# 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 f_{H} C

R2 = R3 = 1 / (2 π x 10^{3} 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 A_{f} = 1 + (R_{f} / 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.