Good evening, Gentlemen.

A thought experiment:

Start with a regular 1/4-wave monopole ground plane. The literature says it
looks like half the value of a dipole, about 35 Ohms, when resonant. It
would be nice to have the resistance at the terminals be a bit higher, and I
very much value a grounded element anyway, so let's let it evolve into a
folded monopole. The literature says it should now have about 4 times the
terminal resistance of the original 1/4-wave we started with (about 140
Ohms). Huh. Now it's a bit high.

They tell me that shortening the antenna below resonance will lower the
resistance and introduce capacitance. But I think I have also seen in the
literature that the antenna can be viewed as a transmission line. A shorted
portion of parallel conductor transmission line (the folded monopole) less
than 1/4-wave long looks inductive. But wait! Which will win? Will the
shortness of the antenna look capacitive or will the transmission line
dominate and the antenna will look inductive?

Even better, is there some choice of the folded section wire diameters and
spacing that will give an inductance that will exactly offset the
capacitance due to shortness? So, then, is there a folded monopole of such
dimensions that the resistance is 50 Ohms (due to being shorter than 1/4
wave) with no terminal reactance (due to the inductive design of the
"transmission line" cancelled by the shortness of the antenna's
capacitance)?

Brain hurts.

John, KD5YI

Sorry!


"The other John Smith" wrote in message
nk.net...

I've recently done some NEC-2 (MultiNEC) modeling of folded dipoles
which might help answer some of your questions. Translating the
results to folded monoploes should be fairly straightforward.

The model is a half-wave folded dipole for 14.2 MHz in freespace,
resonant at 33.15 feet using #18 wire with 2 inch spacing. The
center-fed input impedance is 289 - j0.01, which is 4 times the
resonant impedance of 72 ohms for a conventional dipole. A folded
1/4-wavelength monopole would have half that impedance, or about 144
ohms.

Examining the R-X curves for this dipole shows that it has
characteristics very similar to a 3/2-wavelength dipole, operating at
its third harmonic, and on a relatively low-slope part of the curves,
indicating a low Q and good bandwidth, similar to a fat dipole.

Shortening the antenna increases capacitive reactance, as might be
expected. However, input resistance *increases* as the length
decreases, which is contrary to our experience with common
1/2-wavelength dipoles. This is because we're on the high side of
full-wave resonance, where very high resistance values exist at its
peak. As we shorten the antenna, we're climbing the full-wave
resistance curve, which peaks when the antenna length is 22 feet. If
we further shorten the antenna past full-wave resonance, we now begin
experiencing a "normal" decrease in resistance as we "slide" back down
the low side of the full-wave resistance spike. However, capacitive
reactance has now quickly changed to inductive reactance as we crossed
full-wave resonance.

If we continue to shorten the folded antenna length, we come to a
length of about 17 feet where the input impedance is 50 + j2000 ohms.
Notice that the impedance is *inductive*, not capacitive as we are
accustomed to seeing with ordinary short dipoles. The inductive 2000
ohms can be cancelled with a series capacitor (or other suitable
matching network). Q has increased (because we're on a relatively
steep part of the R-X curves) and bandwidth has narrowed considerably
from the resonance at 33.15 feet.

So, by reducing the length of the 1/2-wavelength folded dipole from
33.15 feet to 17 feet, we have a 50 ohm resistive impedance by
matching the inductive reactance with a capacitor (or split capacitor)
instead of the usual lossy, low-Q loading coils. Gain and patterns
appear to be the same as a conventional dipole.

Translated to a monopole, the length would be a little more than half
the dipole's 17 feet, to boost feed point resistance from 25 ohms to
50 ohms. My guess is (I haven't modeled it) that this antenna
functions much like a 3/8-wavelength monopole, although much shorter.
Actually building this antenna and placing it the real world will
obviously change the above values.

Unfortunately, it doesn't appear that any combination of element size
and spacing will offset the need for impedance matching with the
shortened folded dipole or monopole.

