How can calculate radiation resistance? resistance at 7MHz is
skin effecthttp://circuitcalculator.com/wordpress/2007/06/18/skin-effect-calcula...
0.028mm.
for copper pipe 15mm in diameter:
octave:1> 15*3.14*0.028
ans = 1.3188mm^2
0.0155Om/m * 3.14m
ans = 0.048670 Om.
I think it's not bad.
See
http://personal.ee.surrey.ac.uk/Personal/D.Jefferies/antennexarticles/cloops.htm
for info on calculating the radiation resistance. I believe that will
be helpful. I assume Reg's RJELoop1.exe uses essentially the formula
you'll find there. For your loop, that program says inductance is
2.69uH, conductor loss resistance is 46.2 milliohms, and radiation
resistance is 5.7 milliohms. Although the conductor loss resistance
(essentialy the same as you calculated above is "not bad," you need to
consider it with relation to the very low radiation resistance. A
current in the loop will dissipate far more power in the resistive
loss than in the radiation resistance. You ideally will keep the loss
resistance small compared with the radiation resistance, though for
receiving (because of the very high level of atmospheric noise on HF
and lower frequencies) it matters not nearly so much as for
transmitting. When transmitting, you want your power to go into radio
waves, not heat. When receiving you only need signal greater than
noise, and it is relatively easy to make an amplifier with low enough
noise figure that even an inefficient antenna will result in an
amplifier output whose noise is dominated by the atmospheric noise
received by the antenna.
You recommended did hot use varistors? I'm thinking about
some kindhttp://
www.toshiba.com/taec/components2/Datasheet_Sync//273/1343.pdf
20 in parallel
So 0.02 ohms sounds like a small amount, but it's almost half as much
as the resistance of the copper loop. This may not be a bad thing,
because the Q is so high that the bandwidth will only be about 3.5kHz
assuming a lossless capacitor, and with the added loss the Q will be
lowered to perhaps 1400, allowing a slightly wider bandwidth. With so
narrow a bandwidth you need to be concerned about the stability of the
varicap diodes' capacitance. Still, I would think a very high Q fixed
capacitor supplying most of the total capacitance would be a good
thing. Use only enough varicap to cover the tuning range you want.
So for example, with 2.69uH inductance, if you want to cover 7.00MHz
to 7.30MHz, you need 192.2pF at the low end and 176.7pF at the high
end, a range of a little less than 16pF. You should be able to do
that easily with two of your suggested varicap diodes, perhaps a
couple of fixed 82pF high Q caps, and a high Q trimmer such as a
piston trimmer to trim the center of the range.
I'm thinking about soldering a box from FR4.
If you want the antenna to also do a
What you mean about balanced? Differential output? I should think
about it.
It has more to do with the symmetry of the way the antenna is
mounted. You want to make sure that the capacitance to ground from
each side is as nearly the same as possible. You need to put the gap
in the loop (the feedpoint) either at the top or at the bottom of the
antenna, and for mounting it's often easier to put it at the top.
That way you can clamp onto the middle of the bottom of the loop to
mount to a pole... But if you have a balanced amplifier at the
bottom and bridge the gap symmetrically across the box that amplifier
is mounted in, it should also work well. I recommend to you the
discussion about small loop antennas in King, Mimno and Wing's
"Transmission Lines, Antennas and Waveguides."
Cheers,
Tom