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From: john on 13 Jan 2010 12:00
On 12 Jan, 10:22, Ian Bell <ruffreco...(a)yahoo.com> wrote:
> Suppose you have a simple pot divider consisting of two equal value
> resistors, say 10K each. Across the upper one you connect a series LC
> circuit that resonates at 3KHz or thereabouts. If you drive this network
> from a low impedance source and plot the response across the bottom
> resistor, the Q of the resulting peak is not the Q of the series LC but
> rather is determined by the pair of LC values. For example, choosing
> l=150mH and C=18nF gives a Q of just over 2. Choosing L= 1.5H and
> C=1.8nF gives a much higher Q.
>
> What I need is a simple means of calculating L and C given the pot
> divider resistor value and desired Q and f (assuming the Q of the LC
> itself is much higher).
>
> Cheers
>
> Ian

I'd see it as a Q 'definition' oddity.
If the pot is set at say the 80%, ie 4k/16k position then there's an
output baseline of 0.8V which simply can't be measured using the
normal method of 71% of peak voltage. The skirts are just too high
and the resulting curve shape 'looks' decidedly low Q. But, subtract
the 0.8V offset and remeasure Q as normal, this gives the 'real'
working Q value of XL/R. ('R' in this case appears as 3200ohms hence
the real Q is simply proportional to to the L or C reactance value at
resonance. ).
Set the pot at say the 10% position and the standard Q definition
becomes more and more valid.

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