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Old April 9th 08, 09:22 PM posted to rec.radio.amateur.antenna
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First recorded activity by RadioBanter: Jun 2006
Posts: 395
Default 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

 
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