# Example-10

Example-10

Design a second-order low-pass filter with a gain of 11 and cut-off frequency of 20 kHz.

Solution

Let us arbitrarily select

*C*= 200 pF.For a cut-off frequency of 20 kHz, we need

*R*== 39.789 kÎ©

If we select a standard resistor of 39 kÎ© for

*R*, then the cut-off frequency is about 20.4 kHz.The dc gain for this filter cannot be anything other than

*K*where*K*= 1.586.Thus, for a dc gain of 1.586,

*K*= 1 +*R*/_{f}*R*_{1}= 1.586.This in turn implies that

*R*= 0.586_{f}*R*_{1}.Imposing the dc bias-current balance condition, we obtain

0.586

*R*_{1}= 1.586 (2*R*) = 123.708 kÎ©.Consequently,

*R*_{1}= 211.11 kÎ© and*R*= 123.708 kÎ©._{f}Let us select a standard value of 130 kÎ© for

*R*. Then_{f }*R*_{1}should be about 221.8 kÎ©.We need another amplifying stage to obtain the needed gain of 11. The gain of this stage should be 11/

*K*= 6.936. We have chosen to use non-inverting amplifier for this stage. The output amplifier resistors are calculated as,6.936 = and for

*R*_{A }= 100 kÎ©.,*R*_{2}= 593.6 kÎ©.Thus, the final circuit for the second order low-pass active filter becomes as shown in Fig.

**Circuit of Example (10)**