Is it not true that if the currents on the transmission line are unbalanced
(i.e., unequal on the two conductors) then the transmitter must already be
connected to ground? If not, what is the path of the differential current?

Chuck

Dave Oldridge wrote:

Roy Lewallen wrote in
:

Dave Oldridge wrote:

If the antenna is TRULY balanced and the feedline dressed well away
from it at right angles you should have no common-mode currents on
the feedline.
. . .

That only prevents one of the two ways common mode current can be
created, by coupling. It can also be created by conduction. A common
example is a coax-fed dipole, where the current in the outer feedline
conductor splits between the antenna conductor and the outside of the
coax. An equivalent problem can occur when a dipole is fed with
symmetrical line such as ladder line, and one conductor of the line is
connected to the rig's chassis at the rig end. The current on the
inside of the chassis is equal to the current from the "hot"
conductor, and this splits between the transmission line conductor and
the outside of the chassis. A detailed explanation of conducted common
mode current can be found at
http://eznec.com/Amateur/Articles/Baluns.pdf.

Roy Lewallen, W7EL

Still, if the antenna is TRULY balanced (a situation that only rarely
actually happens), you won't get common-mode currents. I've never had a
problem with them with well-grounded (from an RF standpoint) ground-
mounted verticals either.

Essentially this is why I recommend using open wire or twinlead and
feeding it through a proper balanced-line tuner. Years ago, I built an
amplifier that literally had a balanced line output and fed a 600-ohm
feeder direct off two taps on its output coil. That feedline was only
ten feet long and I worked a TON of 80m DX an the inverted vee that it
connected to. And I could always tap the coil so as to have ZERO RF in
the shack (though my landlady's little 7.5 watt light bulbs used to light
on some frequencies when the house wiring picked up direct from the
antenna).