[Prev][Next][Index][Thread]

Re: Dynamic Q




From: "James" <elgersmad-at-email.msn-dot-com>

    I believe that you're experiancing a condition caused by a bad impedance
match.  If take the Voltage of the output of the primary, and divide it by
the current of the primary's driving circuit, you get the output impedance
of the primary's driver circuit.  Now, when you take the value of the
primary in Henries, times 2, times pi, times the frequency applied by the
diver circuit, you get the AC resistance of the primary.  Now, you take the
capacitor that is in parallel with the primary, and you divide 1 by the
solution of the equation, Taking the value of  the capacitor in Farads,
times 2, times pi, times the frequency of the primary's driver circuit.
Now, one divided by the result of this equation 1divided by the AC
resistance of the Capacitor, added to 1 divided by the AC resistance of the
Primary Coil's inductance.  If the solution to the last equation is less
than the driver circuits Volts/Amps, that is the most likely reason that
your Q calculation is off.  Meaning you need more current from the driver,
or to drive the circuit running at a higher frequency, and swapping out the
capacitor for a smaller value.





>From: "bmack" <bmack-at-frontiernet-dot-net>
>
>Hello again
>
>I saw the the above term and liked it. I use boring terms like loaded
>Q.  Did Bert coin this phrase?
>
>Anyway, what kind of numbers are you getting for dynamic Q?
>
>I find that it's less than a tenth of the calculated unloaded Q
>(undynamic?)
>even before ionization and way before breakout.
>
>
>Can anyone offer an explanation?
>
>Jim McVey
>
>