In the world of HAM radio, there are several myths that survive through generations despite them clashing with laws of physics. However it is quite interesting to see how easy is to debunk some of them with very simple experiments.

The radiating line myth

In HAM radio discussions “experts” frequently advise against antennae tuned by a tuner for a list of “technical” reasons. Often, one of them is that due to increased VSWR, the line begins to radiate, causing any kind of interference.

Let’s try!

Smartest myths are usually founded on unverifiable facts: people are left to choose whether to believe or not. This one, instead, is very easy to verify with a simple experiment.
Let’s start measuring the field radiated by a true antenna:

As we see, as soon as the FT817 goes on-air (red light is on), the RF current meter detects immediately a very strong field radiated from the antenna..

Now, let’s repeat the same experiment on a coaxial cable terminated by a 50Ω resistor:

As expected by everyone, the dummy load is perfectly matched, there are no standing waves and the coaxial cable does not radiate. The VSWR meter, in the green circle, reports “1” (no bars) and the little resistor becomes immediately scorching hot.

It is now time to test the myth. We repeat the same test above but, instead of using a 50Ω resistor, we use a different mismatched value (150Ω) that causes standing waves on the coaxial cable but without triggering the transmitter protection. It the myth is true, the standing waves should cause the coaxial cable to radiate. Let’s try:

As we can see, the VSWR meter shows some bars due to reflected power. The little resistor becomes immediately red-hot, but despite of the experts claims, no radiation whatsoever is detected from the transmission line. Wasn’t VSWR expected to cause the line to radiate?

The explanation

The reason why the coaxial cable does not radiate even on high VSWR is very simple and it is explained by Maxwell laws. They say that a variable current running along a conductor causes a radiating E/M field to be generated. The transmission lines, like the coaxial cable, have the task of transporting energy, not radiating it. So, how do they work?

To avoid radiation, transmission lines employ two conductors that carry equal but opposite currents. These currents do generate an E/M field each, but being one the opposite of the other, they are mutually cancelled.

We can easily verify this statement by modifying a coaxial cable so braid and center core run separated for a few centimeters:

We can now measure the current running on each wire when terminated on a dummy load:

LineOnMeter_eng

As we can see, each wire, if taken separately, does radiate. Also, we see that each wire radiates the same amount power of the other.

With a two channel scope, this is even more evident:

With the scope we can see the two currents of same amplitude and opposite phase.

When the load is not matched, as it happens for our 150Ω resistor, the power can not be transfered completely to the load and it is reflected. This causes the appearance of standing waves on the transmission line: within the line, we will find spots carrying an higher current and other with an higher voltage. However, the standing waves are generated on both conductors: they are symmetrical, with identical amplitude and opposite phase so, once again, they cancel each other.
For this reason, we might experiment line radiation and increased VSWR together, but high VSWR itself can not be the cause of line radiation.

So, when does coax radiate?

In order to cause the coax to radiate, the currents must be unbalanced. This means that the current running on the braid and the center are not equal and opposite anymore. Their sum is not zero and the resulting current (called common mode current) radiates. Note that this can happen also when the system is perfectly matched (i.e. with VSWR=1).

We can see this phenomenon by feeding a single conductor, like the central wire. In this case, there is no opposite current on the braid to cancel it and this forces the coaxial cable, despite of its shielding, to become a wonderful antenna:

Conclusions

There are several reasons that can cause coax line radiation; most of them are related to wrong or missing use of a balun, but none can be imputed to load mismatch. It can happen that a common cause provokes both high VSWR and line radiation, but VSWR can be taken as symptom, not the cause.

 

  1. In a nutshell.
    If I1 and I2 aren’t equal or there is serious mismatch at the antenna, there is a “common mode current” running on the outside of the coaxial cable, and there will be radiation!
    Harmful in most of the cases. (RF bites, distorted voice, key clatters etc)

    • If “I1 and I2 aren’t equal” you have unbalance, the fields don’t cancel and you do have line radiation (common mode radiation).
      If you have “serious mismatch”, you have standing waves on both conductors, they cancel and they do not radiate.
      If I1=I2 (i.e. identical in amplitude and phase), why should be the standing wave ascribed to “I1” be different from the one ascribed to “I2”?
      In other words, your sentence “If I1 and I2 aren’t equal or there is serious mismatch at the antenna” should be rephrased as “If I1 and I2 aren’t equal and not if there is serious mismatch at the antenna“.
      In a nutshell, the “or” conjunction is exactly what the myth is about.
      Obviously, mismatch and unbalance can coexist, but they are not related. This is what this article is about.

