"Bal uhn" or "bayl uhn"?
 

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

"Bal uhn" or "bayl uhn"?
 

On 7/29/2015 5:23 PM, wrote:
rickman wrote:
On 7/29/2015 4:38 PM, wrote:
rickman wrote:
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 energy IN the coax is not carried by either conductor, but in the
field between the conductors.

https://en.wikipedia.org/wiki/Coaxia...al_propagation

Once you connect to coax to something, the outside of the shield looks
like another current path with some impedance of it's own.

Likely easiest to visualize on a vertical antenna as being another radial.

See http://www.eznec.com/miscpage.htm and in particular the article
"Baluns: What They Do and How They Do It".

I'm not at all clear on what you are trying to say. I have no idea why
you are shifting the conversation to the details of the power
transmission. Exactly what was written that you are replying to?

What I am talking about is the current on the outside of a coax, not
"shifting the conversation".

You asked "What exactly is the source..." and I am providing the
answer to that question.

One thing I left out is that the match and reflections have nothing
to do with the current on the outside of the coax.

Try reading the links I provided if you want more details to the
answer to your question of where does the current on the outside of
the coax come from.

Ok, thanks a lot.

--

Rick

"Bal uhn" or "bayl uhn"?
 

On 7/29/2015 6:40 PM, Jeff Liebermann wrote:
On Wed, 29 Jul 2015 15:19:29 -0400, rickman wrote:

I am having trouble forming an image of this. What exactly is the
source of the "returning RF"?

This might help:
http://www.antennex.com/w4rnl/col0606/amod100.html
The first few paragraphs are the applicable parts. Quoting a few
tibits:
"Fig. 1 presents one traditional way to portray the situation
at the dipole feedpoint. Its general purpose is to show why
the insertion of a balun is important as a precautionary measure
in dipole construction."

"However, the current from the braid has 2 paths: the right
leg of the dipole in the figure and the outer side of the
coaxial cable braid."

Yes, I read that, but it doesn't really explain this current. Later
they make the statement, "the current on the braid outside side is the
sum of currents other than transmission line currents on the entire
coaxial cable structure". This is pretty clear, but still does not
explain the source, or maybe I should say "why" the current flows on the
braid and not the antenna.


The rest of the article deals with modeling issues and problems.

When modeling a balun, I use three conductors for the coax. The usual
inner and outer conductors, which are assumed to handle only
differential current and therefore do not radiate, and a mysterious
3rd conductor on the outside, which carries all the common mode
current that does the radiating.

And how is this third wire connected? Why do you see current in it? Is
this just due to the voltage drop across the rest of the coax? If so, I
would expect the current flow to be the same phase as the inner shield
current.

--

Rick

"Bal uhn" or "bayl uhn"?
 

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

"Bal uhn" or "bayl uhn"?
 

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

"Bal uhn" or "bayl uhn"?
 

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

"Bal uhn" or "bayl uhn"?
 

In article , rickman wrote:

Yes, I read that, but it doesn't really explain this current. Later
they make the statement, "the current on the braid outside side is the
sum of currents other than transmission line currents on the entire
coaxial cable structure". This is pretty clear, but still does not
explain the source, or maybe I should say "why" the current flows on the
braid and not the antenna.

Don't ask "Why does current flow on the braid?". Ask "What would *stop*
current from flowing on the braid?".

Current flows on *all* paths that have less than an infinite
impedance. That's its nature.

Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the
braid) will be nonzero, if the voltage at that point is nonzero (E !=
0) and the impedance down the braid at that point is not infinite.

The effect of a balun is to place a high "choking" impedance in series
with the outside of the feedline braid, thus "choking off" the current
flow.

"Bal uhn" or "bayl uhn"?
 

On 7/29/2015 6:31 PM, rickman wrote:
On 7/29/2015 6:40 PM, Jeff Liebermann wrote:
On Wed, 29 Jul 2015 15:19:29 -0400, rickman wrote:

I am having trouble forming an image of this. What exactly is the
source of the "returning RF"?

This might help:
http://www.antennex.com/w4rnl/col0606/amod100.html
The first few paragraphs are the applicable parts. Quoting a few
tibits:
"Fig. 1 presents one traditional way to portray the situation
at the dipole feedpoint. Its general purpose is to show why
the insertion of a balun is important as a precautionary measure
in dipole construction."

"However, the current from the braid has 2 paths: the right
leg of the dipole in the figure and the outer side of the
coaxial cable braid."

Yes, I read that, but it doesn't really explain this current. Later
they make the statement, "the current on the braid outside side is the
sum of currents other than transmission line currents on the entire
coaxial cable structure". This is pretty clear, but still does not
explain the source, or maybe I should say "why" the current flows on the
braid and not the antenna.


