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Design a second-order low-pass filter with a gain of 11 and cut-off frequency of 20 kHz.
Let us arbitrarily select C = 200 pF.
For a cut-off frequency of 20 kHz, we need R = Description: 6701.png
= 39.789 kΩ
If we select a standard resistor of 39 kΩ 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 + Rf /R1 = 1.586.
This in turn implies that Rf = 0.586 R1.
Imposing the dc bias-current balance condition, we obtain
0.586 R1 = 1.586 (2 R) = 123.708 kΩ.
Consequently, R1 = 211.11 kΩ and Rf = 123.708 kΩ.
Let us select a standard value of 130 kΩ for Rf . Then R1 should be about 221.8 kΩ.
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 = Description: 6707.png and for RA = 100 kΩ., R2 = 593.6 kΩ.
Thus, the final circuit for the second order low-pass active filter becomes as shown in Fig.
Description: 2163.png
Circuit of Example (10)

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