Current Flow in Coaxial Transmission Lines

That pretty much sums it up.

A dipole is a different story. Since the coax naturally must lead to the middle of it the coax jacket will be in the field of the dipole and thus couple with that field. A balun helps because it forces symmetry in the dipole behavior. But I found this not to be very substantial. After I blew up a balun back in my high school days (on account of being cheap and having used only one 2" toroid instead of two) I was temporarily out of funds and simply removed the charred remains of the balun. It hardly made a dent in the standing wave ratio. So I groused over not having invested that money into a crate of beer in the first place.

Got a link?

You cannot fix leakage due to non-ideal coax behavior with a balun at the resistor.

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg
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I'm actually done disusing this point with him. He isn't interested in discussing so much as "winning". I've found several references that show how to calculate the currents in both a transmission line and waveguides (which he said also don't have surface currents) as well as animations of the current and he dismisses all that as not being

*exactly* the same.

No one is talking about leakage or other non-idealities. We are discussing first order effects of the coax shield being attached to one element of the dipole. There seem to be two aspects.

You mentioned that the shield is is in the EM field of the antenna, but that would be true for any type of antenna I believe, no? This has only been mentioned peripherally.

The other aspect that we have been discussing is that the connection point of the shield to the antenna element is actually a three way connection; the inner shield surface, the antenna element and the outer shield surface. Since there is voltage on this point the shield outer surface will carry some of the current depriving the antenna element of a current equal to the other element.

Results of this problem include asymmetry in the antenna pattern and having a "hot" receiver from the RF current finding significant impedance between the rig and ground.

I even found some simulations of a yagi antenna with the coax shield at different angles to the antenna and some were rather grossly distorted with a much reduced front to back ratio.

--

Rick
Reply to
rickman

So it's back to discussing Hillary's pant suits and Donald's hairdo, as one guy on Usenet said? :-)

He isn't interested in

Tell him to reduce it to Maxwell's equations. Those haven't been repealed yet :-)

Yes, but one side of the dipole is connected directly to the jacket and they "see" each other. The other side of the dipole and the center conductor can't "see" each other. A balun breaks this imbalance but as I said it isn't as significant as some think. Depends a bit on whether the coax connects perpendicularly (best case) or at an angle towards one side of the dipole. Shallow angles are bad, results in lost of current on the coax shield.

Doesn't matter much because the total current in the shield must in summation roughly be the same as that of the center conductor. It gets a little out of whack if no balun is used but not much.

The standing wave ratio in my case barely increased when I took out the balun.

Yup, an angle with gradually screw things up. The extreme would be a zero angle where the coax shield becomes on side of the dipole and thus "hot". Then you'd essentially have a 1/4-lambda radiator without radials. The antenna becomes ineffective, depends on total coax length and the SWR would not be good. Those are the situations where back in the days we burned our lips on the metal microphones. RF burns took forever to heal.

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

That is a new viewpoint no one has said before, that the inner shield current is balanced against the *total* shield current. Rather they say the inner shield current balances the inner conductor current.

That's fine, but the devil is in the details. One of the points this guy has raised is that the current on the outer shield depends on the length relative to the wavelength. I'm sure this is a very complex function to measure in a real application.

So is this an example of it being a *very* significant problem?

--

Rick
Reply to
rickman

Only in an ideal and balanced situation. When you have a current on the outside that means either the frequency is low enough as to only have an insignificant skin effect or it can mean that you have shield current that shouldn't be there due to an unbalanced load.

One brute force method to muffle currents on the outside of the shield is a ferrite slipped over the coax.

Think about your home electricity: Normally you'd expect 100% of the differential between the two phases to flow through the neutral between street transformer and house. In reality that isn't always the case, sometimes a portion flows via other ground paths. The sum of all those currents must always be zero, whether it's an antenn system or a power system, per Kirchhoff's law.

I ran into that at a client and the boss there (biochemist) said: "Dang! Isn't there some loophole in Kirchhoff? After all, there is one in almost any law".

The impedance of the outer shield path certainly rises with length. Also, even assuming an ideal and lossless coax any current on the outside of the shield will cause radiation and that typically gets worse the longer the cable is. Which is one reason why the EMC guys want you to have full length cables on your system during compliance testing. 1ft of coax won't radiate much at 30MHz but 8ft sings like a bird, meaning more energy is radiated off of the shield instead of conducted along the shield.

When the coax comes off a dipole at a shallow angle, yes, it is serious. Back as a kid I didn't understand that stuff. I still get scared thinking about it: Once I almost screamed because my right arm felt like "cooking" inside after having it parallel to a coax. I could have easily lost an arm there. In hindsight I thought maybe they shouldn't have let us have the highest class ham radio license at such a young age.

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

The presence of the conduction path on the outer shield makes a dipole unbalanced. Some treat it as another element on the antenna for simulation purposes.

The "shouldn't be there" is exactly the issue.

There is also some issue regarding just how to explain the functioning of a balun which is what the ferrite is.

When the inside shield current reaches the balun there is the path on the outside shield for the current to flow. The balun continues the imposition of equal current flow while isolating the shield outer path from the antenna element. Essentially the intended currents are differential which are low impedance through the balun and the unintended currents are common mode which are high impedance through the balun.

With a low impedance path available to the shield inner current, it will not follow the shield outer path.

