Question

Design a wideband band-pass active filter that has a gain of 17dB in the passbandand has corner frequencies at the lower and higher frequency limits of human hearing (i.e. 20Hz to 20kHz).

Answer 1

A wide band pass filter has (Q < 10)

Gain in dB of overall filter = 17 dB

`$F_L = 20 \ Hz $`

`$F_H = 20 \ kHz$`

`$17 = 10 \log A_0$`

`$1.7 = \log A_0 $`

`$A_0 = 5.47 $`

`$F_L = 20 \ Hz $`

`$F_L = (1)/(2 \pi R_1C_1)$`

Let
`$C_1 = 0.1 \mu F$`

`20 = (1)/(2 \pi * R_1 * 0.1 * 10^(-6))$`

`$R_1 = (10^6)/(4 \pi * 0.1) $`

`$R_1 = 79.6 \ k \Omega$`

`$F_H = 20 \ kHz$`

`$F_H = (1)/(2 \pi R_2C_2)$`

`$20 * 10^3 = (1)/(2 \pi R_2 * 0.1 * 10^(-6))$`

`$R_2 = (10^3)/(4 \pi * 10^3) $`

`$R_2 = 79.6 \ \Omega$`

Since the overall gain off filter is 5.47

`$A_0 = 5.47$`

`$A_0 = A_(01) * A_(02)$`

`$A_0 = 3.4 * 1.6$`

`$A_0 \sim 5.47$`

`$A_(01) = 1+ (R_F_1)/(R_i_1)$`

`$3.4 = 1+ (R_F_1)/(R_i_1)$`

`$2.4 = (10 k)/(R_i_1)$`

`$R_i_1 = 4166 \ \Omega$`

`$R_i_1 = 4.1 \ k \Omega $`

`$A_(02) = 1+ (R_F_2)/(R_i_2) $`

`$1.6 = 1+ (R_F_2)/(R_i_2)$`

`$0.6 = (R_F_2)/(R_i_1)$`

`$R_i_2 = (0.1 k)/(0.6)$`

`$R_i_2 = 166 \ \Omega $`