I hope this makes sense. I'm sure Roy, Cecil, Tom, and others might
have comments/corrections that will be helpful to me and others who
are relative neophytes in the wonderful world of antennas.

Al WA4GKQ

Even better, is there some choice of the folded section wire diameters and
spacing that will give an inductance that will exactly offset the
capacitance due to shortness? So, then, is there a folded monopole of such
dimensions that the resistance is 50 Ohms (due to being shorter than 1/4
wave) with no terminal reactance (due to the inductive design of the
"transmission line" cancelled by the shortness of the antenna's
capacitance)?

Modeling a 17 foot folded dipole made from copper #18 wire spaced at 2
inches at 14.2 MHz with EZNEC shows a feedpoint impedance of 46.1 +
j1893 ohms. This can be resonated, as Richard Harrison recently pointed
out, with a series capacitor. There's no free lunch, though -- at 1 kW,
the voltage across the capacitor is almost 9000 V RMS (about 12,000
volts peak), and even at 10 watts, it's almost 900 volts RMS. Besides
concerns about arcing, you'd have to make sure the insulation across the
capacitor is very good, since even a very small leakage current will
cause significant loss. And you end up with a fairly narrow-banded
antenna, with the 2:1 SWR bandwidth of about 130 kHz. The loss due to
finite wire conductivity is 1.9 dB, which might or might not be
acceptable, depending on the particular use. Increasing the wire size
will reduce the loss, but also the bandwidth -- introducing loss nearly
always improves bandwidth, so reducing it narrows the bandwidth. Without
wire loss, and assuming the resulting 29 ohm feedpoint impedance is
transformed to 50 ohms, the 2:1 SWR bandwidth becomes 80 kHz. Like a
great number of variations, this antenna would surely be useful to some
people in some situations, and might well be better than some other
alternatives. But here's an antenna rule you can take to the bank:
Small--broad band--efficient, choose any two. Any time either a modeling
program or an antenna inventor or seller tell you any different, you
should be very, very skeptical.

Roy Lewallen, W7EL

alhearn wrote:
I've recently done some NEC-2 (MultiNEC) modeling of folded dipoles
which might help answer some of your questions. Translating the
results to folded monoploes should be fairly straightforward.

The model is a half-wave folded dipole for 14.2 MHz in freespace,
resonant at 33.15 feet using #18 wire with 2 inch spacing. The
center-fed input impedance is 289 - j0.01, which is 4 times the
resonant impedance of 72 ohms for a conventional dipole. A folded
1/4-wavelength monopole would have half that impedance, or about 144
ohms.

Examining the R-X curves for this dipole shows that it has
characteristics very similar to a 3/2-wavelength dipole, operating at
its third harmonic, and on a relatively low-slope part of the curves,
indicating a low Q and good bandwidth, similar to a fat dipole.

Shortening the antenna increases capacitive reactance, as might be
expected. However, input resistance *increases* as the length
decreases, which is contrary to our experience with common
1/2-wavelength dipoles. This is because we're on the high side of
full-wave resonance, where very high resistance values exist at its
peak. As we shorten the antenna, we're climbing the full-wave
resistance curve, which peaks when the antenna length is 22 feet. If
we further shorten the antenna past full-wave resonance, we now begin
experiencing a "normal" decrease in resistance as we "slide" back down
the low side of the full-wave resistance spike. However, capacitive
reactance has now quickly changed to inductive reactance as we crossed
full-wave resonance.

If we continue to shorten the folded antenna length, we come to a
length of about 17 feet where the input impedance is 50 + j2000 ohms.
Notice that the impedance is *inductive*, not capacitive as we are
accustomed to seeing with ordinary short dipoles. The inductive 2000
ohms can be cancelled with a series capacitor (or other suitable
matching network). Q has increased (because we're on a relatively
steep part of the R-X curves) and bandwidth has narrowed considerably
from the resonance at 33.15 feet.

