Bruce in alaska wrote in news:fast-25B5D6.09195501032008
@netnews.worldnet.att.net:

....
Ed, you would be Much Better Off, if they would allow you to use (2)
runs of Coax, side by side up the mast, and connect only the Center
Conductor of each, to the tuner, with the shield left open on each end
and sealed against water intrusion.

What does this do, what does it achieve?

Owen

Ed, you would be Much Better Off, if they would allow you to use (2)
runs of Coax, side by side up the mast, and connect only the Center
Conductor of each, to the tuner, with the shield left open on each
end and sealed against water intrusion.

What does this do, what does it achieve?

Owen



My expertise is weak in this area, but just guessing.... using twin
coax in the above configuration, if the shields were grounded, would
allow the feedling between the antenna coupler and the feedpoint to be
'balanced' and yet the shields would not radiate as they would with a
single coax run.

Perhaps others, here, will either expand on this, or correct my
misconception.

Ed K7AAT

"Ed_G" wrote in
.91:


Ed, you would be Much Better Off, if they would allow you to use (2)
runs of Coax, side by side up the mast, and connect only the Center
Conductor of each, to the tuner, with the shield left open on each
end and sealed against water intrusion.

What does this do, what does it achieve?

Owen



My expertise is weak in this area, but just guessing.... using
twin
coax in the above configuration, if the shields were grounded, would

But that is not what was said. I read it to say "with the shield left
open on each end".

allow the feedling between the antenna coupler and the feedpoint to be
'balanced' and yet the shields would not radiate as they would with a
single coax run.

Bunk. The only reason the shields would not radiate would be if they
carried equal but opposite currents. That is most unlikely in this case.

Let us just consider a simple example. Assumption is that skin effect on
the coax is fully effective, a reasonable assumption at HF.

Make a quarter wave vertical of a piece of RG213 supported on sky hooks.
Make no connection to the shield at either end, and connect the feed line
between a ground plane / counterpoise / whatever and the centre conductor
of the vertical piece of coax.

What current flows on the outside surface of the vertical coax?

The current on the outside surface of the vertical coax adjacent to the
bottom end of the isolated shield is the same as the current flowing on
the inner conductor adjacent to the same end of the shield.

Does the outer conductor 'shield' the vertical so that it will not
radiate?

No, the outside surface of the shield is the radiatior, it just has a
quarter wave o/c stub in series from the feedline to the radiating
element.



Perhaps others, here, will either expand on this, or correct my
misconception.


Owen

Owen Duffy wrote:
Does the outer conductor 'shield' the vertical so that it will not
radiate?

No, the outside surface of the shield is the radiatior, it just has a
quarter wave o/c stub in series from the feedline to the radiating
element.

Owen, feedline radiation in this particular system might
increase the efficiency which would be a good thing.
--
73, Cecil http://www.w5dxp.com

Bunk. The only reason the shields would not radiate would be if they
carried equal but opposite currents. That is most unlikely in this
case.



If both shields, ( ungrounded ) are tied together, and the two
center conductors are acting as a 'balanced' feedline, how can current
flow on the outsides of the shields, if the interior currents of the two
center conductors are always 180 out of phase?


Ed

"Ed_G" wrote in
. 192.196:


Bunk. The only reason the shields would not radiate would be if they
carried equal but opposite currents. That is most unlikely in this
case.



If both shields, ( ungrounded ) are tied together, and the two

At both ends?

This is the first mention of shields tied together, I certainly didn't
read that into Bruce's "with the shield left open on each end
".

center conductors are acting as a 'balanced' feedline, how can
current flow on the outsides of the shields, if the interior currents
of the two center conductors are always 180 out of phase?

The analysis in this case is different, but if I understand your
scenario, the outer surface of the two coaxes which are tied together at
both ends but connected to nothing else still carries the common mode
current. that exists on the two open wire conductors just prior to
entering the coax assembly. No, you cannot guarantee that those currents
are equal and opposite, ie that there is no common mode current, and the
common mode current will flow entirely on the outside surface of the
outer conductors of the coax assembly when connected as you now propose.

Bruce hasn't explained what his configuration is supposed to do, so we
are still guessing about that one.

There is no answer to this problem, because the problem is ill defined.
You have just added a new element in tying the shields together. Other
questions exist like what other connections exist between tx feed line,
ATU, ant feed line, mast, roofing / rain gutters, any other conductors.

Somethimes knowing how to describe a problem is knowing the answer to the
problem... or conversely, not knowning the answer is the result of not
knowing how to describe the problem.

Owen

If both shields, ( ungrounded ) are tied together, and the two

At both ends?

This is the first mention of shields tied together, I certainly didn't
read that into Bruce's "with the shield left open on each end
".


