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?

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.

--
Roger Hayter

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

On 8/3/2015 5:10 AM, Jeff wrote:

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.


No, a dipole is still balanced, the coax outer does not create a 3rd
element other than by coupling. If the coax is taken off at right angles
the coupling will be low as similar to both elements of the dipole. The
dipole will never be perfectly balanced but can be a very close
approximation.

Ok, don't call the shield an element.

A balanced antenna does not imply balanced current. The load of the
antenna element on the coax will be equal, but the coax also has a
parallel load from the shield outer surface. The two loads in parallel
result in a different voltage on the end of the coax shield than on the
end of the coax inner conductor. This different voltage causes the
different current flow in the antenna element.

A good illustration showing this can be seen near the top of the web
page linked below.


It your contention about the coax acting as a 3rd element were true then
there would be severe distortion of both the impedance of the dipole and
to it radiation pattern. This is not seen in practice and it is also
demonstrable that there is little current flow on the coax outer when
the dipole is well matched to the coax.

I can't say for myself what happens in practice, but others here and on
the web say there *is* severe distortion in the antenna pattern. See
this link about halfway down the page. This is a EZNEC+ simulation. ymmv

http://www.tomthompson.com/radio/EHa...ommonMode.html

--

Rick

Jeff wrote:

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.


No, a dipole is still balanced, the coax outer does not create a 3rd
element other than by coupling. If the coax is taken off at right angles
the coupling will be low as similar to both elements of the dipole. The
dipole will never be perfectly balanced but can be a very close
approximation.

It your contention about the coax acting as a 3rd element were true then
there would be severe distortion of both the impedance of the dipole and
to it radiation pattern. This is not seen in practice and it is also
demonstrable that there is little current flow on the coax outer when
the dipole is well matched to the coax.

Jeff

The coax shield does create a 3rd element but the effect of it highly
depends on on the length of the coax and whether or not the shield is
grounded somewhere along the way.

The effect can be anywhere from negligable and barely measurable to
extremely significant and can be seen with an antenna modeling program.


--
Jim Pennino

On 8/3/2015 1:44 PM, wrote:
Jeff wrote:

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.


No, a dipole is still balanced, the coax outer does not create a 3rd
element other than by coupling. If the coax is taken off at right angles
the coupling will be low as similar to both elements of the dipole. The
dipole will never be perfectly balanced but can be a very close
approximation.

It your contention about the coax acting as a 3rd element were true then
there would be severe distortion of both the impedance of the dipole and
to it radiation pattern. This is not seen in practice and it is also
demonstrable that there is little current flow on the coax outer when
the dipole is well matched to the coax.

Jeff

The coax shield does create a 3rd element but the effect of it highly
depends on on the length of the coax and whether or not the shield is
grounded somewhere along the way.

The effect can be anywhere from negligable and barely measurable to
extremely significant and can be seen with an antenna modeling program.

Let me summarize...

a) No current on shield outer surface

b) Current on shield outer surface

C) Any conceivable combination of the above

Your answer is c). Thanks for playing!

--

Rick

rickman wrote:
On 8/3/2015 1:44 PM, wrote:
Jeff wrote:

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.


No, a dipole is still balanced, the coax outer does not create a 3rd
element other than by coupling. If the coax is taken off at right angles
the coupling will be low as similar to both elements of the dipole. The
dipole will never be perfectly balanced but can be a very close
approximation.

It your contention about the coax acting as a 3rd element were true then
there would be severe distortion of both the impedance of the dipole and
to it radiation pattern. This is not seen in practice and it is also
demonstrable that there is little current flow on the coax outer when
the dipole is well matched to the coax.

Jeff

The coax shield does create a 3rd element but the effect of it highly
depends on on the length of the coax and whether or not the shield is
grounded somewhere along the way.

The effect can be anywhere from negligable and barely measurable to
extremely significant and can be seen with an antenna modeling program.

Let me summarize...

a) No current on shield outer surface

b) Current on shield outer surface

C) Any conceivable combination of the above

Your answer is c). Thanks for playing!

Nope, just yet another knee jerk response.

If you had actually read and understood what I wrote, the only possible
answer is b.

The only thing in question is the magnitude of the current; that there
is a current is a given.


--
Jim Pennino