Re: Tank circuit L/C ratio

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-uswest-dot-net>" <evp-at-pacbell-dot-net>

Tesla list wrote:
> 
> Original poster: "by way of Terry Fritz <twftesla-at-uswest-dot-net>"
<elgersmad-at-fnworld-dot-com>
> 
> The higher the value of L, and the lower the value of C, the higher the
> Q, and ring.  

	No way!  In a passible linear circuit Q is not necessarily related to
the L/C ratio.  In a TC the streamer loading probably results in the
operating Q of a high impedance coil being lower than that of one with
higher C, For distruptive type coils the unloaded Q is of little
importance compared to the Q with streamer loading.
	

> But, in order to derive any usable amoung of energy you
> cannot have an infinite value of L.  Q will increase for lower values
> of L when the number of turns of wire is lower, and Rw is lower.  This
> means that in the equasions that your using increasing the frequency,
> lowering the value of L, and C can result in a stable situation.  For
> example Xc=XL=1ohm at 1Mhz.  In alot of the simulators that I've used
> they crashed on that.  

	They shouldn't if you used them correctly.  Mine don't.

> Consider this in the math as two DC resistances
> in parallel there is a load of only 0.5 ohms, and at ten volts there's
> a demand for a minimum of 20 amperes of current before the voltage you
> can measure across a DC resistance of 0.5 ohms will equal the 10 volts
> provided by the supply.  If the supply has an output impedance of 10
> ohms, then it can only provide one ampere of current at 10 volts,
> therefore, you will only find 5 volts across a DC resistance of 0.5
> ohms.  Now, if that were 10 volts AC with and output impedance from the
> supply of 10 ohms across a parallel tank circuit you might only find
> 3.16 volts or less pending on the Q.  That condition will not change
> because, the starting impedance of the parallel tank circuit is only
> 0.5 ohms, and you will need 20 amperes of current to start the circuit,
> then you'll measure 14.14 volts across the tank and it will demand only
> milliamperes of current from the source.  Low Q, Low Z, Step up, ring
> up, high Q, high Z, step down and out, that's chaotic resonance.  XL in
> parallel with Xc at resonance = surge impedance, or Zmin=1/(1/XL+1/Xc)
> With your equasion if C = 1uf, and L = 1uH, Zsurge=1, and with the
> equation I've suggested using it's 0.5.  For a single pulse it's only
> going to be 0.5 ohms given that it is at the resonant frequency.
> 
> James.

	I have a hard time trying to follow this, but think you're mixed up
between loaded and unloaded circuits.  Note that the Q of a circuit is
proportional to the "per cycle" ratio of the energy stored (in the
capacitor, then the inductor, etc.) to the energy dissipated.  The more
energy stored per unit of energy dissipated (power output, if you will)
The higher the Q.  This leads to maximum Q coming from the lowest L/C
ratio.  No where near that simple, but that's the trend.

ed