Microwave TL coax impedance

[nix85]

@radiolistener Tnx, White Croat brother.

[nix85]

I got another question about coax cable common mode.

https://en.wikipedia.org/wiki/Coaxial_cable#Common_mode_current_and_radiation

I'd like to know why..

The current formed by the field between the antenna and the coax shield would flow in the same direction as the current in the center conductor, and thus not be canceled.

I mean if shield is the return path, in it current flows opposite to inner conductor, why does this unwanted electric field formed between shield and antenna flow in the SAME direction as inner conductor?

[nix85]

Also, does current alternate on the shield or not, according to wikipedia, it does and inner and outer fields cancel out just like in ladder line.

From coax wiki page above:

Most of the shield effect in coax results from opposing currents in the center conductor and shield creating opposite magnetic fields that cancel, and thus do not radiate. The same effect helps ladder line.

However this guy (and others) says there is no oscillation on the shield.

https://youtu.be/GMeOMwf2DJU?t=120

Obviously this is contradictory and confusing, may someone clear this up.

Or does it simply depend on how coax is connected, that is, is shield grounded or is it too connected to the oscillator circuit?

[vk6zgo]

At every other point of the cable, apart from the halfwave points, the impedance looking towards the load is different than 14 Ohms.

You mean 140 Ohm. And so what? I mean that's the point to terminate it exactly at halfwave point to get that load impedance repeated. Are you saying other factors will make those points not exactly where one would expect them to be so this kind of matching is not practical or?

You, in fact, wrote 14OHm, which it would be reasonable to read as 14

The half wavelength points will be as you assumed, & the load impedance will appear at the cable input.

If you read the rest of my posting, you would have realised that, at various points, along the transmission line, the relationships between current through, & voltage across the cable will not be the same as at its input or output.

Your antenna & Transmitter may be happy, although you still need to match 50 to 140 , but remember, the transmission line exists at all these other lengths in between.

At some point, the current will be high, causing I^2R losses, & at others, the voltage will be high, with dielectric losses & the risk of breakdown.,

Losses in a transmission line are specified when correctly matched.
In other situations, all bets are off!

You seem to be ready to argue with every point brought up in response to your OP.
If you feel you already "know it all", why post the query in the first place?

First of all, i have no intention to argue nor i feel "know it all", i came here FOR ANSWER. It's just that people often answer without thinking through or really knowing.

Again, helical antenna radiation resistance at resonance is ~140 Ohm, i don't know where you saw 14.

"If you read the rest of my posting, you would have realised that, at various points, along the transmission line, the relationships between current through, & voltage across the cable will not be the same as at its input or output."

You keep bringing that irrelevant and obvious fact up although it is assumed cable is rated for given current/voltage.

I did not say i will just make cable multiple of halfwavelength, i said i will do it as additional measure to impedance matching.

Do whatever you like, I give up!

[nix85]

Do whatever you like, I give up!

Again, you try to make me look as if i am arguing with you while everyone sees i am respectful and just trying to learn.

[ejeffrey]

Also, does current alternate on the shield or not, according to wikipedia, it does and inner and outer fields cancel out just like in ladder line.

From coax wiki page above:

Most of the shield effect in coax results from opposing currents in the center conductor and shield creating opposite magnetic fields that cancel, and thus do not radiate. The same effect helps ladder line.

However this guy (and others) says there is no oscillation on the shield.

https://youtu.be/GMeOMwf2DJU?t=120

Obviously this is contradictory and confusing, may someone clear this up.

Or does it simply depend on how coax is connected, that is, is shield grounded or is it too connected to the oscillator circuit?

I think you are misinterpreting that video.  Or possibly the video author intends your interpretation, but in that case it is wrong.

Coax absolutely has oscillating charge and current waveforms on both the center conductor and the shield.  The point is that those currents and charges, combined with the currents and charges on the center conductor result in zero field outside the shield and therefore zero voltage on the shield.

In the balanced line configuration the voltage on the lines is equal and opposite.  While the fields fall off rapidly with distance, they are not exactly zero.

