rickman wrote:
On 8/1/2015 4:52 PM, wrote:
rickman wrote:
Here is a more basic question. What are the assumptions to be able to
say the current on the shield inner layer equals the current in the
inner conductor of a coax? I'd be willing to bet I can construct a
circuit where this is not true.

Inside the transmission line the energy is carried in the electromagnetic
field between the conductors, not in the conductors.

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

Nothing external to the transmission line can chage this.

Um, do you want to answer the question about the assumptions required to
assume equal currents in the two conductors of a coax? If not that's
fine.

I did; there is no current in the conductors of a coax where those
conductors physically define a coaxial transmission line.

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

There is current in the conductors only after they no longer physically
define a coaxial transmission line.

Inside the coaxial structure everything is in the field and is balanced.


--
Jim Pennino

On 8/1/2015 8:24 PM, wrote:

There is no current in the shield inner surface, the energy is in the
ELECTROMAGNETIC FIELD between the inner and outer conductors. To be
nit pickingly precise, there is some small current in the inner
surface of the shield and the center wire, but for real coax that
surface current is insignificant.

this is a pretty amazing revelation. So what are the assumptions to
make this true?

If there is no appreciable current flow in the coax, then the resistance
of the wires is of no significance. Funny, when I make a loop antenna
from coax, the Q still seems to be limited by the conductor resistance.
Odd...


You can look at coax as a wave
guide if that makes it easier to understand, though the mode is
different than the mode in what is normally called wave guide.

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

I don't see anything here that says there is no current flow in the coax.


At the end of the coaxial structure, the electromagnetic field
becomes a current flow in any conductors connected to the end
of the coax.

Where do the electrons come from that the current consists of? Does the
wire end act as a capacitor?


One of those conductors is always the outside of the shield because
of the physical structure of coax.

The sum of the currents in the outside of the shield plus all other
conductors connected to the outside shield is equal to the sum of the
currents of all the conductors connected to the center wire.

That seems rather obvious.

--

Rick

On 8/1/2015 8:31 PM, wrote:
rickman wrote:
On 8/1/2015 4:52 PM, wrote:
rickman wrote:
Here is a more basic question. What are the assumptions to be able to
say the current on the shield inner layer equals the current in the
inner conductor of a coax? I'd be willing to bet I can construct a
circuit where this is not true.

Inside the transmission line the energy is carried in the electromagnetic
field between the conductors, not in the conductors.

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

Nothing external to the transmission line can chage this.

Um, do you want to answer the question about the assumptions required to
assume equal currents in the two conductors of a coax? If not that's
fine.

I did; there is no current in the conductors of a coax where those
conductors physically define a coaxial transmission line.

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

There is current in the conductors only after they no longer physically
define a coaxial transmission line.

Inside the coaxial structure everything is in the field and is balanced.

Do you have any references that actually say there is no current flow in
a coax? The reference given above doesn't even come close to saying this.

--

Rick

rickman wrote:

On 8/1/2015 4:31 PM, Roger Hayter wrote:
rickman wrote:

On 8/1/2015 3:29 PM, Roger Hayter wrote:
rickman wrote:

On 8/1/2015 1:38 PM, Roger Hayter wrote:

But your second point is unhelpful in some circumstances. For instance,
if the type of balun is the inductive coil of the feeder with or without
ferrites, then there simply *is no* current path down the outside of the
feeder from the junction of the balun and the feeder, Except from the
outer of the cable in the balun coil, and it is this that is decoupled
by the inductance.

Your description is not clear to me.

I am looking at the junction of the feeder with the balun, and the only
source of current on the outside of the feeder is connected by a very
high inductance to the source of signal at the antenna end.

How is the inner surface of the shield not connected to the outer
surface of the shield?


At the point the balun joins the feeder, the only way RF can get from
inside the coax braid (skin effect deep) to the outside of the coax
braid is to go all the way up to the antenna and down the outer surface
of the balun.

I think this is our first point of disagreement. There is nothing to to
stop the current flowing on the shield inside surface from moving to the
shield outside surface other than a tiny amount of resistance in the
shield wire. Unless the current flow sees a lower impedance path to
follow through the balun, it will travel back on the shield outside
surface.


At RF, the coax braid is an impenetrable Faraday screen. That is what
it is for, after all. The current simply can't go through the
fractional mm of copper from inside to outside.





This is a high inductance path.

