(Mark Keith) wrote in message . com...
"Tom Sedlack" wrote in message The current max is at
the feedpoint. Just guessing on the voltage, probably about 700 volts
potential at the ends or so...x10 for a KW..
I had an E-mail that stated:


As P=V^2/R it follows V is proportional to the square root of P and
not directly proportional.

I was thinking the voltage for 1 kw would be in direct proportion. But
it seems not. 2100v? Thought I would throw this in. My original quote
may cause confusion. :/ MK

Mark Keith wrote:
I was thinking the voltage for 1 kw would be in direct proportion. But
it seems not. 2100v? Thought I would throw this in. My original quote
may cause confusion. :/ MK

Assuming Reg's value of Z0=600 ohms for the wire in a dipole, the RMS voltage
will be about Sqrt(600*1000) = 775v. The net voltage will be double that
value or 1550v making the peak voltage about 2190 volts.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Roy Lewallen wrote in message ...
It's not clear to me what's meant by the voltage (presumably relative to
ground) at the tip of a dipole. Suppose it's a quarter wavelength above
ground. How would you measure it? Or, how would you measure the voltage
at the top of a quarter wavelength vertical?

Roy Lewallen, W7EL

Indeed... Or putting it another way, the potential between two points
does not have a unique value in the presence of a time-varying
magnetic field, which certainly is the case for a radiating dipole.
If you measure the voltage drop along the wire, it's essentially zero,
so along the wire the voltage between the end points of the dipole is
essentially the same as the voltage across the feedpoint. The only
difference between the ends and the feedpoint is due to I*R drop in
the wire. The voltage at the top of Roy's vertical, made out of
fairly large diameter aluminum tubing, is essentially the same as the
voltage at the bottom of that tube, if you measure along the tube.

It would be better to talk about electric field strengths in the
vicinity of the dipole. You could find the potential along a path
from the field if you wished.

(I'm recalling that Roy turned on a lightbulb in my head quite a few
years ago about this. And I'm sad that Kevin, W9CF, doesn't jump in
on things like this very often these days, though I can understand
why.)

Cheers,
Tom

Tom Bruhns wrote:
If you measure the voltage drop along the wire, it's essentially zero,
so along the wire the voltage between the end points of the dipole is
essentially the same as the voltage across the feedpoint.

Brain fart? The feedpoint impedance is the ratio of voltage to current.
The feedpoint impedance of a halfwave centerfed is low, around 70 ohms.
The feedpoint impedance of a halfwave endfed is high, thousands of ohms.

The voltage at the middle of a dipole is low and the current is high.
The same holds true for a 1/4WL monopole feedpoint fed against ground.

The voltage at the ends of a 1/2WL monopole is high and the current
is low. The same is true for the open end of a 1/4WL monopole.

The sum of the forward wave and reflected wave causes standing waves
on an antenna like the above. The voltages, currents, and impedances
vary somewhat akin to an SWR circle on a Smith Chart.
--
73, Cecil, W5DXP

Cecil Moore wrote in message ...
Tom Bruhns wrote:
If you measure the voltage drop along the wire, it's essentially zero,
so along the wire the voltage between the end points of the dipole is
essentially the same as the voltage across the feedpoint.

Brain fart?

Just so we're clear on this, no, certainly not.

If you care why, consider the direction of the electric field adjacent
to the conductor, and integrate the component of that field parallel
to the conductor along the path of the conductor. You will in general
get a different answer than if you integrate along a path from the tip
of the antenna, out say a quarter wavelength, then parallel to the
antenna for a half wave, then back to the other end of the antenna.
There is no such thing as "the voltage" between the ends of your
excited dipole at an instant in time. There are infinitely many
potentials, as there are infinitely many paths you can follow through
the (time-varying) magnetic field.

Cheers,
Tom

Tom Bruhns wrote:
There is no such thing as "the voltage" between the ends of your
excited dipole at an instant in time.

