Re: Math Doodling

```Hi Dave!

I've got to admit, this post got my eyebrows too! :^)

Interesting result - however, this would imply that Vo approaches
infinity as Cp approaches 0, which clearly can't be.

The root of this quandary stems from the fact that Iin is not
independent of Cp:

>From the second equation (below), we get:
Iin = Vin * sqrt(Cp/Lp)

Plugging Iin into the last equation to get it into a form using only Vin
we get:

Vo = Vin * sqrt(Cp/Lp) * sqrt(Ls/Cp)

which then simplifies to the more familiar Vo = Vin * sqrt(Ls/Lp)

As Cp is changed, the impact of the sqrt(Ls/Cp) term is exactly
cancelled by the sqrt(Cp/Lp) term, leaving the output as only a function
of sqrt(Ls/Lp) as expected.

-- Bert --

Tesla List wrote:
>
> Original Poster: David Sharpe <sccr4us-at-erols-dot-com>
>
> Terry, ALL
>
> I've been doing some doodling, and off-line discussion with
> Terry, Barry Benson, and John Freau.  Here is an interesting
> math derivation to try over a cup of coffee...
> --------------------------------------
>
> Here is a simple math analysis situation that blew Richard Hull
> and Alex Tajnsek away.  Based on equations in the Heise paper and
> assuming lossless transfer of power:
>
> Vo = Vin * sqrt ( Ls/Lp )  Where        Vo = max Vout from resonator
>                                         Vin = Vin applied to tank circ.
>                                         Ls = Inductance of resonator
>                                         Lp = Inductance of tank pri.
>
> If the following equation is assumed to be correct in the time domain:
>
> Vin = Iin * sqrt ( Lp/Cp ) Where        Vin = Vin applied to tank circ.
>                                         Iin = peak tank current
>                                         Lp = Inductance of tank pri.
>                                         Cp = Capacitance of tank C
>
> AUTHORS NOTE:  This is RMS tank current times Surge Impedance equals
>                applied voltage to tank circuit.
>
> Then substituting equation 2 into 1 and simplifying results in:
>
> Vo = Iin * sqrt ( Ls/Cp )  Variables as listed above
>
> This suggests that Cp should be made a small as possible, and
> to maximize Vo, as high a Vin as possible should be employed.  This
> makes sense because Iin will go up with higher Vin, and bang energy is
> .5*C*V^2.
>
> Also, if C is made smaller, dielectric losses maybe REDUCED, with a
> given capacitor (since dielectric area and volume are reduced).
> This is the first time that in doodling with the equations, a
> possible mathematical validation of what has been touted by the TCBOR
> all along is derived, make tank capacitors small, and leverage energy
> by the use of very high voltages.
>
> FYI and discussion. Am I full of it or does this make sense???
>
> Regards
>
> DAVE SHARPE, TCBOR
> Chesterfield, VA. USA.

```