the outside of the coax because that current must travel through the balun first somehow ? because this is the only way it can work, as far as i can see ? :( $\endgroup$
    – Andrew
    Apr 13 at 23:39

2 Answers 2

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I have added another picture to further clarify what i am asking - when transmitting, how does the current balun in the drawing below prevent Ishield-outer ?

I'm going to ignore the arrows in your picture and adopt a convention of "positive current sign = upwards", to hopefully avoid confusion. I'm also adding two more measurements, I(balun left side) and I(balun right side) for the currents in each of the two coils.

KCL tells us that I(shield inner) + I(shield outer) - I(balun left side) = 0. [eqn 1]

A quick glance tells us that I(center) = I(balun right side). [eqn 2]

The properties of coax tell us that I(center) = -I(shield inner). [eqn 3]

The balun does its best to ensure that I(balun left side) = -I(balun right side). [eqn 4]

Let's assume that the balun is perfect and that the last equation holds exactly. Then by substituting eqn 4 into eqn 2 we can come up with I(center) = -I(balun left side).

Substituting that into eqn 3 we get I(shield inner) = I(balun left side).

Now substitute that into eqn 1 and we have I(shield outer) + I(shield inner) - I(shield inner) = 0. Or, simplifying, I(shield outer) = 0.

Equations 1 through 3 are another way of stating that "only common-mode current flows on the outside of the shield". If upward current on the outside of the shield makes a "U-turn" and becomes a downward current on the inside of the shield, it induces an upward current on the center conductor in doing so! So perhaps another way of answering your question would be: it does do that, but contrary to your intuition, that doesn't prevent the balun from working — it's exactly what makes the balun work. Any current induced by the outside world on the outside of the shield will be presented to both balun terminals equally, and therefore be subject to the balun's high impedance towards common-mode current.

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  • $\begingroup$ Thank you ! Now i understand, the antenna, both lines of the coax and the transmitter are all in series, any current flowing on the outer of the coax must also be part of an equal and opposite component on the inner conductor, and the balun presents a high impedance to any current which doesn't have an equal and opposite component on it's input terminals. $\endgroup$
    – Andrew
    Apr 16 at 23:10
  • $\begingroup$ To put it another way, current on the outer of the coax cannot flow because it's equal and opposite counterpart on the inner conductor is blocked by the balun, because it's all in series. $\endgroup$
    – Andrew
    Apr 16 at 23:12
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A balun or unun prevents (or reduces) common mode current by providing a low impedance path for differential current and a high impedance path for common mode current.

The pictured balun does this by pairing the wires that you want equal and opposite current in them, while common mode current has to deal with the impedance of the windings around the core. However, if you have high power common mode current with this, the core can saturate, overheat, and possibly crack.

The paired wires form mutual equal and opposite magnetic fields that cancel, so impedance is low. Common mode current flows in the same direction in both wires, so the fields add instead of cancel, and the resulting field interacts with the core. You can think of the core like a huge magnetic field dampener with a high magnetic inertia and resistance, and thus the result of the wire around the core is a high electrical impedance.

The core does not just interact with the shield, it interacts with common mode current in both wires.

Note that balun cores are designed differently than regular transformer cores. Transformer cores typically use laminated plates to have a low magnetic resistance in one direction while having a high resistance in a perpendicular direction to reduce eddy currents. Balun cores use powdered ferrite which has a high magnetic resistance in all directions.

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  • $\begingroup$ Thanks for the reply, how does the balun present a high impedance to the current flowing on the outside of the coax when the coax joins directly to the balun ? there is no part of the balun in series with the outside of the coax, it is a direct connection. What stops the current flowing on the inside of the coax from splitting up at the join and going down the outside of the coax ? $\endgroup$
    – Andrew
    Apr 12 at 4:23
  • $\begingroup$ Is it because the balun forces current of equal amplitude and opposite polarity to flow through itself, and then because the transmission line is in series with the balun. and the same current flows through both, then also the current along the entire length of the line is also forced to be equal and opposite, regardless, and so no current can flow along the outside of the coax ? $\endgroup$
    – Andrew
    Apr 12 at 4:30
  • $\begingroup$ Added a bit more to help answer this. $\endgroup$
    – user10489
    Apr 12 at 11:29
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    $\begingroup$ So, there is a high impedance presented by the balun to any current current coming from the transmitter which does not have an opposing magnetic field, and this is why it can't flow on the outside of the coax ? Is it possible for you to send me a link to a book or similar that explains what you are saying in detail ? Sorry for all the questions, but i really want to understand why a balun prevents the current on the outside of the coax. $\endgroup$
    – Andrew
    Apr 13 at 0:10
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    $\begingroup$ Sorry i disagree with that, due to skin effect for RF current the outside of the coax behaves exactly as a third wire, just the same as if you actually connect a real third wire. $\endgroup$
    – Andrew
    Apr 13 at 23:30

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