ACTIVE_FILTERS_Theory_and_Design

Sallen–Key Filters 25
v i
R
10 K
15.9 nf
C
+
−
Rb
40 K
10 K
R a
ISF = 10 4
v o
A
dB
20
14
10
0
−6
−10
−20
11
0.1 1
a decade
−20 dB/dec
20 dB
10
log f (kHz)
(a)
(b)
FIGURE 2.3 LP Butterworth filter, where f 1 = 1 kHz and K = 5 (14 dB).
Denormalization
ω1 2π
f1 FSF = = = 2π
× 103
∴
ω 1
Figure 2.3 shows the designed filter with its frequency response.
n
Cn
1
C = =
ISF × FSF 2π
× 10
R = ISF × R n
= 10 × 1 Ω=
10
7
= 15. 9 nF
4
k
Ω
R = ISF × R = 10 × 1 kΩ=
10 kΩ
a
an
R = ISF × R = 10 × 4 kΩ=
40 kΩ
b
bn
2.4 FIRST-ORDER HIGH-PASS FILTER
Figure 2.4 shows the first-order high-pass filter with noninverting gain K.
For the node
v 0
/ K,
we have:
K
= 1+
R
R
b
a
− sCV + G + sC V 0
i
( ) = 0 ∴
K
V
Hs ()= 0 KsC Ks
V
=
Ks
= =
i G+
sC 1 s + ω
s +
2
RC
(2.10)