Re: Tank Capacitance: what is the limit

```Subject:  Re: Tank Capacitance: what is the limit
Date:   Thu, 24 Apr 1997 05:28:38 +0000
From:   "John H. Couture" <couturejh-at-worldnet.att-dot-net>
To:    Tesla List <tesla-at-pupman-dot-com>

At 01:58 PM 4/22/97 +0000, you wrote:
>Subject:  Tank Capacitance: what is the limit
>  Date:   Mon, 21 Apr 1997 11:13:22 -0400
>  From:   Brendan Haley <bhaley-at-shore-dot-net>
>    To:   "'Tesla'" <tesla-at-poodle.pupman-dot-com>
>
>
>Hello all-
>
>        I'm sure this question has a simple answer,  and I'm a bit tired
>of
>
>        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?"
>
>        Again,  my intuition tells me that the answer is that you can't
>just
>arbitrarily increase your capacitance to gain current,  but I'm not sure
>exactly
>why.  I have a feeling it has to do with the interaction between the
>capacitor
>and the xfmr,  perhaps related to the impedence of the capacitor, and
>the ideal
>load on the xfmr.  Or perhaps is it a frequency related thing having to
>do with
>the LC circuit formed by the seconday of the xfmr and the capacitor.
>
>        Can anyone shed light on this,  and perhaps bolster or destroy
>my faith
>in my own intuition.
>
>        Also,  could anyone send me the formulas for determining  the
>appropriate capacitance for a transformer, and the appropriate
>inductance for
>the primary,  given a desired freq.,  as this may help me figure this
>out
>as well.  I used to have these,  and they are probably in a book
>somewhere,
>on my bookshelf,  but I am having trouble finding them.
>
>Thanks Again,
>
>        Brendan
>
>---------------------------------------------------------

Brendan -

There are no simple answers when it comes to Tesla coils in spite of
the
fact that the basic circuit appears to be simple.

Theoretically, given a transformer with certain properties there would
be
an optimum tank circuit setup. You are correct in thinking it is a
frequency
related thing. Tesla also was interested in finding the optimum
condition
for his coils and referred to this in his writings.

It's a long and involved story but basically involves a compromize in
operating frequency. If you have a particular power supply and too low
an
operating frequency or too high a frequency the output voltage will not
be
optimum. It is obvious a compromize is required in the tank circuits.

Too low a frequency will give a lower secondary voltage because the Q
factor will be low due to the low frequency   Q = 6.283 FL .  Too high
an
operating frequency will give higher voltages but this parameter will be
reduced because of the secondary coil self capacitance.

Remember, the current thru a capacitor increases as the frequency
increases. In a Tesla coil the current thru the secondary coil self
capacity
increases with the frequency. This current in a Tesla coil is a loss
that
reduces the secondary voltage. To my knowledge no one has ever
researched
and properly tested enough Tesla coils to find these optimum conditions.

Any Tesla coil design computer program must take these problems into
consideration. When I developed the JHCTES computer program I realized
that
because of lack of information on this problem the program would have to
be
designed empirically. That is, the program was based around coils that
been built and tested and found to operate satisfactorially. This is
still
true today and will continue until someone properly tests enough coils
to
determine the optimum conditions.

There is still a lot more engineering design work to do with Tesla
coils
before we can feel that we have coils that are performing at their best.

If you are interested in researching this problem, some of the
equations
you will need are:

Fo = 1/(6.283 sqrt(FL))      Cp = 2W/V^2

Is = Wheeler equation        Cs = Medhurst equation

Amps = Vs/Zs                 Q = 6.283 FL/R

Vs = B Vp sqrt(Qs/Qp)  B = a constant

Note that the log decrement can easily be found by tests and from this
the
Q factors.

You will also need some empirically derived equations obtained by
testing
actual Tesla coils with different tank conditions.

As you can see it is a formidable task but one that is necessary for
solving the optimum tank capacitance problem.

John Couture

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