At frequencies where skin effect is fully developed, and that is a
reasonable assumption for most practical coaxial cables at HF, the
current on the inside surfaace of the outer conductor is equal to but
opposite in direction to the current on the outside surface of the
inner conductor. This is TEM mode propagation.

At the end of the isolated outer conductor, this current must flow
somewhere, and it flows around the end onto the outside surface of the
outer conductor (effectively changing direction as it does so). So, at
that point, the current flowing on the outside of the outer conductor
is exactly equal to the current flowing on the outside of the inner
conductor.

Leaving aside the effects of changing Zo by substitution of coax for
plain conductors:

If you use two coax lines in parallel with the shields isolated, it
makes very little difference, the current that would have flowed on
the two plain conductors now flows on the outer of the coax lines. The
common mode current is the sum of the currents in both coax shields,
as it would be for plain conductors.

If you join the shields together at each end, the sheilds together now
carry the common mode current. A different equivalent circuit, but
almost the same outcome.


The last paragraph above is where I lose you..... when the shields
are joined together. Yes, I understand inside the shield RF
current flowing around the end and to the outer side.... HOWEVER, the
OTHER center conductor is inducing RF current flowing in the opposite
direction. Since both these inside currents are 'shorted' at the ends
of the two shields, I fail to see how you can have any current flowing
on the outer shield since the two opposite currents should cancel....
????


Ed

"Ed_G" wrote in
. 192.196:

of the two shields, I fail to see how you can have any current flowing
on the outer shield since the two opposite currents should cancel....
????

I already said that as I understand your variable configuration, it is most
unlikely that there is zero common mode current, or close to it.
Irrespective of the magnitude of the common mode current, the mechanism is
that the coax doesn't 'shield' it.

Owen

I already said that as I understand your variable configuration, it is
most unlikely that there is zero common mode current, or close to it.
Irrespective of the magnitude of the common mode current, the
mechanism is that the coax doesn't 'shield' it.


No, I wouldn't say "shield" would be a proper term, either. But I
would suggest that a "cancellation" similar to radiation in a balanced
feedline, would be pertinent.


Ed

"Ed_G" wrote in
. 192.196:



I already said that as I understand your variable configuration, it is
most unlikely that there is zero common mode current, or close to it.
Irrespective of the magnitude of the common mode current, the
mechanism is that the coax doesn't 'shield' it.


No, I wouldn't say "shield" would be a proper term, either. But I
would suggest that a "cancellation" similar to radiation in a balanced
feedline, would be pertinent.

Then I will leave you to your view that the system is balanced.

Owen

No, I wouldn't say "shield" would be a proper term, either. But I
would suggest that a "cancellation" similar to radiation in a balanced
feedline, would be pertinent.


Then I will leave you to your view that the system is balanced.

Owen



I would have preferred an explanation of your view on why it wouldn't
be, but I thank you for all your prior discussion.


Ed K7AAT

"Ed_G" wrote in
. 192.196:



No, I wouldn't say "shield" would be a proper term, either.
But I
would suggest that a "cancellation" similar to radiation in a
balanced feedline, would be pertinent.


Then I will leave you to your view that the system is balanced.

Owen



I would have preferred an explanation of your view on why it
wouldn't
be, but I thank you for all your prior discussion.

From my first post on the topic:

"If Ed connects parallel line from the centre of the dipole to the hot
and
common terminals of the ATU, there is likely to be common mode current on
the parallel line adjacent to the ATU. If the only connection on the tx
side of the ATU is the coax, then it will also have a common mode current
adjacent to the ATU and near enough to equal to the common mode current
on the other side of the ATU."

Sure, you can fabricate a parallel line from two coaxial lines, it just
has much more loss than a conventional air spaced line... and although it
is short, you intend operating it at extreme VSWR. The shielded twin line
you synthesise does not have any magic properties in supressing or
shielding feed line radiation.

Note that I am avoiding the term balanced line that some have used.
Balance is not forced by line geometry, but is a result of the
environment, so balanced lines are balanced by external factors, not the
line geometry.

Owen

PS: I wonder if you had considered end feeding the Inverted V with the
ATU. IIRC you had a sheet metal roof. You could just fix the tuner to the
roof, connect the ground terminal to the roof sheet, and take a wire from
the ATU output terminal to the end of the inverted V (which is a
continuous conductor across the apex). This is an unbalanced load
connected to an unbalanced output. Is that too easy?

Owen Duffy wrote in
:

"Ed_G" wrote in
. 192.196:



No, I wouldn't say "shield" would be a proper term, either.
But I
would suggest that a "cancellation" similar to radiation in a
balanced feedline, would be pertinent.


Then I will leave you to your view that the system is balanced.

Owen



I would have preferred an explanation of your view on why it
wouldn't
be, but I thank you for all your prior discussion.

From my first post on the topic:

"If Ed connects parallel line from the centre of the dipole to the hot
and
common terminals of the ATU, there is likely to be common mode current
on the parallel line adjacent to the ATU. If the only connection on
the tx side of the ATU is the coax, then it will also have a common
mode current adjacent to the ATU and near enough to equal to the
common mode current on the other side of the ATU."

Sure, you can fabricate a parallel line from two coaxial lines, it
just has much more loss than a conventional air spaced line... and
although it is short, you intend operating it at extreme VSWR. The
shielded twin line you synthesise does not have any magic properties
in supressing or shielding feed line radiation.

Note that I am avoiding the term balanced line that some have used.
Balance is not forced by line geometry, but is a result of the
environment, so balanced lines are balanced by external factors, not
the line geometry.

Owen

PS: I wonder if you had considered end feeding the Inverted V with the
ATU. IIRC you had a sheet metal roof. You could just fix the tuner to
the roof, connect the ground terminal to the roof sheet, and take a
wire from the ATU output terminal to the end of the inverted V (which
is a continuous conductor across the apex). This is an unbalanced load
connected to an unbalanced output. Is that too easy?




Owen, after reading Roy's explanation, I came back to this one
and what you have been saying is now more clear to me.

I will have to check what is available for end feed on the roof,
but it is a completely rubber covered roof and I doubt there is any
convinent way to access a ground. Tnx.

Ed