[[email protected]]

rickman wrote:
On 8/3/2015 2:25 PM, wrote:
rickman wrote:
On 8/3/2015 1:39 PM, wrote:
rickman wrote:
On 8/3/2015 4:49 AM, Jeff wrote:

This is one of those cases where you have forgotten the details and
underlying premises. A Faraday cage can't stop currents from flowing
through the "cage". It stops fields from penetrating the cage by action
of the resulting currents in the cage. Your car is largely a Faraday
cage but you can still be electrocuted if a live wire is in contact with
the chassis and you touch it while inside. There can be a high
potential across different parts of the car from the current and that
current can pass through you if you touch the cage.

If there are high different potentials between different parts of the
car it is not a Faraday Cage.

You are applying some rule you learned without understanding it. By
ohms law there will be potential differences around the Faraday cage.


Also assuming that it is a Faraday cage then touching it whist the cage
is connected to a live conductor will not cause a shock, as everything
is at the same potential.

Only true if the conductor is perfect with no resistance.

Pedantically true but practically irrelevant; for a real world cage the
voltage differences will be small fractions of a volt.

"Your car is largely a Faraday cage but you can still be electrocuted if
a live wire is in contact with the chassis and you touch it while inside."

Why don't we conduct an experiment. You are saying it will only be a
small fraction of a volt, we'll put you in the car. lol

I notice you did not address what I actually wrote.

Yet another red herring from the red herring master; I would never call
a car a Faraday cage for a number of reasons.

Neither is a coax shield.

I notice you still did not address what I actually wrote.

Yet another red herring from the red herring master; I would never call
a coax shield a Faraday cage.




--
Jim Pennino

[Roger Hayter]

Jeff wrote:

On 04/08/2015 08:30, Jeff wrote:

But there is a difference. With the resistor load the current flowing
from the inner lead of the coax *must* (by Kirchoff's 1st law) balance
the current flowing on the shield inner surface. So there is no current
remaining to flow on the shield outer surface. No diagram needed.

For the case of the dipole, the load is no longer balanced with the
shield outer surface connected to one side and there is no requirement
for the current in the two arms to be equal. So in this case current
flows on the shield outer surface in an amount inversely proportional to
the impedance seen by the current in the antenna element and the shield.


No, the dipole is still balanced, just as it would be if it were fed by
twin wire balanced feeder. The power will ALL be applied to the dipole
and, if perfectly matched, will ALL be radiated (or lost as heat).

The only current that will flow on the coax outer is due to the induced
currents from the radiated field, or that caused if the dipole is not
truly balanced.

Try a NEC simulation of a dipole in free space with say 5m of feeder
attached (wire that simulates the outer for NEC purposes) and look at
the differences in current flow in these 3 scenarios:

1. Coax outer connected to one side of the dipole.

2. Coax outer NOT connected to the dipole outer by a very small
distance. (NEC cfeeder still connected)

3. Coax outer connected to one side of the dipole and an inductance
simulating a choke (balun if you wish to call it such) close to the dipole.

You will find very little difference in the currents flowing in the coax
outer in all 3 cases.

The pattern will be distorted by the currents induced in the coax outer,
and there will be other small differences due to the small lack of
symmetry in the model. What you wont see is a huge difference in the
current in the outer when it is directly connected to one element of the
dipole.

Jeff

To save you the effort here is a nec model for you to try

CM dipole with coax
CE
GW 1 10 -0.489 0 0 -0.05 0 0 1.e-3
GW 2 10 0.05 0 0 0.489 0 0 1.e-3
GW 3 3 -0.05 0 0 0.05 0 0 1.e-3
GW 4 50 0.05 0 0 0.05 0 -5 1.e-3
GW 5 3 0.05 0 -5 -0.05 0 -5 1.e-3
GE 0
EK
TL 5 2 3 2 -50 0
EX 0 5 2 0 1 0
GN -1
FR 0 1 0 0 145 0

Jeff

I confess I haven't looked up the meaning of your model parameters, but
what happens if the feeder comes off the aerial at forty-five degrees in
the plane of the dipole?



--
Roger Hayter

[Roger Hayter]

Jeff wrote:

I confess I haven't looked up the meaning of your model parameters, but
what happens if the feeder comes off the aerial at forty-five degrees in
the plane of the dipole?

In that case the current induced into the outer will be much greater,
and to some degree the feeder will unbalance the dipole which will also
increase the current seen on the outer.

With the feeder at 90 degrees, as in my model, the current in the feeder
is about 1/10 of that in the dipole elements in all 3 of my cases due to
the coupling into the coax and dipole imbalance. The increase in current
in the outer between the cases of a direct connection to that of no
connection or a choke is about 10%. Some of that current is due to the
fact that I have not spent enough time trying to get the swr down to 1:1
ie a perfectly matched dipole.

It helps if you change the coax & system impedance to 70 ohms, but I
could, if I had time, trim the element lengths etc, but it is still
difficult in NEC to make an exactly symmetrical model.

