In message , rickman
writes
On 7/29/2015 3:14 PM, Ian Jackson wrote:
In message , John S
writes
On 7/29/2015 1:16 PM, rickman wrote:

Perhaps someone can explain the issue of current in the coax shield.
Current gives rise to a magnetic field. But the current in the inner
conductor is opposite and would create a magnetic field that would
cancel the field of the outer conductor, no?

What am I missing?


Skin effect. The currents on the inside of the shield and on the
outside of the shield see different things. They each have no idea
what the other is doing.

As for magnetic field, I must step aside. I can only report what the
gurus say (nothing that I've found).

Even though the coax shield is grounded at the shack end, both halves of
the antenna get fed push-pull (in anti-phase) with the RF signal flowing
on the outer skin of the inner conductor and the inner skin of the shield.

However, at the antenna end, the returning RF on the shield side of the
antenna doesn't know that it should stay on the inside of the shield.
Because of the skin effect, it happily makes for the outside, whence it
flows back to shack, and through the shack grounding connections.

I am having trouble forming an image of this. What exactly is the
source of the "returning RF"? Is this reflected RF at the impedance
mismatch at the feedpoint? If so, the situation being discussed has no
impedance mismatch, so no returning RF. Is the returning RF from the
signal being radiated from the antenna inducing current in the shield?
If so, doesn't the inner conductor also pick up the radiated signal?

The RF (which is, of course, an AC signal) doesn't just flow out of the
top end of the coax and into the two halves of the antenna. The fact
that the antenna has a standing wave on it means that some RF is
bouncing off the far ends of the antenna, and back to (and into) the top
end of the coax.

There is no reason why the returning RF current on the shield leg of the
antenna should want to flow back on the inside of the shield - in fact,
a combination of the Faraday shield effect and the skin effect
encourages it to take the easy route on outside of the shield.
--
Ian

On 7/29/2015 4:05 PM, Ian Jackson wrote:
In message , rickman writes
On 7/29/2015 3:14 PM, Ian Jackson wrote:
In message , John S
writes
On 7/29/2015 1:16 PM, rickman wrote:

Perhaps someone can explain the issue of current in the coax shield.
Current gives rise to a magnetic field. But the current in the inner
conductor is opposite and would create a magnetic field that would
cancel the field of the outer conductor, no?

What am I missing?


Skin effect. The currents on the inside of the shield and on the
outside of the shield see different things. They each have no idea
what the other is doing.

As for magnetic field, I must step aside. I can only report what the
gurus say (nothing that I've found).

Even though the coax shield is grounded at the shack end, both halves of
the antenna get fed push-pull (in anti-phase) with the RF signal flowing
on the outer skin of the inner conductor and the inner skin of the
shield.

However, at the antenna end, the returning RF on the shield side of the
antenna doesn't know that it should stay on the inside of the shield.
Because of the skin effect, it happily makes for the outside, whence it
flows back to shack, and through the shack grounding connections.

I am having trouble forming an image of this. What exactly is the
source of the "returning RF"? Is this reflected RF at the impedance
mismatch at the feedpoint? If so, the situation being discussed has
no impedance mismatch, so no returning RF. Is the returning RF from
the signal being radiated from the antenna inducing current in the
shield? If so, doesn't the inner conductor also pick up the radiated
signal?

The RF (which is, of course, an AC signal) doesn't just flow out of the
top end of the coax and into the two halves of the antenna. The fact
that the antenna has a standing wave on it means that some RF is
bouncing off the far ends of the antenna, and back to (and into) the top
end of the coax.

So you are saying that with a perfect match to an antenna with a real
only impedance (the stated condition for this discussion) there will
still be a reflected wave on the feed line?


There is no reason why the returning RF current on the shield leg of the
antenna should want to flow back on the inside of the shield - in fact,
a combination of the Faraday shield effect and the skin effect
encourages it to take the easy route on outside of the shield.

I'm not at all clear on the location of current flow on the shield, but
what about the current flow on the inner conductor? If the antenna
reflects a balanced signal back into the cable isn't there also a
current in the inner conductor which will create an opposing magnetic
field? Maybe that is not the issue as some are talking about the
problems created by the voltage drop to ground on the shield.

--

Rick

On 7/29/2015 4:10 PM, rickman wrote:
On 7/29/2015 4:05 PM, Ian Jackson wrote:
In message , rickman
writes
On 7/29/2015 3:14 PM, Ian Jackson wrote:
In message , John S
writes
On 7/29/2015 1:16 PM, rickman wrote:

Perhaps someone can explain the issue of current in the coax shield.
Current gives rise to a magnetic field. But the current in the inner
conductor is opposite and would create a magnetic field that would
cancel the field of the outer conductor, no?

What am I missing?


Skin effect. The currents on the inside of the shield and on the
outside of the shield see different things. They each have no idea
what the other is doing.

