Owen Duffy wrote in
:

Roy Lewallen wrote in
:


The key to the answer is in two words in the first sentence of your
posting: "immediately adjacent". You're certainly correct that the
current can't abruptly drop to zero at the terminator, because of
Kirchoff's Current Law. But the current doesn't abruptly end, rather
it drops to zero following a sinusoidal distribution curve. It's
quite

My modelling experience is that other things like connection to
ground, and open ends to conductors have more influence on the
location of a standing wave pattern than typical common mode chokes.

I have created a simple model of a Carolina Windom at 7MHz, assuming
that the device at the dipole feedpoint is a 4:1 voltage balun with
negligible common mode impedance, the isolater is 1000+j0 (your
nomination), and a feedline configuration that demonstrates that the
isolator has not caused a minimum in the common mode standing wave
pattern at that point.

A pic of the current distribution is at
http://www.vk1od.net/lost/Clip043.png .
...

Common mode chokes can be made pretty easily to have an impedance of
more than 1k ohm. Both modeling and measurement show this is usually
adequate in typical installations to drop common mode current to very
near zero at the choke location. But you can easily have substantial
current a quarter wavelength on either side of it.

Didn't work for this case, the current minimum is about half wave
between the isolator (left hand blue square) and the dipole, and the
common current entering the shack (right hand blue square) is quite
large.

Apologies, there was an error in the model... I hadn't installed the
source properly. I have replaced the pic at
http://www.vk1od.net/lost/Clip043.png . The situation is a little
different, but the isolator does not force a current minimim at its
location, and the common mode current flowing at the shack is large.

Owen