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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