That's because I added that to the mix. Bruce's comment was a
suggestion for me. I have not done this yet, but my original post
under this thread was solliciting comments on using twin coax to feed a
balanced antenna, or using a single coax feed under the specific set of
circumstances I outlined.


About the common mode current.... please explain how this would be an
issue with the outer shields of two coaxes, shields tied together but
going nowhere ( no ground ), and the balanced antenna fed by the two
center conductors. I do not see common mode current being a factor, but
I'm willing to listen and learn.


Ed

"Ed_G" wrote in
. 192.196:

That's because I added that to the mix. Bruce's comment was a

Ok, well here is a model to shape your thinking and moving the goal
posts.

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.

Most of these 'shielded solutions' arise from a lack of understanding of
how the coaxial transmission line works in TEM mode.

For example, I saw an ham advise someone that station ground connections
were subject to noise pickup and the best improvement he could make was
to shield the ground lead. In his case, his shack was on the first floor
of the building, and his 7m vertical ground lead to the earth stakes etc
was a source of noise, so he used 7m of RG213 with the shield and inner
bonded to the earth stake and the shield left isolated at the top end.

Firstly, this is not a 'shield' at radio frequencies, but what he did
achieve was to insert a s/c stub in series with his station ground
conductor. The impedance of that series stub at 7.1MHz is 3056.20-
j1509.30 ohms... not a good outcome.

It might have 'fixed' his RF feeback problem, but it didn't improve the
station earth at all, it degraded it severely.

Owen

Ed_G wrote:

If both shields, ( ungrounded ) are tied together, and the two center conductors are acting as a 'balanced' feedline, how can current
flow on the outsides of the shields, if the interior currents of the two
center conductors are always 180 out of phase?

The answer is that making the feedline physically symmetrical doesn't
make it "act as a balanced feedline". A feedline is balanced and not
radiating only when the common mode current is zero, i.e., the currents
on the two conductors are equal in magnitude and opposite in phase.
Making a line physically symmetrical doesn't guarantee or cause this.
Nor, for that matter, does making a line physically asymmetrical (e.g.,
coax) necessarily cause a line to become unbalance. There's more about
this at http://eznec.com/Amateur/Articles/Baluns.pdf.

Roy Lewallen, W7EL

Perhaps a little amplification of what Owen has said will help clarify
the situation.

Suppose we have a twinlead transmission line, one conductor of which is
carrying a current of 2 amps at one point, and the other 3 amps at the
same point. For simplicity, we'll assume that the currents are exactly
out of phase. The common mode current (as we'll define it*) is 3 - 2 = 1
amp. (We can directly subtract them due to the assumption that they're
exactly out of phase; otherwise we'd have to do a vector addition.) So
the line will radiate exactly as though there was a single conductor
carrying one amp. This is an unbalanced, radiating feedline.

Now let's replace the line with a coax line of the same impedance so it
doesn't otherwise alter the system. What we'll find is that the center
conductor will carry 2 amps. The inner surface of the outer conductor,
which is always forced to be equal and opposite, carries 2 amps of
opposite polarity, that is, 2 amps going exactly the opposite direction.
On the outside of the shield is one amp, our common mode current. The
inner and outer shield currents combine at the cable ends to become 3
amps. This line will also radiate just like a single conductor carrying
one amp.

Finally let's look what happens when we use two coax lines with the
shields connected but floating. Suppose the 2 amps is on the center of
coax A and 3 amps on the center of coax B. On the inside of the coax A
shield is 2 amps flowing one way (the direction opposite the current on
the center conductor). On the inside of the coax B shield is 3 amps,
flowing the other way. What happens at the ends of the shield? At each
end, the 2 amps flowing one way will add to the 3 amps the other way
(since they're connected at the ends so there's a path from one to the
other), resulting in a 1 amp current which flows down the outside of the
shield. This radiates just the same as the others, like a one amp
current flowing on a single conductor. Using dual coax has accomplished
nothing.

The way to prevent the feedline, whatever the type, from radiating, is
to force the currents on the two conductors to be equal and opposite.
This can be done by making both the antenna and the tuner symmetrical,
in which case any of the three lines will be balanced and not radiate.
Another way is to use one or more common mode chokes (current baluns)
which will also balance any of the three line types. But just changing
from one type of line to another doesn't do it.

I've simplified this analysis to deal only with constant currents, such
as you'd approximately have with an electrically short transmission
line. But the individual currents maintain the same ratio all along
longer lines, so the same result occurs.

(*) Common mode current is sometimes defined as half the vector sum of
the two conductor currents, rather than simply the sum as done here. If
you use the other definition, you assume that the common mode current is
flowing on each of the two conductors to determine the amount of
radiation you'll get. The end result is the same either way.

Roy Lewallen, W7EL