I think you are applying this term without appreciating the full
meaning. It is a high impedance path for common mode currents, but a
low impedance path for differential currents. Since the current in the
shield inner surface balances the current on the center conductor, it is
a very low impedance path for the full current on the shield.

If it were accurate to say the balun was "a high impedance path" without
the qualifications, the balun would prevent the desired signal from
reaching the load.

Which is why I said it is a high impedance path for the current on the
outside of the screen. "Common mode" isn't really quite the right name
for this current, as the inner is totally uninvolved, but it is
certainly not differential, which satisfies your point.





This is easy to see if
the balun is continuous with the feeder, but even if it is not the join
is like the case with the matched load which I have now agreed there is
no 'spare' currrent to cross from the inside to the outside, as the
inside current has to match the inner conductor current.

I won't argue that any of this is correct. It does not conflict in any
way with what I have said.


--
Roger Hayter

rickman wrote:

On 8/1/2015 4:47 PM, wrote:
rickman wrote:
On 8/1/2015 1:38 PM, Roger Hayter wrote:

But your second point is unhelpful in some circumstances. For instance,
if the type of balun is the inductive coil of the feeder with or without
ferrites, then there simply *is no* current path down the outside of the
feeder from the junction of the balun and the feeder, Except from the
outer of the cable in the balun coil, and it is this that is decoupled
by the inductance.

Your description is not clear to me.


Secondly, even if you connect a resistor across the end ot the feeder,
consider that the inner conductor just goes to the resistor, but the
outer conductor sees the resistor and the outer side of the braid in
parallel. So you will get RF (and therefore some radiation) on the
outer of the coax even if you just connect a resistor across the end.

Ok, let's discuss this. You are describing a circuit that is just the
coax and a terminating resistor. You seem to be saying that current
will flow on the outer surface of the shield. If that were true, where
does it come from? In this simple circuit the current on the shield
inner surface matches the current on the inner conductor. So there is
no source for current to flow on the shield outer surface.

The inside and the outside of the shield are connected together at
the point where the resistor connects to them. The source of the
current is the electromagnetic field that propagates inside the coax.

As the shield is another current path, some current will flow down
it. How much depends on the length of the shield in wavelengths which
determines the impedance of that path.

I would like to clarify this point. You are saying that some of the
current that flow to the load on the shield inside surface will flow
back on the shield outside surface. That means the current in the inner
conductor will no longer equal the current in the shield inner surface,
right?

Just to interject a further argument; I think I agree with you on this.
You can't get coax outer current unless some current sink (eg an antenna
element) is connected to the inner conductor side of the load resistor.
But, consider a pefectly symmetrical dipole: if the potential on the
centre conductor and the braid is exactly the same, how can the two
antenna halves have different currents to allow some to flow down the
outside of the braid?

I am beginning to wonder if braid outside current with a symmetrical
antenna *only* occurs when the coax outer interacts with the EM field of
the antenna so as to actually alter the impedance of at least one of the
antenna elements, or alter the two of them to a different extent, so
that a common mode current is "left over" for the braid. It does seem
likely that a long wire coming from the centre of the dipoe and not
being absolutely symmetrical would have this effect. However, on this
argument, you would not need a balun if your feeder was absolutely
symmetrical! This theory seems eminently testable with antenna
simulation programs.

--
Roger Hayter

Jeff wrote:

No, there will be no energy "supplied from the feeder and conducted down
the coax outer" as it will all be conducted into the antenna and
radiated if the antenna is perfectly balanced. The power flowing up the
coax, when perfectly matched, will only 'see' the antenna load, not the
coax outer as and element in parallel.

Jeff

If it was perfectly balanced before you connected the feeder, it will no
longer be perfectly balanced once the shunt impedance to earth of the
coax braid is connected to *one* side of it. Dare I say that that is
why you need a balun? This asymmetrical shunt impedance is the source
of the whole question.


So if what you are saying is correct then terminating a piece of coax
with a 50 ohm resistor will not cause a perfect match and the load will
not be 50 ohms due to the shunt impedance to earth of the coax braid.

I think not.

Jeff

I agree I was probably wrong, and you are probably correct.

