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Re: Tesla Coil Operation - was "Harmonics"
tesla-at-pupman-dot-com on 31.01.99 11:07:43
To: tesla-at-pupman-dot-com
cc: (bcc: Marco Denicolai/MARTIS)
Subject: Re: Tesla Coil Operation - was "Harmonics"
>Original Poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br>
SNIP
>In a Tesla coil, M=k*sqrt(L1*L2), so certainly M<<L2. We can make La=0
>to simplify the circuit. This requires L1=M, or k=sqrt(L1/L2). This
>value for the coupling coefficient k is reasonable for a typical Tesla
>coil. Adding the other elements of the original coil, we have the
>circuit (exchange the gap with C1 if you prefer):
> gap
> o--------+--o o--+---(L2-M)---0(terminal)
> | |
>power supply C1 L1
> | |
> o--------+-------+------------0(ground)
>This is essentially a Tesla coil with the bottom of the secondary
>connected to the top of the primary and the bottom of the primary
>grounded. This configuration may be problematic for the power
>supply, that must be grounded (a normal NST cannot be used), but
<it works exactly as the conventional circuit (with slightly different
>unimportant transmission line effects).
>In old literature, this configuration is known as the Seibt circuit.
>Antonio Carlos M. de Queiroz
Very interesting indeed! I think grounding of one side of the tank supply
cannot be a problem for all of the list members.
- Has anybody on the list tried this configuration?
- Does this provide the same performance of a magnifier but with 2 coils?
If k=sqrt(L1/L2) then L2=L1/(k*k): for L1=13uH and k=0.2 --> L2=0.32mH. On
the other hand, If I just reconnect my secondary I get
k=sqrt(13uH/29mH)=0.02. It looks like the whole secondary needs to be
recalculated or the primary needs to be bigger.
Does this lead to the second coil (tighly coupled to the primary) typical
of 3-coil magnifiers?