Perhaps they meant the voltage 'across' the ends of the dipole. The
ends should always be an electrical half-wave out of phase, right?
There should only be two instants of time during a period when the
difference in potential from end to end is zero.

What are you saying exactly, Tom?

73, Jim AC6XG

On 16 Oct 2003 09:59:07 -0700, (Tom Bruhns) wrote:

Roy Lewallen wrote in message ...
It's not clear to me what's meant by the voltage (presumably relative to
ground) at the tip of a dipole. Suppose it's a quarter wavelength above
ground. How would you measure it? Or, how would you measure the voltage
at the top of a quarter wavelength vertical?

Roy Lewallen, W7EL

Indeed... Or putting it another way, the potential between two points
does not have a unique value in the presence of a time-varying
magnetic field, which certainly is the case for a radiating dipole.
If you measure the voltage drop along the wire, it's essentially zero,
so along the wire the voltage between the end points of the dipole is
essentially the same as the voltage across the feedpoint. The only
difference between the ends and the feedpoint is due to I*R drop in
the wire. The voltage at the top of Roy's vertical, made out of

Ahhh...with the impedance variation I beg to differ.
Otherwise how could I draw a 3 or 4 inch arc off the end of a 160
meter dipole.

The end of the antenna is high impedance while the center (in the case
of a half wave dipole) is low. Low voltage and high current at the
center with high voltage and low current at the ends.

Roger Halstead (K8RI EN73 & ARRL Life Member)
www.rogerhalstead.com
N833R World's oldest Debonair? (S# CD-2)

Tom Bruhns wrote:
There is no such thing as "the voltage" between the ends of your
excited dipole at an instant in time.

Please reference Fig 1, page 2-2, in the 15th edition of the ARRL
Antenna Book. "Current and voltage distribution on a 1/2WL wire.

The RMS (or peak) values of the voltages at the ends of the dipole
are maximum and 180 degrees out of phase. The ratio of net voltage to
net current is the impedance anywhere along the wire.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Cecil wrote,


Tom Bruhns wrote:
There is no such thing as "the voltage" between the ends of your
excited dipole at an instant in time.

Please reference Fig 1, page 2-2, in the 15th edition of the ARRL
Antenna Book. "Current and voltage distribution on a 1/2WL wire.

The RMS (or peak) values of the voltages at the ends of the dipole
are maximum and 180 degrees out of phase. The ratio of net voltage to
net current is the impedance anywhere along the wire.
--
73, Cecil http://www.qsl.net/w5dxp




Cecil, that picture is a gross simplification. In order to show that there's
a unique voltage between the ends of a dipole, you first have to show that
the time-varying electric field between those ends is conservative.
Go ahead. (My money is on Tom, however.)
73,
Tom Donaly, KA6RUH

Jim Kelley wrote in message ...
Tom Bruhns wrote:
There is no such thing as "the voltage" between the ends of your
excited dipole at an instant in time.

Perhaps they meant the voltage 'across' the ends of the dipole. The
ends should always be an electrical half-wave out of phase, right?
There should only be two instants of time during a period when the
difference in potential from end to end is zero.

What are you saying exactly, Tom?

I'm saying that if you measure the voltage between two points on a
good conductor, in a path along that conductor, it will be very small.
The electric field is always perpendicular to a perfect conductor at
the surface of that conductor. For a conductor with resistance, the
drop along it is I*R, and therefore the nearby electric field is in
general not quite perpendicular, but unless it's a darned inefficient
antenna, it's very nearly so.

I'm also saying that the voltage (potential) between two points
depends, in general, on the path you take between the two points. You
should be _especially_ aware of that fact when you're in the presence
of time-varying magnetic fields, such as you have around a powered
antenna.

As I said, if you measure the potential along a line perpendicular to
the antenna, it will be large (when the antenna is excited with some
power). I fully expect the electric field to be high near the wire,
but perpendicular to the wire, NOT parallel to it.

Cheers,
Tom