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Q of passive equaliser



 
 
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Old January 12th 10, 12:43 PM posted to uk.rec.audio
Ian Bell
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Posts: 300
Default Q of passive equaliser

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).

At first sight it might seem the Q should be that of the LC and the 10K
resistor but that gives a 2*pi*f*L/R of about 0.3 for the first example
above rather than the 2 point something you actually get.

Cheers

Ian
  #2 (permalink)  
Old January 12th 10, 04:16 PM posted to uk.rec.audio
Don Pearce[_3_]
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Posts: 1,358
Default Q of passive equaliser

On Tue, 12 Jan 2010 13:43:36 +0000, Ian Bell
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).

At first sight it might seem the Q should be that of the LC and the 10K
resistor but that gives a 2*pi*f*L/R of about 0.3 for the first example
above rather than the 2 point something you actually get.

Cheers

Ian


Why not just Spice model it and play until you get what you want? Or
are you simply trying it out to see if you get something that may be
useful. There are better controlled ways to do this, you know.

Here's a quick analysis for various pot settings from all the way up
to nearly the bottom

http://www.soundthoughts.co.uk/look/ianseq.png

Is that what you were expecting?

d
 




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