ACTIVE_FILTERS_Theory_and_Design

Sallen–Key Filters 35
The radical in the foregoing equations must be positive, hence:
a2
K ≥1−
4 b
(2.35)
For K = 1, Equation (2.33) becomes:
C
1n
a
=
2b
(2.36)
and from Equations (2.34) and (2.31), we have:
C
2n
2
=
a
(2.37)
EXAMPLE 2.3
A second-order low-pass Butterworth filter must be designed with gain 10 and
f 1 = 1 kHz.
Solution
From Butterworth coefficients in Appendix C, for n = 2, we have:
a = 1.414, b = 1.000
From Equation (2.33), we have:
2
C 1
=
and from Equation (2.31):
1. 414 + 1. 414 + 8 × 9 . + . =
4
1 414 8 602
4
= 2. 504 F
1
C 2 = =0. 399 F
2.
504
Finally, we denormalize the resistance and capacitance values. We find the frequencynormalizing
factor FSF and the impedance-scaling factor ISF from Equations (2.4)
and (2.5), respectively.
ISF = 10 4
and
ω1 2π
f1 FSF = = = 2π
× 103
∴
ω 1
n