The main function of filters is to suppress or filter out components from mixed frequency signals for clear communication. As we learned in the **first article** of this series, filters permit a range of frequency to pass that’s known as “pass bend,” rejecting (or suppressing) other frequencies known as “stop bend.”

The cut-off frequency is the parameter that separates these two bends. Depending on the pass and stop bends, there are four types of filters: Active **low-pass**, **high-pass**, band-pass, and **band-reject** filters.

In this tutorial, we’ll cover the band-pass filter (BPF), which allows the pass band of frequencies between the higher and lower cut-off frequencies.

There are two cut-off frequencies for the BPF:

**1. The upper cut-off frequency (f1):** a frequency below where all other frequencies are passed

**2. The lower cut-off frequency (f2):** all of the frequencies above this frequency are passed

From this f1 and f2, let’s define the following parameters, which represent the designing parameters for the BPF.

*The listing of the different parameters for a band-pass filter.*

Based on “Q,” we have two types of band-pass filters…

**1. A wide band-pass filter:** with Q <10, it has a wide flat response over a range of frequencies. The BW is also greater.

**2. A narrow band-pass filter:** with Q>10, it has a sharp bell-type of response. In this case, the BW is much lower.

**The wide band-pass filter
**The wide BPF is a combination of the LPF and HPF as shown here:

*An overview of the wide band-pass filter.*

Here, F2 > F1. This means that we have to design the HPF with F1 and LPF with F2. Now, let’s suppose we want to pass the band of frequencies between 2 to 5 kHz. so F1 = 5000 Hz and F2 = 2000 Hz.

**Designing the HPF section**

**Step 1:**First, assume the required value of the capacitor. It should be less than 0.1 micro Farad. This is necessary for better frequency stability. Let’s assume the value C as the 10 nF (nano farad).**Step 2:**Calculate the value of the resistance from the equation.

*The calculations required to find the resistance of the HPF in a wide band-pass filter.*

Our assumption for the capacitance value as 10 nF is good (or, at least, OK). If we calculate the value of R it’s much less than 1 KOhm. So, we have to assume some other value for the capacitor because the value of R should not be less than 1 K Ohm.

**Step 3:**Next, choose the required pass-band gain. Let’s use 2 for this tutorial.

Here’s the equation…

Af = 1 + (R2 / R1)

2 = 1 + (R2 / R1)

R2 / R1 = 1

R2 = R1

*The** calculations required to find the resistance of the HPF in a wide band-pass filter.*

**Designing the LPF section**

**Step 4:**Now, assume the required value of the capacitor. Let’s suppose we assume the same value C as for10 nF.**Step 5:**Calculate the value of the resistance from the equation.

*The calculations required to find the resistance of the LPF in a wide-band pass filter.*

This value is an odd one for the resistor and may not be available as a fixed value. We might be able to use the potentiometer of 4.7 and tune it to the desired value.

**Step 6:**Choose the required pass-band gain. If we use 2 again, then…

R2 = R1

*The calculations required to find the resistance of the LPF in a wide band-pass filter.*

The final design with the component values is shown here. The op-amp is an active component and it requires +ve and -ve biasing voltages. It’s possible to test the circuit by applying input through the signal generator and observing output on the DSO or oscilloscope and the bode plotter.

*A circuit diagram of the LM741 OPAMP IC-based wide band-pass filter.*

**Note: **I have simulated the above circuit in

**NI’s multisim 11 software**. The schematic design is also prepared using the same software. The software is available as a free one-month trial period from National Instrument’s (

**NI**) website. The below circuits are also prepared using the multisim 11 software and tested in it.

**The narrow band-pass filter **

**Step 1:**For simplicity, let’s assume C1 = C2 = C**Step 2:**Select the centre frequency, FC = 2 KHz, with a pass-band gain of Af = 2, and Q = 10**Step 3:**Next, assume the capacitor value C as 100nF**Step 4:**Calculate the value of the R1 from

*The calculations required to find the resistance for the narrow band-pass filter.*

Here’s the final design:

*The circuit diagram of the LM741 OPAMP IC-based, narrow band-pass filter.*

Filed Under: Tutorials

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