As for magnetic field, I must step aside. I can only report what the
gurus say (nothing that I've found).

Even though the coax shield is grounded at the shack end, both
halves of
the antenna get fed push-pull (in anti-phase) with the RF signal
flowing
on the outer skin of the inner conductor and the inner skin of the
shield.

However, at the antenna end, the returning RF on the shield side of the
antenna doesn't know that it should stay on the inside of the shield.
Because of the skin effect, it happily makes for the outside, whence it
flows back to shack, and through the shack grounding connections.

I am having trouble forming an image of this. What exactly is the
source of the "returning RF"? Is this reflected RF at the impedance
mismatch at the feedpoint? If so, the situation being discussed has
no impedance mismatch, so no returning RF. Is the returning RF from
the signal being radiated from the antenna inducing current in the
shield? If so, doesn't the inner conductor also pick up the radiated
signal?

The RF (which is, of course, an AC signal) doesn't just flow out of the
top end of the coax and into the two halves of the antenna. The fact
that the antenna has a standing wave on it means that some RF is
bouncing off the far ends of the antenna, and back to (and into) the top
end of the coax.

So you are saying that with a perfect match to an antenna with a real
only impedance (the stated condition for this discussion) there will
still be a reflected wave on the feed line?


There is no reason why the returning RF current on the shield leg of the
antenna should want to flow back on the inside of the shield - in fact,
a combination of the Faraday shield effect and the skin effect
encourages it to take the easy route on outside of the shield.

I'm not at all clear on the location of current flow on the shield, but
what about the current flow on the inner conductor? If the antenna
reflects a balanced signal back into the cable isn't there also a
current in the inner conductor which will create an opposing magnetic
field? Maybe that is not the issue as some are talking about the
problems created by the voltage drop to ground on the shield.


Read this:

http://eznec.com/Amateur/Articles/Baluns.pdf

On 7/29/2015 7:13 PM, John S wrote:
On 7/29/2015 4:10 PM, rickman wrote:
On 7/29/2015 4:05 PM, Ian Jackson wrote:
In message , rickman
writes
On 7/29/2015 3:14 PM, Ian Jackson wrote:
In message , John S
writes
On 7/29/2015 1:16 PM, rickman wrote:

Perhaps someone can explain the issue of current in the coax shield.
Current gives rise to a magnetic field. But the current in the
inner
conductor is opposite and would create a magnetic field that would
cancel the field of the outer conductor, no?

What am I missing?


Skin effect. The currents on the inside of the shield and on the
outside of the shield see different things. They each have no idea
what the other is doing.

As for magnetic field, I must step aside. I can only report what the
gurus say (nothing that I've found).

Even though the coax shield is grounded at the shack end, both
halves of
the antenna get fed push-pull (in anti-phase) with the RF signal
flowing
on the outer skin of the inner conductor and the inner skin of the
shield.

However, at the antenna end, the returning RF on the shield side of
the
antenna doesn't know that it should stay on the inside of the shield.
Because of the skin effect, it happily makes for the outside,
whence it
flows back to shack, and through the shack grounding connections.

I am having trouble forming an image of this. What exactly is the
source of the "returning RF"? Is this reflected RF at the impedance
mismatch at the feedpoint? If so, the situation being discussed has
no impedance mismatch, so no returning RF. Is the returning RF from
the signal being radiated from the antenna inducing current in the
shield? If so, doesn't the inner conductor also pick up the radiated
signal?

The RF (which is, of course, an AC signal) doesn't just flow out of the
top end of the coax and into the two halves of the antenna. The fact
that the antenna has a standing wave on it means that some RF is
bouncing off the far ends of the antenna, and back to (and into) the top
end of the coax.

So you are saying that with a perfect match to an antenna with a real
only impedance (the stated condition for this discussion) there will
still be a reflected wave on the feed line?


There is no reason why the returning RF current on the shield leg of the
antenna should want to flow back on the inside of the shield - in fact,
a combination of the Faraday shield effect and the skin effect
encourages it to take the easy route on outside of the shield.

I'm not at all clear on the location of current flow on the shield, but
what about the current flow on the inner conductor? If the antenna
reflects a balanced signal back into the cable isn't there also a
current in the inner conductor which will create an opposing magnetic
field? Maybe that is not the issue as some are talking about the
problems created by the voltage drop to ground on the shield.


Read this:

http://eznec.com/Amateur/Articles/Baluns.pdf

Ok, I think I am getting it. When an unbalanced drive connects to a
balanced line or antenna, the current does not fully flow into the
antenna from the shield. The output of the shield becomes a sneak path
routing it to ground.

I suppose the balun works by giving the shield side of the connection a
virtual zero ohm path so that the sneak path is short circuited.

