Efficiency of 200-ohm hairpin matching
 

Hi Tony,

I would be happy to have a look at the file and add the hairpin if you
wish. But I can also guide you through doing it yourself. It should
be quite easy. Simply click on the " Trans Lines" button, the second
one below Sources in the main window. Specify the first end the
same as the source--perhaps 50% along wire 2, assuming the D.E. is
wire 2. That end will then be in parallel with the source. You can
type "S" in for the other end's wire #, or if you click in the End 2
Wire # box, you should see a list of "open" and "short" and you can
select "short" there. Then enter the length, Z0, velocity factor (1)
and loss. Reverse or normal doesn't matter since only one end is
connected.

Thanks on you guidance. My hairpin is a U of aluminum tube connected directly to
the antenna feedpoints. The U center is connected to the boom.

The hairpin tube diameter is about 0.8 cm. The three sides of the U are
approximately 35 cm, 10 cm and 35 cm. I presume that I shall enter 35 cm as the
transmission line length, but how to calculate Z0? Moreover what about the
velocity factor, perhaps 0.98 for a tube in open air?

Would you have a formula at hand?

Thanks and 73

Tony I0JX

Efficiency of 200-ohm hairpin matching
 

On Apr 9, 11:36 am, "Antonio Vernucci" wrote:
Hi Tony,

I would be happy to have a look at the file and add the hairpin if you
wish. But I can also guide you through doing it yourself. It should
be quite easy. Simply click on the " Trans Lines" button, the second
one below Sources in the main window. Specify the first end the
same as the source--perhaps 50% along wire 2, assuming the D.E. is
wire 2. That end will then be in parallel with the source. You can
type "S" in for the other end's wire #, or if you click in the End 2
Wire # box, you should see a list of "open" and "short" and you can
select "short" there. Then enter the length, Z0, velocity factor (1)
and loss. Reverse or normal doesn't matter since only one end is
connected.

Thanks on you guidance. My hairpin is a U of aluminum tube connected directly to
the antenna feedpoints. The U center is connected to the boom.

The hairpin tube diameter is about 0.8 cm. The three sides of the U are
approximately 35 cm, 10 cm and 35 cm. I presume that I shall enter 35 cm as the
transmission line length, but how to calculate Z0? Moreover what about the
velocity factor, perhaps 0.98 for a tube in open air?

Would you have a formula at hand?

Thanks and 73

Tony I0JX

Hi Tony,

Since the spacing is so large (10cm), it may be better to enter the
hairpin as wires in the model. It might be interesting to compare the
results between wires and a transmission line. First the wires:

I assume that the D.E. is one wire in the model you have now. I
suppose it has an odd number of segments, so that the source can go
just in the middle. I'm assuming the hairpin parallel tube segments
are center-to-center 10cm apart. I will assume the D.E. is currently
250cm long, all at some x value, and y extending from -1.25m to
+1.25m, as one "wire" of some diameter. To be able to connect the
hairpin at the right places, we need wire ends at -.05m and +.05m, so
we can change the D.E. wire to be one segment between y=-.05m and +.
05m, and then add 2 wires the same diameter as the D.E., one from
y=-1.25m to -.05m, and one from y=+.05m to +1.25m, and each with about
half as many segments as the original D.E., all at the original x
value. (Or exchange x and y if I have guessed wrong.) So if the
original D.E. had 11 segments, put 5 segments on each of the two new
wires. Now add three new wires to represent the hairpin U. Assuming
the hairpin is oriented perpendicular to the plane of the antenna, it
will be all at the same x value as the D.E. I'll assume the antenna
is at z=0. So the wires will have ends at x,-.05,0;x,-.05,.35 and x,-.
05,.35;x,+.05,.35 and x,+.05,.35;x,+.05,0 -- and all with diameter 8mm
or .008m.

