160 meter mobile -- Reality Check
 

My bag is VHF, so forgive me if I've bumbled up the calculations. My
numbers show that a standard 108" whip has an input impedance of somewhere
around 0.04 ohms in series with a 22 pf capacitor.

These numbers are so far from what I normally deal with that I'm not sure
that I'm right, and I'd appreciate somebody who actually works down in this
area giving my numbers a reality check.

If they ARE right, how in heaven do most people match to 0.04 ohms? The 22
pf I can resonate out with a 1 millihenry choke (or thereabouts), but how do
most people match 50 ohms to fractional ohms in the homebrew arena -- that
is, without just going out and buying a "magical matching box" of some sort?

Jim

On Sat, 19 Mar 2005 12:45:53 -0800, "RST Engineering"
wrote:

My bag is VHF, so forgive me if I've bumbled up the calculations. My
numbers show that a standard 108" whip has an input impedance of somewhere
around 0.04 ohms in series with a 22 pf capacitor.


Not quite. Over perfect ground the Z is ~ 0.37 -j8170 @ 1.9 MHz for a
12mm diameter radiator.

So you add an inductor of +j8170 with say Q=250. That gives Z =
(0.37+32.68) +j0 = 33.05 +j0. Then you add 16.95 ohm of ground loss
and Z = 50 +j0.

A perfect match with a gain of -17 dBi. :) Easy huh?


These numbers are so far from what I normally deal with that I'm not sure
that I'm right, and I'd appreciate somebody who actually works down in this
area giving my numbers a reality check.

If they ARE right, how in heaven do most people match to 0.04 ohms? The 22
pf I can resonate out with a 1 millihenry choke (or thereabouts), but how do
most people match 50 ohms to fractional ohms in the homebrew arena -- that
is, without just going out and buying a "magical matching box" of some sort?

Jim

RST Engineering wrote:
My bag is VHF, so forgive me if I've bumbled up the calculations. My

numbers show that a standard 108" whip has an input impedance of
somewhere
around 0.04 ohms in series with a 22 pf capacitor.

These numbers are so far from what I normally deal with that I'm not
sure
that I'm right, and I'd appreciate somebody who actually works down
in this
area giving my numbers a reality check.

If they ARE right, how in heaven do most people match to 0.04 ohms?
The 22
pf I can resonate out with a 1 millihenry choke (or thereabouts), but
how do
most people match 50 ohms to fractional ohms in the homebrew arena --
that
is, without just going out and buying a "magical matching box" of
some sort?

Jim
Hi Jim, Your numbers look about right for a 108" whip on 1.8mhz. To
feed a small whip like this you would have to "brute force" it with a
large loading coil at the base. The efficiency will be low and the
bandwidth narrow. To get any efficiency at all, the 2:1 swr bandwidth
will only be around 5khz. Once you tune it up, it will be a single
frequency antenna.
There are things like adding capacitive loading at the top of the
whip, and moving the inductive loading to the center (have to cut the
whip in half). This will make the feedpoint Z a little more managable.
I once loaded an 8' antenna on 160, just to see if I could, but never
got on the air with it. I read somewhere that in the 50's, law
enforcement 2-way radios operated close to our 160m ham band. Wonder
what kind of antennas they used on the police cars. Of course the
distances they were interested in are different from hams.
Gary N4AST

"Wes Stewart" wrote in message
...
On Sat, 19 Mar 2005 12:45:53 -0800, "RST Engineering"
wrote:

My bag is VHF, so forgive me if I've bumbled up the calculations. My
numbers show that a standard 108" whip has an input impedance of somewhere
around 0.04 ohms in series with a 22 pf capacitor.


Not quite. Over perfect ground the Z is ~ 0.37 -j8170 @ 1.9 MHz for a
12mm diameter radiator.

So you add an inductor of +j8170 with say Q=250. That gives Z =
(0.37+32.68) +j0 = 33.05 +j0. Then you add 16.95 ohm of ground loss
and Z = 50 +j0.

A perfect match with a gain of -17 dBi. :) Easy huh?


These numbers are so far from what I normally deal with that I'm not sure
that I'm right, and I'd appreciate somebody who actually works down in
this
area giving my numbers a reality check.

If they ARE right, how in heaven do most people match to 0.04 ohms? The
22
pf I can resonate out with a 1 millihenry choke (or thereabouts), but how
do
most people match 50 ohms to fractional ohms in the homebrew arena -- that
is, without just going out and buying a "magical matching box" of some
sort?

Jim



The following code produces Zin = 0.117 - j2717 ohms at 1.9 MHz. What did
I do wrong? Note the high segmentation to place the feed-point near the
base of the antenna.

Frank

CM 9 ft monopole
CE
GW 1 108 0 0 0 0 0 108 0.25
GS 0 0 0.025400
GE 1
GN 1
EX 0 1 1 00 1 0
LD 5 1 1 108 5.8001E7
FR 0 9 0 0 1.8 0.025
RP 0 181 1 1000 -90 0 1.00000 1.00000
EN

Frank wrote:

The following code produces Zin = 0.117 - j2717 ohms at 1.9 MHz. What did
I do wrong? Note the high segmentation to place the feed-point near the
base of the antenna.

