"Reg Edwards" wrote in message
...
"Tom Sedlack" wrote What kind of voltages and currents are present at
the
ends of dipoles
assuming 100W of RF and the antenna being close to resonant?

-----------------------------------------------

A half-wave dipole is a resonant L & C tuned circuit.

Its lumped equivalent is a series L and C circuit across the feedpoint.

It has a Q value like any other resonant circuit.

As is usual, Q = Inductive reactance of wire divided by resistance.

In the case of a dipole the resistance is the radiation resistance at its
centre, typically around 70 ohms.

Inductance gets smaller as wire diameter increases. So a 2-meter dipole
with 1" diameter tubes has a lower Q (about 4) than a 160-meter dipole
made
with 18 gauge wire. (about 12). Q controls bandwidth.

A 40 meter dipole may have a Q around 9. If it is fed at its centre with
100 watts then the feeding voltage is 84 volts.

So the voltage difference between the ends of the dipole is Q times 84 =
756
volts.

Relative to ground, the voltage at one end of the dipole is half of this,
equal to 378 volts.

This would burn a nice little hole at the tip of your right forefinger if
you touched it. Electrical burns take a long time to heal because the
surrounding flesh is electrocuted. Do not confuse this with 'skin
effect'.
----
Reg, G4FGQ

.................and besides, it SMARTS!!!!!!!!!


Jwerry
K4KWH

Cecil Moore wrote in message ...

Zero current. A voltage estimate would be V^2/600=100w, or V=245V RMS.
The net RMS voltage would be double that value so peak voltage might
be around 700 volts.

V^2/600=100w

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

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

73!

Jeff

On 14 Oct 2003 08:15:58 -0700, (Jeffdeham) wrote:

Cecil Moore wrote in message ...

Zero current. A voltage estimate would be V^2/600=100w, or V=245V RMS.
The net RMS voltage would be double that value so peak voltage might
be around 700 volts.

V^2/600=100w

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

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

73!

Jeff

Hi Jeff,

Reg already explained it. You know the applied voltage to a tuned
circuit whose (presumed for this discussion) opposite leads are at the
far end. You know the characteristic Z and the radiative loss R. You
can guesstimate the Q of this circuit as the multiplier. If you reach
out and touch that element's end (or bring it suitably close to a
conductor or other lossy element) you would necessarily depress that
same Q - by being the load. -Zap!- as the comix used to say. So for
your 5W you can expect that as your hand approaches the tip (how'd you
get up there anyway?) both your body and the proximity of everything
around you (like a ladder, or simply the ground you stand on) has done
far more to depress that same calculated Q to diminish the multiplying
effect, but not without some prospect of surprise. Many shock
injuries come from that alone (falling off the ladder is what kills
you, not the 50V 5mA buzz that rattled your nerves).

When I was in the Navy, we had all sorts of exposed radiators around
the ship. Common lore (certainly untested by me) was that if you
found yourself holding one when transmission was begun, to not let go
because you would draw an arc where formerly the voltage was depressed
by your contact. Now, as to how you could distinguish when the QSO
started or ended to time your unfortunate hand-hold was never offered;
and so I was loath to test this lore. Further, no one asked about why
you would stand there holding it while simmering. I generally carried
an NE-2 neon bulb with me when I went aloft so that I had at least
some method to detect hot surfaces (and scramble back down to find out
what jerk had thrown a tagged breaker).

73's
Richard Clark, KB7QHC

That's why you gotta have a GOOD insulator at the end of a wire antenna.

K9CUN

Whoa! There's gotta be some resistance in there somewhere!

73 de Jack, K9CUN

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



-----= 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! =-----

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



-----= 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! =-----