I'm looking for an ACTUAL RF engineer who understands the EE behind antenna impedance and radiation, to see whether I'm talking rubbish with this understanding or not:

Boost for visibility please?


Honestly, I don't think there are many Motorola, Icom, Yaesu, or Starlink engineers on Mastodon.

Most "actual RF engineers" I've met in the ham radio community give explanations I wouldn't hear in a Physics 400 or Engineering 400 level course.

What circles do RF/EE engineers move in? Google Groups or Reddit better for such questions?

@tayledras I don't know, that's why I'm asking! 😄 I've got a few friends who've helped since I posted that. And I just noticed how much traction this toot got; I've got a lot of reply reading to do now. 🙂


Ping @M0YNG, do you know anyone who might fit the bill?

@smitty You are close, but look at it the wrong way. You are thinking in current and voltage conducted in the wire of the antenna. Instead you should look at it as a transmission line where all the energy is in the field around it and the current/voltage is just an effect (or symptom) of the field. Rick Hartley has given some good presentations on this topic in the context of PCB design. I highly recommend watching those.

@smitty BTW: the feedpoint impedance of a dipole in free space is 73Ω. And that of a folded dipole 292Ω. Both are dependent on the free space impedance of 120π Ω ≈ 377Ω.

But as antennas rarely operate in free space and have some ground, or rather conductive structure, nearby, that shifts the impedance and we are closer to 75Ω and 300Ω. And that's the reason why we use 75Ω coax for TV and radio reception.

@smitty The reason why we don't want common mode current is, because it indicates that we have a field outside of the coax and thus are radiating energy from the cable instead of the antenna.

A coax is not a magical structure that ensures that the shield is at GND potential. It's still a transmission line, not much different than a twin-lead. The reason why the shield is GND is because we make it so, either by grounding both sides or by using methods like CMCC..

@smitty This is the reason why we can use coax cables as transformers. E.g. if you are in the GHz range and need to go from unbalanced to balanced, a short strip of coax is the best balun in terms of performance vs price you can get. Yes, a single piece of coax. No ferrites or anything else needed. Though ferrites can help at lower frequencies.

A nice application of that principle can be found in directional couplers, like described in US patent 4962359 by Dunsmore.

@attilakinali The root of my question is challenging viewing an antenna as a single complex impedance between the conductors of the feedline. I think it's more accurately two separate impedances from each conductor of the feedline into "space" or "ground" or "reference plane" or whatever. When those two impedances are the same, then it can be modeled as a single impedance across the feedline, everything stays differential. When they're different, common mode creeps in.

@smitty Oh..kay.. This is will be long:

If you are using a normal 300-600Ω twin-lead, then the coupling between the two wires will be so small, that they can be ignored. I.e. most of the field around the wires will look like the wires are in free space. I.e. you can look at each wire as a separate conductor in free space.

@smitty If you now go into an antenna without any additional element that would couple the two wires of the twin-lead to each other (e.g. a transformer or balun or anything similar), then each leg of the antenna is driven independently.

Thus, yes, you can separate the two impedances of each leg of the antenna and look at them separately. And this will give you a solution that is very close to what's really going on (remember, we only ignored a small coupling factor).

@smitty Now, if you feed the antenna from a coax cable, you will have to go through a balun or a structure that can drive the antenna from a single ended signal (like a Gamma match).

Here, this separate description of each leg still works, but you need to properly account for that element between the antenna and the coax. It will have a non-negligible influence on the impedance seen by the cable, and where the field will end up. I.e. whether there will be common mode current or not

@smitty The same goes if you feed the antenna with a twin-lead but have a impedance matching element between the cable and the antenna.

How that actually influences the field (and you should really think in fields, not currents and voltages) is not easy to describe in formulas. If you want an accurate answer, use a good field solver (like OpenEMS).

@smitty All this said, at the frequencies most ham radio operate at (i.e. below 30MHz), most of this is quite mood, as the wavelength is so long that any feature of the antenna or cable we look at will be tiny compared to the wavelength and thus will have hardly any influence. Only once we get to feature sizes of >1/20λ will we start to see effects. But even those will be quite small until we get to around 1/8λ to 1/6λ.

@smitty The only reason why baluns and transformers have an effect at all is, because we force the field into an environment (aka the ferrite) where the speed of light is small (remember: c = 1/√(εμ) ) and thus the wavelength is small.

@smitty BTW: Kirchhoff's laws (pronounced [ˈkɪʁçhɔf]) only hold for DC. For any AC system where energy gets radiated they do not hold. In particular the assumption that node currents will add up to zero or that currents going into a wire at one end will leave again at the other end are wrong.

@smitty Well... there are indeed various errors in the text. For a start, a resonant dipole is not necessarily 50Ω. You can't measure the impedance of "half dipoles", you need two points to measure something in electricity. Balun being the most abused word in radio, I am not surprised that you are confused, most amateurs are.