Frank

CM 9 ft monopole
CE
GW 1 108 0 0 0 0 0 108 0.25
GS 0 0 0.025400
GE 1
GN 1
EX 0 1 1 00 1 0
LD 5 1 1 108 5.8001E7
FR 0 9 0 0 1.8 0.025
RP 0 181 1 1000 -90 0 1.00000 1.00000
EN

If you follow NEC guidelines on the minimum recommended segment
length/wire radius ratio, you should have no more than 17 segments; the
model has 127. In this case, though, it doesn't seem to be disturbing
NEC-2 very much. Also, the half inch diameter is somewhat greater than
most CB whips. If you drop the diameter to a quarter inch, you'll get a
substantially greater reactance which would require a larger loading
inductor and hence result in lower efficiency. Otherwise it looks ok to me.

As Wes pointed out, the ground system and loading inductor losses will
be so large as to make the feedpoint resistance (and wire loss)
insignificant. The zero-loss feedpoint resistance is, however, useful in
determining efficiency. If you set the wire loss to zero by removing the
LD "card", you get the more useful value of feedpoint resistance when
losses are zero, 0.098 ohms. Using Wes' value of about 50 ohm feedpoint
resistance when losses are included, you can calculate the efficiency as
0.098/50 = 0.2%, or 2 watts radiated for each kW applied. In American
mobile terms that's 1.5 watts per horsepower. I'm always glad to see
more QRP signals on the band.

Roy Lewallen, W7EL

To obtain obtain some idea of the numerical values involved with
mobile, short, antennas on 160 meter and other low frequency bands,
download programs -

HELICAL3
VERTLOAD
LOADCOIL

from website below. Takes only a few seconds to download. Not zipped
up. Run immediately.

With a very short antenna, to obtain a useful efficiency, a physically
large loading coil is essential. The only space available is up the
antenna. So the best thing to do is extend the coil up the antenna
from near the base, in the form of a long slender helical winding on a
pvc plastic pipe.

The whole thing behaves as a short 1/4-wave resonant vertical. There
is a short rod at the top which is pruned to resonate the antenna in
the required band. It is inevitably a single frequency job. The
bandwidth on 160m is only a few kHz.

With a 100 watt transmitter, when both vehicles are in the low-noise
countryside, it is possible to work 100 miles or more, on groundwave,
in daylight. Much further via skywave in darkness. G3YXM has worked
transatlantic from a car in Scotland to a base station in Canada on
occasions.

Maximum overall antenna heights are about 9 feet above roof of car
with the coil about 2" in diameter, the coil extending to a height of
6 or 7 feet.

The most tedious procedure is pruning all antenas in a group of
mobiles to the same frequency on the 160 meter band. But it has been
done. Mobile to a base station is easy. For many years the standard
UK frequency for mobiles was 1930 kHz.

Matching from antenna base to a 50-ohm transmiier can be accomplished
by a single or bank of mica capacitors of several hundred pF.

For quite a number of years, mobile working on 160m was quite popular
in the UK using helically-wound antennas.
----
.................................................. ..........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. ..........

Frank wrote:
"Wes Stewart" wrote in message
...
On Sat, 19 Mar 2005 12:45:53 -0800, "RST Engineering"
wrote:

My bag is VHF, so forgive me if I've bumbled up the calculations.
My
numbers show that a standard 108" whip has an input impedance of
somewhere
around 0.04 ohms in series with a 22 pf capacitor.


Not quite. Over perfect ground the Z is ~ 0.37 -j8170 @ 1.9 MHz
for a
12mm diameter radiator.

So you add an inductor of +j8170 with say Q=250. That gives Z =
(0.37+32.68) +j0 = 33.05 +j0. Then you add 16.95 ohm of ground
loss
and Z = 50 +j0.

A perfect match with a gain of -17 dBi. :) Easy huh?


These numbers are so far from what I normally deal with that I'm
not sure
that I'm right, and I'd appreciate somebody who actually works down
in
this
area giving my numbers a reality check.

If they ARE right, how in heaven do most people match to 0.04 ohms?
The
22
pf I can resonate out with a 1 millihenry choke (or thereabouts),
but how
do
most people match 50 ohms to fractional ohms in the homebrew arena
-- that
is, without just going out and buying a "magical matching box" of
some
sort?

Jim



The following code produces Zin = 0.117 - j2717 ohms at 1.9 MHz.
What did
I do wrong? Note the high segmentation to place the feed-point near
the
base of the antenna.

