Current Flow in Coaxial Transmission Lines

In a discussion in another group, someone is saying that in a coax, because the energy is carried in the EM field and so there is only an "insignificant" current flow in the conductors.

He also says that if you terminate a coax with a matched resistor some of the current will flow back on the outside of the shield. It seems to me this would result in an imbalance of current in the transmission line which seems to be problematic.

Are either of these correct? He doesn't provide much evidence in either case. He points to wiki pages that don't seem to support his statements.

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Rick
Reply to
rickman
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Don't rely on this person as a mentor >:-}

...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
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I love to cook with wine.     Sometimes I even put it in the food.
Reply to
Jim Thompson

My favorite transmission line textbook is _Electric Transmission Lines_ by Hugh Hildritch Skilling, McGraw Hill, 1951.

I'll try to scan the relevant pages about high frequency behavior in coax a little later if I have a chance.

Reply to
bitrex

Not directly on topic but something related that you maybe interested in, which may help you understand other questions about transmission lines.

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Hers a basic break down of COAX cable.. It also gets into the character impedance issue, which is part of what you were talking about.

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Also look up "Standing Waves".

With all that, you should get a better view of a coax cable

Jmaie

Reply to
M Philbrook

It's true that the energy is carried in the EM field in the waveguide, but without current flowing in the conductors, no EM field could be generated! It's not like, shooting a beam of light down an optical fiber. You need moving charges in the conductor to create an EM field in the coax, but the majority of the power being delivered to the load is being delivered by the EM field in the dielectric, not the moving charges themselves. See:

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So there are two terms in the loss that you'll get with coax, or two wire line, or whatever: there will be a resistance term that is the loss caused by the resistance the moving charges encounter in the conductor, and a dielectric loss term that occurs due to the EM field losing energy because of the power factor of the dielectric.

At high frequency due to "skin effect" the current in the conductors will tend to crowd around the inner surface of the outer conductor and the outer conductor of the inner conductor, increasing the magnitude of the line resistance term.

Reply to
bitrex

So far your's is the closest to the issues in the discussion there. The original question had to do with using a balun between a balanced dipole and a coax relating to current returning on the coax shield outer surface. A lot of red herring showed up in this conversation and one line of reasoning referred to the current in the coax being negligible with the end of the coax being a "generator" of current from the EM field.

He won't really explain the line of thought, rather just keeps referring to wiki pages which don't actually say what he says they say.

I'm pretty over the discussion at this point. The group seems to have people who want to discuss and understand, but there are always one or two who seem to be there mostly for the drama or maybe just have their own way of looking at the universe but can't really explain it for any number of reasons.

He really got my goat when he suggest that even in the case of a purely resistive load matched to the coax there would be current leaking out onto the shield outer surface. I thought Kirchoff's laws would show this to be wrong, but he pulls anything he wants out of the hat when he wants.

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Rick
Reply to
rickman

When you have AC current in a transmission line, you are generating propagating electromagnetic fields that carry energy. The Poynting vector, E cross H divided by mu_0, or whatever, has units of energy density. It's the same thing with power lines - their energy is being delivered by the EM field, not the moving charges. You're actually burning energy overcoming the resistance of the wire to the charges to get them moving and get the EM field going.

So in a coaxial line, or a power line, or whatever, you will have a Poynting vector inside the conductor(s), and inside the dielectric. As the frequency increases, due to the skin effect, the magnitude of the Poynting vector inside the conductor(s) will decrease, until the majority of the power is being transmitted through the dielectric. As well, as frequency increases the current will begin to crowd on the outer surface of the conductor(s), causing the apparent resistane of the conductor(s) to increase, requiring that your generator expend more power in the process of keeping those charges in motion.

If the current in a coaxial line is in opposite directions in each conductor, then the current in the conductors will migrate _towards_ the axis of the propagating EM field in the dielectric, that is towards the inner surface of the outer conductor and the outer surface of the inner conductor.

As far as I know, that's all that classical non-relativistic electrodynamics can say about things. The issue is also that to "fully" describe classical electrodynamics one needs special relativity (an electric field and a magnetic field are the same from different reference frames), and to "fully" describe resistance you need quantum mechanics. So I guess in a sense to "fully" describe what is going on here according to our current understanding of physics, one would essentially need quantum field theory.

