rickman wrote:
On 8/1/2015 4:47 PM, 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.


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.

The inside and the outside of the shield are connected together at
the point where the resistor connects to them. The source of the
current is the electromagnetic field that propagates inside the coax.

As the shield is another current path, some current will flow down
it. How much depends on the length of the shield in wavelengths which
determines the impedance of that path.

I would like to clarify this point. You are saying that some of the
current that flow to the load on the shield inside surface will flow
back on the shield outside surface. That means the current in the inner
conductor will no longer equal the current in the shield inner surface,
right?

There is no current in the shield inner surface, the energy is in the
ELECTROMAGNETIC FIELD between the inner and outer conductors. To be
nit pickingly precise, there is some small current in the inner
surface of the shield and the center wire, but for real coax that
surface current is insignificant. You can look at coax as a wave
guide if that makes it easier to understand, though the mode is
different than the mode in what is normally called wave guide.

https://en.wikipedia.org/wiki/Transm...#Coaxial_cable

At the end of the coaxial structure, the electromagnetic field
becomes a current flow in any conductors connected to the end
of the coax.

One of those conductors is always the outside of the shield because
of the physical structure of coax.

The sum of the currents in the outside of the shield plus all other
conductors connected to the outside shield is equal to the sum of the
currents of all the conductors connected to the center wire.



--
Jim Pennino