AN231K04-DVLP3 – AnadigmApex Development Board
UM023100-K001h
Page 11 of 16
Figure 5: Suggested Rauch filter circuits
For low pass response
:
H(s) = 1/((R1/R2)+ (sC2(R1+R3+(R1*R3/R2)))+
(s
2
R1R3(2.C1)C2))
R1 = R2 = 2R3 = 2R
And
C1 = 2C2 = 2C
Fc = 1/(4
π
RC(SQRT 2))
Re-arranging these equations for low pass filter
R1 = Rin;
R2 = G*Rin;
R3 = G*Rin;
C1 = Q.(G+1)/(G.4.
π
.Fo.Rin);
C2 = 1/(G.4.
π
.Fo.Q.R1)
Examples
Rauch filter settings:
For Fc = 30kHz:
R1=R2=10k, R3=5k, C1=0.75nF, C2=0.375nF
For Fc = 40kHz:
R1=R2=10k, R3=5k, C1=0.56nF, C2=0.28nF
For Fc = 10kHz
R1=R2=10k, R3=5k, C1=2.2nF, C2=1.1nF
Alternative math.
C1=2.C2.(G+1).Q^2 R1=1/(Q.4.G.
π
.F.C2) R2=G*R1 R3=1/(Q.4.(G+1).
π
.F.C2)
from these : C1=Q.(G+1)/(2.G.
π
.F.R1) and C2 = 1/(G.
π
.F.R1.4.Q)
Example,
Starting from a requirement for a corner frequency of 10kHz and we want a gain of 1
Choose to use 10kohm resistors
so that R1=R2=10k,
also if we choose a Q=0.707 for flat response then:
C1=1.414/(2 x
π
x 10k x 10k) = 2.25nF
and C2=1/(1 x
π
x 10k x 10k x 4 x 0.707) = 1.1nF
and R3=1/(0.707 x 4 x 2 x
π
x 10k x 1.1n) = 5.1kohm