So, by reducing the length of the 1/2-wavelength folded dipole from
33.15 feet to 17 feet, we have a 50 ohm resistive impedance by
matching the inductive reactance with a capacitor (or split capacitor)
instead of the usual lossy, low-Q loading coils. Gain and patterns
appear to be the same as a conventional dipole.

Translated to a monopole, the length would be a little more than half
the dipole's 17 feet, to boost feed point resistance from 25 ohms to
50 ohms. My guess is (I haven't modeled it) that this antenna
functions much like a 3/8-wavelength monopole, although much shorter.
Actually building this antenna and placing it the real world will
obviously change the above values.

Unfortunately, it doesn't appear that any combination of element size
and spacing will offset the need for impedance matching with the
shortened folded dipole or monopole.

I hope this makes sense. I'm sure Roy, Cecil, Tom, and others might
have comments/corrections that will be helpful to me and others who
are relative neophytes in the wonderful world of antennas.

Al WA4GKQ

Roy Lewallen wrote:
"Modeling a 17 foot folded dipole made from copper #18 wire spaced at 2
inches at 14.2 MHz with EZNEC shows a feedpoint impedance of 46.1 +
j1893. This can be resonated as Richard Harrison recently pointed out
with a series capacitor. There`s no free lunch though---at 1kW, the
voltage across the capacitor is almost 9000 V RMS (about 12000 volts
peak) and even at 10 watts its almost 900 volts RMS.

I agree that at 1KW input to Roy`s folded dipole the power-correction
capacitor has 8466 volts across it. That`s close enough to 9 KV for me.

No single antenna fits all applications and alterations may adapt an
antenna for more than one application.

Antennas have a voltage to current ratio (Zo) which is a function of
position along along the conductor. Zo is also a function of conductor
length to diameter ratio. Fat wires have lower Zo than do thin wires.
Low Zo means low voltage (relatively).

Also spacing the folded antenna conductors farther apart lowers
impedance and Q. This helps bandwidth.

Raising the current by lowering Zo is no panacea as the volts across the
capacitor are Amps x XC.

The capacitance of 1893 ohms at 14.2 MHz is about 0.000006 pF. If the
plate size is kept significant, the spacing should be good for 12 KV
with no problem.

The Andrew Corporation folded monopoles I am familiar with were usually
working with 500-watt VHF FM transmitters in our land-mobile operations.
Bandwidth required was 2f + 2d, if I recall, and the (f) was maximum
modulation frequency, and the (d) was the peak deviation. Bandwidth was
less than 20 KHz. Half-duplex was the communications mode so we needed
the antenna only to work at one carrier frequency.

It was a cakewalk. Antennas only flashed over on lightning strikes and
the 50-ohm Heliax saw most of the lightning as a common-mode disturbance
and rejected its passage through the coax (via counter-emf from coax
distributed inductance).

The VHF Andrew folded monopole element was similar to the slide pipe on
a trombone only made of stainless steel. It had clamps to hold its
position once set. Andrew set its length for 50-ohms at our frequency, I
suppose, and adjusted the reactance for a net zero. When we set it atop
our tower we always had about 500 watts forward and nearly zero
reflected power. Some of these are surely operating well at this moment
after 50 years or more, though they`ve surely accumulated many small
pits from countless lightning strikes.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

. . .
The capacitance of 1893 ohms at 14.2 MHz is about 0.000006 pF. If the
plate size is kept significant, the spacing should be good for 12 KV
with no problem.
. . .

By my reckoning, a capacitive reactance of 1893 ohms at 14.2 MHz is 5.9 pF.

Roy Lewallen, W7EL

"Roy Lewallen" wrote in message
...
Modeling a 17 foot folded dipole made from copper #18 wire spaced at 2
inches at 14.2 MHz with EZNEC shows a feedpoint impedance of 46.1 +
j1893 ohms.