--
Roger Hayter

On 8/2/2015 4:37 AM, Roger Hayter wrote:
rickman wrote:

On 8/1/2015 4:31 PM, Roger Hayter wrote:
rickman wrote:

On 8/1/2015 3:29 PM, Roger Hayter wrote:
rickman wrote:

On 8/1/2015 1:38 PM, Roger Hayter wrote:

But your second point is unhelpful in some circumstances. For instance,
if the type of balun is the inductive coil of the feeder with or without
ferrites, then there simply *is no* current path down the outside of the
feeder from the junction of the balun and the feeder, Except from the
outer of the cable in the balun coil, and it is this that is decoupled
by the inductance.

Your description is not clear to me.

I am looking at the junction of the feeder with the balun, and the only
source of current on the outside of the feeder is connected by a very
high inductance to the source of signal at the antenna end.

How is the inner surface of the shield not connected to the outer
surface of the shield?


At the point the balun joins the feeder, the only way RF can get from
inside the coax braid (skin effect deep) to the outside of the coax
braid is to go all the way up to the antenna and down the outer surface
of the balun.

I think this is our first point of disagreement. There is nothing to to
stop the current flowing on the shield inside surface from moving to the
shield outside surface other than a tiny amount of resistance in the
shield wire. Unless the current flow sees a lower impedance path to
follow through the balun, it will travel back on the shield outside
surface.


At RF, the coax braid is an impenetrable Faraday screen. That is what
it is for, after all. The current simply can't go through the
fractional mm of copper from inside to outside.

This is one of those cases where you have forgotten the details and
underlying premises. A Faraday cage can't stop currents from flowing
through the "cage". It stops fields from penetrating the cage by action
of the resulting currents in the cage. Your car is largely a Faraday
cage but you can still be electrocuted if a live wire is in contact with
the chassis and you touch it while inside. There can be a high
potential across different parts of the car from the current and that
current can pass through you if you touch the cage.


This is a high inductance path.

I think you are applying this term without appreciating the full
meaning. It is a high impedance path for common mode currents, but a
low impedance path for differential currents. Since the current in the
shield inner surface balances the current on the center conductor, it is
a very low impedance path for the full current on the shield.

If it were accurate to say the balun was "a high impedance path" without
the qualifications, the balun would prevent the desired signal from
reaching the load.

Which is why I said it is a high impedance path for the current on the
outside of the screen. "Common mode" isn't really quite the right name
for this current, as the inner is totally uninvolved, but it is
certainly not differential, which satisfies your point.

Actually, I quoted you accurately above. You omitted the qualifier and
that was my point.

--

Rick

On 8/2/2015 4:37 AM, Roger Hayter wrote:
rickman wrote:

On 8/1/2015 4:47 PM, wrote:
rickman wrote:
On 8/1/2015 1:38 PM, Roger Hayter wrote:

But your second point is unhelpful in some circumstances. For instance,
if the type of balun is the inductive coil of the feeder with or without
ferrites, then there simply *is no* current path down the outside of the
feeder from the junction of the balun and the feeder, Except from the
outer of the cable in the balun coil, and it is this that is decoupled
by the inductance.

Your description is not clear to me.


Secondly, even if you connect a resistor across the end ot the feeder,
consider that the inner conductor just goes to the resistor, but the
outer conductor sees the resistor and the outer side of the braid in
parallel. So you will get RF (and therefore some radiation) on the
outer of the coax even if you just connect a resistor across the end.

Ok, let's discuss this. You are describing a circuit that is just the
coax and a terminating resistor. You seem to be saying that current
will flow on the outer surface of the shield. If that were true, where
does it come from? In this simple circuit the current on the shield
inner surface matches the current on the inner conductor. So there is
no source for current to flow on the shield outer surface.

The inside and the outside of the shield are connected together at
the point where the resistor connects to them. The source of the
current is the electromagnetic field that propagates inside the coax.

As the shield is another current path, some current will flow down
it. How much depends on the length of the shield in wavelengths which
determines the impedance of that path.

I would like to clarify this point. You are saying that some of the
current that flow to the load on the shield inside surface will flow
back on the shield outside surface. That means the current in the inner
conductor will no longer equal the current in the shield inner surface,
right?

Just to interject a further argument; I think I agree with you on this.
You can't get coax outer current unless some current sink (eg an antenna
element) is connected to the inner conductor side of the load resistor.
But, consider a pefectly symmetrical dipole: if the potential on the
centre conductor and the braid is exactly the same, how can the two
antenna halves have different currents to allow some to flow down the
outside of the braid?

