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. This is a high inductance path. 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.






Secondly, even if you connect a resistor across the end ot the feeder,
consider that the inner conductor just goes to the resistor, but the
outer conductor sees the resistor and the outer side of the braid in
parallel. So you will get RF (and therefore some radiation) on the
outer of the coax even if you just connect a resistor across the end.

Ok, let's discuss this. You are describing a circuit that is just the
coax and a terminating resistor. You seem to be saying that current
will flow on the outer surface of the shield. If that were true, where
does it come from? In this simple circuit the current on the shield
inner surface matches the current on the inner conductor. So there is
no source for current to flow on the shield outer surface.

I am beginning to think you may be right! Sorry. But it doesn't affect
the argument if there is an antenna, however symmetrical, connected to
both conductors.

Actually it does. With an antenna connected currents can flow from the
inner and shield conductors unequally. It will only be unequal if there
is an alternate path for the current. That alternate path will be the
outer surface of the shield.


This would actually be quite a simple lab experiment, at UHF or higher.
Compare the amount of RF on the outer with a bare surface mount 50ohm or
with one of the screened 50 ohm terminations (which does not allow any
signal to get to the outer). Or compare the SWR which will be near 1.0
with the screened load and might be very different with the unscreened,
sim,ply because the coax outer is shunting one side of the load.

Ok, can anyone do this?

It would be very interesting!


--
Roger Hayter