Cecil Moore wrote:

Many examples exist for current drops in distributed networks. That's
one thing that makes circuit analysis invalid for distributed network
problems. The series current is NOT the same value everywhere in a
distributed network. Asserting that there is no such thing as "current
drop" in distributed networks simply indicates an invalid choice of
models.

How much current drop is there at 440 MHz in 100 feet of RG-58 between
the source and a 50 ohm load? Answer: A 20 dB power drop equates to a
40 dB current drop.

Cecil,

You seem to like the "Roach Motel" theory of current flow. The electrons
check in, but they don't check out.

Here's a clue.

Conservation of charge is every bit as fundamental as conservation of
energy. Current does not just disappear. So what happens in your abused
RG-58 case?

Answer: this is not a simple series circuit.

At every point along the line the current splits between continuing down
the line and shunting to the other half of the transmission line. When
the line is lossless, the shunting is purely reactive, and no net
current flows. However, when there is loss in the line, there is a small
phase shift along with the attenuation, and net current is shunted.

The "circuit" model, as you like to call it, is every bit as valid as
the "distributed network" model. However, due to the distributed time
and space considerations in a transmission line, the "circuit" model is
mathematically intractable for many applications.

The physical reality remains the same even if we cannot easily do the math.

Oh, by the way, in a constant impedance environment the current change
corresponding to a power reduction of 20 dB is also 20 dB, not 40 dB.

73,
Gene
W4SZ

On Mon, 25 Oct 2004 15:10:22 GMT, Gene Fuller
wrote:

However, due to the distributed time
and space considerations in a transmission line, the "circuit" model is
mathematically intractable for many applications.

AKA Violation of Kirchhoff

Richard,

I am not sure why you think there is a violation of Kirchhoff.

The current law is unchanged. The analysis of the voltage law is much
more complicated, but not incorrect.

The situation is not really any different than the use of retarded
potentials for radiation. One must carefully keep track of the loop
voltages with consideration for time and space differences, but there is
no fundamental difference in the physics.

For distributed networks, the Kirchhoff voltage calculation is difficult
(intractable). It is not impossible, but it is unnecessary due to the
existence of the much more friendly transmission line formulations.

73,
Gene
W4SZ

Richard Clark wrote:
On Mon, 25 Oct 2004 15:10:22 GMT, Gene Fuller
wrote:


However, due to the distributed time
and space considerations in a transmission line, the "circuit" model is
mathematically intractable for many applications.


AKA Violation of Kirchhoff

Tom Donaly, KA6RUH wrote:
"I know there are people in the past who have attempted to characterize
antennas as transmission lines."

True. You can include Terman among them.

Terman discusses antenna current distribution on page 866 of his 1955
edition:

"Under most circumstances the losses are sufficiently low and the ratio
of wire length to diameter sufficiently great so that to a first
approximation the current distribution can be taken as that for a line
with zero losses; it then has the characteristics discussed in Sec,
4-5."

Sec.4-5 is found on page 95 and is titled: "The Effect of Attenuation on
Voltage and Current Distribution - Lossless Lines".

Sec. 4-5 is in Terman`s chapter on Transmission lines. Obviously an
open-circuit antenna has the same current distribution as an
open-circuit transmission line, and for the same reason.

Best regards, Richard Harrison, KB5WZI

"Tom Donaly" wrote:
I expect someone on this newsgroup will start talking about
"voltage flow" next.

The E-field and the H-field are inseparable in an EM RF wave.
If the H-field (current component) is flowing, then so is the
E-field (voltage component). If the E-field (voltage component)
is dropping due to radiation, then so is the H-field (current
component). That's one of the differences between a distributed
network RF wave and a DC bench circuit.

And how could we possibly have a Power Flow Vector without the
voltage flowing along with the power and the current? :-)
--
73, Cecil http://www.qsl.net/w5dxp

Tom Donaly wrote:

If you make clear what you mean and why you think it makes sense
there's nothing wrong with using non-standard terminology if it
aids communication. If you just pull it out of the air as if it
were a normal technical term that all technical people use, then
you're just gassing nonsense. I expect someone on this newsgroup
will start talking about "voltage flow" next.
73,
Tom Donaly, KA6RUH

Probably best we distinguish between "flow", and propagation, before we
digress into discussions of "standing wave propagation". ;-)

73, Jim AC6XG

Richard Harrison wrote:
Tom Donaly, KA6RUH wrote:
"I know there are people in the past who have attempted to characterize
antennas as transmission lines."

True. You can include Terman among them.

Terman discusses antenna current distribution on page 866 of his 1955
edition:

"Under most circumstances the losses are sufficiently low and the ratio
of wire length to diameter sufficiently great so that to a first
approximation the current distribution can be taken as that for a line
with zero losses; it then has the characteristics discussed in Sec,
4-5."

Sec.4-5 is found on page 95 and is titled: "The Effect of Attenuation on
Voltage and Current Distribution - Lossless Lines".

Sec. 4-5 is in Terman`s chapter on Transmission lines. Obviously an
open-circuit antenna has the same current distribution as an
open-circuit transmission line, and for the same reason.

Best regards, Richard Harrison, KB5WZI


You don't even have to go as far back as Terman. Reg Edwards bases his
antenna programs on the transmission line behavior of antennas, and
his programs do as well as you'd want. That's one way to look at
antenna behavior. It's not the only way, and it doesn't mean that
Cecil knows what he's talking about when he espouses the theories
he made up in his head out there in the hot Texas sun.
73,
Tom Donaly, KA6RUH

In message , Cecil Moore
writes
How much current drop is there at 440 MHz in 100 feet of RG-58 between
the source and a 50 ohm load? Answer: A 20 dB power drop equates to a
40 dB current drop.

Uh? Are you really REALLY sure?
Do you want to change your mind?
Ian.
--

It's not the only way, and it doesn't mean that
Cecil knows what he's talking about when he espouses the theories
he made up in his head out there in the hot Texas sun.
73,
Tom Donaly, KA6RUH


We get the picture of you Tom, you can stop with that crap, quite enough!
Chipster will get jelous for you to take away his crown as troller
extraordinaire.

Yuri

Gene Fuller wrote:
Conservation of charge is every bit as fundamental as conservation of
energy. Current does not just disappear. So what happens in your abused
RG-58 case?

Answer: this is not a simple series circuit.

Thanks Gene, that is exactly my point. 17th century simple series
DC concepts don't work on RF distributed networks.
--
73, Cecil http://www.qsl.net/w5dxp


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