For the transmission line model: the diameter (8mm) and spacing
(100mm center to center) tells me the line impedance is about 380
ohms. It should be OK to assume VF=1.00, if there is no insulation
around the tube. One problem is that the short across the end is not
really a short--it is more an inductance. We could go to the trouble
of calculating the inductance and putting that as a load on the end of
the transmission line instead of just a short, but it should be a good
approximation to just lengthen the line by half the length of the
"short"; I would enter the line length just as 40cm instead of 35. (I
won't be surprised if Owen pops in here with either a confirmation or
a better suggestion!)

Cheers,
Tom

Efficiency of 200-ohm hairpin matching
 

I assume that the D.E. is one wire in the model you have now. I
suppose it has an odd number of segments, so that the source can go
just in the middle. I'm assuming the hairpin parallel tube segments
are center-to-center 10cm apart. I will assume the D.E. is currently
250cm long, all at some x value, and y extending from -1.25m to
+1.25m, as one "wire" of some diameter. To be able to connect the
hairpin at the right places, we need wire ends at -.05m and +.05m, so
we can change the D.E. wire to be one segment between y=-.05m and +.
05m, and then add 2 wires the same diameter as the D.E., one from
y=-1.25m to -.05m, and one from y=+.05m to +1.25m, and each with about
half as many segments as the original D.E., all at the original x
value. (Or exchange x and y if I have guessed wrong.) So if the
original D.E. had 11 segments, put 5 segments on each of the two new
wires. Now add three new wires to represent the hairpin U. Assuming
the hairpin is oriented perpendicular to the plane of the antenna, it
will be all at the same x value as the D.E. I'll assume the antenna
is at z=0. So the wires will have ends at x,-.05,0;x,-.05,.35 and x,-.
05,.35;x,+.05,.35 and x,+.05,.35;x,+.05,0 -- and all with diameter 8mm
or .008m.

Thanks Tom,

I did the EZNEC model exactly as you suggested, but I now have another doubt (I
must re-read the EZNEC help one of these days...)

I would say that I shall still feed the driven element at its center, and not
separately feed the two junctions between the hairpin and the driven element

Is that right?

73

Tony I0JX

Efficiency of 200-ohm hairpin matching
 

K7ITM wrote in
:

On Apr 9, 2:08 am, Owen Duffy wrote:
K7ITM wrote

om:

...

Yes, EZNEC has some nice features. I exercised a couple of them
tonight, doing frequency sweeps with inductive and capacitive
matching. I did cheat: since I do not know the dimensions of the
hairpin match, I elected to just use a pure lumped inductance. I
suppose the error compared with a transmission line stub (hairpin)
won't be great. I also used for the first time ASCII file import
for

It probably isn't.

However, you could get an idea from my Two Wire Line Loss Calculator
(http://www.vk1od.net/tl/twllc.htm). It looks like a hairpin made of
4mm dia aluminium 50mm spacing and 150mm in length give an impedance
of 0.02 +j61... so the Q is probably mainly determined by the end
connections.

a o/c stub made of 4mm dia aluminium 50mm spacing and 1300mm in
length give an impedance of 0.07-j80... so, it is not quite as good
electrically, it is unweildly and end connection resistance will
probably still be significant. This is no doubt why people use a
hairpin in preference to an o/c stub!

Owen

Thanks, Owen. Actually, the error I was thinking of was not the Q,
since both a good hairpin and a good helical coil will have Qu's very
much higher than the loaded Q of the matching network -- but rather of
the slope of reactance versus frequency, since the hairpin is a
transmission line and will presumably show a sharp resonance at a
different frequency from the coil's self resonance. But that error
should also be small--negligible as Wim notes.

Hi Tom,

Warning bells sound to me when applications call for reactors fabricated
from TL sections. Not to say that are always bad, but they aren't always
good, and they bear examination.

I think that may have been in Tony's mind over the matching network.

In the case of the hairpin, fabricated from substantial material, it
looks good in this application. Because the line section is so short,
inductive reactance is almost linearly proportional to length, and self
resonance of the line section itself is nearly a decade higher.

