Re: Measured Q

["Barton B. Anderson" <tesla123-at-pacbell-dot-net> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)]

Hi Malcolm, 

Tesla list wrote: 
>
> > The skin depth of my coil is .0104 inches using the equation 
> > Sd(inches) = (66 / sqrt(Fo)) / 25.4 
> > 
> > If I calc the resistive value of .0104 inch wire diameter using the same 
> length 
> > of wire the coil is wound at, I end up with a large resistance of 335 ohms.
>
> > 
> > If I then take XL/R = 34,543 / 335 = Q of 103 
> > 
> > Obviously eddy currents and hysterisis are not accounted for and the only 
> > proper way to find Q is measurement, but this approach seems to get close
> (or 
> > does it?). 
>
> Have you measured the Q of your coil? I think it would be 
> somewhat lower if the coil is closewound because although skin 
> effect would be pretty well absent there would still be some 
> proximity effect. Please let me know. I'm curious about this 
> case. 
>
> Regards, 
> Malcolm




You know, I have never measured the Q (yes, this is embarassing. I keep letting
other things get in the way. I'll look through some of the posts on Q
measurement and see if there is some method that will work for me (I don't have
a generator). Any suggestions? 

I agree, that the Q is probably lower. The coil is closewound (12.75 x 44.5, 18
awg .0418 total wire diameter, 23.4 turns per inch, Loaded Fo is 86.2kHz
measured). I actually used a calc'd Fo in the Sd above. With the measured Fo,
the Q here would work out to 96.4 (a little lower). I have "a lot" of top load
shielding going on. Unloaded Fo would work out to 87.3 (due to the higher Fo).
So proximity effects should make it even lower. I wonder what it really is. 

I've got the secondary sitting by itself as I'm in process of some RSG work, so
it would be a good time to make the measurement. 

Take care, 
Bart