Hi every one

Please I am trying to design band pass filter using sallen & key filter for the sum of the left audio the frequancy range of which is 23KHz to 53KHz

Please I need to understand the calculation baesed on Fc=1/2piRC as I ahve to use equal values for Rs and equal values for C

*pass band gain = 0
*Gain flatness error = 1dB
*Attenuation at twice the cut-off frequancy 12dB

the sallen key filter should be secound order filter and I dodnt know how many secound order filter I'll need

Many thanks

 

Hello,

On this page some links with information on sallen key filters can be found.
http://www.educypedia.be/electronics/analogfil.htm
Scroll down the page for some PDF's.

Greetings,
Bertus

 

thx alot
but the website is very complecate for me
I coulden anderstand most of the things so please can you expline it fro me

 

Hello,

This comes from one of the links of the page I gave you.
http://www.ecircuitcenter.com/Circuits/opsalkey1/opsalkey1.htm

Sallen-Key Low-Pass Filter

CIRCUIT

OPSALKEY1.CIR Download the SPICE file​

Suppose you had a large interfering signal you needed to get rid of. To get lots of attenuation, you could cascade several RC filters. Unfortunately, the impedance of one RC section affects the next. This means that the “knee” or transition between the pass and stop bands won’t be very sharp. A sharp knee helps you reduce the interfering signal without degrading your desired signals. In this situation, the Sallen-Key active filter can save the day. This circuit implements a 2-pole filter. Cascading several stages can give you a steep attenuation curve with a very sharp knee.

LOW-PASS FILTER DESIGN
Although there are many filter types and ways to implement them, here’s an active low-pass filter that’s greatly simplified if R1=R2 and the op amp stage is a unity gain follower (RB=short and RA=open). Designing a 2-pole Butterworth filter requires just a few steps.
1. Choose a cutoff frequency fo (Hz).

As an example, select fo=10 kHz to reduce a noise signal at 50 kHz and pass your desired signals below 5 kHz.
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2. Pick a convenient cap value C2 between 100pF and 0.1 uF.

Suppose you’ve got plenty of 1000pF caps in stock, select this value for C2.
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3. Make C1 = 2 x C2

C1 = 2 · C2 = 2000pF
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4. Calculate R1 = R2 = 0.707 / (2 · π · fo · C2)

R1 = R2 = 0.707 / (2 · π · 10kHz · 1000pF) = 11.2 K ohms


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Greetings,
Bertus

 

Many thank this is very helpfull