[T3sl4co1l]

The shield is, well, a shield.  It supports two independent current flows: currents inside the line (the image current of the signal line's current flow), and currents outside the shield (image current carried by surroundings, or free space).

(When that current reflects a current in free space, that's also known as an antenna.)

I think it makes more sense when described in this way.  It is equivalent to talking about directions of current flow, but that may be confusing to the beginner who only sees alternating currents everywhere -- the sign (or more generally, the phase) is easy to get confused about.

For further reading:look up the image current, displacement current, path of least impedance, stuff like that.

Tim

[radiolistener]

coax cable has low radiation loss due to skin effect in the braid conductor. High frequency current flows mainly at skin surface. In case of coax cable it means that RF current flows on the inner side of the braid surface. So, RF current cannot go from the inner side of braid to the outer side of braid due to skin effect. This is why bad braid quality leads to higher radiation loss for coax cable.

Actually RF energy flows as electromagnetic wave inside insulator (between center conductor and braid). Conductors here just keep this wave inside insulator.

The same thing happens with symmetrical transmission line. The RF energy flows as electromagnetic wave inside insulator between two wires. But since there is no shield, the part of this electromagnetic wave flying away. This is why it has radiation loss.

[nix85]

I don't think i am misinterpreting the video, nor i think he is referring to cancelation of voltage on the shield, you can hear in the vid he makes it clear there is no current oscillating on the outer shield 'cause it is "just ground".

If that is how coax is connected THEN HE IS RIGHT, except for the secondary induced currents due to transformer effect.

If however both inner conductor and shield are connected to the balanced source then HE IS NOT RIGHT.

So, like i assumed, it depends to which type of source coax is connected.

Yea, i know about skin effect, that is, energy flowing around the conductors at higher frequencies as EM wave (poynting vector) as described by Oliver Heaviside.

As for image current, i assume that's the 1:1 transformer effect between inner and outer conductor which is secondary effect and not really an answer to my question if shield carries oscillating current fed directly from the signal source.

I also know what displacement current is.

[radiolistener]

not really an answer to my question if shield carries oscillating current fed directly from the signal source.

Shield don't carries oscillating current and center conductor don't carries it.
Insulator carries it in the form of EM waves

When EM wave falls to a conductor surface, it makes oscillating current and this oscillating current emits back EM wave. So, conductor works like mirror for the light and keeps EM wave inside insulator of the coax cable by reflecting it back when it trying to fly away. In such way EM waves traveling through coax cable in it's insulator. Current oscillations on conductor surface is just an effect of traveling EM wave in the insulator

So, technically, current oscillations are present on both - on center conductor surface and on inner surface of the shield. But they are carried by EM wave in the insulator

The signal source just initiating this process by making first current oscillation, which leads to emit EM wave. And then coax cable carries this EM wave in the insulator between conductors. When you consume energy of current oscillations at the end of cable, this energy will not be emitted as EM wave again, because it was consumed by load. And the path of RF energy will ends. If you will consume just a part of energy of these current oscillations, the rest will be emitted as EM wave and will travel back to the source.

The average speed of electrons in the conductor is very-very small, they cannot carry current oscillations with a speed of light. EM wave doing it.

[ejeffrey]

I don't think i am misinterpreting the video, nor i think he is referring to cancelation of voltage on the shield, you can hear in the vid he makes it clear there is no current oscillating on the outer shield 'cause it is "just ground".

That is wrong, and I still think you are misinterpreting the video.  There is charge and current oscillation on the shield.  At the ends of the transmission line those currents flow into the ground connection at either end.  The electrons don't care whether they are "driven by the source" or "induced by transformer effect".  Those are distinctions that only exist in your mind.  That is like saying the neutral wire in your electrical supply doesn't carry current because it is just ground.

So, like i assumed, it depends to which type of source coax is connected.

If you already "know" the answers and are going to ignore the answers, please don't bother asking.

[vk6zgo]

I don't think i am misinterpreting the video, nor i think he is referring to cancelation of voltage on the shield, you can hear in the vid he makes it clear there is no current oscillating on the outer shield 'cause it is "just ground".

