Current through coils
 

Richard,

After that response all I can say is, this newsgroup is sure good for
entertainment.

73,
Gene
W4SZ

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"Now explain how you`d do it with a box having only two terminals--"

I`ll give the mathematician`s answer: "It`s of no interest. It`s already
been solved." Cecil said he would put a coil in the box. I agree.
Retardation between incident and reflected waves in each direction would
in most cases cause a current difference between the two ends of the
coil. Unlike the usual transmission line, the wire is coiled to get
reactance into a small space. The effect is the same in that phase shift
is distributed along the length of the wire. There is just more of it
and and the intervals between maxima and minima are short. Impedance and
therefore voltage along the vire is a function of site along the wire.
There will be a standing wave pattern throughout the coil.

Best regards, Richard Harrison, KB5WZI

Current through coils
 

Cecil,

This is getting more interesting by the moment. Have you now removed
some of the well-known physical attributes of wire and transmission
lines? Specifically, what happened to the L and C of the wire?

I have no issue with the use of network theory, reflection coefficients,
standing waves, or any other commonly used descriptions. However, none
of these mathematical conveniences change the fundamental physical laws.
If current, and therefore charge, appears to be unbalanced, then there
must be charge storage somewhere.

As Reg pointed out, the charge is stored in the capacitance of the coil.
No need to invoke any magic incantations about networks and standing waves.

In principle any of these problems can be solved with very basic
equations found in any E&M text. In practice it is extremely cumbersome
to do so, and that is why all of the network formulations have been
developed. Just don't fall into the trap of thinking that any new
physical behavior is created by the reflections and standing waves. I
believe in previous messages you have referred to that thinking as
"seduced by the math models."

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:

Your response makes no sense at all. Unequal currents into and out of
a passive black box implies charge storage, which generally means
capacitance.


Boundary condition: There's nothing but wire inside the black box.

Current through coils
 

Roy Lewallen wrote:
Now explain how you'd do it with a box having only two terminals -- and
assuming the box is very small compared to a wavelength.

Assume a one-wavelength dipole off-center fed 1/4WL from
one end. Using EZNEC with 60 segments, feeding at segment
number 15 is 24.2% from one end and that's close enough
for this example. This is actually done in EZNEC with a
130 ft. dipole on 7.2 MHz and I'll email out that file
upon request. I'm going to describe the current
distribution in the following diagram with 60 segments
running from left to right in *fixed font*. Eash dash
corresponds to a segment in EZNEC and F is the feedpoint.

seg L L seg
1 v v 60
--------------F---------------------------------------------
^ ^ ^
N N N

The current distribution is sinusoidal. N stands for 'node'
which is a current minimum point. L stands for 'loop' which
is a current maximum point. Since I'm limited to ASCII, the
reader will need to imagine a current envelope drawn from
seg 1 up to 'L', down to seg 30, back up to 'L', and back
down to seg 60. I'll follow this posting up with actual
EZNEC graphics posted to my web page.

Now we are going to replace part of that wire with a 6" long
coil. A 6" long coil on 7.2 MHz is about 1/3 of one percent
of a wavelength so that should qualify as 'very small'. And,
to illustrate another fact, I'm going to make the coil from
1/4 wavelength of wire, 33' on 40m, and try to model that
using the helical coil feature of EZNEC. That may or may not
violate an EZNEC design rule - I just don't know yet. But
it doesn't change the concepts being presented here.

Let me say this is a very rough approximation to what happens
in the real world. The concepts are accurate. The values may
be off by a relatively large percentage. The coil certainly
distorts the current away from that near-perfect sinusoid and
certainly doesn't radiate like the wire it replaces. But roughly,
here will be the results of placing the bottom of the coil at
seg 30:

seg L L seg
1 v v 46
--------------F---------------////----------------
^ ^ ^
N N N

The current at the left end of the coil will be low because
that is roughly the location of a current node (minimum). The
current at the right end of the coil will be high because that
is roughly the location of a current loop (maximum). If one
considers the current flowing from left to right, more current
will be flowing into the coil than is flowing out of it, like
the current at:

http://www.qsl.net/w5dxp/qrzgif35.gif

This is a standing-wave antenna so the standing-wave current
displayed by EZNEC is flowing hardly at all. That standing-
wave current consists of two component phasors, rotating in
opposite directions. That's why the phase of the standing-wave
current is relatively constant. The standing-wave phasor, the
superposition of the forward and reflected current phasors,
rotates hardly at all, usually by just a few degrees from end
to end in a 1/2WL dipole. If the dipole is made of 'thin wire',
the phase of the standing-wave current is fixed at zero degrees.
(Can a phasor that doesn't rotate be called a phasor?)

