# Re: Couples Therapy

```Some stuff to add to what Antonio wrote.  For circuits with equal
primary and secondary Q's, the "flat-top"response occurs at critical
coupling, where kQ = 1.  For high-Q coils the condition is met with very
small coupling, for low-Q ones with correspondingly more.  Any talk
about critical coupling in connection with a Tesla coil is probably
meaningless, as the loading of the secondary due to the impedance of the
discharge is is non-linear and variable.  The whole object is to pump as
much power as possible into the discharge, however that is accomplished.

Ed

Tesla List wrote:
>
> Original Poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br>
>
> Tesla List wrote:
> >
> > Original Poster: "Ruud de Graaf" <rdegraaf-at-daxis.nl>
>
> > Thank you for your comprehensive answer. At first I was a little surprised
> > with your first comment: 'I don't see any relation between the theory of
> > coupled bandpass filters and Tesla
> > transformers.', but apparently you didn't meant it like it sounded. I
> > understand the need for resistor(s) in a bandpass filter to make it
suitable
> > for its job, but do you mean with 'this condition' the moment of critical
> > coupling? I don't think resistors can have any influence on the moment a
> > bandpass filter has the transition from one peak to two peaks in the dB/f
> > graph.
>
> Hi Rudd:
>
> The resistors determine totally if there are two peaks or just one.
> Without them the bandpass has always two peaks, ideally of infinite
> amplitude, or limited only by losses in a real circuit (except if the
> coupling coefficient between the coils is exactly 1, what is impossible
> in practice). A Tesla coil operates in this condition.
>
> Imagine, for example, what happens in the filter with just one resistor:
>                k
>   o------C1--+   +--+--+---o
>   +          |   |  |  |   +
>  Vin         L1  L2 C2 R  Vout
>   -          |   |  |  |   -
>   o----------+   +--+--+---o
>
> If R is large, the circuit is almost lossless, resonates at two
> different frequencies, and there are two sharp peaks in the frequency
> response.
> If R is small, it practically short-circuits C2 and L2, taking them
> out of the circuit, but leaves C1 connected to an almost purely
> inductive impedance seen at the input of the transformer. The
> remaining circuit has only one resonance frequency and the
> frequency response has only one peak, sharp too.
> Try this circuit in a simulator:
>
> C1=C2=1/sqrt(2)
> L1=2*sqrt(2)
> L2=sqrt(2)
> k=1/sqrt(2)
> R=1
>
> This is a normalized Butterworth design, with a passband geometrically
> centered at 1 rad/s, with 1 rad/s of bandwidth, and unity gain. It is
> exactly at the limit between one and two peaks. R>1 produces two peaks.
> R<1 just one. (sqrt=square root.)
>
> Antonio Carlos M. de Queiroz

```