Active 2nd order LPF on breadboard

Here second order active low pass filter design and test is illustrated. The active filter is build on a breadboard and the input and output signal amplitude waveform to and from the filter is tested. Also the frequency spectrum and frequency response graph is shown for the designed filter. This test can be performed easily at home requiring just PC/laptop and PC soundcard software.

Recommended Prerequisites

• Active 1st Order Low Pass Filter Design on Breadboard
• First order Active HPF Design and Test
• Active 2nd order HPF with LM358
  

Contents

1. 2nd order Active LPF on breadboard
2. Circuit diagram
3. Calculation
4. Results

2nd order Active LPF on breadboard

For the construction of the active low pass filter LM358 operational amplifier is used. Other operational amplifier like TL072 can be used as well. The following shows the active low pass filter on a breadboard.

Circuit diagram

The following shows the circuit schematic diagram of the above constructed active high pass filter.

The active 2nd order low pass filter consist of two RC low pass filter at the input. The first LPF is formed by the resistor R1 and capacitor C1 and the second LPF is formed by the resistor R2 and capacitor C2. The resistor Rf and R3 provides the passband gain for the filter. For Butterworth response with flat passband, the passband gain must be 1.586. 

Calculation

The cutoff frequency for the filter is given by the mathematical expression below.

$$f_c=\frac{1}{2 \pi \sqrt{R_1 R_2 C_1 C_2}}$$

 For simplicity we can select the resistors R₁ = R₂ = R and select the capacitor values C₁ = C₂ = C. Then the above formula for the cutoff frequency becomes, 

$$f_c=\frac{1}{2 \pi R C}$$

 

In the filter circuit above, we have used R₁ = R₂ =15kohm and C₁ = C₂ = 0.01uF. This gives cutoff frequency of 1khz.

 

The passband gain  A_F get is set by the resistors R_F and R₃ which is given by the following equation.

$A_F = 1 + \frac{R_F}{R_3}$

For Butterworth response A_F = 1.586 so we can calculate the value of R_F by setting R₃.

 $R_F = R_3(A_F - 1)$  

By using R₃ =10kohm and with A_F = 1.586 we get R_F =5.86 KOhm. 

The above values can be directly calculated using the online 2nd Order Active LPF Calculator.

Results

The following consecutive waveform diagram obtained by passing signal into the filter illustrates how the signal waveform of the output decreases as the frequency is increased. The yellow waveform is the reference input signal to the filter and the blue waveform is the output signal from the filter. 

 

The first waveform graph shows signal at 315Hz which is below the cutoff frequency 1kHz. At this frequency of 315Hz we can see that the output signal waveform is larger in magnitude than the input signal magnitude.

 

The next waveform graph shows the input and output signal waveform and their amplitude at 630Hz. As can be observed, as magnitude of the output signal has decreased from previous frequency of 315Hz. The magnitude of both the input and output is almost equal.

At 800Hz the output signal magnitude is further decreased as shown below.

As we increase the frequency the magnitude decreases further due to low filter action. At 1Khz the waveform is as shown.

At 3.15kHz,

The frequency response graph of the filter is shown below.