My experience with one of the HyGain (now MFJ) 2m Yagis was that VSWR was
stable with weather and over time. I had taken some care to exclude air +
water from key connections, in fact I replaced all the fastners used for
electrical connection with SS when the beam was less than a year old
because of corrosion. Tweny years later the remaining fastners used only
for mechanical retention required replacement due to corrosion. But,
after the first rework, the matching system seemed robust (bird proof)
and reliable.

I have used them since on home made antennas with good success.

Owen

Efficiency of 200-ohm hairpin matching
 

On Apr 9, 2:59 pm, Owen Duffy wrote:
...
Hi Tom,

Warning bells sound to me when applications call for reactors fabricated
from TL sections. Not to say that are always bad, but they aren't always
good, and they bear examination.

....

:-) Yes, indeed. I just this afternoon fabricated a 500MHz LPF that
I wanted to give about 40dB attenuation up to several GHz. My design
was a 5th order elliptical, so I had two parallel tanks for the series
paths and three shunt capacitors. Because the coils (10nH and 15nH)
are so small, I though about using a shorted stub instead, but then
realized that my shorted stub would look like a short when it was 1/2
wave long, not a good thing to have as a series element at a frequency
you want to block. Probably would have worked OK since the two stubs
were different and wouldn't have the same resonance frequency; and by
the time you reached that freq., the shunt caps would be pretty
effective on their own. Anyway, it works fine with tiny coils. Thank
goodness for good microscopes.

Cheers,
Tom

Efficiency of 200-ohm hairpin matching
 

On Apr 9, 1:22 pm, "Antonio Vernucci" wrote:
I assume that the D.E. is one wire in the model you have now. I
suppose it has an odd number of segments, so that the source can go
just in the middle. I'm assuming the hairpin parallel tube segments
are center-to-center 10cm apart. I will assume the D.E. is currently
250cm long, all at some x value, and y extending from -1.25m to
+1.25m, as one "wire" of some diameter. To be able to connect the
hairpin at the right places, we need wire ends at -.05m and +.05m, so
we can change the D.E. wire to be one segment between y=-.05m and +.
05m, and then add 2 wires the same diameter as the D.E., one from
y=-1.25m to -.05m, and one from y=+.05m to +1.25m, and each with about
half as many segments as the original D.E., all at the original x
value. (Or exchange x and y if I have guessed wrong.) So if the
original D.E. had 11 segments, put 5 segments on each of the two new
wires. Now add three new wires to represent the hairpin U. Assuming
the hairpin is oriented perpendicular to the plane of the antenna, it
will be all at the same x value as the D.E. I'll assume the antenna
is at z=0. So the wires will have ends at x,-.05,0;x,-.05,.35 and x,-.
05,.35;x,+.05,.35 and x,+.05,.35;x,+.05,0 -- and all with diameter 8mm
or .008m.

Thanks Tom,

I did the EZNEC model exactly as you suggested, but I now have another doubt (I
must re-read the EZNEC help one of these days...)

I would say that I shall still feed the driven element at its center, and not
separately feed the two junctions between the hairpin and the driven element

Is that right?

73

Tony I0JX

Hi Tony,

I can understand your confusion and doubt. I hope this says it in
clear terms:

In the revised model with the hairpin done as three wires, the
original D.E. single wire was replaced by three wires: a 10cm long
center section which should have the source in its center, and a
"left" and a "right" outer section. Those three sections are three
separate wires. The reason to break the original single wire into
three wires is so that there will be wire "ends" to connect the
hairpin wires to. You can only connect wires at their ends, even
though they have multiple segments. So yes, in the new model the
source is just in the middle of the wire which is the middle section
of the D.E.

