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Author Topic: current and voltage along a non resonant dipole  (Read 18082 times)
AE5EK
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Posts: 53




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« on: November 11, 2013, 07:07:25 PM »

When  dipole is resonant the voltage is highest at the ends and current is highest as the middle.
But what happens when the dipole is not resonant, say for example and OCFD or a random wire at any frequency where it is not a half wavelength?  Where is the max current and what is the voltage at the ends?

tks
Dennis
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WD4ELG
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Posts: 875




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« Reply #1 on: November 11, 2013, 09:05:54 PM »

It depends upon the length of the wire and the frequency of operation.  EZNEC can give you a good detailed answer.
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AE5EK
Member

Posts: 53




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« Reply #2 on: November 11, 2013, 09:30:00 PM »

I have heard it said that the ends have to be low or no current and high voltage.  How can that be when the antenna is not 1/2 wavelength.
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WB6BYU
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Posts: 13335




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« Reply #3 on: November 11, 2013, 09:36:14 PM »

Highest voltage is always at the ends, and current is minimum, because there is no
place for current to flow to.

To analyze the current flowing on a wire, start at the ends and mark it as a voltage
maximum.  Go 1/4 wavelength in from the end and that will be a voltage minimum
and current maximum.  Another wave puts you back at a voltage maximum, etc.

At adjacent current maxima the currents will be 180 degrees out of phase.

So, for example, consider the current on a 1 wavelength wire:  voltage will
be maximum at the ends, currents maximum at 1/4 wave in from each end,
and voltage maximum at the center.  The currents in the two half wave
sections are out of phase.  I often find it easiest to draw a picture of the
antenna and mark it, for example, with dots at the voltage maxima and
arrows denoting direction (phase) at the current maxima:  in the case
of the 1 wavelength wire there would be a dot in the middle and one at
each end, and between the dots one arrow would point right and the
other left.

Then from looking at the paper (broadside to the antenna, such that the
arrows are equidistant from you) you would see equal numbers of arrows
pointing right and left, the radiation from which would cancel each other
in your direction, so there is a null in the pattern broadside to the antenna.

The feedpoint does make a difference, however:  if fed in the center such
a wire would have maximum radiation broadside to the antenna, while
fed 1/4 wave from the end you would have an "X" pattern with a null
broadside to the wire.  How to tell the difference?  Start by marking
the voltage and current nodes along the wire from each end towards
the feedpoint.  At the feedpoint you know that current flowing out
of one side of the feedpoint flows into the other side at the same
phase.  This means there is no phase shift across the feedpoint, whether
it is at a current maximum (low impedance) or voltage maximum (high
impedance, where otherwise there would be a phase reversal in the
currents.)

Once you draw out the currents on a few wires (especially those several
half wavelengths long) then it becomes easier to imagine the current
distributions on them.


The same approach works with a closed loop antenna, but in that case you
start from the midpoint of the loop at the far side from the feedpoint with
maximum current and work back around the loop from there.
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KA4POL
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Posts: 2028




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« Reply #4 on: November 11, 2013, 09:48:00 PM »

Similar to DC resistance you need to consider impedance. Where there is a low impedance you have a higher current. At the antenna ends you find high impedance and therefore no current. Keep in mind that the impedance is not constant over the complete length of the antenna.
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W5DXP
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Posts: 3613


WWW

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« Reply #5 on: November 12, 2013, 05:03:50 AM »

A dipole is a standing wave antenna. The SWR on a typical 1/2WL dipole is in the ballpark of 20:1 at the feedpoint. The SWR is nearly infinite at the tip ends of the dipole. For a pure standing wave, the voltage envelope is always 90 degrees out of phase with the current envelope, i.e. where a voltage maximum exists, a current minimum will exist and vice versa. The maximum points are called "loops" and the minimum points are called "nodes". The voltage and current distribution on a dipole is essentially the same pattern as the voltage and current distribution on an open-circuit transmission line stub.
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73, Cecil, www.w5dxp.com
The purpose of an antenna tuner is to increase the current through the radiation resistance at the antenna to the maximum available magnitude resulting in a radiated power of I2(RRAD) from the antenna.
G3RZP
Member

Posts: 4713




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« Reply #6 on: November 12, 2013, 06:28:42 AM »

Cecil,

Don't you mean 180 degrees out of phase, not 90?
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WB6BYU
Member

Posts: 13335




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« Reply #7 on: November 12, 2013, 06:38:51 AM »

No, 90 degrees is correct.

Current is maximum when voltage is minimum.  1/4 wave down the line the current is minimum while
the voltage is maximum.

