Hummm...Should be about .336 mh for a grounded coil.
About .182 mh for a insulated coil. In the real
world will usually amount to about 5 turns of coil
average on say a .5 to 1 inch form. Trim coil for
best match. MK

In article ,
john doe wrote:

Is that 58.5 inch figure a typo? Your NEC model says 51 inches, and
my quickie spreadsheet calculation says 50.9 inches for a 145 MHz
center of band.

Actually, the 51 you're seeing in the NEC model is probably the number of
segments. One end is at 0,0,36 and the other is at 0,0,94.5; so yes it's
really 58.5 inches.

Whups... my bad.

I came to this number by playing with the model until
the REAL component of the impedance got as close to 50 as I could get it.

Hmmm. What did that do to the pattern? You no longer have a 5/8-wave
antenna. Adding about 8 inches has brought it very close to being a
3/4-wave radiator. As such, it's going to have a substantially lower
amount of towards-the-horizon energy in its pattern, and a big lobe
aiming upwards at roughly 45 degrees above the horizon.

This is the classic problem with running a 2-meter J-pole on 440 -
it'll load up and radiate, but a lot of its radiation is aimed at
airplanes rather than repeaters :-(

I also wonder about the coil - it calculates out to be just over 1
microHenry, or about j910 ohms at 145 MHz. That seems like quite a
bit too much, based on jgboyles's posting earlier today indicating a
feedpoint Z of about 80-j300.

My model comes up with a feedpoint impedance of 5.4485E+01-j2.8560E+03

So I tried to build a coil with an inductive reactance to cancel that .. I
came up with 3.13 microhenries.

Is my model way off?????

I think you might want to take two looks at it:

- Check the radiation pattern. By lengthening it to get a 50-ohm
resistive component in the feedpoint, I suspect you've given up
much of the gain benefit of a true 5/8-wave radiator. You may
actually have less towards-the-horizon power and sensitivity than
you'd get with a 1/4-wave groundplane or a 1/2-wave J-pole.

- Check the formula and actual inductance for your coil.

With so much capacitive reactance from the radiator to cancel out with
the coil, I suspect that you may also find that you've calculated out
an antenna which is going to be rather narrow-banded. Even slight
frequency shifts, or errors in the coil winding (a fraction of a turn)
could leave you with a lot of residual reactance and an unacceptable
SWR.

--
Dave Platt AE6EO
Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

On Mon, 14 Mar 2005 14:26:57 GMT, "john doe"
wrote:

Can anyone point me to a good site for construction a 5/8 wave 2M antenna???

I tried home-brewing one already with poor results .... a conductor up a
piece of PVC .... same conductor wound into a coil at the bottome
connected in series to the center of the coax. 4 1/4 wave radials
connected to outer conductor.

Thanks,

-- Rob
ka2pbt


Can the 5/8 wave antenna be matched with a tuning stub such as the
J-Pole uses?


--
73 for now
Buck
N4PGW

Can the 5/8 wave antenna be matched with a tuning stub such as the
J-Pole uses?...

Probably, but seems easier to to use a 5 turn coil at
the base...But I agree...A 5/8 GP is not a very good 2m
antenna. Better off building a high quality 1/4 wave GP.
Seriously... By high quality, I mean instead of 3-4 sloped
radials, use 6-8 radials, and even better , use a 2nd set
for decoupling. You can also build "sleeve" versions.
5/8's are ok on HF, where the angles are not critical, and on
10m, will usually be the best choice. But on 2m, they are usually
lame.
But, don't take my word. Build a 5/8 GP, and then build a 1/4
GP, and see which is best. I bet the 1/4 wave wins. For a
5/8 to work well on VHF/UHF, it needs to be a collinear dual
5/8 design, or at the least, use sloped 5/8 or 3/4 wave radials.
5/8's with 1/4 wave radials have pretty lousy patterns for 2m use.
MK

Can the 5/8 wave antenna be matched with a tuning stub such as the
J-Pole uses?


--
73 for now
Buck
N4PGW

Yes, A 5/8 radiator with an input Z of 80-j300 with series 19cm of 450
ohm line will be about 55-j0 ohms. Of course all of this is highly
dependent on the antenna environment on 2m.
Gary N4AST

Ok,

I have not followed all of this thread.

One of the pdf's shows a shunt tapped inductor
as the base loading.

This is an auto-transformer match.

The entire inductor is set to cancel the capacitive reactance
of the aerial, then the coax feed is tapped to the 50R point,
The result is 50R J0.


For 5/8 aerials you may want to look at a physical short radiator
with a capacity hat to lower the angle of radiation.



wrote in message
ups.com...




Can the 5/8 wave antenna be matched with a tuning stub such as the
J-Pole uses?


--
73 for now
Buck
N4PGW

Yes, A 5/8 radiator with an input Z of 80-j300 with series 19cm of 450
ohm line will be about 55-j0 ohms. Of course all of this is highly
dependent on the antenna environment on 2m.
Gary N4AST

I'm trying to understand the shunt tapped inductor as a circuit.

How does it work??? As long as the inductor cancels out the reactance of
the radiator you just need to find the 50 ohm point on the coil ... or for
that matter any feedline impedance???

Pardon my lack of knowledge.

de ka2pbt

john doe wrote:
I'm trying to understand the shunt tapped inductor as a circuit.

How does it work??? As long as the inductor cancels out the reactance
of
the radiator you just need to find the 50 ohm point on the coil ...
or for
that matter any feedline impedance???

Pardon my lack of knowledge.

de ka2pbt

Hi John, Is the shunt tapped inductor you are refering to is a coil
with one end connected to the 5/8 radiator, the other end to the
radials, the coax shield to the radials, and the center conductor
tapped up the coil from the radials? If this is the case, according to
the Smith Chart, you can only get a perfect 50 ohm match if the real
part of the impedance you are trying to match less than 50 ohms. Our
models indicate this is not the case with a 5/8 radiator.
The shunt tapped inductor is actually a 2 element matching system,
with both elements being inductors. As you stated, the inductor in
series with the radiator cancels the reactance, and the tapped shunt
inductor provides a 50+j0 point. Works only if the real part of the
impedance is 50 ohms.
Gary N4AST

Let's suppose you have a vertical antenna with a base feedpoint
impedance of, say, 75 - j300 ohms (75 ohms resistance in series with 300
ohms of capacitive reactance), typical of a thin 5/8 wave vertical. A
network consisting of 1275 ohms in *parallel* with 319 ohms of
capacitive reactance has the same impedance. (Note how the Xc isn't much
different from the Xc of the series circuit in this case.) If we put an
inductor with 319 ohms of inductive reactance in parallel with the
antenna (that is, from the base feedpoint to ground), the reactance of
the inductor cancels out the capacitive reactance of the antenna, and
we're left with 1275 ohms of pure resistance from the antenna base to
ground (that is, across the inductor).

We can use the inductor as an autotransformer. If we tap up on the
inductor some fraction k of the whole way up, the impedance we'll see at
that tap will be very nearly 1275 * k^2, and it'll be purely resistive
(no reactance) because the impedance across the whole coil is purely
resistive. For example, half way up the coil we'll see 1275 * (.5)^2 =
1275 / 4 = 319 ohms. So to get 50 ohms, we tap up sqrt(50/1275) = 20% of
the way up the coil.

Roy Lewallen, W7EL

john doe wrote:
I'm trying to understand the shunt tapped inductor as a circuit.

How does it work??? As long as the inductor cancels out the reactance of
the radiator you just need to find the 50 ohm point on the coil ... or for
that matter any feedline impedance???

Pardon my lack of knowledge.

de ka2pbt