BESSEL 4TH ORDER BAND PASS FILTER

In this post we will use two multiple feedback (MFB) 2nd order band pass filters (see the previous post) to build a 4th order band pass filter. See Fig. 1 for the schemtic and the figure below for the SPICE AC analysis results (click on the figures for a larger view.) This method uses two filters with staggered tunning so that a flat band pass can be obtained. We will use Bessel parameters so that we obtain linear phase shift performance. Refer to Texas Instruments Active Filter Design Techniques Literature Number SLOA088 for a good description of the complete theory for this approach.

Lets design the 4th order band pass Bessel filter with the following specifications:

Center frequency Fm = 1000
Bandwidth B = 100
Q Q = Fm / B = 10
Center gain Km = 1 absolute value

The Bessel coefficients for a Q of 10 are

a1 = 1.3617
b1 = 0.6180

A quantity called alpha is also associated with the Bessel coefficients for a Q of 10. We will call alpha the quantity "a" in the following.

a = 1.0324

The tuning frequencies of the 2nd order sections is

Fm1 = Fm / a = 968.6

Fm2 = Fm * a = 1032

The Q or quality factor for both 2nd order sections is the same:

Q12 = Q * ( 1 + a^2 ) * b1 / ( a * a1) = 9.08

The gain of both sections is the same and the absolute value is

K12 = ( Q12 / Q ) * ( Km / b1 )^0.5 = 1.155

We will define the same capacitor value for all the capacitors in both sections as

C = 0.01 * 10^-6 Farad

So now we can compute the resistor values shown in Fig. 1.

For filter section no. 1 (see Fig. 1)

R2 = Q12 / ( Pi * Fm1 * C ) = 298.4K

R1 = R2 / (2 * K12 ) = 129.2K

R3 = K12 * R1 / ( 2 * Q12^2 - K12 ) = 911 ohms

And for filter section no. 2 we have


R6 = Q12 / ( Pi * Fm2 * C) = 280K

R5 = R6 / ( 2 * K12 ) = 121.2K

R7 = K12 * R12 / ( 2 * Q12^2 -K12 ) = 855 ohms

We have now specified the filter, assigned the value for the capacitors, and computed the resistor values as shown in Fig. 1 (click on the figure for a larger view.) The simulation results are shown below Fig. 1. The simulation results show that the filter does indeed have a center frequency of 1KHZ and a bandwidth of 100HZ at -3db points according to our specifications. Note that a high Q analog filter like the one we designed will not in general be stable for step or pulse transient signals but it will work fine for steady state or slowly varying sine wave input signal.

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.

2ND ORDER NARROW BAND FILTER WITH GAIN

A high Q narrow band filter with gain is shown in Fig. 1. This filter can be used to filter steady state signals but it is not suitable for audio use or with signals that have frequent transient signals. Its high Q and gain will cause it to ring or be unstable with such transients. The high Q, high gain circuit is with R5 not populated. The circuit can be used with audio signals if Q and gain is reduced with the " Q spoiler" resistor R5 set to approximately 12.5K maximum. With R5 in place the Q is approximately 1.3 and the gain is approximately 22.5db at the pass frequency, in the case shown, of 2KHZ. Without R5 populated, the Q is set to 10 and the gain is approximately 40db or 100 at the pass frequency. A 0.05 volt peak 2KHZ input signal will produce a 5 volt peak output signal.

The design shown has been simulated for AC response and transient response using the LTC6244 rail-to-rail 5MHZ op-amp. Fig.1 is the actual simulation schematic.

To design the circuit, we must first decide on the pass band center frequency Fo.
Next set the Q value. Then set a practical value for C1 and C2, e.g., 0.01UFD. Next we will calculate a value called Beta which is the ratio of the center frequency in radians to the Q value. Set Fo to 2KHZ and set Q to 10.

Beta = 2 * Pi * Fo / Q

Beta = 1256.6 with an Fo of 2KHZ and a Q of 10.

In the case of the circuit shown we can now calculate the value of R4 as

R4 = 2 / ( Beta * C ) = 159.16K with C=C1=C2= 0.01UFD

The closest standard value would then be selected from the standard resistance decade table.

Also we can compute the bandwidth of the filter measured at the -3db points as

BW = Beta / ( 2 * Pi ) = Fo / Q = 200. The bandwidth for this circuit is 10% of the center frequency with Q set at 10.

It is also known for the circuit of Fig. 1 that the circuit component values are related to the center frequency as

Fo = 1 / ( 2 * Pi * ( R2||R3 * R4 )^0.5 * C )

where R2||R3 is the parallel value of R2 and R3. We can re-write the equation as


R2||R3 = 1 / ( Wo^2 * C^2 * R4 ) = 397.9

where

Wo = 2 * Pi * Fo

Assuming that R2 = R3, we can then say that R2 and R3 are 2 * (R2||R3) = 795.8 each as shown in Fig. 1. The closest standard value would be selected from the standard resistance decade table.

