On 6 abr, 23:43, "Antonio Vernucci" wrote:
I have recently mounted a 6-meter long Yagi. The driven element is a
(non-folded) dipole whose impedance is brought up to 200 ohm by significantly
shortening it (so introducing a high series capacitive reactance) and using a
(rather small) hairpin to resonate the residual reactance. The system is fed by
a 4-to-1 balun (200-to-50 ohm) made of a half-wavelength cable. The SWR at
resonance is almost perfect, but the SWR response vs. frequency is very sharp,
this meaning that the system has a high Q. I am wondering whether such a 200-ohm
feed technique could cause non negligible losses caused by the higher current
circulating in the hairpin (due to its low reactance as well as to the high
voltage on the high-impedance feed point, that is 200 ohm + reactance of the
shortened driven element). 73 Tony I0JX

Hello Antonio,

When I understand you correctly, the series component of the dipole
input impedance is 200 Ohm? I would expect value lower then 70
Ohms.

When 200 Ohms real component of Zi, you used a series hairpin to
cancel the capacitive component. You can guess the loss by calculating
the AC resistance of the pin.

Assuming homogeneous current distribution around the circumference,
you can calculate the AC resistance with (assuming length of hairpin
0.15 lambda):

Rac = 80*10^-9*sqrt(f)*le/d , Rac in Ohm, f in Hz, le (wire length)
in m, d (diameter of wire) in m, valid for copper. For other
materials (non magnetic), when conductance is factor 4 less (so 4
times higher spec. resistance), Rac doubles.

When the result for your hairpin 200 Ohm, the loss is negligible.

In case of a parallel pin, you can use same technique, but in that
case you have to calculate the inductance of the pin. With the
inductance you know the reactance and the current that runs through it
(for example with 100Vrms across it). With Rac you can calculate the
power loss.

Hope this helps a bit.

Best regards,

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