Recent discussions have highlighted the need for good transformer balance
and low interwinding capacitance, for the matching transformers used for
pennants and flags. Further examination shows that the problem is not trivial.
The problem stems from common mode excitation of the antenna, in which it
behaves like a bent monopole with an elevated feedpoint:
*
* *
* *
* * X * X = feedpoint
* * * (bent monopole mode)
* * *
* *
*
*
*
-----------*-----------
Ground
K6SE reports the gain of the pennant and flag as about -36 and -30 dBi
respectively, over average ground. My own modelling of these antennas
confirms these figures and indicates a front-to-back ratio of 20 dB,
meaning that the rear lobes are at -56 and -50 dBi respectively.
To ensure that the front-to-back ratio is not degraded by more than 1 dB,
it is necessary to keep the "bent monopole" gain at least 6 dB further
down, i.e. at -61 and -56 dBi for the pennant and flag antennas respectively.
Considering that the gain of a quarter wave monopole over similar earth is
about +1 dBi, one begins to see the magnitude of the problem. In other
words, to avoid filling in the rear null to any significant degree, it is
necessary to attenuate the common-mode gain by some 50 to 60 dB!
To determine the transformer capacitance needed to produce this
attenuation, I modelled the pennant as a bent monopole, as shown above. I
also modelled the flag on the assumption that it is supported by a single
mast at the centre, with the feedline travelling horizontally back to the
mast, and then vertically down to the ground.
Over average ground, the common-mode impedance of the pennant was
0.76-j7507 ohms, and that of the flag was 0.68-j19175 ohms (calculations
were done with NEC4WIN95, which uses the Mininec engine). Assuming a 50 ohm
source, this produces mismatch losses of 43.5 and 51.7 dB respectively.
Although the flag is within ballpark, the interwinding capacitance of the
pennant transformer needs to contribute a further 17.5 dB loss (1).
This is a simple potentiometer problem. To obtain a loss of 17.5 dB,
Xc must be at least 7507*10^(17.5/20) = 56295 ohms. From this, we can
calculate the maximum allowable interwinding capacitance as 1.54 pF at a
frequency of 1.83 MHz.
This is the effective capacitance, averaged over the whole winding. If each
turn contributes equally to the capacitance, the maximum allowable total
capacitance is twice the above or 3.08 pF. However practical windings are
less than uniform, so the maximum allowable capacitance will be (say) 2 pF.
This order of isolation is likely to be marginal or impossible to achieve
with the fairly large balun core mentioned by W8JI, because of the close
proximity of the windings. In my opinion it would be much better to use a
toroid, with the primary and secondary well separated, and coupling through
the ferrite only. I concur with W8JI on the absolute need to use the
correct ferrite, and based on the curves published by Fair-Rite and Amidon,
believe the best material to be #43 (ui = 850). This will produce a core Q
of 4.25 and consequent loss of about 1 dB, which is reasonable for the
application.
I would avoid the use of a manganese-zinc ferrite like 73, as it appears
far too lossy for this sort of transformer at 1.8 MHz. The fact that it
works well for transmission line transformers up to 30 MHz (or beyond) is
not really relevant, because the application here relies on magnetic
coupling, whereas transmission line transformers do not.
At this point, it must be emphasized that a transmission line transformer
SHOULD NOT be used for this application, as their end to end isolation is
much too low. They are really only suitable for low impedance circuits.
My choice is an FT140-43 toroid with 8 turns on the secondary, and 34 or 35
turns on the primary (for 900 or 950 ohms respectively). Sorry to specify
the larger and more expensive core, but one needs the higher AL value to
get sufficient inductance (2). With this core, the shunt reactance will be
+j689 ohms, leading to a nice conservative design. This larger core will
also make it easier to keep the windings separated.
This transformer should bring everyone's pennants and flags "to life", as
reported by W7IUV.
I would advise against stacking cores to increase the AL value, because the
primary to secondary capacity increases in proportion to the number of
cores. This happens because the wires run parallel for a longer distance.
If a higher AL value is needed, choose a core with a larger outside
diameter. Although this approach is more expensive, the interwinding
capacity is not significantly increased.
A final touch, although probably not necessary, would be to balance the low
impedance winding by interposing a second transformer. The result would be
to cancel the remaining capacity of the first transformer, as far as common
mode coupling is concerned. Suggested details are: 10+10+10 turns trifilar,
FT50-43 core, coax to one winding, the other two windings in series, and
their centre tap grounded. Since this transformer steps the impedance up to
200 ohms, the main matching transformer will need twice the number of turns
on the low impedance winding, i.e. 16 turns instead of 8. Losses should be
low.
Finally, with the antenna floating, it is imperative to provide a DC path
for static discharge. I suggest grounding the opposite side of the antenna,
at the centre. Don't be tempted to take the easy way and add a centre tap
to the 900/950 ohm winding, because the first nearby strike will
disaccomodate the ferrite and greatly increase its losses.
73,
Peter VK3APN
(1) The figures assume the bent monopole has the same directivity as a
straight quarter wave monopole. It doesn't, but the difference is minor
(only one or two dB). We are seeking ballpark figures here, not fractional
dB precision.
(2) AL = inductance of one turn in nH. This equals the mH per 1000 turns as
quoted by Amidon.
pnesbit@melbpc.org.au
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