So the current is offset from the voltage by 1/4 of a cycle, or 90 degrees.
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AE5EK
Member

Posts: 53




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« Reply #8 on: November 12, 2013, 06:50:34 AM »

But what about when the dipole is not a multiple of 0.5 wavelengths because it is NOT resonant? What would be the current distribution when the dipole is 3/4 wavelength? How can both ends be a high voltage? If one end is a high voltage, the other end would need to be ZERO volts because it is 3/4 wavelength away?

tks
Dennis
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WB6BYU
Member

Posts: 13335




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« Reply #9 on: November 12, 2013, 08:29:31 AM »

Quote from: AE5EK

But what about when the dipole is not a multiple of 0.5 wavelengths because it is NOT resonant? What would be the current distribution when the dipole is 3/4 wavelength? How can both ends be a high voltage? If one end is a high voltage, the other end would need to be ZERO volts because it is 3/4 wavelength away?



But if you have any current flowing at the end of the wire, where is it flowing from/to?
If you don't have current flowing, then the impedance is high.

(Or, if you look at it another way, all the current flowing outwards at that point is reflected
from the open circuit at the end of the wire so it flows back the opposite direction.  The
net current is zero.)


The answer to your conundrum is what happens at the feedpoint.

But let's look at the antenna first.  Starting at the right end, we have high impedance at
the end, low impedance (maximum current) 1/4 wave in from the end, and 1/8 wave
later we hit the feedpoint at a point in the cycle where both the voltage and current
are at intermediate values.  If we repeat the process from the left end, we have the
same current distribution, and end at the same intermediate point in the cycle.

If we were to plot the voltage distribution we'd see it high at the ends, dropping to zero
1/4 wave in from the end, then rising part way back up to the maximum at the feedpoint.
The current would be minimum at the ends, maximum 1/4 wave back, then dropping back
somewhat by the time you hit the feedpoint.

What happens at the feedpoint?  We know that the current has to be in phase across
the feedpoint, so if we are plotting the current along the wire we can make the
sections in phase on each side of it.  Since there is no point of minimum current
between the feedpoint and the end of the wire (which is where the current changes
phase along the wire) then all the current on the wire will be in the same phase.


Basically, the fact that the current must be 0 at the open end of a wire antenna, and
the sine / cosine distribution along the wire, force a particular condition at the feedpoint
regardless of the length of the wire.  That's where you have any discontinuity in the
voltage or current distributions, and that's why the impedance can vary wildly in the
center of a doublet as you change frequency.  If the feedpoint is at one end of the
wire rather than in the middle the current distribution will be different, but the same
rules apply.  (Except that you wouldn't say that the current is zero there because
it isn't a free end of the wire - it is connected to something, namely the feedline.)
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W5WSS
Member

Posts: 1744




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« Reply #10 on: November 12, 2013, 09:48:21 AM »

Dale, you are very good at technical writings Smiley

Appreciate you!

73 Smiley
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G3TXQ
Member

Posts: 1523




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« Reply #11 on: November 12, 2013, 10:54:18 AM »

On a related issue ......

Don't get confused between: on the one hand the phase relationship between the time-varying voltage and current at a particular point on the wire; and on the other hand the phase relationship between the voltage and current standing waves along the wire - they are two quite different things!

For example, we know at any point along a resonant half-wave dipole the time-varying voltage and current would be in-phase with each other; at the same time we know there is a 90 degree phase shift between the voltage and current standing wave patterns along that wire. They are different things, and it's easy to confuse the two.

Steve G3TXQ
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G3RZP
Member

Posts: 4713




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« Reply #12 on: November 12, 2013, 02:42:21 PM »

Right Steve, because if the feed voltage and feed current at the feed point have a 90 degree phase difference, the impedance must be a pure reactance - which, by definition, is not resonance.
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KC8Y
Member

Posts: 247




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« Reply #13 on: November 12, 2013, 02:46:45 PM »

AE5EK:

I use an OCF dipole, that seems to work on 80/75, 40, 20, 17, 12, 10 & 6-meters; also just use end-fed 1/2 wave for 30-meters.  Have a tuner that does fine tuning for each band.  I never go above 50-watts & operate digital modes.

Ken KC8Y
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WB6BYU
Member

Posts: 13335




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« Reply #14 on: November 12, 2013, 02:58:49 PM »

A further source of confusion is between the magnitude and polarity of the standing
wave distribution along the wire, where there are points of minimum and maximum
voltage and current, and the actual plot of voltage vs. time at any single point, where
the voltage (and current) will show the normal alternation between positive and negative
values over the course of a single cycle of RF.


This is a confusing and difficult concept for a lot of people, especially with all
of the potential misunderstandings such as those that we have noted.  Take your
time, read through the various resources again, try plotting some distributions on
a drawing of an antenna, or whatever works best for your preferred learning style.

And kudos to you, Dennis, for asking questions and jumping in to try to understand
it.  Keep asking questions until you get it:  at that point things will make much
better sense, though it may not feel like that while you're still muddling in the middle.


And your reward (or, perhaps, penance might be a better term) will be then
to explain it to others...
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