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Old 10th September 2010, 11:30 AM   #1
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Default 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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Old 10th September 2010, 11:51 AM   #2
Mooly is offline Mooly  United Kingdom
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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.
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Old 10th September 2010, 12:45 PM   #3
DF96 is offline DF96  England
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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.
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Old 10th September 2010, 05:05 PM   #4
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Quote:
Originally Posted by DF96 View Post
The value given by the RC formula is a corner frequency, not a resonance.
What DF96 said. Corner frequency (-3dB), as in a RC filter. Close, but no cigar.

For resonance you need XL=XC.

XL=2*pi*f*L
XC=1/(2*pi*f*C)

w
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Old 10th September 2010, 07:03 PM   #5
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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 AN-67.
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File Type: pdf Res_Freq_Op_Amp.pdf (9.6 KB, 16 views)
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Old 10th September 2010, 08:16 PM   #6
DF96 is offline DF96  England
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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.
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Old 10th September 2010, 08:19 PM   #7
Elvee is offline Elvee  Belgium
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Quote:
Originally Posted by johngalt47 View Post
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?
Depends on the opamp: if it's a micropower one with 0.05MHz GBW, it's unlikely.
If it's the latest LTCxxx or ADzzz, your chances are better.
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