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Low pass Active filters

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fala

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Hello, I have a question about active filters. with reference to the image I understand the role of R2 & C2 they are actual low pass filter, but what I don't understand fully,is the role of R1 & C1 why they are necessary? and why R1 should be equal to R2 and C1 equal to C2? I want gain of my filter be one at all frequencies( frequencies less than -3db cut off) can I remove R3 & R4 and directly connect pin 6 OpAmp to pin 2 ?
thanks.
 

jeffttan

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Hello, the image you have posted does not show up. thanks:)
 

Kral

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fala,
The filter is a 2-pole low pass filter. This filter topology is know as the Sallen Key topology. If you derive the circuit transfer function you will find that the denominator is a 2nd order polynomial of the form.
K/(s2 + 2ZetasW0 + w0^2).
Where Zeta is the damping factor (=.707 for a Butterworth filter), W0 is the undamped natural frequency.
~
You can, indeead connect pin 6 directly to pin 2 to obtain a unity DC gain. One reason for using a gain > 1 is that it allows you to select C1 = C2. With a gain of 1, setting C1 = C2 results in a negative value for R2. Radio Shack does not stock many negative resistors. If you are willing to use unequal value capacitors, then setting the gain = 1 is not a problem. The usual practice is to select one of the capacitors to be a standard value, and calculate the other capacitor, which will almost always turn out to be a non-standard value.
~
See any book on anmalog filter design for more information, or Google on "active filter".
~
Regards,
Kral
 

    fala

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fala

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hello jeffttan, image dose not show automatically(I don't know why) but I could download it myself so you should be able to. but you have to log in to be able to see images or download them, thanks.
thanks kral, your reply was helpful. I'm looking the web for articles on this subject. but as I understood from your response I have two options first remove R1 &C1 and use the active filter as one pole then I don't need amplification for dc. second because I'm going to fed output of the filter to a 16bit ADC, I should find a way to calibrate ADC at each frequency. and because even if I use one pole filter then I will have less than unity gain anywhere between DC and -3db cutoff and this gain dose not seem linear or easily computable then I still have to calibrate ADC at each frequency(even if I use one pole). so why not use second pole that has sharper attenuation because I have to calibrate my ADC for each frequency anyway. can you please confirm this that my understanding so far is right? thank you very much.
 

IanP

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    fala

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