--

Rick

rickman wrote:
On 7/29/2015 7:13 PM, John S wrote:
On 7/29/2015 4:10 PM, rickman wrote:
On 7/29/2015 4:05 PM, Ian Jackson wrote:
In message , rickman
writes
On 7/29/2015 3:14 PM, Ian Jackson wrote:
In message , John S
writes
On 7/29/2015 1:16 PM, rickman wrote:

Perhaps someone can explain the issue of current in the coax shield.
Current gives rise to a magnetic field. But the current in the
inner
conductor is opposite and would create a magnetic field that would
cancel the field of the outer conductor, no?

What am I missing?


Skin effect. The currents on the inside of the shield and on the
outside of the shield see different things. They each have no idea
what the other is doing.

As for magnetic field, I must step aside. I can only report what the
gurus say (nothing that I've found).

Even though the coax shield is grounded at the shack end, both
halves of
the antenna get fed push-pull (in anti-phase) with the RF signal
flowing
on the outer skin of the inner conductor and the inner skin of the
shield.

However, at the antenna end, the returning RF on the shield side of
the
antenna doesn't know that it should stay on the inside of the shield.
Because of the skin effect, it happily makes for the outside,
whence it
flows back to shack, and through the shack grounding connections.

I am having trouble forming an image of this. What exactly is the
source of the "returning RF"? Is this reflected RF at the impedance
mismatch at the feedpoint? If so, the situation being discussed has
no impedance mismatch, so no returning RF. Is the returning RF from
the signal being radiated from the antenna inducing current in the
shield? If so, doesn't the inner conductor also pick up the radiated
signal?

The RF (which is, of course, an AC signal) doesn't just flow out of the
top end of the coax and into the two halves of the antenna. The fact
that the antenna has a standing wave on it means that some RF is
bouncing off the far ends of the antenna, and back to (and into) the top
end of the coax.

So you are saying that with a perfect match to an antenna with a real
only impedance (the stated condition for this discussion) there will
still be a reflected wave on the feed line?


There is no reason why the returning RF current on the shield leg of the
antenna should want to flow back on the inside of the shield - in fact,
a combination of the Faraday shield effect and the skin effect
encourages it to take the easy route on outside of the shield.

I'm not at all clear on the location of current flow on the shield, but
what about the current flow on the inner conductor? If the antenna
reflects a balanced signal back into the cable isn't there also a
current in the inner conductor which will create an opposing magnetic
field? Maybe that is not the issue as some are talking about the
problems created by the voltage drop to ground on the shield.


Read this:

http://eznec.com/Amateur/Articles/Baluns.pdf

Ok, I think I am getting it. When an unbalanced drive connects to a
balanced line or antenna, the current does not fully flow into the
antenna from the shield. The output of the shield becomes a sneak path
routing it to ground.

Or radiating it.

I suppose the balun works by giving the shield side of the connection a
virtual zero ohm path so that the sneak path is short circuited.

Actually a balun works by giving the shield side of the connection
a high impedance path so that the current is minimized.



--
Jim Pennino

In message , rickman
writes
On 7/29/2015 4:05 PM, Ian Jackson wrote:




The RF (which is, of course, an AC signal) doesn't just flow out of the
top end of the coax and into the two halves of the antenna. The fact
that the antenna has a standing wave on it means that some RF is
bouncing off the far ends of the antenna, and back to (and into) the top
end of the coax.

So you are saying that with a perfect match to an antenna with a real
only impedance (the stated condition for this discussion) there will
still be a reflected wave on the feed line?

Maybe 'standing wave' is the wrong description. What I'm referring to is
the approximately sinusoidal current and voltage distribution along the
length of the antenna (high voltage at the ends, high current at the
centre feedpoint).

However, this does result from the outgoing AC wave meeting the wave
bouncing back from the ends of the antenna. If you accept that both legs
of the antenna have this 'waveform' (although where there's no balun,
they are probably unequal), and the shape of the ;waveform; is the
vectorial summation of the go-and-return RF signals, then it's pretty
easy to see why a fair proportion of the returning signal should head
down the outside of the shield. It's probably easier to visualise this
than trying to work out why some of the forward-going RF signal (on the
inside of the shield) should chose to do an about-turn at the antenna
feedpoint, and immediately come back down on the outside of the shield -
instead of flowing into the antenna wire.

There is no reason why the returning RF current on the shield leg of the
antenna should want to flow back on the inside of the shield - in fact,
a combination of the Faraday shield effect and the skin effect
encourages it to take the easy route on outside of the shield.

I'm not at all clear on the location of current flow on the shield, but
what about the current flow on the inner conductor? If the antenna
reflects a balanced signal back into the cable isn't there also a
current in the inner conductor which will create an opposing magnetic
field? Maybe that is not the issue as some are talking about the
problems created by the voltage drop to ground on the shield.


--
Ian