# Active op amp high pass filter circuit

### Op amp high pass filters are very easy to design and have straightforward equations for the Butterworth response

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Op amp high pass filters are easy to implement using a few components and they are required in a variety of different circuits to eliminate hum and other noise.

Although any form of filter response can be chosen, the Butterworth response simplifies the equations and it is possible to calculate the values required in seconds.

Despite the ease of design, op amp high pass filters are able to provide a high level of performance from just a relatively small number of components.

## What is a high pass filter

As the name implies, a high pass filter is a filter that passes the higher frequencies and rejects those at lower frequencies.

This can be used in many instances, for example when needing to reject low frequency noise, hum, etc. from signals. This may be useful in some audio applications to remove low frequency hum, or within RF to remove low frequency signals that are not required.

The shape of the curve is of importance. One of the most important features is the cut-off frequency. This is normally taken as the point where the response has fallen by 3dB.

Another important feature is the final slope of the roll off. This is generally governed by the number of 'poles' in the filter. Normally there is one pole for each capacitor inductor in a filter.

When plotted on a logarithmic scale the ultimate roll-off becomes a straight line, with the response falling at the ultimate roll off rate. This is 6dB per pole within the filter.

## Single pole op amp high pass filter

The simplest circuit high pass filter circuit using an operational amplifier can be achieved by placing a capacitor in series with one of the resistors in the amplifier circuit as shown. The capacitor reactance increases as the frequency falls, and as a result this forms a CR low pass filter providing a roll off of 6 dB per octave.

The cut off frequency or break point of the filter can be calculated very easily by working out the frequency at which the reactance of the capacitor equals the resistance of the resistor. This can be achieved using the formula:

Where:

**Xc** is the capacitive reactance in ohms

**Π** is equal to 3.142

**f** is the frequency in Hertz

**C** is the capacitance in Farads

## Two pole active high pass filter

Although it is possible to design a wide variety of filters with different levels of gain and different roll off patterns using operational amplifiers, the filter described on this page will give a good sure-fire solution. It offers unity gain and a Butterworth response (the flattest response in band, but not the fastest to achieve ultimate roll off out of band).

The calculations for the circuit values are very straightforward for the Butterworth response and unity gain scenario. Critical damping is required for the circuit and the ratio of the resistor vales determines this.

$C}_{1}=2{C}_{2$

$f=\frac{\sqrt{2}}{4\pi R{c}_{2}}$

When choosing the values, ensure that the resistor values fall in the region between 10 kΩ and 100 kΩ. This is advisable because the output impedance of the circuit rises with increasing frequency and values outside this region may affect the performance.

When using the op amp active high pass filter higher levels of attenuation and steeper roll-of can be achieved by cascading a number of circuits. The design can also be altered to accommodate different forms of filter, although the calculations do become more difficult.

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