There are 2 cut off frequencies for band pass filter.

1. upper cut off frequency (f1) – the frequency below which all the frequencies are passed

2. lower cut off frequency (f2) – all the frequencies above this frequency are passed

From this f1 and f2 we shall define following parameters which are designing parameters for band pass filter.

*Fig.1: Screenshot listing different parameters for band pass filter*

Based on Q we have **two types of band pass filter**

1. **Wide band pass filter –** with Q <10 it has wide flat response over range of frequencies. BW is more

2. **Narrow band pass filter –** with Q>10 it has sharp bell type response. BW is very less.

**Wide band pass filter**

Itâ€™s actually combination of LPF and HPF as shown in figure

*Fig. 2: Overview of Wide Band Pass Filter*

Here F2 > F1. So we have to design HPF with F1 and LPF with F2. Suppose we want to pass the band of frequencies between 2 KHz to 5 KHz. so F1 = 5000 Hz and F2 = 2000 Hz.

__Designing HPF section__

** Step 1:** assume the required value of capacitor. It should be less than 0.1 micro Farad. This is required for better frequency stability. Suppose we assume value C as 10 nF (nano farad)

** Step 2:** calculate the value of resistance from equation

*Fig. 3: Screenshot of calculations required to find resistance of HPF in Wide Band Pass Filter*

*Fig. 4: *

so our assumption for capacitance value as 10 nF is good (or OK). Otherwise if the calculated value of R is much less than 1 KOhm, we have to assume some other value of capacitor. Because the value of R should not be less than 1 K Ohm.

** Step 3:** choose required pass band gain. Suppose we take it as 2. So form equation

Af = 1 + (R2 / R1)

2 = 1 + (R2 / R1)

R2 / R1 = 1

R2 = R1

*Fig. 5:** Screenshot of calculations required to find resistance of HPF in Wide Band Pass Filter*

__Designing LPF section__

** Step 4:** assume the required value of capacitor. Suppose we assume same value C as 10 nF

** Step 5:** calculate the value of resistance from equation

*Fig. 6: Screenshot of calculations required to find resistance of LPF in Wide Band Pass Filter*

This value is odd value of resistor. May not be available as fixed value. So we may use potentiometer of 4.7 K? and tune it to desire value.

** Step 6:** choose required pass band gain. If we take it as 2 then again

R2 = R1

*Fig. 7: Screenshot of calculations required to find resistance of LPF in Wide Band Pass Filter*

Final design with component values are as shown. Op-amp is active component and it requires +ve and -ve biasing voltages. One can test the circuit by applying input through signal generator and observing output on DSO or oscilloscope as well as bode plotter as shown in figure.

*Fig. 8: Circuit Diagram of LM741 OPAMP IC based Wide Band Pass Filter*

__Note: –__ The schematic design is prepared in NI’s multisim 11 software. Software is available as free for 1 month trial period from NI’s website. Following all the circuits are also prepared in multisim 11 software.

## Narrow Band Pass Filter

**Step 1:** for simplicity assume C1 = C2 = C

**Step 2:** select centre frequency fc = 2 KHz, pass band gain Af = 2, and Q = 10

** Step 3:** assume capacitor value C as 100nF

** Step 4:** calculate value of R1 from

*Fig. 9: Screenshot of calculations required to find resistance for Narrow Band Pass Filter*

Final design is as shown below.

*Fig. 10: Circuit Diagram of LM741 OPAMP IC based Narrow Band Pass Filter*

## Questions related to this article?

đź‘‰Ask and discuss on EDAboard.com and Electro-Tech-Online.com forums.

Tell Us What You Think!!

You must be logged in to post a comment.