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Follow-up on traps as loads

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Subject: Follow-up on traps as loads
From: (L. B. Cebik)
Date: Tue, 18 Feb 1997 09:46:40 -0500 (EST)
Antenna Element Losses

I thought I might follow up on the discussion of "losses" due to traps with
some quantification via modeling.  This will graphically illustrate the
fact that the insertion of loads in an antenna element results in lesser
gain just from the reconfiguration of the element, even when there are no
ohmic losses, and also give an indication of additional loss of gain
occasioned by ohmic losses related to the Q of the load.

To do this, I created a model of a 20 meter dipole.  Then I shortened it
from roughly 34' to 24' (about 70% normal length).  To reresonate the
dipole, I inserted various types of loads.  I then added, where relevant,
some ohmic losses to the load elements.  All models are in free space.  The
full size model used 21 segments; the 70% models used 15 segments so that
distance between current check points is about the same as in the full size

The data can be looked at in at least two very useful ways.  First, there
are the standard numbers associated with gain and feedpoint impedance. 
Second, we can examine the pattern of current levels along the elements.  I
shall list the current levels from the feedpoint to the end for only one
side of the dipole, since they are the same on the other side.  Note the
abrupt changes from the normal virtually sinusoidal curve of currents
wherever loads are present.

Use caution in interpreting these numbers.  First, gain figures are in dBi,
which means that you should never think such things as "This version has
only 50% of the gain of that one," if the gains are 2 dBi and 1 dBi,
respectively.  Rather, version 2 is down 1 dB from version 1, and would be
so over real ground of any sort (where the numbers might be 8 dBi and 7
dBi, respectively).  Second, when applied to multi-element arrays, the
differentials are roughly--but not exactly, additive.  For dipoles with a
0.5 dB differential, a 3 element beam might show about 1.5 dB differential,
but with due allowance for other design factors that can modify this rough
measure.  Third, the models use a fairly radical element shortening:  30%. 
Traps in a 10-15-20 meter beam, on 20 meters, do not approach this level of
shortening.  On 15 and 10 meters, the situation is more complex, since part
of the gain reduction may stem from a trap appearing as an inductive load
and part from trap inefficiency in terminating the element.

Finally, note that currents are at the center of segments of these NEC
models; hence, the current closest to the feed point will be less than 1.0
and the current in the last segment will be greater than zero.  Currents
are for a set power level to the feedpoint of 100 Watts.  You will have to
mentally add the line beneath the readings to create the linear element
along which these current levels exist.

Time for some numbers.  All models are of copper wire (#12) at 14.174 MHz. 
"Current" = the current with a feedpoint power of 100 watts; "C (I=1)" =
the current with a preset feedpoint current of 1A.

1.  Full size dipole:  gain 2.09 dBi; Z = 73.0 + j1.0 ohms
Current:  1.16 1.12 1.06 .98  .88  .75  .61  .46  .29  .10
C (I=1):   .99  .96  .91 .84  .75  .64  .52  .39  .25  .09

2.  Capacity hat loading at 70% point:  gain 1.97 dBi; Z = 59.7 + 0.0 ohms
Current:  1.28 1.25 1.18 1.09 .98  .86  .72  (load begins here)
C (I=1):   .99 .96   .91  .84 .76  .66  .56
Note that the current level for 100 watts is higher due to the lower
feedpoint impedance.  Relative current distribution is shown by the "C
(I=1)" line where the current is set to 1 A at the feedpoint.  Otherwise
note that the current remains at normal levels until those levels have
decreased to the point of adding little to the antennas radiation.  The
current continues to decrease in the hat elements, but radiation from those
elements cancels itself due to being of equal magnitudes but opposite
phase.  Gain is down by 0.12 dB from the full size dipole.

3.  Center loading coil of infinite Q (no ohmic losses).  Load = j458 ohms. 
Gain 1.86 dBi; Z = 29.2 + j1.3 ohms
Current:  1.70 1.51 1.29 1.04 .78  .49  .18
C (I=1):  .92  .82  .70  .56  .42  .27  .10
Note:  the gain reduction is 0.23 dB just from the missing high current
portion of the antenna element unavailable for radiation.  (Modeling
software treats inductive loads as non-radiating.  They do radiate a
little, but too little to affect the results here.)

