Re: Tank circuit L/C ratio

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

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.  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.  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.


-- Original Message As Follows --

Subject: Re: Tank circuit L/C ratio
From: "Tesla list" <tesla-at-pupman-dot-com>
To: tesla-at-pupman-dot-com
Date: Thu, 11 Jan 2001 11:21:49 -0700

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

> Can you define what would be "surge impedance"? High L, low C
certainly
> reduces the current in the primary by rising its impedance level, but
> I don't see how an special kind of impedance comes from this.
> 
> Antonio Carlos M. de Queiroz

	I learned the term "surge impedance" in college, almost 60 years ago. 
The definition was 

	Zs = sqrt(L/C)

This is equal to the reactance of either L or C at the resonant
frequency of the referenced circuit, whatever it may be.  (Work it
out.)  Convenient for Q calculations and stuff like that.  Higher L/C
ratios result in higher impedances, nothing more than that, and nothing
magic at all.

Ed








---------------------------------------------------------------
This Message was Powered by Xcel Communications
Sign up for your FREE EMAIL account today at http://www.mailroom-dot-com
Give your FAX machine an email address http://www.faxroom-dot-com