Taking 1/4WL of the antenna wire and winding it into a high-Q
coil above replaces *roughly* 90 degrees of the antenna. The
radiation pattern certainly changes because the coil doesn't
radiate much. But we are not concerned about radiation patterns
in this discussion. We are concerned about the current at each
end of the coil, the same current that we measure and the same
current reported by EZNEC. That current is certainly not constant
through the coil and THE DIFFERENCE IN THE MAGNITUDE OF THE
CURRENT AT EACH END OF THE COIL DEPENDS UPON WHERE IT IS PLACED
IN THE STANDING-WAVE SYSTEM.

The traveling-wave current through a coil is close to equal at
each end. The standing-wave current at each end of a coil is
NOT equal unless we locate the center of the coil at a current
node or at a current loop. In a bottom-loaded mobile antenna,
the coil is located very near a current loop where the slope
of the current is near zero. In fact, the net current peaks
inside the bottom-loading coil.

So the concept that net current at each end of a coil installed
in a standing-wave environment is equal is just a myth, an old
wives' tale that needs to be banned from ham radio.

The coil does indeed cause considerable distortion away from the
perfect cosine current wave exhibited by a thin wire. But the
macro effects of that cosine wave still exist when a coil is
installed. The current at each end of a coil installed in a
standing-wave antenna depends upon its location in the system.
--
73, Cecil http://www.qsl.net/w5dxp

Current through coils
 

Richard Harrison wrote:

Unlike the usual transmission line, the wire is coiled to get
reactance into a small space. The effect is the same in that phase shift
is distributed along the length of the wire.

W8JI measured a 60 degree phase shift through a 100uH coil
at 1 MHz. Consider that at one end of that coil the
forward and reflected currents may be:

Ifor = 0.55 amps at zero deg, Iref = 0.45 amps at zero deg.

Inet = Ifor + Iref = 1 amp at zero degrees.

At the other end of the coil, the forward and reflected
currents may be:

Ifor = 0.55 amps at +60 deg, Iref = 0.45 amps at -60 deg

Inet = Ifor*cos(60) + Iref*cos(-60)

Inet = 0.275 + 0.225 = 0.5 amps at zero deg

Some items of note:

1. The forward current magnitude is the same at both ends
2. The forward current phase is shifted by 60 degrees
3. The reflected current magnitude is the same at both ends
4. The reflected current magnitude is shifted by -60 deg
5. The forward and reflected current phasors rotate in
opposite directions
6. The net current phase is unchanged through the coil
7. The net current magnitude is changed by 100% from 0.5
amps to 1.0 amps.
--
73, Cecil http://www.qsl.net/w5dxp

Current through coils
 

Gene Fuller wrote:
I have no issue with the use of network theory, reflection coefficients,
standing waves, or any other commonly used descriptions. However, none
of these mathematical conveniences change the fundamental physical laws.
If current, and therefore charge, appears to be unbalanced, then there
must be charge storage somewhere.

Gene, the flaw is in your misunderstanding of the
fundamental physical laws, not in those laws.

We measure the net current at one end of a coil at 0.1
amp and we measure net current at the other end of the
coil at 0.7 amps (my web page example). The net current
*APPEARS* to be unbalanced, but appearances can be
deceiving. THERE IS NO STEADY-STATE CHARGE STORAGE
ANYWHERE IN THE SYSTEM. Does this violate any fundamental
physical laws? Of course not. Here's why (neglecting losses):

V*I*cos(theta) equals the same power at both ends of the
coil. That proves there is no steady-state energy storage.

V1*(0.7)*cos(theta1) = V2*(0.1)*cos(theta2)

This is a distributed network. A lumped circuit analysis
fails miserably when you try to use it in a standing-
wave environment and you have just proved it. That is
also the same mistake that W8JI and W7EL have been
making.

The forward current at the 0.7 amp point is 0.4 amps at
zero deg. The reflected current at the 0.7 amp point is
0.3 amps at zero degrees. The net current is the phasor
sum of those two component currents.

Inet = (0.4 amps at zero deg) + (0.3 amps at zero deg)
Inet = 0.7 amps at zero deg

The forward current at the 0.1 amp point is 0.4 amps at
82 degrees. The reflected current at the 0.1 amp point is
0.3 amps at -82 degrees. The coil causes an 82 degree
phase shift in both forward and reflected currents and
since their phasors are rotating in opposite directions,
the sign of their phase shifts are opposite.

The net current at this end of the coil is:

Inet = Ifor + Iref
Inet = (0.4 amp at 82 deg) + (0.3 amp at -82 deg)
Inet = 0.057 amps + 0.043 amps = 0.1 amp at zero deg

The fundamental physical laws are perfectly valid as has
been demonstrated here. It is your understanding of them
that seems to be the problem. You seem to have been fooled
by appearances and as a result, you chose the wrong model
with which to try to solve the problem. The distributed
network analysis was developed because the lumped circuit
analysis falls apart under certain circumstances. One of
those circumstances is the presence of standing waves like
the ones that exist in a 75m mobile antenna.
--
73, Cecil http://www.qsl.net/w5dxp

Current through coils
 

Gene Fuller, W4SZ wrote:
"This is getting more interesting by the moment."

There are plenty of coils in boxes which have different currents into
and out of their two ends.