We have now completely designed the circuit of Fig. 1.

The performance of the circuit was verified using a SPICE circuit simulator. The center frequency was at 2KHZ. Bandwidth was 200 HZ. Gain was 40db or 100. With a 50 millivolt peak 2KHZ input signal, the output signal was 5 volts peak. AC analysis showed that symmetry was excellent.

2ND ORDER BAND PASS FILTER

Fig. 1 shows a second order band pass filter that is designed to have a center frequncy of 2KHZ and is stable with transient signals or noise inputs. The schematic is the actual simulation schemtaice used including the biasing circuit as the basic circuit. Note that a practical value of 0.01UFD is set for the capacitors to start with but the R values are calculated as shown below and the calculated values are shown to 5 or 6 digits for use in the simulation but standard resistor decade values that are close will provide acceptable results.

We want this filter to have a maximally flat response in the pass band so the damping factor d is set to the square root of 2, as

d = ( 2 )^2 = 1.1412...

If we wanted a maximally flat time delay instead, such as might be required for an audio cross-over network, we would set d to the square root of 3 instead.

Once we have decided on d we can then calculate a gain that will be stable for the circuit. For an ordinary wide band pass filter, like the circuit in Fig. 1, we don't need a lot of gain and we need to be careful not to use too much gain as the filter may become unstable and ring on any kind of input transient, making it useless. We will use a standard formula to calculate the gain K that will work. We will treat the gain K as a positive value although the real gain is actually negative due to the inverting amplifier circuit we are using.

K = 3 - d = 1.5857

We can round off the gain to 1.59.

Now there is a criteria for setting the minimum value of the quality factor of the filter Q. Again, we don't want too high a Q for this simple wide band filter as the circuit may not then be time stable. The criteria is:

Q > ( K / 2 )^0.5 = 0.89

So what ever Q we choose it needs to be greater than 0.89. So lets try a Q = 1.

Also we set the center of the pass band to 2KHZ for demonstration.

Fo = 2000

Now we can calculate the resistor values for the filter:

R3 = Q / ( 2 * Pi *K * C * Fo ) = 1 / ( 2 * 3.14159 * 1.5857 * 0.01*10^-6 * Fo )

R3 = 5004.9

R4 = Q / (( 2 * Q^2 - K ) * 2 * Pi * C * Fo ) = 19409

From the equation for R4 we can see that 2Q^2 must be greater than the absolute value of K, and 2Q^2 cannot equal K, or R4 must be removed from the filter and then we have a different circuit and performance. I have found that including R4 makes a better filter.

R2 = 2 * Q / ( 2 * Pi * C * Fo ) = 15915.5

R1 is simply a dummy load for the op-amp and has no theoretical effect on calculations.

Simulating the circuit as shown in Fig. 1 with the Linear Technology LTC6244 rail-to-rail 5MHZ op-amp we have the following results:

Quite symmetrical band pass curve on a log of frequency vs voltage db scale with

Fo = 2KHZ

F low at -3db = 1.24KHZ approx.

F high at -3db = 3.3KHZ approx.

-20db at 20KHZ approx.
-20db at 200HZ approx.

Phase shift at Fo is -90 degrees.

With an input signal of 0.62893, the output signal is 1 volt as predicted from the gain K calculation.

Note that

Fo = ( F low * F high )^0.5 = ( 1240 * 3300)^0.5 = 2023HZ , very close to Fo = 2KHZ.

You will note also that the band pass shape for the rates of attenuation is proportional to a percentage of the frequency, not an absolute value of frequency delta from Fo.

2ND ORDER LOW OR HIGH PASS FILTER

Fig. 1 shows a circuit that can be configured as either a high pass or low pass filter depending on which components are populated. The values shown are for a low pass filter with a cut-off frequency of approximately 200HZ.

In the case shown R2 and R3 are set to 8.1K, and C1 and C4 are set to 0.1UFD. The component parts parallel to these are set to parasitic values (not populated) but would be populated for the high pass version and R2, R3, C1, and C4 would not be populated for the high pass configuration.

We can calculate the cut-off frequency value for either the high pass or low pass filter from the equation

Fo = 1 /( 2 * Pi * ( R2 * R3 * C1 * C4 ) ^ 0.5 ) = 196.48 HZ

R2 and R3 are usually set equal to each other and C1 an C4 are usually set equal to each other. We can then simplify the formula to:

Fo = 1 / ( 2 * Pi * R * C)

which we should recognize as the equation for a simple RC filter, but in this case we have a 2nd filter instead of a 1st order filter with the circuit of Fig. 1.

