MFB 2ND ORDER BAND PASS FILTER

Fig. 1 shows a schematic of a multiple feedback type 2nd order band pass filter. Click on the figures for a larger view. For a complete description, transfer function and other information you can refer to Texas Instrument Literature Number SLOA088, "Active Filter Design Techniques". In this post we will show the equations and information that is usually needed for a practical design.

To start with the design engineer should know or specify the center frequency, the bandwidth, the Q of the filter, and the gain at the center frequency. This circuit is very good for this application because of its low parts count and the ability to separately adjust the center frequency, the gain, and the Q of the filter. Note that Q is always the ratio of center frequency to bandwidth for a bandpass filter.

Q = Fm / (F2 - F1)

where Fm is the center frequency, F2 is the -3db high side cut-off and F1 is the -3db cut off on the low frequency side of the response curve. See the simulation response curve above of the filter we will design here.

Although the center frequency is specified, it can be adjusted according to its relation to the component values:


Fm = (1 / (2 * Pi * C)) * ( ( R1 + R3) / (R1 * R2 * R3 ) )^0.5

Note that the actual center frequency is inversely proportional to the value of the capacitors C.

The gain of the filter at the center frequency is

Ko = R2 / (2 * R1)

The Q is related to the center frequency as

Q = Pi * Fm * R2 * C

and the bandwidth is

BW = 1 / ( Pi * R2 * C )

Both capacitors in the circuit are usually set to the same value.

To design the filter we pick both capacitors to be equal to a practical value. Then since we have already picked the center frequency Fm, the Q, and the gain Ko at Fm, we can now calculate the three resistors values R1, R2, and R3 shown in Fig. 1.

R2 = Q / ( Pi * Fm * C )

R1 = R2 / ( 2 * Ko )

R3 = Ko * R1 / ( 2 * Q^2 + Ko )

Note that Ko is the absolute value of the gain even though the op-amp is an inverting configuration.

A design example of a fairly high Q MFB filter is as follows:

Fm = 10KHZ
Ko = 1.0
Q = 10
BW = 1000HZ

C = 0.01ufd ( set for both capacitors as a designer's choice. )

Using the above equations to calculate the resistor values we have

R2 = 31.83K
R1 = 15.92K
R3 = 79.18 ohms

I am using the exact computed resistor and capacitor values here in order to test the accuracy of the design by SPICE simulation. Fig. 1 is the actual simulation schematic.

The simulation results are shown for the AC analysis run in the plot above. The simulation provided the following results:

Upper -3db cut off frequency 10.549KHZ

Lower -3db cut off frequency 9.554KHZ

From above the band width is 995HZ. ( The design band width is 1000HZ.)

The Q is also then found to be 10.05. The design value was 10.0.

So the simulation results are within 1% of the design targets ( as long as the exact component values are used.)

With the high Q of the design, the filter is only conditionally stable in the presence of transient inputs, but for constant sine wave inputs the circuit would function according to its specifications. If a transient step is applied with a 0.1 microsecond rise-time, the circuit will ring about 1000 uS as observed at its output. If transient stability is required, a lower Q would have to be used with of course less frequency selectivity or a wider band width.

4TH ORDER BESSEL HIGH PASS FILTER

A 4th order Bessel high pass filter can be constructed as shown in Fig. 1 (click on the image to see a larger view.) The design example is for a 1KHZ cut off high pass filter. The transient and AC analysis SPICE plots are shown also. Note the smooth variation in phase in the AC analysis and the stability shown by lack of ringing after a pulse edge in the transient analysis.

For convenience we will let the value of the complex operator s be transformed to a substitute complex operator as

u = 1 / s

Then we can write the transfer function for the high pass filter as

A(u) = K / (( 1 + a1 * u + b1 * u^2) * ( 1 + a2 * u + b2 * u^2 )...)

A(u) is valid as long as the filter is a time-invariant linear circuit.

The coefficients for various filters are already known and we will use the Bessel coefficients for a 4th order filter. We choose the Bessel coefficients because it will give us a linear phase response and good pulse response. The Bessel coefficients for 4th order filters that fit the above transfer function are

a1 = 1.3397
b1 = 0.4889
a2 = 0.7743
b2 = 0.3890

To design the filter we will first pick a standard capacitor value that we will use for both 2nd order stages and then calculate the corresponding resistor values using the four Bessel coefficients.

R1 = 1 / ( Pi * Fo * C* ai ) and R2 = ai / (4 * Pi * Fo * C * bi)

where ai and bi are a1, a2 and b1, b2 coefficients depending on which stage the resistors are calculated for, that is a1 and b1 for the first stage and a2 and b2 for the second stage.

Doing the calculations we find

R1 = 2.376K
R2 = 2.181K
R3 = 4.111K
R4 = 1.584K

with C = 0.1UFD and Fo = 1KHZ

Note that R1 and R2 are for the first stage, and R3 and R4 represent the R1 and R2 for the second stage. See Fig. 1.

