Re: Big Coils (was DOLLINGER)

To: tesla-at-pupman-dot-com
Subject: Re: Big Coils (was DOLLINGER)
From: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> (by way of Terry Fritz <twf-at-verinet-dot-com>)
Date: Mon, 14 Dec 1998 20:43:38 -0700
Approved: twf-at-verinet-dot-com

Marco Denicolai wrote:

> I have been reading dolnger1.jpg through dolnger8.jpg and I must say it is
> a remarkable piece of infomation! Very useful! I suggest it to everybody.
> 
> I also made a Matlab program based on the equation presented in those
> papers and I experimented with different values of k and L1, L2.

I have a program that solves the problem exactly in the general case and
plots everything. Take a look at it at
ftp://coe.ufrj.br/pub/acmq/teslasim.zip

> According to the equation, top voltage rise at the secondary depends mainly
> on the square root of the secondary/primary inductance ratio (accordingly
> to the formula usually reported by other textbooks). That would give a
> LOWER secondary voltage using big coils (e.g. the new one I am designing)
> compared with small coils (e.g. my old one).

No. The equation predicts a higher rise: V2max=V1max*sqrt(L2/L1)
This is simply energy conservation. The input energy is
E1=0.5*C1*V1max^2
The maximum output energy would be E2=0.5*C2*V2max^2. If the system is
tuned,
L1*C1=L2*C2, or L2/L1=C1/C2. With no losses and complete transfer,
E2=E1,
and V2max/Vinmax=sqrt(C1/C2)=sqrt(L2/L1).

> We all know this doesn't hold, but I don't believe that it can be all
> explained saying that you have lower losses with big coils... 
> So how it can be explained that (physically) bigger coils provide higher
> secondary voltages, although they have a lower secondary/primary inductance
> ratio?

They have -higer- secondary/primary inductance ratio.

> using the paper's data, the top voltage rise at the secondary should have
> been about 130 times the primary voltage. Still those guys used 50V at the
> primary, measured about 2 kV at the secondary (factor is 40:1) and
> concluded that the computer simulation gave the same results (!).
> How's that?

I didn't check, but probably the effect of resistive losses. Try with my
program.

Antonio Carlos M. de Queiroz