@smitty Anyway, a bit more on the subject. You need to start the discussion with transmission lines (t.l.), it is simpler that way. Any t.l. has an impedance, which is what you get if the line is very long and you measure V and A when feeding it. That impedance is a pure resistance and the number depends on the shape and distance between the wires. It can be 50, 75, 300Ω, whatever.

@smitty If what is at the other end (e.g. the antenna, but it can be something else) has the same impedance as the line, there are no reflexions. So we try to make our antennas the same impedance as the line. For that we may need an impedance transformer (e.g. two coils on a ferrite), often improperly called a balun.

@smitty A real balun is something else (even if it can also be made with coils on a ferrite, but wound differently). A balun is to connect something BALanced (symmetric as a dipole or a loop) to something UNbalanced (asymmetric as a coax cable).

@smitty The reason we want a balun (or symmetriser, a better word) is to ensure that the outside of the coax is not part of the system. Because of the skin effect, the coax has actually 3 conductors: center, inside of the shield and outside of the shield. You want the outside of the shield to be insulated from the system. Why in the next toot.

@smitty If the outside of the shield is part of the system, it will be part of the antenna (at the antenna end). So your nice resonant dipole has a third conductor with a length which is what you randomly chose for your coax. It can just work (by chance, if the random length happens to be good), or not. That explains why amateurs report different experiences and some say one does not need a balun. That third conductor will also be influenced by nearby objects and radiate RF.

@smitty If you use a ladder line, the line and antenna are symmetric, so you don't need a balun (symmetriser). You may still need an impedance transformer (which some people wrongly call a balun but should not). But this kind of line is dependent on being symmetric, so you can't have nearby objects on a side, etc... And it needs to be connected symmetrically to your amplifier.

@dl2jml What I'm challenging is the "single complex impedance across the feedline" model of an antenna. If that were the case, there's no way to REFLECT CM currents from the antenna. There are other sources of CM current, but there's no RLC you can put at the end of DM feed line that will result in CM current.

I think the antenna is actually two different impedances, from each feed line conductor, to "free space" or "ground" or "reference plane" or whatever. (more)

@dl2jml If those two separate impedances are the same as would be the case with a perfect antenna, then you can model it as a single impedance across the feedline. Everything remains DM. But when they aren't the same, you start getting CM reflections from the antenna.

I'm trying to verify whether this understanding is correct. It makes sense, but that doesn't mean its correct. 🙂

@smitty You seem to use a nonstandard definition of "impedance". The standard definition is something very simple: you need two points and you connect a voltmeter and an ampere-meter. Then R=U/I. (For the complex impedance you also measure the phase difference between U and I, but let us make it simple). So you always need 2 points to connect your instruments.

In your post, I don't see where is the second point. You measure one half of the dipole, why not, but you still need a second point to connect to. You hint that this second point could be the ground. Why not again, by why the ground and not something else? An if it is the ground, taking into account that both arms of the dipole are identical, I don't see why measuring identical pieces of metal to the ground would give different values.

@smitty Then, there is the idea of "reflected common mode currents". Common mode currents don't care whether they are reflected or not. They happen at the coaxial line for a simple reason: at the end the inside and the outside of the shield are connected. Thee CM current happen with anything, not only with antennas. They happen at the radio side as well. And the balun (symmetriser) simply serves one purpose: make sure that the outside current, at this point, is zero.

@dl2jml If both sides of the dipole are exactly identical, then there would be no current flow into this mythical third node I'm talking about, and it would be as if it didn't exist. Which leads to the simple "single complex impedance across the feedline" model working. It's the case where the two halves of the antenna are NOT the same that I'm trying to understand.

@smitty @smitty I think that what you believe is that CM current are created out of nothing and that when a dipole is connected to a coax, you still have equal currents inside the coax, and an extra CM current. This is not what is actually happening. If you connect the dipole directly to the coax, you will have i1 in the center conductor, i2 in the inside of the braid and i3 in the outside of the braid, with i1=i2+i3. The current is split between both sides of the braid.

@dl2jml I get that. The outside of the shield and the half of the dipole connected to the shield, are effectively two different wires connected to that feed point and will share the currents based on their relative impedances. If your feed line is an even multiple of 1/4 wavelength, the shield is a high impedance and the antenna gets most of the current. If it's an odd multiple, it's low, and your feed line and antenna share it.

@smitty Yes, this is about the idea. You can actually make a balun with a quarterwave transmission line. This is a standard technique at 70cm and above (for longer waves, the balun would be very large and this is not pratical).

@dl2jml @smitty I love reading this type of back and forth. Lots to learn. I don't think your equation is correct though. The current on the pin and inside of the coax will always be equal and opposite. Any current on one, will induce an equal and opposite current on the other. Common mode current can exist on the outside, so i1=i2, and i3 can be any value unrelated to the first two. As hams, we try to reduce i3 with various devices, and those devices should aim to minimally effect i1 and i2.

