Q-factor measurements of different LW-Antennas
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- เผยแพร่เมื่อ 5 ก.พ. 2025
- Q-factor comparison measurements for different AM magnetic antennas starting with LW-coils.
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3 different ferrite-rod LW-coils, 1 spiderweb-coil, 1 frame-antenna
![Resonant Magnet-Field Antennas [Pt.5]: Dielectric & Magnetic Losses](/img/n.gif)
Excellent video and info , makes a little more sense for me ! Thanks !
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Another great video. Question: does your measurement setup compromise the accuracy of the results due to the antenna location breaking the 1meter proximity rule you spoke of in previous videos? I imagine it does but that the effects at relatively low frequencies must be slight.
A basket weave coil is wound like a cylinder and is different than a spider web coil
Hi Roger, have you done another video explaining the origin of the x5 multiplier to convert cycles to Q-factor? (For my own calculations, I use pi *1.44 which gives a factor x4.5).
In this video the method is introduced:
th-cam.com/video/gJlkWt3VYxM/w-d-xo.html
From my memory the factor of 5 of results from e^(-0.5)*pi = 5,18
The general formula should be Q = n(x) * pi * e^(-x)
where n(x) is the number of cycles at a point x, where x is the fraction of amplitude relative to a reference amplitude (e.g. the maximum amplitude)
Hi Roger,
If you express exponential decay of a variable U(t) as:
U(t) = U(0)*e^(-t/tau)
where tau is the time-constant at which the variable decays to 1/e of its initial value, then the time it takes to decay to half its value is:
t(half) = tau * (-ln(0.5)) = 0.693 * tau
In other words, if you *measure* the half-life as t(half) on an oscilloscope, then
tau = t(half) * (1/0.693) = 1.44 * t
Then, if Q-factor is defined as _"Pi times nCycles in one decay-time-constant tau"_ then it may be calculated as n*1.44* Pi, i.e. about a factor 4.53*n (where n is the nCycles in one half-life)
One of us is making a mistake ;)
Did you see the FEMM model, btw?
Cheers!
You might be right. I haven´t the book at hand and where I got the factor 5 from.
Anyway we are both less than 10% off from the rounded value of 5.
And the slightly "better" Q-values that result from multiplying by 5 instead of 4.5 compensate for the non-negligible loading of the tank-circuit with the 10MOhm resistors.
During the measurements I recognized a strange effect that there is a big difference if I use 1 MOhm coupling-resistors or 10 MOhms. That shouldn´t be the case because the calculated resonance-resistances are in the range of 100 kOhms.
I still have to make a comparison with 100 MOhm coupling-resistors to see if 10 MOhms is really enough.
And yes I got your email with the FEMM-file.
Still have to find the time to work myself through the software :-)
Hi Aditya, you are correct that the factor is 4.5 and it's easy to confirm this using a simple LTSpice or similar model.
@@wd8dsb Cool. Thanks for checking :)
Can you make a video with nanovna q mesure
Yes, if you send me a Nano-VNA.
Great video very interesting thanks
hello
Is the vidéo série about radio and antenas is over ?
It was so interesting
No, it ain´t over yet :-) I still have some topics to show (like e.g. mythbusting crystal radios). In the last 6 months I didn´t have time to shoot new videos but I will probably continue in January.
thank you fpr sharing .. very good vidéo
Would love a copy of your notes on this good video big thumbs up 👍
Sorry, but I don´t make notes or a script. It´s all prepared beforehand in my head :-)
Just freely speaking.
KainkaLabs thank you for your reply 👍