4.  Center loading coil:  Q=100.  Load = 4.58 + j458 ohms
Gain 1.23 dBi; Z = 33.8 + j1.3 ohms
Current:  1.58 1.40 1.20 .97  .72  .46  .16
5.  Center loading coil:  Q=50.  Load 9.16 + j458 ohms
Gain 0.68 dBi; Z = 38.4 + j1.3 ohms
Current:  1.49 1.32 1.12 ..91 .68  .43  .15
Note that the current distribution remains the same for a present feedpoint
value of 1.0.  However, element gain is down significantly due to
additional power being lost in the center load resistive losses, as
indicated by the decreased current levels in 4. and 5.  Again, remember
that relative to a trap element, this is a very high loading value.  Modern
traps may (or may not) achieve coil Q values higher than 100.  However, a
more accurate reflection of their action come from models of mid-element
loading coils.

6.  Mid-element loading coil:  infinite Q (no ohmic losses).  Loads = j478
ohms (x2).  Gain 1.89 dBi; Z = 46.1 + 0.47 ohms
Current:  1.46 1.42 1.36 1.24 .93  .59  .21
C (I=1):  .99  .96  .92  .84  .64  .40  .14
   Coil Position:         ^
Note that although mid-element loading coils yield a slightly higher gain
(0.03 dB) than a center loading coil, the difference is not significant. 
Each coil must actually be larger than a single center loading coil for
resonating the same length of element.  Compare the current distribution to
that of the full size dipole and the rapid drop of current past the loading

7.  Mid-element loading coil:  Q=100.  Loads = 4.78 + j478 (x2)
Gain 1.29 dBi; Z = 53.0 + 0.1 ohms
Current:  1.36 1.32 1.27 1.16 .88  .55  .20
8.  Mid-element loading coil:  Q=50.  Loads = 9.56 + j478 (x2)
Gain 0.76 dBi; Z = 59.8 - 0.2 ohms
Current:  1.28 1.25 1.19 1.10 .82  .52  .19
Note that again, the current distribution is the same as in the lossless
load case, although the gain is down significantly (1.33 dB for Q=50; 0.80
for Q=100).  For a constant power, current levels are down along the
element compared to a set of lossless loads.  Again, the much smaller
values of trap coils may occasion far lower losses.

Note that we have been focusing on current distribution and on currents
from a constant power level.  We did this to sort out the questions of gain
related to load configuration and reactance from the questions of gain
related to resistive losses.  Constant-power current figures are comparable
only within a model type, where feedpoint impedances are close enough to
make the current levels meaningfully similar.

Plugging these numbers into a spreadsheet and making simple line graphs can
display the discontinuities occasioned by loading elements even more
vividly.  That however, and in the words of all those pretentious
textbooks, I leave to the reader as an exercise.

If you use a trap beam, you can estimate for 20 meters a total equivalent
midelement loading coil by seeing what value of inductive reactance is
necessary to resonate the element on 20 at its manufactured length. 
Actually performance will additionally depend on the Q of the coil(s)
functioning as coils rather than as part of a tuned circuit, since the
capacitor has little effect.  (However, ultra precision would also take
into account signal coupling through the capacitor--of very high Q but of
significant capacitive reactance.)

Loosely, you can do the same for 15 meters by measuring the element from 15
meter trap to 15 meter trap.  On 10 meters, the element should be full
length.  However, trap efficiencies come into play for these bands, and
that is another topic entirely.



L. B. Cebik, W4RNL         /\  /\     *   /  /    /    (Off)(423) 974-7215
1434 High Mesa Drive      /  \/  \/\     ----/\---     (Hm) (423) 938-6335
Knoxville, Tennessee     /\   \   \ \   /  / || /      (FAX)(423) 974-3509
37938-4443     USA      /  \   \   \ \       ||    

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