On 8/2/2015 4:37 AM, Roger Hayter wrote:
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?
Just to interject a further argument; I think I agree with you on this.
You can't get coax outer current unless some current sink (eg an antenna
element) is connected to the inner conductor side of the load resistor.
But, consider a pefectly symmetrical dipole: if the potential on the
centre conductor and the braid is exactly the same, how can the two
antenna halves have different currents to allow some to flow down the
outside of the braid?
This has been explained previously. A dipole is not balanced when it is
connected to the coax. The shield outer surface presents a third
element which makes the shield side of the dipole different from the
center conductor side. In the case of the resistor the current flowing
in one side must flow out the other, so it is balanced no matter what.
The dipole has no such requirement. If you restrict the current running
into one side and not the other it can do nothing about it.
I am beginning to wonder if braid outside current with a symmetrical
antenna *only* occurs when the coax outer interacts with the EM field of
the antenna so as to actually alter the impedance of at least one of the
antenna elements, or alter the two of them to a different extent, so
that a common mode current is "left over" for the braid. It does seem
likely that a long wire coming from the centre of the dipoe and not
being absolutely symmetrical would have this effect. However, on this
argument, you would not need a balun if your feeder was absolutely
symmetrical! This theory seems eminently testable with antenna
simulation programs.
You seem to be thinking the antenna elements present some sort of
current sink that *must* extract some amount of current no matter what.
They are just loads no different from the shield outer surface. The
current will flow according to the impedance seen by the current. There
certainly will be some interaction, but if you use a triaxial cable and
only connect the outer shield to ground at the transmitter, the antenna
won't "see" the inner shield and the current will still flow on the
outside of that shield.
The symmetrical feeder argument does not make sense to me. Why wouldn't
the impact on both antenna elements be identical? I haven't heard
anyone say you *do* need a balun if the feed line is symmetrical. That
would be a balbal transformer.
--
Rick