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Design >> Analog Design >> 2-pole active filter question
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Message started by Giorgos on Jun 30^th, 2013, 6:35am

Title: 2-pole active filter question
Post by Giorgos on Jun 30^th, 2013, 6:35am

Hello! I am desiging in PSpice a 2-pole active filter. the values are taken from a wikipedia example to be sure about the fcut.

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this is the circuit http://en.wikipedia.org/wiki/File:Sallen-Key_Lowpass_Example.svg

and the fcut=16KHz. but when i am doing it in pspice the fcut is at 10khz.

i tried with other values and again problem. i read about FREQUENCY SCALING FACTOR accoring to Butterworth filter but i don't understand
why for example wiki has wrong fcut.

please someone to explain.thanks a lot :)

Title: Re: 2-pole active filter question
Post by rfidea on Jun 30^th, 2013, 12:29pm

There is nothing wrong at Wiki. You are mixing up the 3dB-frequency with the resonance frequency. The resonance frequency of your frequency is 15.9 kHz but that is not the 3dB frequency, which seems to be about 10.5 kHz.

The Q=0.5 means that your filter is NOT a Butterworth filter. A second order Butterworth filter has Q=1/sqrt(2)=0.707. A Butterworth filter has the same resonance frequency as 3dB-frequency, valid for all orders.

Title: Re: 2-pole active filter question
Post by Giorgos on Jun 30^th, 2013, 1:34pm

rfidea wrote on Jun 30^th, 2013, 12:29pm:

hi and many thanks for the reply. So this formula:fcut=1/2π(sqrt(R1R2C1C2) is valid only about butteworth filters .

so the formulas of fcut, depends on Q of the type of filter design (bessel,butterchebysec etc)....

So the Q=0.5 what is this? how can i know ?

Is there any rule of thumb?

Title: Re: 2-pole active filter question
Post by rfidea on Jul 1^st, 2013, 10:37am

Yes, you are right. I think the relationship between fcut/f0 and Q is pretty complex.