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Re: Measured Q



Hi Malcolm, 
Question below: 

Tesla list wrote: 
>
> Original poster: "Malcolm Watts" <M.J.Watts-at-massey.ac.nz> 
>
> Hi Mark, 
>
> On 12 Aug 00, at 19:26, Tesla list wrote: 
>
> > Original poster: "Mark Broker" <broker-at-uwplatt.edu> 
> > 
> > > You can calculate         Q = XL / R 
> > > 
> > > Note:       XL = 2*pi*f*L 
> > > 
> > > Where L is the inductance of the coil in Henries. 
> > 
> > Hello, list 
> > 
> > I finally decided to measure the Q of my coil.  I have a measured DC 
> resistance 
> > of 26.7 ohms.  According to E-T5, my coil's operating frequency is about 
> > 125kHz.  The inductance is calculated to be 33mH by WinTesla.  Putting this
>
> > into 
> > the equations gives me a Q of 970! 
> > 
> > This number seems about 5 times too high, and my primary tuning isn't 
> terribly 
> > finicky.  Is this high of a Q possible? 
>
> Not in your coil unfortunately. You cannot use DC resistance 
> in those calcs since frequency at DC = 0Hz.  In practice you 
> have to use what is known as ESR (effective series resistance) 
> which lumps all losses (skin, radiation etc.) at the frequency 
> of interest into a single resistance. You can quantify the ESR 
> of your coil by doing an accurate Q measurement and deriving 
> ESR = wL/Q. 
>  





Doesn't w*L = XL? Also, the skin depth at the operating frequency will cause
the ohmic value of the winding to increase. I'm curious if the method I used
below will hold water. 

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?). 

Bart