Lol

--

Rick
Reply to
rickman

That's almost always the case with RF. Tsssk ... PHUFF ... *BAM* ... "Hey, why did that happen? How can a capacitor explode?"

RF can be like a water bed. You squish it here and it comes back out over there. It can do really weird stuff. As a kid I found out that it can turn white ceramic into bubbly green glass in a spectacular fireworks. Oh, and afterwards I had to lay new carpet.

Yup. You should have become an RF engineer. The world doesn't have enough of those.

To most people this stuff is voodoo. Which is mostly how I make a living. Want to slowly retire but the clients won't let me.

[...]
--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

Precisely! Indeed, it's so innate, it's in the math. You can't just take a complex polynomial expression and add, multiply and divide terms at will: there's a very strict algebra you must use, which corresponds to the application of components (of whatever impedance you happen to have, but they're generally one of only three things: constant resistance, inductive reactance proportional to frequency, or capacitive reactance inverse to frequency), in series and parallel, to the network in question.

So one also must take a stroll through analytical calculus, if one wishes to attempt solutions of component values, particularly in regards to phase matching (pesky arctangents, and all).

In a broader sense, it connects with the general Heisenberg uncertainty principle: one cannot have a dual-domain wave phenomenon (e.g., time and space; voltage and current; electric and magnetic; etc.), perfectly localized in both domains at the same time. There is a necessary and constant tradeoff.

The matter, when it comes to EMC, and filter design, and baluns and antennas and all that, is: there is energy. Energy is conserved (duh?). You can't use reactances to absorb energy (duh!), only push it around (sausage effect!). You must either divert it (often, the obvious conclusion is: reflection, back to the transmitter -- high SWR, and potential psst-BANG!), or absorb it (lossy ferrite beads, or a duplexer into something else, even if it's a termination resistor).

Send some my way ;-)

Tim

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

We've had the discussion before about products vs. consulting. You are still working 40 hours a week (or something like that) and I am working

40 hours a year. You may make somewhat more than I do, but my hourly rate is a bit higher, lol

My job literally involves sending emails for a few days to get a quote from the assembly house, then placing my order in response to the PO from my customer. Then some weeks later I pick up a box full of tested and bubble wrapped board to drop off at FedEx. This time it won't even be international. The new middle man (international assemblers who build my customer's systems) has their own middle man in Texas to receive orders which they than carry across the border. My markup is around 250%. Of course it took some time to get to this point and I have no idea of when the next order will come in. But I don't care a lot either.

Oh yeah, I have to get them to accept my terms and conditions which makes them indemnify me for any liability.

--

Rick
Reply to
rickman

Didn't you forget radiating it?

--

Rick
Reply to
rickman

On Mon, 03 Aug 2015 22:14:00 -0400, rickman Gave us:

Fire that pebble into the pond (at a perpendicular angle)and watch the little waves!

We used to drop a bowling ball into the river off the end of our dock. Early in the morning when the water was "like glass".

The barge mud (that's what we called it) was so thick, we always knew where to look for the ball afterward. It was in the mud!

Reply to
DecadentLinuxUserNumeroUno

Radiation is just loss. Well, 99.99994% of it. :^)

Tim

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

currents must always be zero, whether it's an >antenna system or a power system, per Kirchhoff's law.

There sure is in Kirchhoff--it's a low frequency approximation that neglects radiation and displacement current.

Cheers

Phil Hobbs

Reply to
Phil Hobbs

The asymmetric feed will also affect the radiation pattern.

Not a big deal for a single dipole at low HF, but a nasty thing with multielement antennas (Yagi etc.) on VHF/UHF, so you definite need some proper matching, such as gamma- or delta-match to have a decent radiation pattern.

Reply to
upsidedown

Actually I am down to four day weeks. Half a day goes to volunteering and the other half day to mountain biking and road biking.

I like the sizzle of always designing new stuff or re-designing existing stuff to make it work better. Once things are in production I kind of get bored.

I had a client this year where their new legal folks balked at that. So the company currently isn't a client :-)

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg
[...]

Or like Rick said, radiate it off. Dithering helps to get it under the limit. But that's not a nice way to do it.

Want to stay self-employed? In that case maybe flesh out the resume on Linkedin a bit. For example, if you have RF and regulatory experience (important for my kind of jobs) there would have to be some more detail about that so potential clients can see. They usually want to see the sort of projects someone did in that field and then familiarity with standards such as UL60601 and RTCA/DO-160.

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

But those loopholes get you into trouble like most other do. Radiation can cause a botched EMC test and displacement current either dioes loop back of the magnetic field couples into something else and then trouble brews.

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

If Kirchhoff were the whole story, van de Graaf generators wouldn't work, and neither would antennas.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Is displacement current not the field embodiment of Kirchoff itself?

For example, a dipole antenna's current flows are closed by the electric fields (carrying charge) that span the rods. That field, in turn, produces a closed loop magnetic field, which produces a closed loop electric field, which... At all points, the fields are either complete loops, or the electric fields terminate in a radiator (lacking magnetic monopoles, the converse is absent).

You'll have a hard time convincing people that current doesn't, in fact, flow through capacitors...

Tim

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

Phil has a point. Much of the particles and EM radiation directed into space will never loop back. So theoretically you can have a (large, as in whole system) node where the sum of all currents isn't zero.

[...]
--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

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