A coil in a box used to be a common way to resonate a too-short 1/4-wave
(90-degree) whip. A company I worked for had many Land Rovers, trucks,
boats, and ships on and around the Argentine side of the island of
Tierra del Fuego.These were equipped with H-F SSB tranceivers. Mobile
amntenna was a stainless whip mounted atop a substantial fiberglass box.
The box contained the loading coil which was accessible for preselecting
the right coil tap to resonate the whip with the vehicle for a
particular operating frequency. The box also contained a motor-driven
band-switch to automatically change taps on the coil when the frequency
was changed on the radio.

I am well aware of the ability to resonate a 90-degree whip with no more
than the proper coil in series with the short whip on a base insulator.
I tuned every one of those coils for each of the frequencies we used in
Argentina with my own hands.

ON4UN has a graph, Fig 9-22 on page 9-15 of "Low-Band DX-ing" which
shows current distribution of a base-loaded whip, In his example, the
whip is 45-degrees long.. The loading coil provides the extra 45-degrees
required for resonance.

Current at the base of ON4UN`s whip is one amp times the cosine of
45-degrees, or 0.707 amp. The loading coil has an input of one amp.

With 1 amp into the loading coil and 0.707 amp out of the loading coil,
the coil definitely does not have the same current at both ends.

Best regards, Richard Harrison, KB5WZI

Current through coils
 

Richard Harrison wrote:
ON4UN has a graph, Fig 9-22 on page 9-15 of "Low-Band DX-ing" which
shows current distribution of a base-loaded whip, In his example, the
whip is 45-degrees long.. The loading coil provides the extra 45-degrees
required for resonance.

Current at the base of ON4UN`s whip is one amp times the cosine of
45-degrees, or 0.707 amp. The loading coil has an input of one amp.

With 1 amp into the loading coil and 0.707 amp out of the loading coil,
the coil definitely does not have the same current at both ends.

It's not that perfect in the real world but the basic concept
still applies. The actual current at the top of the coil is
somewhat higher than 0.707 amp because the current inside
the coil is greater than 1 amp. EZNEC says the current about
1/3 of the way up from the bottom of the coil is about 1.15
amp. The inductance of the coil forces the phase between
the voltage and current to increase. To maintain the same
V*I*cos(theta) power, the current must also increase.

The high flux fields developed inside the coil somewhat distort
the perfect current cosine wave found in a thin wire dipole so
it is not quite as black and white as ON4UN indicates.
--
73, Cecil http://www.qsl.net/w5dxp

Current through coils
 

Cecil Moore wrote:
If one
considers the current flowing from left to right, more current
will be flowing into the coil than is flowing out of it, ...

Correction! Should say: If one considers the current flowing from
left to right, more current will be flowing out of the coil than
is flowing into it, ...
--
73, Cecil http://www.qsl.net/w5dxp

Current through coils
 

Roy Lewallen wrote:
And now things must
be getting boring again out in the Texas prairie, so here it comes
again. Have fun, folks.

Roy, you are like the novice who uses a DC ohm-meter to measure
the feedpoint impedance of an antenna. You are using a lumped
circuit analysis in the presence of standing waves, not
realizing what an extreme technical blunder that really is.
Worse yet, you belittle people who point out your blunder.
--
73, Cecil http://www.qsl.net/w5dxp

Current through coils
 

Gene Fuller wrote:
I have lots of flaws, most of them unrelated to RRAA. But a lack of
understanding of fundamental physical laws is not one of them.

Gene, you proved beyond any doubt, during the rest of your posting,
that you even though may quote the fundamental laws, you are
misunderstanding those fundamental laws. I think Richard Harrison
does understand. If you refuse to listen to me, then please listen
to him.

Current is generally accepted as the flow of charge.

Standing wave current is a net charge flow of zero. Standing wave
current is DIFFERENT from traveling wave current. At any and every
point, the standing wave current is NOT moving. Since it is not
moving, there is NO net charge flow. Please read, understand, and
respond to the following simple example and questionai


Lossless Transmission Line, SWR is infinite
-------------------------------------------------
Ifor-- 1 amp --Iref 1 amp


1. Is there any net flow of charge? ______

2. Is the current at a current maximum point equal to 2 amps? ______

3. Is the current at a current minimum point equal to 0 amps? ______

4. Is there any net flow of charge between the 2 amp point and the
zero amp point? ________

5. If I replace the 1/4WL section of wire with a 90 degree phase
shifting coil, have I changed very much? _________

6. If the current at one end of that coil is 0 amps and the current
at the other end of that coil is 2 amps, is there any net flow of
charge through the coil? ________

When the current is different at the ends of a simple two terminal
device then the charge flow is different as well.

That is definitely NOT true for standing wave current. There
is ZERO net charge flow for standing wave current. And standing wave
current is exactly what you are measuring in a standing wave antenna
like a 75m mobile antenna.
--
73, Cecil http://www.qsl.net/w5dxp