It's sort of moot in that as far as I know, in most cases when a certain coax is used for the RF frequency range that it was designed for, runs of a few dozens or hundreds of meters are so short that the lines can essentially be treated as lossless for analysis purposes.

Reply to
bitrex

If that's true at all it's only going to apply when the diameter of the line gets to be a significant fraction of a wavelength.

Unless he's terminating it in some unique (and probably stupid) way, this is just plain BS. As long as the current on the center conductor matches the current on the shield, the current is "inside" the coax.

If it's an radio group, well, there's a lot of profoundly astute people in that hobby, and there's a lot of severely misinformed, self-important wackos. Sorting them out is up to you.

--
www.wescottdesign.com
Reply to
Tim Wescott

Yeah, I often just ignore this particular guy. But sometimes he makes a post that is reasonable and actually has some analysis in it. Then he will make some absolute statement like this one and won't back down or modify it or even *explain* it in any way. I've asked him for the underlying assumptions under which his premise might be true and he keeps referring to the same wiki section on coax as transmission lines.

I thought I'd ask here to see if anyone had some idea of where he was getting this.

The discussion is actually about the current flow on the outside of the shield that happens when driving a balanced antenna and he keeps wanting to turn it into a different topic altogether. Whatever...

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Rick
Reply to
rickman

You can run DC through a coax. Works fine.

At low frequencies, the entire cross-section of all the conductors carries current.

--

John Larkin         Highland Technology, Inc 
lunatic fringe electronics 

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com
Reply to
John Larkin

Another way to think of it, is in terms of basis sets.

  1. You can have independent currents on core and shield. Period. They don't even need to be coaxial anymore!
  2. Or you can subtract the shield current from the core current, and call them common and differential mode. Or more typically, normal current and shield current.
  3. Or you can take the sum and difference, and call them differential and common mode (in the conventional sense, as with twisted pairs).

Where signals meet the circuit, however, you'll find that (2) is the most consistent and easiest applied for cable of that construction, and (3) is best for symmetrical construction like twisted pair. (1) is only ever used where "IDGAF" is in effect, like automotive wiring (where everything is a ground, because it's bolted together) ... usually with disastrous results when it comes to RF accidentally flowing in those wires (i.e., automotive EMC). :)

Tim

--
Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

Yes, of course you can have many types of current in a coax. The issue is what you get under given situations.

I have no idea what IDGAF means. Or... IHNIWIM

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Rick
Reply to
rickman

Yeah. Thanks

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Rick
Reply to
rickman

If you put a step voltage on one end, then the pulse propagates along the cable, charging he capacitance between the centre conductor and the shield as it goes. That's going to require the transfer of electrons, which implies a current.

Sylvia.

Reply to
Sylvia Else

On Sun, 02 Aug 2015 14:51:47 -0400, rickman Gave us:

Notice how cell towers no longer have a "repeater station" at their base piping the RF up to the cell antenna arrays.

The RF amp is placed locally right behind each antenna, and the base shack is a mere gateway to the trunk line/Internet backbone. Likely fiber linked up to the amps.

Notice how modern PCB designs also use DC power modules local to the actual load devices.

It's the new thing. Been around a long time though, actually.

Reply to
DecadentLinuxUserNumeroUno

On Mon, 3 Aug 2015 12:08:53 +1000, Sylvia Else Gave us:

and a reflection.

Reply to
DecadentLinuxUserNumeroUno

Whether there's a reflection depends on what's at the other end. If the cable is terminated by a resistance that will carry that current at that voltage (that is, if it equals the characteristic impedance), then there will not be a reflection.

Sylvia.

Reply to
Sylvia Else

Nice Answers Biertx...

Reply to
kenny.sourcing

There are two types of baluns used between balanced antennas and unbalanced feed lines:

- balanced tansformer with center tap - common mode choke on unbalanced line

If the antenna in really balanced (including its near-field environment), both methods lead to similar results. For real-life situations, where the antenna is seldom balanced, the common-mode choke often leads to better results.

The discussion of outside currents on the coax may come from observed currents due to the antenna element coupling to the feed line. In practice, balanced dipoles have been working well even when connected directly to coax (no balun, with all the outside currents).

--

Tauno Voipio (OH2UG)
Reply to
Tauno Voipio

"I Don't Give A ..." :-)

Tim

--
Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

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