The Q or quality factor can also be calculated for each stage from


Q1 = ( b1 )^0.5 / a1

Q2 = ( b2 )^0.5 / a2

SALLEN-KEY BESSEL FILTER 4TH ORDER

The transfer function for filters constructed from cascading a 2nd order filter can be calculated with the product of the 2nd oder polynomials from the equation

A(s) = Ao /(( 1 + a1 * s + b1 * s^2)( 1 + a2 * s + b2 * s^2) ... )

We are going to cascade two 2nd order filters that are not identical to obtain a 4th order filter as shown in Fig. 1. The result produces the Bessel filter characteristics overall as long as we use the Bessel coefficients in the calculations of R and C values. The Ac response curve below the schematic shows the results of simulating the circuit on SPICE with the ouput levels and the phase shifts vs. frequency of the first stage and the second stages combined. The circuit is designed to be a Bessel type filter so that the phase shifts are linear and pulse response is good. The circuit values shown produce a 1KHZ low pass filter. We calculate the values below.

First note that since we are not cascading two identical filters, we must use a1, a2, b1, and b2 coefficents. a1 and b1 are used for the first stage, and a2 and b2 are used for the second stage to conform to the theory.

We choose the coefficents in accordance with known coefficients for the major filter types as tabulated here:

4th order Bessel coefficients:

a1 1.3397
b1 0.4889
a2 0.7743
b2 0.3890

Q is calculated from:

Q1 = ( b1 )^0.5 / a1 for the first stage = 0.52

q2 = ( b2 )^0.5 / a2 for the second stage = 0.805

using the Bessel coefficients.

In Fig. 1 we will rename the first resistor on the left as R1 and the next resistor R2. The lower capacitor we will call C1 and the capacitor in the feedback loop as C2 for both stage calulations for reasons of simplicity.

Next we need to assign a value for C1 so that we can calculate a value for C2 and the resistor values R1, R2 The value of C2 must be set so that the R values are positive real numbers. First,

C1 = 0.1ufd (defined as a starting point). Then

C2 > = 4* C1 * b1 / a1^2 = 0.1333ufd

We will set C2 to 0.15ufd for the first stage and we will need a hihger value of 0.33ufd for C2 in the second stage as an available standard value. Now we can calculate the resistor values from the solutions to the quadratic equation for the first stage

R1, R2 =( a1 * C2 -/+ (a1^2 * C2^2 - 4 * b1 * C1 * C2)^0.5) / (4 * Pi * Fo * C1 *C2)

The solutions are

R1, R2 = 508, 1.624K ohms (find the closest standard values.)

In the theory it is correct to use R1 as the value resulting from the calculation with the negative of the square root of the above equation, and R2 as the result of the calculations from the positive square root. It is a good idea to always verify your design with a simulation and testing a prototype circuit.

Similar calculations are done for the second stage to get resistor values of R1 = 331 ohms and 901 ohms.

The 2nd order stages have now been completely designed and in Fig. 1 we show two stages cascaded to achieve a 4th order filter.

The AC response at the output of each stage is shown if the figure below the schematic above. The advantage of the linear phase shift is that the group delay is constant and the pulse response is excellent for the Bessel filter. Simulation with a pulse input shows that the pulse response at the first stage output is faster than that of the 2nd output stage. So the trade off of the faster cut-off of the 4th order vs. the 2nd order filter is a slower pulse response of the higher order filter.

Note that you cannot apply the same identical stage for both stages as the results will not be accurate and it does not conform to the theory of the correct transfer function.

60HZ TWIN-T NOTCH FILTER

A Twin-T RC filter can be buffered with one or two op-amps to provided a usable notch filter for audio circuits or other applications.

Fig. 1 shows the AC response of the circuit of Fig. 2, a Twin-T filter with LTC6244 rail-to-rail 5MHZ op-amps. We used the following equations to calculate the values in the circuit elements R1 - R3 and C1 - C3. R4 and R5 could be set as shown or a 50K potentiometer could be used to adjust the Q of the circuit. We set the values shown to give us a high Q notch with approximately 60db of attenuation at 60HZ.

The following calculations were used:

Fo = 1 / ( 2 * Pi * R1 * C1 )

For this design we will set Fo to 60HZ and calculate the basic RC time constants of the filter using the normal value relationships for this type of filter

R1 = R2 = 2 * R3

C1 = C2 = C3 / 2

R1*C1 = 1 / ( 2 * Pi * Fo ) = 0.0026526

Let C1 = 0.1 * 10^-6

Then,

R1 = 0.0026526 / ( 0.1 * 10^-6 ) = 26526

R2 = 26526 C2 = 0.1 * 10^-6

C3 = 0.2 * 10^-6 R3 = 13263

In regard to setting values for R4 and R5, we found that better Q and response was obtained by setting R4 less than R5. Fig. 1 shows the results of simulation of the circuit of Fig. 2.

Note that the source has to be able to drive the input impedance of the filter which is approximately 160 ohms at 20KHZ. If a higher input impedance is needed you must reduce the value of C1 and C2 which mainly control the input impedance of the filter. Then you will have to calculate new values for R1 - R3.