So if you know the frequency cut-off you want you can set the capacitor values and calculate the R values. For example, say you want a cut-off of 500 HZ and you will set the C values to 0.1UFD as before. The R values are then

R = 1 / ( 2 * Pi * Fo * C ) = 3.183 K

Or if we know the R values and the cut-off frequency, we can calculate the capacitor values;

C = 1 / ( 2 * Pi * Fo * R ) = 1.000 * 10^-7

or 0.1UFD.

The circuit in Fig. 1 was simulated with the components shown using +5V and -5V power supplies on a SPICE simulator and verified to have the calculated cut-off frequency.

What is a class T amplifier?

A class T amplifier is a type of class D amplifier except that it has some characteristics based on advanced technology in switching modulation and feedback techniques, such as very high switching frequencies, sigma-delta type pulse width modulation, and other usually proprietary features that are not published as generally available information. What might be some of the advantages of this type of design? We can expect that the higher switching frequency would help to reduce distortion and lower noise. Sigma-delta modulation spreads out the switching frequency over a range of frequencies which helps to reduce EMI testing problems.

What would we expect to be the disadvantages of the class T system? The variable switching frequency has some disadvantages over a fixed frequency system, especially in testing, and filtering out the switching frequency in the final output signal. The circuitry is usually more complicated and harder to design than the classical fixed frequency ramp generated pulse width modulation.

Should you buy a class T system? This is a question of personal taste, but I can tell you that the class T system has been very well accepted by large companies that are including the system in TV’s and home theatre systems. If the system sounds good and you do not want to spend a lot of money on high end equipment, class T may be good for you. If you are looking for a high end system you may want to shop carefully and do your research first. You may still find that a class T system is great and just right for you. Good hunting and good listening.

what are differences of full bridge vs. half-bridge class d

What are the advantages of full-bridge vs. half-bridge Class D amplifiers?

The following is a quick summary:

Half-Bridge
PWM Two-level switching (high and low logic levels)
Current Limited to the rating of each mosfet, assuming 1 mosfet high side, and 1 low.
No. of mosfets 2 minimum
Gate drive One dual gate driver possible
DC offset Critical -- must be virtually zero
Supply voltages Need positive and negative voltages
Distorsion Less than 1% possible
Bandwidth 20-20KHZ possible depending on design
Power regulation A pumping effect makes regulation more difficult

Full bridge
PWM Three-level possible
Current capacity Twice the half-bridge rating for the same mosfet types
No. of mosfets 4 minimum
Gate drive More complex than half-bridge
DC offset Self-canceling
Supply voltages One voltage can be used. Also can work from positive and negative supplies
Distorsion Much less than 1% possible
Bandwidth 20-20KHZ practical
Power regulation Easier than half-bridge

One comment on the gate drive system for the full-bridge circuit is that the logic is straight-forward but it may be necessary to add gate drive buffers to increase the strength of the gate drives.

WHAT IS A FIRST ORDER ACTIVE FILTER?

What is an active low pass filter? Fig. 1 shows a simple first order low pass filter. Since it is a linear time invariant amplifier, we can write the transfer function in the frequency domain as

G = Vo / Vi = K * Wo / (s + Wo)

where K is the basic DC gain (see the link below for more information on op-amps), and we define s and Wo as

s = j * 2 * Pi * f, (j is the square-root of minus one, Pi = 3.14159…, and f is the frequency of computation.)

Wo = 2 * Pi * fo . fo = 1 / (2*Pi*R2*C)

K = - R2 /R1

where fo is the “cut-off frequency” of the filter. Since the filter is a first order filter, a Bode plot of gain vs. frequency would show that at the cut-off frequency, fo, the gain will decline at the rate of -20db per decade of increase of frequency.

A more useful and practical way to compute gain of this filter is to compute its absolute (non-complex) value as

Gabs = K * Wo / (W^2 + Wo^2)^1/2

In words, the above equation would read absolute gain is the product of K, and Wo, divided by the square root of the sum of the squares of W and Wo.

We can intuitively see that if W >> Wo the impedance of C will be a lot less than R2 and the gain will be very low, essentially blocking high frequencies from passing. In the high frequency range the circuit acts as an integrator. On the other hand, if W is much less than Wo, the impedance of C becomes very large compared to R2 so the low frequencies are passed all the way down to and including DC signals.

WHAT IS A FIRST ORDER HIGH PASS FILTER?

What are the basic equations for the first order high pass filter shown in Fig. 1?

