On 7 abr, 23:15, "Antonio Vernucci" wrote:
Hello Tom,



First, the matching is being done essentially with an "L" network (or
rather the balanced version of an "L" network), where there is a load
resistance (the resistive part of the feedpoint impedance, which
includes radiation resistance and element loss resistance reflected to
the feedpoint), the series capacitive reactance of the feedpoing
element, and a shunt inductive reactance, provided by the hairpin.
Because shortening the driven element causes a decrease in resistance
and an increase in capacitive reactance, it's possible to find a
length that allows matching to any of a wide range of resistances.
But the higher the resistance to which you match, the shorter you need
to make the element and the lower the feedpoint resistance. The ratio
of feedpoint resistance to matched resistance determines the loaded Q
of the matching network; as you make the matched resistance higher,
the loaded Q goes up rather quickly. If you know the loaded Q and the
unloaded Q of the hairpin, you have a good handle on the amount lost
to heat in the hairpin: if the hairpin Q is two times the loaded Q,
half the power is dissipated in the hairpin, for example.

However, unless you build an antenna with a very low feedpoint
resistance at resonance, there almost certainly won't be an efficiency
problem: the reactance changes quickly enough with changes in driven
element length that the resistance won't drop much by the time you
reach enough reactance to get a match to 200 ohms. It appears that
the loaded Q of the match to 200 ohms for your case will be less than
4. I would think unless you really messed up badly, the hairpin
unloaded Q should be well in excess of 100, and if that's the case,
the power lost in the hairpin would correspond to well under 0.1dB
signal level change.

All OK. I however reckon that, due to the parasitic elements effect, the
radiation resistance of the driven element (before shortening it) would be in
the order of 20 ohm. So, impedance gets brought up by a factor of 10 or so.

On the other hand, I'm surprised by the comment from the manufacturer
about difficulties making a 1:1 balun. I have had good luck using
ferrites and/or self-resonant coils of feedline and/or coils of
feedline specifically resonated with additional capacitance. A 4:1
balun from a "hairpin" of 1/2 wave of coax, arranged symmetrically, is
easy enough to make, but I would not rule out using a 1:1, if it has
advantages for you.

His argument is that, for high-power operation (say 1500W), it is more
convenient for him to provide a quarter-wavelength 4:1 balun made of RG-142
teflon cable than a sealed box with a 1:1 coil on an ironpowder toroidal core. I
mentioned him that, just using some extra length of RG-142 cable, he can easily
build a 1:1 balun, and hence design the antenna matching system for 50 ohm
instead of 200 ohm.

I hope he will listen to me, because that antenna is really narrowband!

73

Tony I0JX

Hello,

Imagine 10 Ohms radiation resistance + capacitive component after
shortening the open dipole radiator. When you convert that to 200 Ohms
via a parallel inductance (hairpin), the Q factor of such a network is
about 4.4. At 50 MHz that will result in a bandwidth (VSWR=2) of 8
MHz.

A 200 Ohms to 50 Ohms coaxial balun will have a lower Q factor. So the
combination of balun and L-network will certainly have a useful
bandwidth 340 kHz.

Maybe the radiation resistance is less (you can derive that from the
hairpin inductance) and/or the antenna is by nature (very) narrow
band.

Best regards,

Wim
PA3DJS
www.tetech.nl
remove abc from the address.