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 andimpedance model are shown in Figure 15.25.

Let us focus on sinusoidal inputs and use the impedance method to obtainits input-output relationship. First, notice that the portion of the circuit withinthe dashed box is a non-inverting connection of the Op Amp with gain:

G = 1 + R1

R2. (15.94)

+

-

C

C

RR

R1R2

vi

vov2

v1

(a) Circuit

+

-

Vi

VoV2

V1

(b) Impedance model

1/Cs

R

R1

R2

1/C

s

R

F IGURE 15.25 The Sallen-Keycircuit.

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Thus, for the purpose of analysis, we can replace the circuit within the dashedbox with an amplifier whose gain is G. Therefore, we can write

Vo = GV1. (15.95)

Applying KCL for node V1

V2 − V1

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

GVo. (15.96)

Now, KCL for node V2 yields,

Vi − V2

R= V2 − V1

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+1G

Vo

R=(

RCs+1G

)Vo − 1

GVo

R+(

RCs+1G

)Vo − Vo

1Cs

. (15.98)

We can simplify Equation 15.98 and obtain the following expression relatingthe output voltage to the input voltage:

H(s) = Vo(s)

Vi(s)= G

R2C2s2 + RCs(3 − G) + 1. (15.99)

As a specific example, let us draw the frequency response for the filter trans-fer function for RC = 1 and R1 = R2. For these values, G = 2 and the

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-1.5 -1 -0.5 0 0.5-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real Axis

Imag

Axi

s

F IGURE 15.26 Pole-zero plot ofthe Sallen-Key filter.

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 thedenominator has a pair of complex conjugate roots: −1/2 + j

√3/4 and

−1/2 − j√

3/4. In terms of the pole-zero nomenclature introduced inSection 13.4.3, the transfer function has two poles and no zeros. The polesare 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. (Whenzeros exist, their locations are depicted using circles.)

We can now plot the frequency response as shown in Figure 15.27 bysubstituting s = jω in the transfer function:

H( jω) = 2

( jω)2 + jω + 1(15.102)

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F IGURE 15.27 Frequencyresponse for the Sallen-Key filter.

10-1 100 101 -200

-150

-100

-50

0

Frequency (radians)

Pha

se (

degr

ees)

10-1 100 10110-2

10-1

100

101

Frequency (radians)

Mag

nitu

de

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 thephase of H( jω) versus the frequency ω.

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