"Bal uhn" or "bayl uhn"?
 

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.

--

Rick

"Bal uhn" or "bayl uhn"?
 

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.




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.

I am beginning to think you may be right! Sorry. But it doesn't affect
the argument if there is an antenna, however symmetrical, connected to
both conductors.




This would actually be quite a simple lab experiment, at UHF or higher.
Compare the amount of RF on the outer with a bare surface mount 50ohm or
with one of the screened 50 ohm terminations (which does not allow any
signal to get to the outer). Or compare the SWR which will be near 1.0
with the screened load and might be very different with the unscreened,
sim,ply because the coax outer is shunting one side of the load.

Ok, can anyone do this?

It would be very interesting!


--
Roger Hayter

"Bal uhn" or "bayl uhn"?
 

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?


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.

I am beginning to think you may be right! Sorry. But it doesn't affect
the argument if there is an antenna, however symmetrical, connected to
both conductors.

Actually it does. With an antenna connected currents can flow from the
inner and shield conductors unequally. It will only be unequal if there
is an alternate path for the current. That alternate path will be the
outer surface of the shield.


This would actually be quite a simple lab experiment, at UHF or higher.
Compare the amount of RF on the outer with a bare surface mount 50ohm or
with one of the screened 50 ohm terminations (which does not allow any
signal to get to the outer). Or compare the SWR which will be near 1.0
with the screened load and might be very different with the unscreened,
sim,ply because the coax outer is shunting one side of the load.

Ok, can anyone do this?

It would be very interesting!

--

Rick

"Bal uhn" or "bayl uhn"?
 

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. This is a high inductance path. 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.






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.

I am beginning to think you may be right! Sorry. But it doesn't affect
the argument if there is an antenna, however symmetrical, connected to
both conductors.

Actually it does. With an antenna connected currents can flow from the
inner and shield conductors unequally. It will only be unequal if there
is an alternate path for the current. That alternate path will be the
outer surface of the shield.


This would actually be quite a simple lab experiment, at UHF or higher.
Compare the amount of RF on the outer with a bare surface mount 50ohm or
with one of the screened 50 ohm terminations (which does not allow any
signal to get to the outer). Or compare the SWR which will be near 1.0
with the screened load and might be very different with the unscreened,
sim,ply because the coax outer is shunting one side of the load.

Ok, can anyone do this?

It would be very interesting!


--
Roger Hayter

"Bal uhn" or "bayl uhn"?
 

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.


This would actually be quite a simple lab experiment, at UHF or higher.
Compare the amount of RF on the outer with a bare surface mount 50ohm or
with one of the screened 50 ohm terminations (which does not allow any
signal to get to the outer). Or compare the SWR which will be near 1.0
with the screened load and might be very different with the unscreened,
sim,ply because the coax outer is shunting one side of the load.

Ok, can anyone do this?

Anyone with appropriate test equipement.

--
Jim Pennino

"Bal uhn" or "bayl uhn"?
 

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.



--
Jim Pennino

"Bal uhn" or "bayl uhn"?
 

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.


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.


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.

--

Rick

"Bal uhn" or "bayl uhn"?
 

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?

--

Rick

"Bal uhn" or "bayl uhn"?
 

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.

--

Rick

"Bal uhn" or "bayl uhn"?
 

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?

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. 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

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

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.



--
Jim Pennino