Re: Tank Capacitance: what is the limit

To: tesla-at-poodle.pupman-dot-com
Subject: Re: Tank Capacitance: what is the limit
From: Tesla List <tesla-at-stic-dot-net>
Date: Wed, 23 Apr 1997 23:48:16 -0500
Approved: tesla-at-stic-dot-net
Subject:  Re: Tank Capacitance: what is the limit
  Date:   Wed, 23 Apr 1997 13:55:04 -0500 (CDT)
  From:   c604313-at-showme.missouri.edu
    To:   Tesla List <tesla-at-pupman-dot-com>
    CC:   tesla-at-poodle.pupman-dot-com


> 
> 
> Hello all-
> 
>         I'm sure this question has a simple answer,  and I'm a bit tired
> of 
> thinking about it,  so I will ask the list.
> 
>         My intuition is that,  given a transformer with certain
> properties (output
> voltage and current),  there is an ideal tank capacitance at which the
> circuit 
> (transformer -> cap. -> inductor(primary)) will run optimally.  My
> friend however
> asks me:  "Why can't you increase the capacitance indefinitely,  and
> retune
> your primary appropriately,  in order to achieve more current in the
> tank
> circuit?"
> Thanks Again,

      
   Brendan,
   Yeah, what your talking about is transformer / capacitance matching.
And as you say, there is a optimal configuration.  The simple answer 
basically goes like this:

When dealing with AC components, notably capacitors and inductors, an 
interesting phenomena appears known as reactance. Reactance is a kind of
"apparent" resistance and is measured in ohms just like resistance.
Reactance, however, is frequency dependent unlike your straight forward
resistors.
If I remember correctly, it is calculated this way:


                     1 
         Xc = ----------------                Eq #1
              2 * Pi * Hz * C 

Where,
      Pi = 3.14
      Hz = frequency in Hertz
      C  = Capacitance in Farads
      Xc = Reactance in Ohms
      * means multiplication

To make sure were speaking the same language.
 If you have a .01 microfarad capacitor (= 0.00000001 farads) and a
perfect, ideal 60 Hertz SINE wave hooked across the capacitor it would
give an apparent resistance of 265,393 Ohms.

Now, if either the capacitance OR the frequency were to increase, the 
Reactance would decrease and vice-a-versa.

This is all important because of the relation:
 
     V = I * R
where,
      V = Volts
      I = Amps
      R = Resistance

Just because a transformer is rated at 10,000 volts 30 mili amps does
NOT mean that it will not or can not supply more amperage, indeed it
can!!
But it may not be DESIGNED to supply that much current. If you try to
take more current from a transformer than it was designed for you will
end up with burned up windings either primary windings or secondary 
windings or both. (but of course transformers can usually be pushed
safely
past there rated values).

If we rearange the above equation we get:
     V
    --- = R                 Eq #2
     I
Given a constant voltage, if we increase R the only way to make this
Equation work is to decrease the Amperage (thus decreasing power).
Conversly if we lower R the Amperage must also rise to make this 
equation balance risking burning out the windings in the transformer.
With Neon transformers, if too much current is drawn from them they
start
to shunt themselves to save their own hide. They start shutting down by 
lowering their voltage output resulting in lower power.

By setting Xc in Eq #1 equal to R in Eq #2  and solving for C we get.



                                1
                    C = -------------------
                         2 * Pi * Hz * (V/I)

and this is the standard eq for matching the transformer with the
Capacitor.


So, for a 15,000 V 30 miliamp =(.03 amp) operating at 60 hertz
(house frequency)
we get:
                                      1
          0.0000000053 =   ----------------------------- 
                           2 * 3.14 * 60 * (15000/0.03)

or 0.0053 microfarads.
  
Hope this helps.