Answer to Question #194292 in Electrical Engineering for John Gly

Question

Electronics engineering: FILTERS

Design a 2nd order crossover network with the following specifications:  

• High Pass cutoff frequency = 3100 Hz
• Low Pass cutoff frequency = 310 Hz
• Bandpass frequency = 310Hz to 3100Hz   

Answer 1

An Op-Amp based bandpass filter is as shown below

The transfer function $\frac{V_0(s)}{V_{in}(s)}=-\frac{Z_0(s)}{Z_{in}(s)}$

$\frac{V_0(s)}{V_{in}(s)}=-[\frac{R_2}{(1+sC_2R_2)}]*[\frac{sc_1}{(1+sC_1R_1)}]=\frac{sc_1R_2}{(1+s \tau_1)(1+s \tau_2)} \implies \tau =RC$

The circuit will behave as BPF if $\tau _1> \tau_2$

$f_L=\frac{1}{2 \pi R_1C_1}$

$f_H=\frac{1}{2 \pi R_2C_2}$

$R_1=\frac{1}{2 \pi *310*47*10^{-9}}=10.924k \Omega$

Commercial available resistance nearest to calculated R₁ is 11k$\Omega$

$\implies f_L=\frac{1}{2 \pi R_1C_1} =\frac{1}{2 \pi 11*10^3*47*10^{-9}} =307.84 Hz$

Therefore, $R_1=11k \Omega; C_1=47nF$

$C_2=\frac{1}{2 \pi R_2f_H} =\frac{1}{2 \pi *3100*20*10^3} = 2.567 nF$

Commercial available capacitor nearest to calculated C₂ is 2.7nF

$f_H=\frac{1}{2 \pi R_2C_2}=\frac{1}{2 \pi*20*10^3*2.7*10^{-9}}=2947.31 Hz$

Therefore, $R_2=20k \Omega; C_2=2.7nF$