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Re: Tank Capacitance: what is the limit

Subject:  Re: Tank Capacitance: what is the limit
  Date:  Fri, 02 May 1997 14:43:00 -0500 (EST)
  From:  Benson_Barry%PAX5-at-mr.nawcad.navy.mil
    To: tesla-at-pupman-dot-com


Hi Malcolm, All,
I agree that the resistance goes down as the current
goes up and the power dissipation goes up.  Increasing
the primary inductance increases Q.  Increasing the
primary capacitance decreases the Q but increases
the energy available to feed the secondary limited
by the input power feed rate.  As the primary and
secondary exchange energy the current amplitude
and number of energy exchanges before streamer
breakout determine the amount of energy given
to the secondary.  The current amplitude is limited
by the inductance of the primary and the resistance
of the spark gap and skin effect resistance of
the conductors.  Increasing the primary Q will
decrease the energy loss per beat except in the
spark gap.  The question I am purplexed by is
whether it is more efficient to transfer energy
with many beats from a high Q primary system,
incurring spark gap losses which are highest
at the transitions, or with one or less beats
from a low Q, big capacity, low turn, primary
system which incurs less spark gap transition
losse but bigger spark gap power loss from
the higher current?

The secondary is a classic resonator with a variable
resistance (streamer), skin effect resistance,
other interturn capacitive esr losses, and the
variable capacitance of the streamer.
With many beats the energy lost to heating the
wire and ionizing the air will be greater.  With a
single pulse would the energy loss be less with
the result that more energy goes to the streamer?
Or could it be that the streamer requires multiple
pulses to reach maximum length?

Have you ever tried feeding a single pulse at
full power into the primary with a trigatron
or field distortion swinging cascade switch?

Reference equation from the Radiotron Designers
Handbook, fourth edition, page 410:

I = E * r / ( r**2 + (w**2)(L**2))  at resonance

Q = w * L / r = 1 / (w * C)

I = current in parallel resonant circuit (primary)
E = voltage applied to capacitor
r = resistances in primary (nonreactive)
w = 2 * Pi * frequency
L = inductance
Q = energy input per unit time / energy lost per unit time
C = capacitance


Barry

 ----------
From: "tesla"-at-pupman-dot-com-at-PMDF-at-PAXMB1
To: Benson Barry; "tesla"-at-poodle.pupman-dot-com-at-PMDF-at-PAXMB1
Subject: Re: Tank Capacitance: what is the limit
Date: Monday, April 28, 1997 6:39PM

<<File Attachment: 00000000.TXT>>
Subject:       Re: Tank Capacitance: what is the limit
       Date:   Mon, 28 Apr 1997 16:29:46 +1200
       From:   "Malcolm Watts" <MALCOLM-at-directorate.wnp.ac.nz>
Organization:  Wellington Polytechnic, NZ
         To:   tesla-at-pupman-dot-com


Hi all,
        A comment on Fr McGahee's post:

> While it is generally true that the primary is more than one turn of
> wire,
> as the secondary becomes LARGER in DIAMETER, it is often advantageous to
> use larger capacitors and fewer turns in the primary.

I usually run at lower frequencies and keep primary Q up by keeping
Lp large. I hear things like "lower frequency implies lower Q" a lot.
That no doubt comes from observing that Q = X/R. Let's have a closer
look at that:
                  Q = (2 x PI x f x L)/R

and we know that f = 1/(2 x PI x SQRT(L x C))

     Let's halve f in three ways, first by making C four times larger.
That leaves L unaffected which indeed causes Q to dive in half since
f has halved.
     Now lets achieve the same thing by doubling L and doubling C. Q
is now the same (as it should be because the L/C ratio is preserved.
     Finally, let's quadruple L. Despite f having halved, Q is
actually doubled.

     With reference to the primary, we know that "R" is a non-linear
resistance. In fact I maintain it goes down as gap current goes up.
If the voltage across the gap when it is conducting is reasonably
constant, then gap losses double as the current is doubled. Despite
gap resistance doubling as current is halved, it loses less power
because it roughly follows a VxI dissipation law.

Malcolm

> That is just what
> Tesla did on many occasions. As you change one parameter, such as the
> coil
> diameter or the size of the tank capacitor, it is sometimes necessary to
> break some of the rules that NORMALLY work quite well. Never be afraid
> to
> experiment to find out what ACTUALLY works best in THIS particular
> setup.
> The common wisdom is often all too common and not much wisdom.
>
> In reading Tesla's Colorado Springs Notes I am kept contantly chuckling
> at
> the way he did things that EVEN went against his OWN recommendations (of
> just a page or two before!) You HAVE to experiment and improvise in this
> business. What at first appears to be an absolutely minor change can
> have
> an absolutely major effect on overall circuit performance. I have
> learned
> not to just blindly accept what others say is the BEST way of doing
> anything, but to experiment on my own and try and understand WHY some
> things work and some things don't *in this particular case*.
>
> Fr. Tom McGahee

>
>