I didn't know I could model loops in EZNEC. But now I see that it has
problems only with small loops. I guess a 1/4-wave loop is not considered
small. I'll go back and try it.

My fall-back plan is to make a simple 1/4-wave resonant folded monopole and
feed it with a 1/4-wave length of 75 Ohm coax. There will probably be some
mismatch, but I think it will be tolerable.

Thanks, guys.

John

Roy Lewallen, W7EL wrote:
"By my reckoning, a capacitive reactance of 1893 ohms at 14.2 MHz is 5.9
pF."

Dang me and my sliderule. Neither of us keeps track of decimals very
well. 6 pF can be obtained with wide spacing and high breakdown volts
with small plates. I don`t see much of a hurdle to clear because 6 pF is
a small capacitance.

As a practical matter, Andrew Corporation used another method to tune
its folded monopoles, I believe, because they had d-c continuity. They
supplied these antennas for decades to work at VHF. To move from 10m to
20m brings problems of scale, mechanical and electrical.

The original question said that the open-circuit ground plane has a
35-ohm feedpoint (at some elevation), and a folded ground plane has
about 140 ohms as a feedpoint. Neither ground plane matches the usual
coax at the antenna resonant frequency. Commercial antenna makers
advertise and deliver open-circuit and folded radiator ground plane
antennas which are nearly 50 + j0 ohms feedpoint impedance at a
specified frequency when mounted high and in the clear.

The folded radiator offers more lightning protection than the
open-circuit radiator. The folded radiator contains the ability to
step-up feedpoint impedance in cases where an open-circuit radiator
would have an inconveniently low feedpoint. Most TV yagis, for example,
use a folded dipole as the driven eleement due to the low feedpoint
impedance caused by mutual coupling with the parasitic elements.

Most energy in a lightning strike is at lower frequencies. Tune the
bands during thunderstorm season and notice where the static crashes are
worse, though much of this is due to propagation, some is due to the
shape of the transient. Where the folded antenna loop is small in terms
of wavelength, the loop is nearly a short-circuit and differential
energy is small.

I saw lightning problems solved by replacing open-circuit antennas with
folded-element antennas. As lightning is an interference problem taken
to an extreme, folded elements are also useful in solving some other
interference problems. But there are cautions. A folded dipole has a
resonance where it is only 1/4-wave from tip to tip. Its circumference
is 1/2-wave and resonates. This gives a folded dipole twice as many
resonances as an open-circuit dipole.

I make arithmetic mistakes more frequently when I don`t know for sure
that the number I calculate is reasonable or not. I do know that 20-kV
to 40-kV sparkplug voltage does not ordinarily leap many feet through
the air. I also have a formula for capacitance:

CpF = 0.225 K A / S

CpF = capacitance in pF

K = dielectric constant

A = area of one of the 2-plate capacitor plates
(sq. in.)

S = spacing between the plates in inches

For air, K = 1.0006

For a vacuum, K = 1

6 pF is not much so it should be easy to create.

Best regards, Richard Harrison, KB5WZI

"Roy Lewallen" wrote in message
...
Modeling a 17 foot folded dipole made from copper #18 wire spaced at 2
inches at 14.2 MHz with EZNEC shows a feedpoint impedance of 46.1 +
j1893 ohms.


Now I'm confused again.

I modeled a monopole at 434 MHz. Varying the vertical element showed the
terminal impedance to get lower in resistance and become increasingly
capacitive as the vertical element is shortened.

I thought it was supposed to be backwards from the usual unfolded
monopole.?.


John

John wrote:

I didn't know I could model loops in EZNEC. But now I see that it has
problems only with small loops. I guess a 1/4-wave loop is not considered
small. I'll go back and try it.
. . .

Because EZNEC uses NEC-2 for calculations, it has the same problems with
small loops that NEC-2 does. It's able to model any kind of antenna that
NEC-2 can, within its segment limitation.

Roy Lewallen, W7EL