The rest of the article deals with modeling issues and problems.

When modeling a balun, I use three conductors for the coax. The usual
inner and outer conductors, which are assumed to handle only
differential current and therefore do not radiate, and a mysterious
3rd conductor on the outside, which carries all the common mode
current that does the radiating.

And how is this third wire connected? Why do you see current in it? Is
this just due to the voltage drop across the rest of the coax? If so, I
would expect the current flow to be the same phase as the inner shield
current.

Consider a half-wave dipole fed by coax.

Pretend that the feed point of the antenna has the RF applied (which it
does when you consider that the coax is just a medium to bring the RF up
to that point). What does the the feed point see? It sees one antenna
element on the coax center wire and two antenna elements on the shield
(element + shield). The outside of the shield is now an antenna element
going down toward the earth while the wire attached to the shield is the
other actual element. So the 'balanced' antenna is now unbalanced due to
the coax becoming one of the elements.

This is easy to model in EZNEC. And, you can see the results of changing
the length of the coax as well as the effects of installing impedances
in the coax.

"Bal uhn" or "bayl uhn"?
 

On 7/30/2015 2:01 PM, Dave Platt wrote:
In article , rickman wrote:

Yes, I read that, but it doesn't really explain this current. Later
they make the statement, "the current on the braid outside side is the
sum of currents other than transmission line currents on the entire
coaxial cable structure". This is pretty clear, but still does not
explain the source, or maybe I should say "why" the current flows on the
braid and not the antenna.

Don't ask "Why does current flow on the braid?". Ask "What would *stop*
current from flowing on the braid?".

Current flows on *all* paths that have less than an infinite
impedance. That's its nature.

Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the
braid) will be nonzero, if the voltage at that point is nonzero (E !=
0) and the impedance down the braid at that point is not infinite.

The effect of a balun is to place a high "choking" impedance in series
with the outside of the feedline braid, thus "choking off" the current
flow.

I can't say I agree with your "choking" impedance idea. The coax
connects to the balun in the same way it connects to the antenna. The
balun can have no effect on the impedance of the coax shield. Just as
you ask, "What would *stop* current from flowing on the braid?" when
connected to the antenna what will stop the current from flowing on the
braid when connected to the balun?

The only thing that will stop the current from flowing on the outside of
the shield when connected to the balun is if the balun presents a much
lower impedance path for the current than does the shield.

The only way your suggestion makes sense is if the current actually
comes *from* the antenna and the balun prevents that current from
returning to the feed line.

Maybe this is one of those pointless distinctions and both ways of
looking at it are correct.

--

Rick

"Bal uhn" or "bayl uhn"?
 

rickman wrote:
On 7/30/2015 2:01 PM, Dave Platt wrote:
In article , rickman wrote:

Yes, I read that, but it doesn't really explain this current. Later
they make the statement, "the current on the braid outside side is the
sum of currents other than transmission line currents on the entire
coaxial cable structure". This is pretty clear, but still does not
explain the source, or maybe I should say "why" the current flows on the
braid and not the antenna.

Don't ask "Why does current flow on the braid?". Ask "What would *stop*
current from flowing on the braid?".

Current flows on *all* paths that have less than an infinite
impedance. That's its nature.

Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the
braid) will be nonzero, if the voltage at that point is nonzero (E !=
0) and the impedance down the braid at that point is not infinite.

The effect of a balun is to place a high "choking" impedance in series
with the outside of the feedline braid, thus "choking off" the current
flow.

I can't say I agree with your "choking" impedance idea. The coax
connects to the balun in the same way it connects to the antenna. The
balun can have no effect on the impedance of the coax shield. Just as
you ask, "What would *stop* current from flowing on the braid?" when
connected to the antenna what will stop the current from flowing on the
braid when connected to the balun?

You fail to understand the difference between a choke balun and a
voltage balun.

A voltage balun forces the voltages at the output to be equal.

A choke balun provides a very high impedance to the shield path.

The only thing that will stop the current from flowing on the outside of
the shield when connected to the balun is if the balun presents a much
lower impedance path for the current than does the shield.

Nope, it is just the opposite.

A high impedance path prevents the flow of current.

I = E/Z

From high school algebra, as Z appoaches zero, I approaches infinity,
and as Z approaces infinity, I approaches zero.

The only way your suggestion makes sense is if the current actually
comes *from* the antenna and the balun prevents that current from
returning to the feed line.

Nope, the current comes from the end of the coax and the outside
of the shield is a separate current path from the inside of the
coax and the antenna.

Maybe this is one of those pointless distinctions and both ways of
looking at it are correct.

Nope, there is a very big point to it.

Read www.eznec.com/Amateur/Articles/Baluns.pdf for a detailed discussion
with pictures showing the current flow.



--
Jim Pennino