On 8/1/2015 4:31 PM, Roger Hayter wrote:
rickman wrote:

On 8/1/2015 3:29 PM, Roger Hayter wrote:
rickman wrote:

On 8/1/2015 1:38 PM, Roger Hayter wrote:

But your second point is unhelpful in some circumstances. For instance,
if the type of balun is the inductive coil of the feeder with or without
ferrites, then there simply *is no* current path down the outside of the
feeder from the junction of the balun and the feeder, Except from the
outer of the cable in the balun coil, and it is this that is decoupled
by the inductance.

Your description is not clear to me.

I am looking at the junction of the feeder with the balun, and the only
source of current on the outside of the feeder is connected by a very
high inductance to the source of signal at the antenna end.

How is the inner surface of the shield not connected to the outer
surface of the shield?


At the point the balun joins the feeder, the only way RF can get from
inside the coax braid (skin effect deep) to the outside of the coax
braid is to go all the way up to the antenna and down the outer surface
of the balun.

I think this is our first point of disagreement. There is nothing to to
stop the current flowing on the shield inside surface from moving to the
shield outside surface other than a tiny amount of resistance in the
shield wire. Unless the current flow sees a lower impedance path to
follow through the balun, it will travel back on the shield outside
surface.


This is a high inductance path.

I think you are applying this term without appreciating the full
meaning. It is a high impedance path for common mode currents, but a
low impedance path for differential currents. Since the current in the
shield inner surface balances the current on the center conductor, it is
a very low impedance path for the full current on the shield.

If it were accurate to say the balun was "a high impedance path" without
the qualifications, the balun would prevent the desired signal from
reaching the load.


This is easy to see if
the balun is continuous with the feeder, but even if it is not the join
is like the case with the matched load which I have now agreed there is
no 'spare' currrent to cross from the inside to the outside, as the
inside current has to match the inner conductor current.

I won't argue that any of this is correct. It does not conflict in any
way with what I have said.

--

Rick