Jeffdeham wrote:
Cecil Moore wrote:
V^2/600=100w

Where did the value of V and 600 come from in this formula?

V is the unknown. 600 ohms is the approximate ball park feedpoint
impedance for a traveling-wave antenna.

I'd like to be able to calculate the voltages also for let's say my 5 watt QRP rig.

Remember, it is a really rough estimate. For 5w, V^2/600=5w, so
V ~ 55v RMS. The peak voltage at the end of a dipole fed with 5w
would be very roughly 150 volts.
--
73, Cecil http://www.qsl.net/w5dxp



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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

Cecil Moore wrote:
Jeffdeham wrote:

Cecil Moore wrote:

V^2/600=100w


Where did the value of V and 600 come from in this formula?


V is the unknown. 600 ohms is the approximate ball park feedpoint
impedance for a traveling-wave antenna.

I'd like to be able to calculate the voltages also for let's say my 5
watt QRP rig.


Remember, it is a really rough estimate. For 5w, V^2/600=5w, so
V ~ 55v RMS. The peak voltage at the end of a dipole fed with 5w
would be very roughly 150 volts.

The ball-park impedance, Zo, of any isolated reasonable length of wire such
as part of a radio antenna, is 600 ohms.


It is first necessary to force yourselves to accept the uncomfortable idea
that any length of wire is a single-wire transmission line.


Zo = 60*(Log(4*Length/dia) - 1) is near enough for most purposes. Jot it in
your notebooks. Sorry I'm unable to provide a reference but you can quote ME
if you like. I found it jotted in MY tattered notebook. ;o)


Even Terman knew that. But perhaps not having sufficient confidence in the
matter he may never have said so explicitly. In which case, unless somebody
can work it out from the information he cribbed from Grover et al, hardly
anybody knows it. ;o)


In the case of a 40 meter dipole of length 20 meters, using 14 awg wire, the
wire Zo = 588 ohms.


To calculate matched-loss in dB/100 feet of single-wire lines would be more
complicated. It is akin to calculating what happens in the case of a
Beverage. It is non-linear versus length.


Cecil's 600 ohms was correct but his description could be confusing. He
need not have mentioned travelling-waves or any other sort of waves because
that depends on whether or not a line or anenna wire is terminated. A
line's termination has no effect on its Zo. Frequency does not enter the
argument. Like any other sort of line it's just a matter of Sqrt(L/C).


A single-wire non-resonant transmission line is used to feed the original
1920's (?) Windom. The line's input impedance is around 600 ohms. It is
correctly terminated by tapping into the resonant antenna at the appropriate
off-center point. The line is of course low loss but it does radiate a bit.
But then, who loses sleep about a bit of radiation from a feedline to a
multi-directional antenna - it's not wasted.

I'm on Chinese vinyards' "Greatwall" white tonight. They have certainly
woken up. ;o)
----
Reg, G4FGQ

Before I find myself inundated with invitations to attend tea-parties, in
the formula for Zo replace "Log" with "Ln".

Reg.

Roy Lewallen wrote:
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?

Who said anything about measuring it? We know it exists and can cause
corona in moist/salty circumstances. But I assume it could be measured
using something like an artificial ground at the tip of the monopole.

A dipole is akin to a leaky unterminated transmission line. The forward
wave travels out to the ends of the dipole where it is reflected by the
open circuit. Just as there is a large voltage at the end of an unterminated
transmission line, there is a large voltage at the ends of an unterminated
dipole (or at the end of a monopole). And just as we can make some assumptions
and estimate the magnitude of the voltage at the end of an unterminated
transmission line, we can make some assumptions and estimate the magnitude of
the voltage at the end of an unterminated dipole. The voltage anywhere along
a center-fed dipole is (Vfwd+Vref). The current anywhere along a
center-fed dipole is (Ifwd+Iref). The feedpoint impedance of a dipole is
(Vfwd+Vref)/(Ifwd+Iref) at the feedpoint. A CF dipole is a standing-wave
antenna with the voltages in phase and maximum at the tips. The voltages
are out of phase and minimum at the center feedpoint.

All we need is an estimate of the feedpoint impedance if the dipole was
terminated at each end thus turning it into a traveling-wave antenna.
I estimated about 600 ohms which put the tip voltage in the same ballpark
as Reg's estimate based on 'Q'.
--
73, Cecil http://www.qsl.net/w5dxp



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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

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