That is wrong, and I still think you are misinterpreting the video.  There is charge and current oscillation on the shield.  At the ends of the transmission line those currents flow into the ground connection at either end.  The electrons don't care whether they are "driven by the source" or "induced by transformer effect".  Those are distinctions that only exist in your mind.  That is like saying the neutral wire in your electrical supply doesn't carry current because it is just ground.

So, like i assumed, it depends to which type of source coax is connected.

If you already "know" the answers and are going to ignore the answers, please don't bother asking.

I rest my case!

[nix85]

not really an answer to my question if shield carries oscillating current fed directly from the signal source.

Shield don't carries oscillating current and center conductor don't carries it.
Insulator carries it in the form of EM waves

When EM wave falls to a conductor surface, it makes oscillating current and this oscillating current emits back EM wave. So, conductor works like mirror for the light and keeps EM wave inside insulator of the coax cable by reflecting it back when it trying to fly away. In such way EM waves traveling through coax cable in it's insulator. Current oscillations on conductor surface is just an effect of traveling EM wave in the insulator

So, technically, current oscillations are present on both - on center conductor surface and on inner surface of the shield. But they are carried by EM wave in the insulator

The signal source just initiating this process by making first current oscillation, which leads to emit EM wave. And then coax cable carries this EM wave in the insulator between conductors. When you consume energy of current oscillations at the end of cable, this energy will not be emitted as EM wave again, because it was consumed by load. And the path of RF energy will ends. If you will consume just a part of energy of these current oscillations, the rest will be emitted as EM wave and will travel back to the source.

The average speed of electrons in the conductor is very-very small, they cannot carry current oscillations with a speed of light. EM wave doing it.

Ok, sure it travels outside of conductor due to skin effect, we got it but you didn't address the difference if both inner conductor and shield are connected to oscillating circuit aka balanced feed, or if just inner conductor is fed oscillating signal and shield is connected only to ground.

[nix85]

I don't think i am misinterpreting the video, nor i think he is referring to cancelation of voltage on the shield, you can hear in the vid he makes it clear there is no current oscillating on the outer shield 'cause it is "just ground".

That is wrong, and I still think you are misinterpreting the video.  There is charge and current oscillation on the shield.  At the ends of the transmission line those currents flow into the ground connection at either end.  The electrons don't care whether they are "driven by the source" or "induced by transformer effect".  Those are distinctions that only exist in your mind.  That is like saying the neutral wire in your electrical supply doesn't carry current because it is just ground.

You are wrong, i am not misinterpreting, listen again to what he says in the vid "all the going back n' forth is being done JUST by center conductor".

"The electrons don't care whether they are "driven by the source" or "induced by transformer effect"." Hah, there you go, you didn't get my question at all cause that is exactly what i was pointing to, not if current forms on the shield as a transformer effect but if shield is DIRECTLY driven on not. And the answer is, of course, it can be, but as an unbalanced line it is usually not.

If you already "know" the answers and are going to ignore the answers, please don't bother asking.

Please, from now on, if you don't really know the answer, refrain from answering.

[nix85]

I rest my case!

You had no case. Chill.

[EEVblog]

If you already "know" the answers and are going to ignore the answers, please don't bother asking.

Please, from now on, if you don't really know the answer, refrain from answering.

That's not how forum's work. You don't get to tell people to stop providing their answers or opinions.
You won't make friends here with that sort of approach.

[EEVblog]

I been researching how to impedance match microwave antenna and i am not sure about (characteristic) impedance of coax TL. Namely, we all know CI is not length dependent as all units determining it are per unit length and thus cancel out, but what about frequency?

Yesterday i stumbled upon this article from IetLabs and they clearly state:

Although it can be represented in terms of inductors, capacitors and resistors, characteristic impedance is a complex number that is highly dependent on the frequency of the applied signal. Zo is not a function of the cable length. At high frequencies (> 100kHz), the characteristic impedance is almost purely resistive. At mid-range frequencies (1kHz), Zo is affected by capacitance (ωC) and at low frequencies (DC – 100Hz), Zo is influenced by conductance (G). Refer to Figure 2.

https://www.ietlabs.com/pdf/application_notes/5-Characteristic%20Cable%20Impedance-Digibridge.pdf

What does this mean, that coax at 100KHz+ behaves purely like a resistor? If so is this because reactance of the coax is mainly due to capacitive reactance which drops with frequency?