Cheers,
Tom

Efficiency of 200-ohm hairpin matching
 

I can understand your confusion and doubt. I hope this says it in
clear terms:

In the revised model with the hairpin done as three wires, the
original D.E. single wire was replaced by three wires: a 10cm long
center section which should have the source in its center, and a
"left" and a "right" outer section. Those three sections are three
separate wires. The reason to break the original single wire into
three wires is so that there will be wire "ends" to connect the
hairpin wires to. You can only connect wires at their ends, even
though they have multiple segments. So yes, in the new model the
source is just in the middle of the wire which is the middle section
of the D.E.

Thanks for your clarification that confirms my expectation that I shall put the
generator in the middle of the 10 cm-long center section of the D.E..

Everything works fine and, for a reference 200-ohm resistance, I get an SWR of
almost 1 not too far away from the frequency at which it actually occurs.

If I may take a bit more of your time, I have noted two inconsistencies:

- with the hairpin EZNEC gives an antenna gain of 12.76 dB but, if I just delete
the three hairpin segments leaving everything else unchanged, the gain grows up
to 13.59 dB. I am surprised about that result, as I had understood that, in
EZNEC, the antenna gain has nothing to do with the way the antenna is fed.

- with the hairpin, at the antenna resonant frequency EZNEC shows an impedance
of 200.5 +j2.1 ohm which obviously corresponds to an SWR of almost 1. If I
remove the hairpin, at the same frequency EZNEC shows an impedance of 42.4 - 39
ohm which could seem reasonable, but it is actually not. As a matter of fact, by
applying the appropriate series-to-parallel conversion formula, that impedance
corresponds to the parallel of a 78.2 ohm resistance and a 85.1 ohm capacitive
reactance. Resonating the reactive capacitance with an equal (though opposite)
reactance hairpin, the remaining resistance would be just 78.2 ohm, that is far
away from the 200 ohm EZNEC shows with the hairpin installed.

May I have your opinion?

73

Tony I0JX

Efficiency of 200-ohm hairpin matching
 

On Apr 10, 1:22 pm, "Antonio Vernucci" wrote:
I can understand your confusion and doubt. I hope this says it in
clear terms:

In the revised model with the hairpin done as three wires, the
original D.E. single wire was replaced by three wires: a 10cm long
center section which should have the source in its center, and a
"left" and a "right" outer section. Those three sections are three
separate wires. The reason to break the original single wire into
three wires is so that there will be wire "ends" to connect the
hairpin wires to. You can only connect wires at their ends, even
though they have multiple segments. So yes, in the new model the
source is just in the middle of the wire which is the middle section
of the D.E.

Thanks for your clarification that confirms my expectation that I shall put the
generator in the middle of the 10 cm-long center section of the D.E..

Everything works fine and, for a reference 200-ohm resistance, I get an SWR of
almost 1 not too far away from the frequency at which it actually occurs.

If I may take a bit more of your time, I have noted two inconsistencies:

- with the hairpin EZNEC gives an antenna gain of 12.76 dB but, if I just delete
the three hairpin segments leaving everything else unchanged, the gain grows up
to 13.59 dB. I am surprised about that result, as I had understood that, in
EZNEC, the antenna gain has nothing to do with the way the antenna is fed.

- with the hairpin, at the antenna resonant frequency EZNEC shows an impedance
of 200.5 +j2.1 ohm which obviously corresponds to an SWR of almost 1. If I
remove the hairpin, at the same frequency EZNEC shows an impedance of 42.4 - 39
ohm which could seem reasonable, but it is actually not. As a matter of fact, by
applying the appropriate series-to-parallel conversion formula, that impedance
corresponds to the parallel of a 78.2 ohm resistance and a 85.1 ohm capacitive
reactance. Resonating the reactive capacitance with an equal (though opposite)
reactance hairpin, the remaining resistance would be just 78.2 ohm, that is far
away from the 200 ohm EZNEC shows with the hairpin installed.

May I have your opinion?

73

Tony I0JX

Hi Tony,

Hmm...very interesting. I hope some others will check in on this with
ideas too.

