Active Butterworth Filter Design

prerequisite basis

   As we all know, filters are divided into active filters and passive filters, the difference between the two is whether additional power is required. Among them, passive filters are divided into RC filters and LC filters. The simplest RC low-pass filter is shown in the figure below.

   Due to the presence of load resistance, the cutoff frequency and passband magnification of the filter will change when there is no load or load, and the specific changes are as follows: No load:
A
$$A_{up}=1f_p=\frac{1}{2\pi RC}$$
$$A_u=\frac{1}{1+j\frac{f}{f_p}}$$
带载:
$$A_{up}=\frac{R_L}{R+R_L}$$
$$f_p=\frac{1}{2\pi (R//R_L)C}$$
   It can be seen that the RC filter has a certain loss, and the other LC filter will not be described in detail here.

Active Butterworth Filter Structure

2nd and 3rd order active filters

   The basic structure of the active Butterworth filter is an RC filter, but the op amp is used to eliminate the cutoff frequency change and loss caused by the load. The op amp can be connected into two types of amplifying circuits in the same direction and in the opposite direction. Here, only the connection in the same direction is considered. The basic structure is a second-order and third-order section composed of an op amp. As shown in the figure below (single gain), let all the resistors in the circuit be 1. The transfer function is calculated by the current loop theorem and voltage loop theorem in circuit analysis: T ( S ) 2 = 1 C 1 C 2 S 2 + 2 C 2 S + 1 T(S)_2=\frac{1}{C_1 C_2S^2+2C_2S+

1
$$T(S{)}_2=\frac{1}{C_1{C}_2{S}^2+2C_{2}S+1}$$
$$T(S{)}_3=\frac{1}{C_1{C}_2{C}_3{S}^3+2C_2(C_1+C_2)S^2+(C_2+3C_3)S+1}$$
These two transfer functions are directly obtained from the circuit. When designing the Butterworth filter, the normalized transfer functions of the second and third orders are:
$$T(S{)}_2=\frac{1}{S^2+\sqrt{2}S+1}$$
$$T(S{)}_3=\frac{1}{S^3+2S^2+2S+1}$$
By comparing the coefficients, for the second order:
$$\begin{array}{rl}& C_1{C}_2=1\\ & 2C_2=\sqrt{2}\end{array}$$
For the third order:
$$\begin{array}{rl}& C_1{C}_2{C}_3=1\\ & 2C_2(C_1+C_2)=2\\ & C_2+3C_3=2\end{array}$$
Solving the system of equations gives:
For the second order:
$$\begin{array}{rl}& C_1=\sqrt{2}=1.414\\ & 2C_2=\frac{1}{\sqrt{2}}=0.7071\end{array}$$
For third order:
$$C_1=3.546,C_2=1.392,C_3=0.2024$$
   It should be noted that this equation is solved by the normalized transfer function, so the calculated capacitance value is also a normalized value, the unit is F, and R in the circuit is 1 ohm, which is also a normalized value. When designing the actual parameters, it must be denormalized, and the denormalized method will be given in the actual design later.

High Order Active Filter

   Active filters above 4th order can be combined with the 2nd order and 3rd order active filters given above. According to the relationship between the system transfer functions, the final system transfer function should be the product of the combined filters, namely: H ( S ) = H 1 ( S ) H 2 ( S ) H 3 ( S ) … H n ( S ) H(S)=H_1(S)H_2(S)H_3(S)\dots H_n(S
$$H(S)=H_1(S)H_2\left(S\right)H_3(S)\dots H_n\left(S\right)$$
The cascaded configuration of the 4th order to 7th order filters is given below:

1. 4th order (=2+2)
   

2. 5th order (=3+2)
   

3. 6th order (=2+2+2)
   

4. 7th order (=3+2+2)
   

The normalized low-pass Butterworth filter denominator polynomial is directly given here:

This picture comes from the link: Butterworth filter design

In parentheses is the transfer function of the 2-stage section or 3-stage section active filter of each stage.

denormalization

   Solution 1: Let R=1ohm, and then divide all reactance components (L, C) by a frequency scaling factor FSF, so that a known filter response index can be calibrated to different frequency ranges, that is, denormalization. FSF=required frequency/current reference frequency, such as the current reference frequency is ${oh}_c=1rad/s$ , and the required reference frequency is$2\pi f_c$($f_c$is the frequency required in the actual topic.

   Solution 2: The actual value of the components calculated in solution 1 is very unrealistic, the capacitance value is too large, the inductance value is too small, and R=1om is even more unrealistic. This has to be resolved with the impedance scale Z. For any linear active or passive network, if all resistor and inductor values ​​are multiplied by Z, and all capacitors are divided by the same factor Z, the transfer function remains unchanged. The reason is the Z cancellation in the transfer function. This is not proven here. Using solution 2 for denormalization, the denormalized value is given by the following formula:
$$\begin{array}{rl}& R^{{}^{\prime }}=R\times Z\\ & L^{{}^{\prime }}=\frac{L\times Z}{FSF}\\ & C^{{}^{\prime }}=\frac{C}{FSF\times Z}\end{array}$$

design example

Requirements:
   active low-pass filter;
   attenuation of 3dB at 75Hz;
   minimum attenuation of 30dB at 150Hz;
   gain of 1;

Design steps:
   (1) Determine the filter order according to the steepness coefficient: because $A_S=150/75=2$ , that is, the filter is required to have a minimum attenuation of 30dB at 2rad/s. It can be seen from the magnitude-frequency response diagram of the Butterworth filter as follows:

   The 5th-order Butterworth filter meets the requirements.
   (2) Determine the circuit form, and select the appropriate circuit according to whether there is a gain requirement.
   (3) Obtain the normalized values ​​of R and C, and obtain R and C according to the comparison between the normalized low-pass Butterworth filter denominator polynomial given above and the 2nd-order section or 3rd-order section form of the Butterworth filter. The method has been given above.
   (4) Denormalization; first determine the impedance scaling coefficient Z, select according to the frequency band where the filter is located, and choose a larger Z for the low frequency band, otherwise vice versa. The cut-off frequency of this question is as low as 75Hz, Z should be larger,$Z=R=50Kohm，FSF=2p.m\times f_c=471.2$ , so the actual C value can be obtained:
   3rd order node:
$$\begin{array}{rl}& C_1^{{}^{\prime }}=\frac{C_1}{FSF\times Z}=\frac{1.753}{471.2\times 5\times 10^4}=0.074uF\_\\ & C_2^{{}^{\prime }}=0.057uF\_C_3^{{}^{\prime }}=0.018uF\end{array}$$
   2nd order node:
$$C_1^{{}^{\prime }}=0.137uF{C}_2^{{}^{\prime }}=0.013uF$$
   (5) The component value is substituted into the normalization circuit to obtain the final filter circuit as follows;

postscript

   Other types of active filters can be designed in the same way according to their transfer functions of different orders.