In the frequency domain for a time invariant linear network we can write the transfer function of gain from input to output as

G = Go * s / (s + Wo)

where s = j *2 *Pi *f, and j is the square root of -1, Pi = 3.14159… , and f is the frequency variable. Go is the gain in the pass band and Wo = 2*Pi*fo. fo is the frequency at filter cut-off. G is the complex gain and the expression shows us that the transfer function has what is called a “zero” in its response at f=0, and a “pole” in its response at Wo (the cut-off frequency in radians.) A more practical equation has been derived from taking the absolute value of the gain and expressing it as

Gabs = Go*W / (W^2 + Wo^2)^1/2

where W= 2*Pi*f. From the above equation we see that as W becomes much greater than Wo, the response approaches Go = -R2 / R1. We can also see that as f approaches zero, the gain also approaches zero.

Lastly, we can easily compute the cut-off frequency fo as

fo = 1 / (2*Pi*R1*C)

The first order high pass filter is useful for removing very low frequency noise from a signal. But due to its first order characteristic, its roll-off below the cut-off frequency fo is only -20db per decade of frequency. So for example if the cut-off frequency is 10KHZ, the signal would be down to -80db at 1 HZ, assuming we had zero db in the pass band, or a gain of 1.0.

WHAT IS A SALLEN-KEY FILTER?

The Sallen-Key filter can be configured in a number of different ways. We will discuss a low pass filter shown in Fig. 1. (at bottom of post.) This is a second order filter and provides 40db attenuation per decade of frequency above the cut-off frequency. This is a useful circuit for applications where we wish to obtain a fairly rapid roll-off with a one-stage filter.

I won’t go into the transfer function equation here. If you want to study it, you can find the equation on Wikipedia. But to explain my design procedure (which is not found on Wikipedia), we do need to know the equations for the cut-off frequency, fo, and the quality factor, Q, of the circuit which are as follows:

fo = 1 / ( 2 * Pi * (R1*R2*C1*C2)^1/2)

The inner parenthetical quantity is the square-root of the product of the R’s and C’s multiplied together.

Q = ((R1*R2*C1*C2)^1/2) / (C2*(R1+R2))

where again the numerator is the square-root of the product of R’s and C’s. These are the equations we need to design the circuit.

First we combine the two equations to get the product of fo and Q:

Q*fo = 1 / (2*Pi*(R1+R2)*C2)

A Butterworth maximally flat filter would have a Q equal to the reciprocal of the square-root of two. For ease of calculation here, I will set Q = 1 to obtain a simple equation for fo:

fo = 1 / (2*Pi*(R1+R2)*C2)

where in all of the above Pi = 3.14159…

Assume that we want a filter with a 10KHZ cut-off and we are satisfied with a Q = 1 which will provide a reasonably flat response in its pass band. Then we know that the quantity important to the cut-off frequency is

(R1+R2)*C2 = 1 / (2*Pi*fo) = 1.5915 X 10^-5

The units of the above are reciprocal radian frequency. Let C2 be set to a reasonable value of say 1 nano-farad. (Don’t make C2 too large a value because C1 will need to be even larger.) Now we have that

R1+R2 = 15910

So as a quick approximation let us set R1 = R2 = 8K ohms each so the total is 16K ohms. Then we can calculate C1 from the square of our first equation above as

C1 = 1 / (4*Pi^2*R1*R2*fo^2*C2)

C1 = 3.957 X 10^-9 ~ 4 nano-farad

Now let’s check the design by calculating fo and Q:

fo = 1 / (2*Pi*(R1*R2*C1*C2)^1/2) = 9947 Hz

That is close enough to our original goal of a 10KHZ cut-off frequency.

Now for Q:

Q = (R1*R2*C1*C2)^2 / (C2*(R1+R2)) = (8000^2*4X10^-9*1X^-9)^1/2 / (1X^-9*16000) = 1.0

The calculated value of Q agrees with our original decision to set Q = 1.

I have now completed the design of the filter so that we have the following component values:

R1 = R2 =8K

C1 = 1X10^-9

C2= 4X10^-9

Q = 1.0

fo = 10KHZ

In fact I simulated the design using a 1MHZ JFET input op-amp and had the following results:

fo ~ 12.7KHZ (-3db down frequency)

flatness 1.3 db DC to 7KHZ

40 db per decade attenuation above 10KHZ

-54 db at 300KHZ then rising slowly to -46db at 1MHZ.

The response was very smooth with no ring or ripple in the pass-band, with a steady 40db per decade roll-off after cut-off.

The Butterworth Filter

The Butterworth filter is a class of filters that meet certain mathematical characteristics and performance characteristics that can be met by certain filter types both passive and active. The Butterworth filter theory was developed by Stephen Butterworth in 1930. The theory is still applied today.
A Butterworth filter is maximally flat in the pass band-- that is no ripples, and has a monotonic shape in its pass band and in its stop band. The Butterworth filter is the only type that is monotonic in the stop band as opposed to Bessel, Chebyshev, and elliptic filter types which are not. In the stop band, the Butterworth filter rolls off at a rate of -20db per decade for each order. First order is -20db slope. