# Active 2nd order HPF with LM358

Here an active 2nd order high pass filter(HPF) design using LM358 operational amplifier is illustrated. The HPF is build on a breadboard. It has a cutoff frequency of 1kHz. The active filter designed is tested and filtering action is illustrated. The filtering action shows that the amplitude of the input signal at the output increases as the frequency is increased from lower 0Hz to upper frequency such as 10kHz. The frequency response graph of the designed filter is also illustrated.

Recommended Prerequisites

### Video

The following video shows the designed filter and testing of the filter.

### 2nd order HPF on breadboard

The following picture shows the active high pass filter build on a breadboard.

### Circuit Diagram of 2nd order active HPF

The following shows the circuit drawing of the 2nd order active high pass filter that was build on the breadboard shown above.

The 2nd order high pass filter shown above consist of two RC filters. The first is made up of R1 and C1 and the second is made up of R2 and C2. The cutoff frequency of the HPF is given by the following equation.

$f_c=\frac{1}{2 \pi \sqrt{R_1 R_2 C_1 C_2}}$
For simplicity we can set the resistors R1 and R2 equal. Let R1 = R2 = R. Similarly we can set the capacitors C1 and C2 equal. That is let C1 = C2 = C. With this simplification, the above cutoff frequency formula can be written as,
$f_c=\frac{1}{2 \pi R C}$

We have set R1 = R2 =15kohm and C1 = C2 = 0.01uF which gives cutoff frequency of 1khz.

The formula to calculate the passband gain is,
$$A_F = 1 + \frac{R_F}{R_3}$$
Thus the passband gain  AF get is set by the resistors RF and R3
If we want Butterworth response then passband gain AF should be 1.586. Doing so we can calculate the value of RF by selecting R3.
$$R_F = R_3(A_F - 1)$$
By using R3 =10kohm and with AF = 1.586 we get RF =5.86 KOhm.

### Waveform and Frequency Response

The following shows the input signal and output signal waveform from the high pass filter at 630Hz.

In the above oscilloscope waveform graph, the yellow waveform is the output and the blue is the input signal waveform. At 630Hz, the amplitude of the filter output signal is lower than than the reference input signal amplitude.

The following shows the signal waveform at 2KHz.

The above graph show that at 2KHz as the frequency has increased, the output signal amplitude is higher than that of the original reference input signal amplitude. This is because in high pass filter, with increase in frequency above cutoff frequency of 1KHz the signal amplitude at the output will be lesser attenuated.

The following shows the frequency response graph of the 2nd order high pass filter designed with LM358 operational amplifier.