Designing components for designing second order low pass filter at a high cut-off frequency

a. R2, R3 = 33k Ω, C1 = C2 = 0.0047μ F, R1 = 27k Ω & Rf = 15.8k Ω

b. R2 = R3 = 66k Ω, C1 = C2 = 0.0047μ F, R1 = 35k Ω & Rf = 60k Ω

c. R2 = R3 = 51k Ω, C1 = C2 = 0.0047μ F, R1 = 67k Ω & Rf = 70k Ω

d. None of the above

 

Since the value of high cut- off frequency is mentioned. Below enlisted are the steps to evaluate filter components of second order low pass filter for designing purpose.

Step 1 : Assume R2 = R3 = R & C1 = C2 = C.

Step 2 : Select the value of C = 1μ F
Suppose that C1 = C2 = C = 0.0047μ F

Step 3 : Determine R by using formula R = 1 / 2 π f_H C
R2 = R3 = 1 / (2 π x 10³ x 47 x 10^-10)
= 33.86k Ω
(consider the round figure value i.e. 33k Ω
As we know that Rf should be equal to 0.586 x R1, then R1 = 27k Ω)
Thus, Rf = 0.586 x 27 = 15.86k Ω
By taking into consideration the pot value of about 20k Ω, estimated resistances are
R2 = R3 = 33k Ω
C2 = C3 = 0.0047μ F
R1 = 27k Ω & Rf = 15.8k Ω

Step 4 : Since the values of resistors and capacitors are equal , the passband voltage gain must be equal to A_f = 1 + (R_f / R1) 1.586
However, the gain must satisfy the Butterworth response & hence, the value of R1 must be selected = 100k Ω in order to estimate the value of Rf.

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