0
7.1kviews
Design a second order KRC low pass filter with a cut off frequency $f_O$ = 2KHz and Q=5.
1 Answer
0
326views

Given :

$f_O$ = cut off frequency = 2KHz

Q = 5

Solution :

Figure 1 shows the circuit diagram and frequency response of second order KRC low pass filter.

enter image description here

Step 1 : Determine the values of $R_2$,$R_3$ and $C_2$,$C_3$

$$f_o = \frac{1}{2π\sqrt{R_2 R_3 C_2 C_3}}$$

If $R_2$,$R_3$ = R and $C_2$,$C_3$ = C then $f_O$ becomes

$$f_o = \frac{1}{2πRC}$$

$$ 2KHz = \frac{1}{2πR×0.1µf}…………….Assume C=0.1µf$$

$$\boxed{R = 795.57Ω = 820Ω(Std)}$$

$R_2$ = $R_3$ = 820Ω and $C_2$ = $C_3$ = 0.1µf,

Step 2 : Determining Gain

$$A = 3-\frac{1}{Q}$$

$$A = 3-\frac{1}{5}$$

$$A = 2.8$$

Step 3 : Selection of an amplifier component

$$A = 1 + \frac{R_f}{R_1}$$

$$2.8 = 1 + \frac{R_f}{R_1} $$

Let $R_1$ = 10K $$1.8 = \frac{R_F}{10K}$$ $$\boxed{R_F = 18K}$$ Use 20K potentiometer at $R_F$.

Step 4 : Designed Circuit Diagram

enter image description here

Please log in to add an answer.