My first thought is that the hairpin conductors are really more a part
of the antenna than you may have thought. The current in that 10cm
section of the hairpin, especially, is in a direction parallel to the
D.E. and to all the directors and to the reflector(s). It will couple
to them, and could be changing the gain and impedance some.

I am also surprised by the high resistance you are seeing with the
hairpin wires removed. From what you posted before, and from the 3-
element NBS and the 6-element designs I ran in EZNEC, I would expect a
lower feedpoint resistance than that. If I put a pure inductance in
parallel with your 42.4-j39 ohms, I get the same answer as you did,
that it moves to 78 ohms resistive.

My offer to look at your EZNEC file is still open, of course. Maybe I
could see something in it--I know that I can look and look at
something and not see an obvious problem, even when I know perfectly
well what to look for.

Also--in the model before you split the D.E. into three wires, did you
get a more reasonable feedpoint impedance (with no hairpin, just
looking at the D.E. feedpoint)? I think we were expecting something
like 19-j58 ohms. If you get that, then I would look for the reason
things change (so very much!) when you split the D.E. into three
separate (but connected, end-to-end) wires. I can understand a small
change but not so large a change. Is the total number of segments for
the three wires making up the D.E. still about the same as it was in
the original version of the model, with just one wire for the D.E.?

Cheers,
Tom

Efficiency of 200-ohm hairpin matching
 

I am also surprised by the high resistance you are seeing with the
hairpin wires removed. From what you posted before, and from the 3-
element NBS and the 6-element designs I ran in EZNEC, I would expect a
lower feedpoint resistance than that. If I put a pure inductance in
parallel with your 42.4-j39 ohms, I get the same answer as you did,
that it moves to 78 ohms resistive.

My offer to look at your EZNEC file is still open, of course. Maybe I
could see something in it--I know that I can look and look at
something and not see an obvious problem, even when I know perfectly
well what to look for.

If your personal mail on the newsgroup is correct, I could send you the EZNEC
file of my antenna.

Also--in the model before you split the D.E. into three wires, did you
get a more reasonable feedpoint impedance (with no hairpin, just
looking at the D.E. feedpoint)? I think we were expecting something
like 19-j58 ohms. If you get that, then I would look for the reason
things change (so very much!) when you split the D.E. into three
separate (but connected, end-to-end) wires. I can understand a small
change but not so large a change. Is the total number of segments for
the three wires making up the D.E. still about the same as it was in
the original version of the model, with just one wire for the D.E.?

I am also surprised as I would have expected a much lower resistance (as you
suggested)

I did the test you have suggested and I noted some change, but not much. With a
"solid" D.E., i.e. not broken in three pieces (and without the hairpin of
course), impedance is 43.27 - j 40.97 ohm, not terribly away from 42.4 - j39
ohm. In that test I configured EZNEC for 17 segments, whilst when the D.E. was
broken in three parts I had used 17 segments for the two outer arms and only 5
segments for the 10-cm center section, due to its much shorter length (I see
that result depend somewhat on the number of segments).

73

Tony I0JX

Efficiency of 200-ohm hairpin matching
 

Antonio Vernucci wrote:
. . .
- with the hairpin EZNEC gives an antenna gain of 12.76 dB but, if I
just delete the three hairpin segments leaving everything else
unchanged, the gain grows up to 13.59 dB. I am surprised about that
result, as I had understood that, in EZNEC, the antenna gain has nothing
to do with the way the antenna is fed.

What you're probably seeing is a numerical problem in the NEC
calculating engine. It's very fussy about the region near a source, and
doesn't like small loops which include a source. You should run an
Average Gain check (see "Average Gain" in the EZNEC manual index), which
will reveal whether this is the problem.

The double precision calculating engine in EZNEC+ is considerably more
tolerant of small loops, but can still have problems with average gain
for other reasons.

Roy Lewallen, W7EL