      Vy 73 de Davide IZ2UUF

  2. The measurement suggests wrong results. Coax radiation has several ingredients. One is common mode. The other one is caused by the attenuation of the shield. The shield IS NOT TIGHT. It leaks. And it leaks more when VSWR is higher because voltages and currents are much higher every quarter wavelength. THERE it leaks more upon high SWR. The measurement method is simply not sensitive enough to show the additional radiation. Use a vector network analyser and a parallel cable for pick-up. You will see. If not, attend my lectures. It is one of my demonstrations.

    • Hello Fritz.

      Obviously you are right. Load mismatch causes greater current within the line, which provokes higher Joule attenuation due to resistance and, of course, a greater radiation due to leakage.
      However, as happens in engineering, the problem is not understanding whether an event occurs, but its relevance according to the application being considered: the word “zero” is a synonym of “negligible”.
      Let’s take the example above: a 150 ohm resistive load (VSWR=3) implies an increased leakage ranging somewhere from <1dB to 2dB according to cable length. An average cable leakage is between -70dB and -100dB (see http://www.w8ji.com/coaxial_cable_leakage.htm or http://www.arrl.org/files/file/Technology/tis/info/pdf/8104028.pdf). Adding 2dB to this leakage can be considered negligible. Even with a 500 ohm load (VSWR=10) leakage is increased by only 6dB. And whatever the leakage is, adding a few dB to it, in normal HAM radio operations it will not change its nature: if leakage represents a real problem, it does so with or without few additional dB’s.

      Instead, even a little unbalance causes a huge increase of common mode radiation. If I attach a short wire to the center conductor of my dummy load, the RF ammeter on the coax goes to full scale: I have to select the x10 scale and use minimum power (0.5W) to have readings that are not permanently somewhere over the top. I recorded the experiment above in the video below: the field generated by unbalance was so strong that not only the RF ammeter went full scale, but I’ve been able to call CQ and be copied by a CW reverse beacon:

      http://www.youtube.com/watch?v=UHVGLbGOKZs

      The difference between radiation due to unbalance and increased leakage is not comparable and so they are their effects on real world HAM radio applications.

      The simplicity of my RF ammeter represents the needs of HAM radio application: if the needle doesn’t move (or even it moves a little) when feeding the line with 100W, whatever radiation it is, it can be considered negligible: because resulting pattern distortion is negligible, resulting RF in the shack is negligible and resulting common-mode noise is negligible.

      In my opinion, comments like your, coming from a very authoritative person like you, are very precise under a scientific point of view, but without underlining the real figures, simply help spreading the myth. If a friend of yours is coming to you asking for help because with its SWR=3 end-fed antenna is receiving electric shocks from his microphone, would you advise him to better tune his antenna to solve the problem, as we read on many HAM radio forums? This is what this post is about.

      vy 73 de Davide IZ2UUF

    • Hello Anthony.

      Thank you for your appreciation!
      By the way, I see you are from Parma (OH): I graduated in 1988-1989 at the Berea High School, I guess you know what I’m talking about! 🙂

      Vy 73, Davide IZ2UUF

  3. Brilliant way to explain it. It looks like you used a 1/4 watt 50 ohm resister and I am surprised it did not turn black, smoke and fail in seconds, even on low power.

    • Hello Tim.

      I used 0.25W resistors with the TX power set to its minimum, which is 0.5W.
      To make a measurement or take a picture you need just one second TX or so. The resistor doesn’t have even the time to become warm.
      If you look at attenuators datasheets, they are rated for a given average power and a much higher low duty cicle power. For example, my Weinschel 3200 attenuator is rated for 1W average, but it can handle up to 50W with 1uS pulses and duty cycle of 1%.

      Vy 73 de Davide IZ2UUF

  4. I think this is a great article showing the truth, but I am more interested. In your test meter and how to build one. All the best and 73’s. Don. VA3DDW

  5. Using current and magnetic field cacellation is good way to explain,but what if opposite voltage on the center and shield emitted slice different electric field,as shield may not be perfect.

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