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

## What will be the designing components for designing second order low pass filter at a high cut-off frequency of 1kHz assuming C=0.0047 μ F along with consideration of standard pot values ?

a. R2,R3 =33 k Ω ; C1 = C2 = 0.0047 μ F , R1 = 27 k Ω & Rf = 15.8 k Ω
b. R2 = R3 = 66 k Ω ; C1= C2 = 0.0047 μ F , R1 = 35 k Ω & Rf = 60 k Ω
c. R2 = R3 = 51 k Ω ; C1 = C2 = 0.0047 μ F , R1 = 67 k Ω & Rf = 70 k Ω
d. None

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

Explanation :

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 = 1 μ F
Suppose that C1= C2 = C = 0.0047 μ F

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

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