SLOA049 Application note

SLOA049D July 2000 – February 2023

B.1 Fifth-Order Low-Pass Butterworth Filter

Referring to Table 9-1, for a fifth-order Butterworth filter, the required circuit transfer function can be written as:

Equation 28.

H_{LP} (f) = \frac{K}{(\frac{j f}{f_{c}} + 1) (- {(\frac{f}{f_{c}})}^{2} + \frac{1}{0 .6180} \times \frac{j f}{f_{c}} + 1) (- {(\frac{f}{f_{c}})}^{2} + \frac{1}{0 .6180} \times \frac{j f}{f_{c}} + 1)}

Figure 15-1 shows a Sallen-Key circuit implementation and the required component values. $f_{c}$ is the –3-dB point. The overall gain of the circuit in the pass band is $K = K_{a} \times K_{b}$ .

Figure 15-1 Fifth-Order Low-Pass Filter Topology Cascading Two Sallen-Key Stages and an RC

Stage	$f_{c}$	$Q$	$K$
1	$\frac{1}{2 πRC}$	$N / A$	$1$
2	$\frac{1}{2 π \sqrt{R_{1 a} R_{2 a} C_{1 a} C_{2 a}}}$	$\frac{\sqrt{R_{1 a} R_{2 a} C_{1 a} C_{2 a}}}{R_{1 a} C_{1 a} + R_{2 a} C_{1 a} + R_{1 a} C_{2 a} (1 - K_{a})} = 0.618$	$K_{a} = \frac{R_{3 a} + R_{4 a}}{R_{3 a}}$
3	$\frac{1}{2 π \sqrt{R_{1 b} R_{2 b} C_{1 b} C_{2 b}}}$	$\frac{\sqrt{R_{1 b} R_{2 b} C_{1 b} C_{2 b}}}{R_{1 b} C_{1 b} + R_{2 b} C_{1 b} + R_{1 b} C_{2 b} (1 - K_{b})} = 1.618$	$K_{b} = \frac{R_{3 b} + R_{4 b}}{R_{3 b}}$