Frank

CM 9 ft monopole
CE
GW 1 108 0 0 0 0 0 108 0.25
GS 0 0 0.025400
GE 1
GN 1
EX 0 1 1 00 1 0
LD 5 1 1 108 5.8001E7
FR 0 9 0 0 1.8 0.025
RP 0 181 1 1000 -90 0 1.00000 1.00000
EN

Hi Frank, Well, I got different numbers than both of you. Must depend
on the ground type of the models, as well as the # of segments.
If you are trying to match an antenna with fractions of an ohm
resistive and thousands reactive, the differences we are all observing
will not make a whole lot of difference in the matching network you
decide to use. It will be big and bulky, and very inefficient.
Gary N4AST

On Sat, 19 Mar 2005 23:23:30 GMT, "Frank"
wrote:

[snip]

The following code produces Zin = 0.117 - j2717 ohms at 1.9 MHz. What did
I do wrong? Note the high segmentation to place the feed-point near the
base of the antenna.

Frank

CM 9 ft monopole
CE
GW 1 108 0 0 0 0 0 108 0.25
GS 0 0 0.025400
GE 1
GN 1
EX 0 1 1 00 1 0
LD 5 1 1 108 5.8001E7
FR 0 9 0 0 1.8 0.025
RP 0 181 1 1000 -90 0 1.00000 1.00000
EN

My bad. I hate it when that happens. A typo on the length on my
part. 101 instead of 108.

It's more like 0.12 -j2890 in EZNEC. I'm not real fluent in NEC decks
but I think you are using too many segments.

"Roy Lewallen" wrote in message
...
Frank wrote:

The following code produces Zin = 0.117 - j2717 ohms at 1.9 MHz. What
did I do wrong? Note the high segmentation to place the feed-point near
the base of the antenna.

Frank

CM 9 ft monopole
CE
GW 1 108 0 0 0 0 0 108 0.25
GS 0 0 0.025400
GE 1
GN 1
EX 0 1 1 00 1 0
LD 5 1 1 108 5.8001E7
FR 0 9 0 0 1.8 0.025
RP 0 181 1 1000 -90 0 1.00000 1.00000
EN

If you follow NEC guidelines on the minimum recommended segment
length/wire radius ratio, you should have no more than 17 segments; the
model has 127. In this case, though, it doesn't seem to be disturbing
NEC-2 very much. Also, the half inch diameter is somewhat greater than
most CB whips. If you drop the diameter to a quarter inch, you'll get a
substantially greater reactance which would require a larger loading
inductor and hence result in lower efficiency. Otherwise it looks ok to
me.

As Wes pointed out, the ground system and loading inductor losses will be
so large as to make the feedpoint resistance (and wire loss)
insignificant. The zero-loss feedpoint resistance is, however, useful in
determining efficiency. If you set the wire loss to zero by removing the
LD "card", you get the more useful value of feedpoint resistance when
losses are zero, 0.098 ohms. Using Wes' value of about 50 ohm feedpoint
resistance when losses are included, you can calculate the efficiency as
0.098/50 = 0.2%, or 2 watts radiated for each kW applied. In American
mobile terms that's 1.5 watts per horsepower. I'm always glad to see more
QRP signals on the band.

Roy Lewallen, W7EL

I should have read the manual before modeling, but I was so concerned that
the feed-point be near the bottom. Changing the segmentation to 17,
removing the LD card, and reducing the radius to 0.125 ", calculates the
feed-point impedance as: 0.123 - j3286. I considered Jim's calculation of
0.37 - j8170 to be far too reactive. The fact is these points are so close,
when observed on the Smith chart, as to be virtually identical.

Analyzing a shunt L/series C matching network, with inductor Q at 300,
indicates a loss of 21.5dB with 30 kV RMS at the base of the antenna (1.5 kW
into the matching network). (In the case of 0.37 - j8170 the loss is 21.8
dB, with 42 kV at the base). These voltages are so large, I am beginning to
wonder if I goofed again! The losses, at least, are the same order of
magnitude as Roy's calculation.

73,

Frank

If you had made this model with EZNEC, you would have seen a warning
that the segment length is shorter than recommended. And you could have
entered the dimensions directly in inches (or immediately converted
between meters and inches). The free demo program is adequate for this
model, providing you don't want more than 20 segments.

Roy Lewallen, W7EL

Frank wrote:

I should have read the manual before modeling, but I was so concerned that
the feed-point be near the bottom. Changing the segmentation to 17,
removing the LD card, and reducing the radius to 0.125 ", calculates the
feed-point impedance as: 0.123 - j3286. I considered Jim's calculation of
0.37 - j8170 to be far too reactive. The fact is these points are so close,
when observed on the Smith chart, as to be virtually identical.

Analyzing a shunt L/series C matching network, with inductor Q at 300,
indicates a loss of 21.5dB with 30 kV RMS at the base of the antenna (1.5 kW
into the matching network). (In the case of 0.37 - j8170 the loss is 21.8
dB, with 42 kV at the base). These voltages are so large, I am beginning to
wonder if I goofed again! The losses, at least, are the same order of
magnitude as Roy's calculation.

73,

Frank