15.6.5 S A L L E N - K E Y F I L T E R
This section introduces a lowpass filter called the Sallen-Key filter. Its circuit and
impedance model are shown in Figure 15.25.
Let us focus on sinusoidal inputs and use the impedance method to obtain
its input-output relationship. First, notice that the portion of the circuit within
the dashed box is a non-inverting connection of the Op Amp with gain:
G=1+
R1
R2
.
(15.94)
C
vi
v1
R
v2
+
vo
R
-
C
R1
R2
(a) Circuit
F I G U R E 15.25 The Sallen-Key
circuit.
1/Cs
R
R
V1
+
V2
Vo
-
1/Cs
Vi
R1
R2
(b) Impedance model
866a
Thus, for the purpose of analysis, we can replace the circuit within the dashed
box with an amplifier whose gain is G. Therefore, we can write
Vo = GV1 .
(15.95)
Applying KCL for node V1
V2 − V 1
=
R
V1
1/Cs
,
which simplifies to:
V2 = (RCs + 1)V1 .
Substituting for V1 in terms of Vo from Equation 15.95, we get
V2 =
RCs + 1
G
Vo .
(15.96)
Now, KCL for node V2 yields,
Vi − V 2
V2 − V1
=
R
R
V2 − Vo
+
1/Cs
.
(15.97)
Substituting for V1 and V2 in terms of Vo from Equations 15.95 and 15.96,
we get
Vi −
RCs+1
Vo
G
R
=
RCs+1
G
Vo −
1
V
G o
R
+
RCs+1
G
Vo − Vo
1
Cs
.
(15.98)
We can simplify Equation 15.98 and obtain the following expression relating
the output voltage to the input voltage:
H(s) =
Vo (s)
Vi (s)
=
G
R 2 C 2 s2
+ RCs(3 − G) + 1
.
(15.99)
As a specific example, let us draw the frequency response for the filter transfer function for RC = 1 and R1 = R2 . For these values, G = 2 and the
866b
1
0.8
0.6
Imag Axis
0.4
0.2
0
F I G U R E 15.26 Pole-zero plot of
the Sallen-Key filter.
-0.2
-0.4
-0.6
-0.8
-1
-1.5
-1
-0.5
Real Axis
0
0.5
transfer function is given by:
H(s) =
2
s2
+s+1
(15.100)
or, factoring the denominator,
H(s) =
2
.
(s + 1/2 + j 3/4)(s + 1/2 − j 3/4)
(15.101)
The transfer function represents a second-order filter. The expression
in the
3/4 and
denominator
has
a
pair
of
complex
conjugate
roots:
−1/2
+
j
−1/2 − j 3/4. In terms of the pole-zero nomenclature introduced in
Section 13.4.3, the transfer function has two polesand no zeros. Thepoles
are a complex conjugate pair located at −1/2 + j 3/4 and −1/2 − j 3/4.
Figure 15.26 depicts the pole locations using X’s in the complex plane. (When
zeros exist, their locations are depicted using circles.)
We can now plot the frequency response as shown in Figure 15.27 by
substituting s = jω in the transfer function:
H( jω) =
2
( jω)2
+ jω + 1
(15.102)
866c
Magnitude
101
100
10-1
F I G U R E 15.27 Frequency
response for the Sallen-Key filter.
Phase (degrees)
10-2
10-1
100
Frequency (radians)
101
100
Frequency (radians)
101
0
-50
-100
-150
-200
10-1
or, in terms of the factored transfer function,
H( jω) =
2
.
( jω + 1/2 + j 3/4)( jω + 1/2 − j 3/4)
(15.103)
As before, the frequency response in Figure 15.27 plots the magnitude and the
phase of H( jω) versus the frequency ω.
866d