This has been explained previously. A dipole is not balanced when it is
connected to the coax. The shield outer surface presents a third
element which makes the shield side of the dipole different from the
center conductor side. In the case of the resistor the current flowing
in one side must flow out the other, so it is balanced no matter what.
The dipole has no such requirement. If you restrict the current running
into one side and not the other it can do nothing about it.


I am beginning to wonder if braid outside current with a symmetrical
antenna *only* occurs when the coax outer interacts with the EM field of
the antenna so as to actually alter the impedance of at least one of the
antenna elements, or alter the two of them to a different extent, so
that a common mode current is "left over" for the braid. It does seem
likely that a long wire coming from the centre of the dipoe and not
being absolutely symmetrical would have this effect. However, on this
argument, you would not need a balun if your feeder was absolutely
symmetrical! This theory seems eminently testable with antenna
simulation programs.

You seem to be thinking the antenna elements present some sort of
current sink that *must* extract some amount of current no matter what.
They are just loads no different from the shield outer surface. The
current will flow according to the impedance seen by the current. There
certainly will be some interaction, but if you use a triaxial cable and
only connect the outer shield to ground at the transmitter, the antenna
won't "see" the inner shield and the current will still flow on the
outside of that shield.

The symmetrical feeder argument does not make sense to me. Why wouldn't
the impact on both antenna elements be identical? I haven't heard
anyone say you *do* need a balun if the feed line is symmetrical. That
would be a balbal transformer.

--

Rick

On 8/2/2015 4:20 AM, Jeff wrote:

No, there will be no energy "supplied from the feeder and conducted down
the coax outer" as it will all be conducted into the antenna and
radiated if the antenna is perfectly balanced. The power flowing up the
coax, when perfectly matched, will only 'see' the antenna load, not the
coax outer as and element in parallel.

Jeff

If it was perfectly balanced before you connected the feeder, it will no
longer be perfectly balanced once the shunt impedance to earth of the
coax braid is connected to *one* side of it. Dare I say that that is
why you need a balun? This asymmetrical shunt impedance is the source
of the whole question.


So if what you are saying is correct then terminating a piece of coax
with a 50 ohm resistor will not cause a perfect match and the load will
not be 50 ohms due to the shunt impedance to earth of the coax braid.

I think not.

Maybe you can draw a diagram and show yourself where the current flows.
The earth connection on the shield outside is in parallel with the
earth connection on the shield inside. You can't have any current flow
on the outside because any current flowing on the shield inside balances
with the current flowing in the center conductor. The resistor requires
the two currents are equal with none left to travel down the shield
outside.

Although I am still waiting for jim to explain his idea that there is no
current flow in the coax. If that is true it will make this a *very*
simple picture.

--

Rick

rickman wrote:
On 8/1/2015 8:24 PM, wrote:

There is no current in the shield inner surface, the energy is in the
ELECTROMAGNETIC FIELD between the inner and outer conductors. To be
nit pickingly precise, there is some small current in the inner
surface of the shield and the center wire, but for real coax that
surface current is insignificant.

this is a pretty amazing revelation. So what are the assumptions to
make this true?

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

Do you think there is significant current in the walls of a wave guide?

If there is no appreciable current flow in the coax, then the resistance
of the wires is of no significance. Funny, when I make a loop antenna
from coax, the Q still seems to be limited by the conductor resistance.
Odd...

Not at all, it is just that you don't understand that in that case it
is not a transmission line.

You can look at coax as a wave
guide if that makes it easier to understand, though the mode is
different than the mode in what is normally called wave guide.

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

I don't see anything here that says there is no current flow in the coax.

What it says is where the energy IS, not where it is NOT.

If you want to understand the effect of ohmic resistance, read
and understand the differential equations described he

https://en.wikipedia.org/wiki/Transm....27s_equations

At the end of the coaxial structure, the electromagnetic field
becomes a current flow in any conductors connected to the end
of the coax.

Where do the electrons come from that the current consists of? Does the
wire end act as a capacitor?

Are you really asking this?

Where do the electrons come from at the terminals of a receiving antenna?


One of those conductors is always the outside of the shield because
of the physical structure of coax.

The sum of the currents in the outside of the shield plus all other
conductors connected to the outside shield is equal to the sum of the
currents of all the conductors connected to the center wire.

That seems rather obvious.

Then why do you keep asking about it?

--
Jim Pennino