@ve3mal @smitty As you wrote "Any current on the pin of the coax, will induce an equal and opposite current on the inside of the coax." But "inducing currents" takes a few wavelength and says little about what happens right at the entrance (or exit) of the coax... and I was only discussing the entrance of the coax.

@ve3mal @dl2jml this is correct Jason. i1=i2 necessarily. i3 is independent, and we try to minimize it. Kirchov’s laws get weird at RF when dealing with antennas.

@dl2jml Ohms law states that complex impedance is the ratio of voltage to current, and the phase between the two. Voltage requires two points to measure.

You're correct that I'm implying a third measurement point that represents free space. Part of the problem is that I don't know what to call it. Is it ground? Is it free space? It's what a ground plane is trying to tap into in a vertical unbalanced antenna.

@smitty Free space is not a single point. The measurement you are taking will be different depending where in "free space" you place that second point...

@dl2jml Now the part you've got me chewing on is Kerchov's current laws. The energy may radiate into free space, but the currents still have to sum to zero in the feedline/antenna system. I'm comfortable with "free space" having a potential relative to the antenna, but I'm less comfortable with current flowing from half my antenna into free space.

I need to think about this more, and I've already spent more time on this today than I should. 🙂

@dl2jml Thank you VERY MUCH for working through this with me today. I really appreciate you taking the time.

@smitty I should say that you are not the first person who was confused. I am regularly surprised by the number of people who make a complete mess of what are actually very simple concepts. All what is needed is to go the the basics: you have pieces of wires and can measure U and I (and you need the knowledge that RF cannot go into metal, the so called skin effect). People like Ampere, Coulomb, Hertz... had nothing more and they were able to explain everything.

@smitty Time spent to understand the laws of physics is always time well spent...
Here again, it is quite simple. The currents don't disappear in free space, just like currents don't disappear in a resistor. Voltage gets smaller, so that power (which is voltage times current) gets smaller. The missing power is what is sent to space (or heats the resistor).

@smitty I'm not an actual RF engineer(tm), but I recently spent quite a lot of time researching why common mode chokes work as baluns for a friend. I might have some pointers for you.

In the end we went with an explanation that is similar to the in chapter "9.7.6 Baluns and Transformers" of "Constantine A. Balanis Antenna Theory - Analysis and Design". It slightly simpler than your explanation, but I think they are at least partially equivalent.

Quick copy of the relevant part of the book:
[...] coaxial cable is inherently unbalanced. Because the inner and outer (inside and outside parts of it) conductors of the coax are not coupled to the antenna in the same way, they provide the unbalance. The result is a net current flow to ground on the outside part of the outer conductor. This is shown in Figure 9.25(a) where an electrical equivalent is also indicated.


@smitty The amount of current flow I_3 on the outside surface of the outer conductor is determined by the impedance Z g from the outer shield to ground. If Z_g can be made very large, I_3 can be reduced significantly. Devices that can be used to balance inherently unbalanced systems, by canceling or choking the outside current, are known as baluns (bal ance to unbalance).


@smitty In the end I think voltages and currents might not the best abstraction/model to reason about those things. It might be better to think about it in terms of fields propagating along a transmission line. Unfortunately I haven't found a good (as in it gives you an intuitive understanding before vomiting all the math at you) introduction into that topic yet.

When I explain this stuff in amateur radio exam classes (so on a non-math level), I start with applying voltage to a line. The first few nanoseconds, whatever you do has not reached the other end. So what do you get in that brief time frame? An ohmic resistor of the line impedance. Thinking about what happens when the signal comes back from the other end gives you the idea of reflection. Sine wave signals then produce the usual shortcut of λ/4 line with C before and L after.

@sebastian @smitty
Then unfold the line to become a dipole. When you vary the dipole length, you get qualitatively the same behavior as you do in a transmission line, with a bit of loss superimposed that accounts for the radiated energy.

@dj3ei @smitty I always find that that part of the explanation clashes with the math.
To my mind the better approximation would be that the antenna is a special section of transmission line, providing a coupling with ideally least reflection between the fields in the transmission line formed by the cable in the field forming in free air.

@dj3ei @smitty btw. if anyone else is looking for a starting point I can recommend
which also has a nice quick primer on smith charts

@sebastian @dj3ei I'm challenging the "single complex impedance across the transmission line" model of an antenna. If that were the case, all reflections would be differential mode currents. There's no RLC you can put across any feed line that results in CM current, without some other connection.

I think antennas are TWO impedances, one from each side of the feed line, to "space" or "reference plane" or "ground." When those impedances match, they can be modeled as a single impedance. (cont)

@sebastian @dj3ei But when those impedances do NOT match, then reflections start getting common mode components to them.

@sebastian "Vomiting Math" is my new Einstürzende Neubauten cover band. 🙂

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