Applications of Operational Amplifier: Filters

Filters:

A filter is a frequency selective circuit that, passes a specified band of frequencies and blocks or attenuates signals of frequencies out side this band. Filter may be classified on a number of ways.

1. Analog or digital
2. Passive or active
3. Audio or radio frequency

Analog filters are designed to process only signals while digital filters process analog signals using digital technique. Depending on the type of elements used in their consideration, filters may be classified as passive or active.

Elements used in passive filters are resistors, capacitors and inductors. Active filters, on the other hand, employ transistors or OPAMPs, in addition to the resistor and capacitors. Depending upon the elements the frequency range is decided.

RC filters are used for audio or low frequency operation. LC filters are employed at RF or high frequencies.

The most commonly used filters are these:

1. Low pass filters
2. High pass filter
3. Band pass filter
4. Band reject filter.
5. All pass filter

Fig. 1, shows the frequency response characteristics of the five types of filter. The ideal response is shown by dashed line. While the solid lines indicates the practical filter response.

Fig. 1

A low pass filter has a constant gain from 0 Hz to a high cutoff frequency f_H. Therefore, the bandwidth is f_H. At f_H the gain is down by 3db. After that the gain decreases as frequency increases. The frequency range 0 to f_H Hz is called pass band and beyond f_H is called stop band.

Similarly, a high pass filter has a constant gain from very high frequency to a low cutoff frequency f_L. below f_L the gain decreases as frequency decreases. At f_L the gain is down by 3db. The frequency range f_L Hz to ∞ is called pass band and bleow f_L is called stop band.

First Order Low Pass Filter:

Fig. 2, shows a first order low pass Butter-worth filter that uses an RC network for filtering, opamp is used in non-inverting configuration, R₁ and R_f decides the gain of the filter.

According to voltage divider rule, the voltage at the non-inverting terminal is:

Fig. 2

Thus the low pass filter has a nearly constant gain A_f from 0 Hz to high cut off frequency f_H. At f_H the gain is 0.707 A_f and after f_H it decreases at a constant rate with an increases in frequency. f_H is called cutoff frequency because the gain of filter at this frequency is reduced by 3dB from 0Hz.

Filter Design:

A low pass filter can be designed using the following steps:

1. Choose a value of high cutoff frequency f_H.
2. Select a value of C less than or equal to 1 µF.
3. Calculate the value of R using .
4. Finally, select values of R1 and RF to set the desired gain using .

Example - 1

Design a low pass filter at a cutoff frequency of 1 kH z with a pass band gain of 2.

Solution:

Given f_H = 1 kHz. Let C = 0.01 µF.

Therefore, R can be obtained as

A 20 kΩ potentiometer can be used to set the resistance R.

Since the pass band gain is 2, R₁ and R_F must be equal. Let R₁ = R₂ = 10 kΩ.

Low pass filter with adjustable corner frequency

One advantage of active filter is that it is often quite simple to vary parameter values. As an example, a first-order low-pass filter with adjustable corner frequency is shown in fig. 3.

Fig. 3

The voltage at the opamp inputs are given by

Setting v + = v -, we obtain the voltage, v₁, as follows:

where 

The second opamp acts as an inverting integrator, and

Note that we use upper case letters for the voltages since these are functions of s. K is the fraction of V₁ sent to the integrator. That is, it is the potentiometer ratio, which is a number between 0 and 1.

The transfer function is given by

The dc gain is found by setting s = 0 (i.e., jω =0)

The corner frequency is at KA₂ / RC. Thus, the frequency is adjustable and is proportional to K. Without use of the opamp, we would normally have a corner frequency which is inversely propostional to the resistor value. With a frequency proportional to K, we can use a linear taper potentiometer. The frequency is then linearly proportional to the setting of the potentiometer.

Example - 2

Design a first order adjustable low-pass filter with a dc gain of 10 and a corner frequency adjustable from near 0 t0 1 KHz.

Solution:

There are six unknowns in this problems (RA, RF, R1, R2, R and C) and only three equations (gain, frequency and bias balance). This leaves three parameters open to choice. Suppose we choose the following values:

C = 0.1 µF
R = 10 KΩ
R₁ = 10KΩ

The ratio of R₂ to R₁ is the dc gain, so with a given value of R₁ = 10KΩ, R₂ must be 100 kΩ. We solve for A₁ and A₂ in the order to find the ratio, R_F / R_A.

The maximum corner frequency occurs at K = 1, so this frequency is set to 2π x 1000. Since R and C are known, we find A₂ = 6.28. Since A₂ and A₁ are related by the dc gain, we determine A₁ / A₂ = 10 and A₁ = 62.8. Now, substituting the expression for A₂, we find

and since 

we find R_F / R_A = 68. R_A is chosen to achieve bias balance. The impedance attached to the non-inverting input is 10 KΩ || 100 KΩ = 10 KΩ.

Fig. 4

If we assume that R_F is large compared with R_A ( we can check this assumption after solving for these resistors), the parallel combination will be close to the value of R_A. We therefore can choose R_A = 10 KΩ. With this choice of R_A, R_F is found to be 680KΩ and bias balance is achieved. The complete filter is shwn in fig. 4.

Second Order Low-Pass Butterworth filter:

A stop-band response having a 40-dB/decade at the cut-off frequency is obtained with the second-order low-pass filter. A first order low-pass filter can be converted into a second-order low-pass filter by using an additional RC network as shown in fig. 1.

Fig. 1	Fig. 2

The gain of the second order filter is set by R₁ and R_F, while the high cut-ff frequency f_H is determined by R₂, C₂, R₃ and C₃ as follows:

Furthermore, for a second-order low pass Butterworth response, the voltage gain magnitude is given by

where, 

Except for having the different cut off frequency, the frequency response of the second order low pass filter is identical to that of the first order type as shown in fig. 2.

Filter Design:

The design steps of the second order filter are identical to those of the first order filter as given bellow:

1. Choose a value of high cutoff frequency f_H.
2. To simplify the design calculations, set R₂ = R₃ = R and C₂ = C₃ = C. Then choose a value of C less than 1 µF.
3. Calculate the value of R using .
4. Finally, because of the equal resistor (R₂ = R₃) and capacitor (C₂ = C₃) values, the pass band voltage gain A_F has to be equal to 1.586. This gain is necessary to guarantee Butterworth response. Therefore, R_F = 0.586 R₁. Hence choose a value of R₁= 100 kΩ and calculate the value of R_F.

First Order High Pass Butterworth filter:

Fig. 3, shows the circuit of first order high pass filter.This is formed by interchanging R and C in low pass filter.

The lower cut off frequency is f_L. This is the frequency at which the magnitude of the gain is 0.707 times its pass band value. All frequencies higher than f_L are pass band frequencies with the highest frequency determined by the closed loop bandwidth of the OPAMP.

The magnitude of the gain of the filter is

Fig. 3

If the two filters (high and low) band pass are connected in series it becomes wide band filter whose gain frequency response is shown in fig. 4.

Fig. 4