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.
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: ruclips.net/video/gJlkWt3VYxM/видео.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 :-)
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.
![Resonant Magnet-Field Antennas [Pt.5]: Dielectric & Magnetic Losses](/img/1.gif)
Excellent video and info , makes a little more sense for me ! Thanks !
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.
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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:
ruclips.net/video/gJlkWt3VYxM/видео.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 :)
Great video very interesting thanks
Can you make a video with nanovna q mesure
Yes, if you send me a Nano-VNA.
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 👍