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10th September 2010, 11:30 AM  #1 
diyAudio Member

Resonant circuit design?
I wasn't sure which area this question belongs so I guess this one is as good as any...
When designing a capacitive resonant circuit using the formula f=1/2*pi*R*C does it matter about the relative impedances of the resistors and capacitors? I mean, as long as the combination arrives at the resonant frequency you choose, can you just put any combination of resistor and capacitor you want?
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10th September 2010, 11:51 AM  #2 
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There's a bit more to it than that... not least the practicalities and properties of real world components.
Look up "Q" with regard to resonance... it's a measure of the "goodness" of a resonant circuit and is affected by real world properties of components. To answer the question, yes, if the impedance of a cap deviates from the ideal then it affects the resonant circuit. This is particularly true of coils as well. The formula doesn't take these into consideration, it assumes ideal. 
10th September 2010, 12:45 PM  #3 
diyAudio Member
Join Date: May 2007

RC is not a resonant circuit, you need LC or an active circuit for resonance. The value given by the RC formula is a corner frequency, not a resonance.
The impedances will give the stated frequency whatever their individual values  but you have to realise that the resistance/capacitance which appears in the formula is not the value of a component which you have just soldered in, but the effective value after the loading of the rest of the circuit is considered. People often forget this. 
10th September 2010, 05:05 PM  #4 
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10th September 2010, 07:03 PM  #5 
diyAudio Member

Ok, I see that a picture is worth a thousand words. In the attachment, if I just change the capacitors by an order of magnitude, will I get approximately an order of magnitude change in the oscillation frequency?
This was taken from Linear Technology's app note AN67.
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Any technology sufficiently advanced is indistinguishable from magic 
10th September 2010, 08:16 PM  #6 
diyAudio Member
Join Date: May 2007

Yes, because they are already significantly larger than any likely circuit stray capacitances. However, if you dropped them by a factor of 100 then the frequency might not change by quite that much as a few hundred pF is starting to get near stray values.

10th September 2010, 08:19 PM  #7  
diyAudio Member
Join Date: Sep 2006

Quote:
If it's the latest LTCxxx or ADzzz, your chances are better. 

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