Coax's are designed to act as transmission lines with a constant impedance across a wide bandwidth, that's their job.

[radiolistener]

What does this mean, that coax at 100KHz+ behaves purely like a resistor? If so is this because reactance of the coax is mainly due to capacitive reactance which drops with frequency?

Technicaly thats  not correct. Coax cable is not resistor, this is transmissionline. Resistor consumes energy and transforms it into heat. Coax cable transfer energy, it doesn't consume it.

From source feed point side coax line is really looks like resistor equivalent, but this is just for some limited period of time. When EM wave in coax cable runs to the end of cable and will not be consumed, it will be reflected back and when it runs back to the source the things will be changed. The source will see that moment like coax line input resistance was suddenly changed. The change will depends on the load impedance at the second end of the coax cable.

If load impedance on the second end of cax line will be equal to coax line impedance, all energy will be consumed by load and reflection will not happens. So the source will continue to see the same resistor equivalent on the coax cable input. But when the load impedance will not be match with coax line impedance, a part of EM wave will be reflected back and the source will see the change of input impedance at coax cable input. It will depends on frequency and cable length.

[nix85]

Coax cable transfer energy, it doesn't consume it.

Technicaly thats not correct : ) cause there will always be losses.

[nix85]

Just to add, i did not misinterpret that video, guy is clearly showing no current on the shield and this is of course wrong as everyone here agrees. And i know electrons hardly move, that energy is carried by EM wave, poynting vector, but that is beyond the scope of this thread.

Some food for thought... Ask yourself how does flux running through the core of a transformer induce voltage in the secondary altho flux is totally confined in the core and coil is outside of it, no secondary wire cuts a single line of flux.

There is much for you to learn, to expand upon.

[TheUnnamedNewbie]

Some food for thought... Ask yourself how does flux running through the core of a transformer induce voltage in the secondary altho flux is totally confined in the core and coil is outside of it, no secondary wire cuts a single line of flux.

Some food for thought.... Ask yourself why you feel the need to phrase this as some riddle you solved and the rest of the world hasn't, even though this was all solved in the 1800's.

[nix85]

Some food for thought.... Ask yourself why you feel the need to phrase this as some riddle you solved and the rest of the world hasn't, even though this was all solved in the 1800's.

Since you think it was solved in 1800's, explain how is voltage induced in the secondary altho not a single line of flux cuts a single wire.

[Kalvin]

Since you think it was solved in 1800's, explain how is voltage induced in the secondary altho not a single line of flux cuts a single wire.

The operation principle of a transformer can be explained using Lentz's law and Faraday's law. So, you may be wrong stating "not a single line of flux cuts a single wire".

Like others have already written, you can measure transmission line impedance vs frequency with a network analyzer, for example. At very low frequencies (like below 100 kHz or so), a transmission line has frequency dependent impedance, but at higher frequencies (1 MHz or so) the characteristic impedance remains pretty constant although the attenuation increase with the frequency.

[T3sl4co1l]

Lines of flux aren't a terribly great explanatory tool.  They lead to confusion with the spinning magnet-disk problem (doesn't matter if the magnet spins, it's radially symmetric -- but you might be tempted to think the lines are bonded to it like a brush), and the transformer just looks weird.

Consider the de-energized transformer.  Where are the flux lines?  There aren't any inside.  But there must be flux lines somewhere, if we imagine them to be some kind of conserved physical material.  The answer is: they are at infinity.  In infinite supply, unreachably far away, so their presence there can't possibly matter otherwise.  When the transformer is energized, lines of flux are brought in, from infinity, into the local loop -- thus cutting wires and inducing voltage, as the story goes.

Far better to just stick with enclosed area and field intensity...

Tim

[nix85]

Since you think it was solved in 1800's, explain how is voltage induced in the secondary altho not a single line of flux cuts a single wire.

The operation principle of a transformer can be explained using Lentz's law and Faraday's law. So, you may be wrong stating "not a single line of flux cuts a single wire".

False and false.