Jeff

The conclusion I would draw is that generally there is no need to for a
balun with a well set up coax fed dipole unless you get problems with RF
in the shack, or local RFI, more likely at high power. But if the
geometry is not very symmetrical a balun may improve the aerial
performance. Which, by a coincidence, is what the books generally say.
But the discussion has been useful, and the modelling you have kindly
done tends to confirm the textbook version. (10% is quite a lot if
running kWs, even if it doesn't affect the aerial much.)


--
Roger Hayter

[John S]

On 8/4/2015 2:30 AM, Jeff wrote:

But there is a difference. With the resistor load the current flowing
from the inner lead of the coax *must* (by Kirchoff's 1st law) balance
the current flowing on the shield inner surface. So there is no current
remaining to flow on the shield outer surface. No diagram needed.

For the case of the dipole, the load is no longer balanced with the
shield outer surface connected to one side and there is no requirement
for the current in the two arms to be equal. So in this case current
flows on the shield outer surface in an amount inversely proportional to
the impedance seen by the current in the antenna element and the shield.


No, the dipole is still balanced, just as it would be if it were fed by
twin wire balanced feeder. The power will ALL be applied to the dipole
and, if perfectly matched, will ALL be radiated (or lost as heat).

The only current that will flow on the coax outer is due to the induced
currents from the radiated field, or that caused if the dipole is not
truly balanced.


If the dipole is truly balanced (and the coax is perpendicular to the
antenna), how does current get induced into the coax? My understanding
is that perpendicular wires have little current induction due to this
orientation.

Try a NEC simulation of a dipole in free space with say 5m of feeder
attached (wire that simulates the outer for NEC purposes) and look at
the differences in current flow in these 3 scenarios:


Is EZNEC sufficient for this?


1. Coax outer connected to one side of the dipole.

2. Coax outer NOT connected to the dipole outer by a very small
distance. (NEC cfeeder still connected)

3. Coax outer connected to one side of the dipole and an inductance
simulating a choke (balun if you wish to call it such) close to the dipole.

You will find very little difference in the currents flowing in the coax
outer in all 3 cases.


Free space is hard to come by for us amateurs. In your model, can you
attach the end opposite the antenna to ground and repeat the test?


The pattern will be distorted by the currents induced in the coax outer,
and there will be other small differences due to the small lack of
symmetry in the model. What you wont see is a huge difference in the
current in the outer when it is directly connected to one element of the
dipole.

Jeff


I don't remember that the possible pattern changes are in debate. But,
my memory is flawed and the thread is long.

[rickman]

On 8/4/2015 11:19 AM, John S wrote:
On 8/4/2015 2:30 AM, Jeff wrote:

But there is a difference. With the resistor load the current flowing
from the inner lead of the coax *must* (by Kirchoff's 1st law) balance
the current flowing on the shield inner surface. So there is no current
remaining to flow on the shield outer surface. No diagram needed.

For the case of the dipole, the load is no longer balanced with the
shield outer surface connected to one side and there is no requirement
for the current in the two arms to be equal. So in this case current
flows on the shield outer surface in an amount inversely proportional to
the impedance seen by the current in the antenna element and the shield.


No, the dipole is still balanced, just as it would be if it were fed by
twin wire balanced feeder. The power will ALL be applied to the dipole
and, if perfectly matched, will ALL be radiated (or lost as heat).

The only current that will flow on the coax outer is due to the induced
currents from the radiated field, or that caused if the dipole is not
truly balanced.


If the dipole is truly balanced (and the coax is perpendicular to the
antenna), how does current get induced into the coax? My understanding
is that perpendicular wires have little current induction due to this
orientation.

Current does not need to be "induced" in the shield outside. There is
a voltage on the shield at the antenna feed point and a path for the
current on the shield outside. Current flows. Why would the current
prefer the antenna element over the shield outside?

Yes, the antenna has not changed and it is still "ballanced", but the
antenna is not the only load seen by the feed line. The current in the
two antenna elements are not equal because some of the current flows on
the shield having nothing to do with induction from the antenna.

--

Rick

[rickman]

On 8/5/2015 5:43 AM, Jeff wrote:
If the dipole is truly balanced (and the coax is perpendicular to the
antenna), how does current get induced into the coax? My understanding
is that perpendicular wires have little current induction due to this
orientation.

Current does not need to be "induced" in the shield outside. There is
a voltage on the shield at the antenna feed point and a path for the
current on the shield outside. Current flows. Why would the current
prefer the antenna element over the shield outside?

Yes, the antenna has not changed and it is still "ballanced", but the
antenna is not the only load seen by the feed line. The current in the
two antenna elements are not equal because some of the current flows on
the shield having nothing to do with induction from the antenna.


Well NEC and reality do not agree with you!!

