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https://designers-guide.org/forum/YaBB.pl Design >> Analog Design >> 2-pole active filter question https://designers-guide.org/forum/YaBB.pl?num=1372599302 Message started by Giorgos on Jun 30th, 2013, 6:35am |
Title: 2-pole active filter question Post by Giorgos on Jun 30th, 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. [url] ![]() 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 30th, 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 30th, 2013, 1:34pm rfidea wrote on Jun 30th, 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 1st, 2013, 10:37am Yes, you are right. I think the relationship between fcut/f0 and Q is pretty complex. |
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