In article , rickman
wrote:

I can't say I agree with your "choking" impedance idea. The coax
connects to the balun in the same way it connects to the antenna.

No, it's really rather different.

The balun can have no effect on the impedance of the coax shield.
Just as you ask, "What would *stop* current from flowing on the
braid?" when connected to the antenna what will stop the current
from flowing on the braid when connected to the balun?

Because, properly done, there's nowhere for that current to come from.

The only thing that will stop the current from flowing on the outside of
the shield when connected to the balun is if the balun presents a much
lower impedance path for the current than does the shield.

No, Kirchhoff's current law comes to the rescue. In fact you want the
(current) balun to have a very high impedance ot common mode current.

The only way your suggestion makes sense is if the current actually
comes *from* the antenna and the balun prevents that current from
returning to the feed line.

Now you're close. In one way of looking at it, if you connect a coaxial
feedline directly to a dipole antenna, that's EXACTLY where the current
on the outside of the coax braid comes from -- one side of the antenna!

I'll try to explain the whole thing in a different way.

First, the current flowing "up" the inner conductor of the coax equals
the current flowing "down" the inside surface of the outer conductor of
the coax, no matter what. This assumes the coax shielding is 100%. The
field between the outside surface of the inner conductor and the inner
surface of the outer conductor is completely contained insisde the coax,
and this requires that those two currents be identical. This much
happens no matter what (if anything) is connected to the top end of the
coax.

Now connect something to the top of the coax, and what happens to those
two (equal) curents? It depends upon what is connected.

Connect a physically small resistor (resistance of the resistor doesn't
matter, and if you want, you can put a physically small inductor or a
physically small capacitor in series with that resistor). Kirchhoff's
current law requires the current into the one end of this load to equal
the current out of the other end of the load. The current into the end
of the load connected to the inner conductor of the coax is the same as
the current flowing up that inner conductor (Kirchhoff's law again). The
same amount of curent flows out of the other end of the load, which is
connected to the shield. So the current flowing down the inside of the
shield equals the current flowing out of the shield's end of the load,
and there is none left to flow down the outside of the braid
(Kirchhoff's law once more). If some current did flow down the outside
of the braid, the curent into one end of the load couldn't equal the
current out of the other end of the laod and Kirchhoff would be unhappy.

Now connect an antenna to the coax, instead of a resistor/reactor load.
There isn't a Kirchhoff's law for antennas. Since the two halves of the
dipole aren't perfectly coupled together, it is quite possible for the
current flowing into the left half of the dipole to be different from
the current flowing out of the right half. Should that happen, the
current flowing out of the right half of the antenna down into the coax
shield (inner plus outer surfaces, some on each) will be different from
the current flowing down just the inside of the shield. (Remember the
current down the inside of the shield HAS to equal the current up the
inner conductor.) Where does the difference come from? The difference is
the current flowing down the OUTSIDE of the shield.

Put a choke consisting of some bifilar turns of parallel conductor
feedline, or better yet consisting of a piece of small diameter coax,
wound around a toroid, between the top of the coax feedline and the
antenna feedpoint, and the current flowing up one conductor of the choke
(one wire, or the inner conductor if you use coax in the choke) has to
equal the current flowing down the other conductor of the choke (the
other wire, or the inner surface of the shield of the piece of coax in
the choke). That means the the currents out of one conductor at the top
of the choke and into the other conductor at the top of the choke have
to be equal. If they were different, the difference (common mode
current) would have to flow through the common mode impedance of the
choke. If the choke is a good choke, that common mode impedance will be
high. So such common mode current has to be low. Only to the extent that
it isn't zero, can the current out of the one conductor differ from the
current into the other conductor (at the top end of the choke). So the
choke does its best to "force" the currents at its top end to be equal.
That's why it's called a current balun (not a voltage balun). Even if
the impedances of the two halves of the antenna are different, the
current balun will do its best to make the current into one side of the
antenna equal the current out of the other side.

Since the currents at the BOTTOM of the choke in the two conductors are
equal, the stupid coaxial feedline thinks it is feeding a resistor (or
maybe a resistor and a reactor in series), not an antenna, and behaves
accordingly; there is no current flowing down the outside of the coax
shield because there is nowhere for such a current to come from.

David, VE7EZM and AF7BZ

--
David Ryeburn

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