If the feeder sees an impedance that is identical to its own
characteristic impedance that all of the power will go into that load,
that is the situation with a 50 ohm resistor and the radiation
resistance of a dipole is no different. The feeder is not clairvoyant so
that it knows that it is a dipole is attached, and a perfectly matched
dipole will not reflect any power back down onto the coax outer, the
only load seen will be the dipole's radiation resistance not some
mythical addition load in parallel. In a perfectly matched situation the
current in the dipole element will be equal, it is only imbalance that
will cause currents other than those caused by induction to flow on the
outer. This is clearly demonstrable in NEC. NEC also shows that the
currents induced on the outer by radiation are much larger than any
currents caused by small imbalances in the dipole elements when
resonant. Those induced currents are there even if a balun or choke is
used.

Jeff, I am tired of discussing this. Any time you need to invoke
"clairvoyance" I think you have missed the mark. The simple matter is
that a dipole is *not* like a resistor load. The resistor load is a
closed path and does not accumulate charge (except for non-idealities
which we are purposely ignoring), so the current in one side is the same
as the current out the other side. A dipole does *not* have this
limitation and the currents into the two sides does not have to be the
same.

There are many ways to use simulation and only one will be correct. If
you simulate an antenna with a balanced feed you will not get the
results for an unbalanced feed. I found such simulations on the
Internet and posted the links somewhere, not sure which thread or if it
was even this group. I was discussing this elsewhere with people who
deal with EMC and RF for a living and they agree with me. If you have
done such simulations, please show us results. Meanwhile, I'll try to
dig up the info I found on the Internet.

--

Rick

[Sal M. O'Nella]

"Jeff" wrote in message ...

On 27/07/2015 21:45, Wayne wrote:
Just today I got a question from a new ham on the pronunciation of balun.

He has been around the scientific community a lot (physicists, etc.) but
not many RF types such as engineers or hams.

He claims that he rarely has ever hear the pronunciation "bal uhn", and
I've rarely heard "bayl uhn".

Anybody want to weigh in on this, heh heh


Definitely 'Bal-Un' only. Never heard it said any other way.

Jeff

================================================== ===============
Sadly, when I worked as a tech, one of the engineers I routinely went out
with said "BAIL-em" 100% of the time. Ouch. (I was not in a position to
correct him but he was top-notch on the job, so the hurt was minor.)

Sal

[Sal M. O'Nella]

"rickman" wrote in message ...

On 7/29/2015 2:32 PM, John S wrote:
On 7/29/2015 1:16 PM, rickman wrote:
On 7/29/2015 1:42 PM, John S wrote:
On 7/29/2015 11:56 AM, Wayne wrote:

Lots of questions. A large area for investigation.

The reason for my balun question (other than to generate meaningful
technical banter on the newsgroup) is that some years ago in the age of
sliderules, a widely known and respected antenna guru told me that a
balun was unnecessary at resonance ( j=0 ).

I lost touch with him and don't know if his views changed over the
years.

I disagree with him.

Please see Roy Lewallen's (W7EL, author of EZNEC) site for some good
reading. He is a superb writer of easy to understand technical tidbits.
He has some balun stuff among other stuff.

http://eznec.com/misc/

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).

I don't follow what skin effect has to do with the issue. The current
flowing on the outside of the shield is the only current flowing in the
shield. What's your point?

================================================== ============

At RF, the current flows almost entirely at the surface of the conductors
Since the shield has a discrete thickness, it can carry two unrelated
currents, one around the outer surface of the shield and a simultaneous one
within the inner surface of the shield. The current along the inner
circumference of the shield is approximately the equal of the current in the
center conductor, while the current on the outside of the shield represents
a portion of the current at the feed point that is not delivered to the load
because the outer part of the shield represents a parallel path for the
flow.

"Sal"

[rickman]

On 8/11/2015 1:27 AM, Sal M. O'Nella wrote:


"rickman" wrote in message ...

On 7/29/2015 2:32 PM, John S wrote:
On 7/29/2015 1:16 PM, rickman wrote:
On 7/29/2015 1:42 PM, John S wrote:
On 7/29/2015 11:56 AM, Wayne wrote:

Lots of questions. A large area for investigation.

The reason for my balun question (other than to generate meaningful
technical banter on the newsgroup) is that some years ago in the
age of
sliderules, a widely known and respected antenna guru told me that a
balun was unnecessary at resonance ( j=0 ).

I lost touch with him and don't know if his views changed over the
years.

I disagree with him.

Please see Roy Lewallen's (W7EL, author of EZNEC) site for some good
reading. He is a superb writer of easy to understand technical tidbits.
He has some balun stuff among other stuff.

http://eznec.com/misc/

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).

I don't follow what skin effect has to do with the issue. The current
flowing on the outside of the shield is the only current flowing in the
shield. What's your point?

================================================== ============

At RF, the current flows almost entirely at the surface of the
conductors Since the shield has a discrete thickness, it can carry two
unrelated currents, one around the outer surface of the shield and a
simultaneous one within the inner surface of the shield. The current
along the inner circumference of the shield is approximately the equal
of the current in the center conductor, while the current on the outside
of the shield represents a portion of the current at the feed point that
is not delivered to the load because the outer part of the shield
represents a parallel path for the flow.

"Sal"

You posted two seemingly contradictory paragraphs separated by a line.
I am confused about what you are trying to say.

--

Rick