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Question on Modelling a Magnifier




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From:  Antonio C. M. de Queiroz [SMTP:acmq-at-compuland-dot-com.br]
Sent:  Friday, March 13, 1998 11:13 PM
To:  Tesla List
Subject:  Re: Question on Modelling a Magnifier

Andrew Chin wrote:

>I've just recently started following the thread on Modelling a Magnifier and found
> it very interesting that a >magnifier operatates in the same manner as a conventional coil.

>Just one thing I want to know though.  It was probably mentioned way back in this
>thread but what were the >formulas for output voltage and coupling coefficient, and 
>how were they derived?

A quick resume:

A conventional capacitor-discharge Tesla coil is composed by two resonator
LC tanks tuned for the same frequency, with the coils magnetically coupled
with a low coupling coefficient.
The lumped model for a conventional Tesla coil after the firing of the
spark gap, and before any breakout of sparks in the secondary, ignoring
resistances, is:

+-----+     +-----+
|     | <k> |     |
C1    L1    L2    C2
|     |     |     |
+-----+     +-----+

k is the coupling coefficient, k=M/sqrt(L1*L2) (a).

A Tesla magnifier has a transformer with higher coupling coefficient, and
a separate "third coil" resonator mounted some distance away.
An ideal lumped model for a magnifier, assuming as negligible the capacitances
C2' in parallel with L2' , would be:

+-----+      +--L3-+
|     | <k'> |     |
C1    L1     L2'   C3
|     |      |     |
+-----+      +-----+

k' is the coupling coefficient, k'=M'/sqrt(L1*L2') (b).

This model is exactly equivalent to the model of the conventional coil
with the same primary tank, if:
M=M'
k=M'/sqrt(L1*(L2'+L3))=k'*sqrt(L2'/(L2'+L3)
L2=L2'+L3
C2=C3

Why: 
The differential equations that model a linear transformer are:
v1 = L1*di1/dt + M*di2/dt
v2 = M*di1/dt  + L2*di2/dt  
where v1 and i1 are the voltage and current at the primary and v2 and
i2 the same for the secondary.
If L2 is split in two parts, L2' and L3, with the same current i2 and
the same voltage v2 across the combination:
v1 = L1*di1/dt + M'*di2/dt
v2 = M'*di1/dt + (L2'+L3)*di2/dt
The equations are identical, and their solution too, if L2=L2'+L3
and M=M'. From (a) and (b), the value of the effective coupling
coefficient k is obtained.

The maximum output voltage is given by energy conservation:
VC2max=VC1max*sqrt(C1/C2) for the conventional coil.
VC3max=VC1max*sqrt(C1/C3) for the magnifier.

Why:
In both systems, the initial energy in the system, just before the 
conduction of the primary spark gap, is the energy in a capacitor C1 
charged to VC1max: E1=0.5*C1*VC1max^2.
With the system properly tuned, after a few cycles all this energy
(ignoring losses) will be in the outpur capacitance: 
E2=0.5*C2*VC2max^2, or E2=0.5*C3*VC3max^2.
E1=E2 gives the relations.
This assumes that the break rate is low enough, so there is no
significant energy left from the previous break in the circuit.

This is valid as long as L2'<<L3. If the two coils are similar, the
presence of a significant capacitance C2' may result in different
behavior, with possibly higher output voltage for the same total 
inductance, because C3 can be smaller and the system can be tuned
in a way that periodically concentrates the energy in C3 only,
after the opening of the primary circuit. (I am still not sure if
it is possible to remove all the energy from the primary circuit
in this condition, but probably is.)

Antonio Carlos M. de Queiroz
http://www.coe.ufrj.br/~acmq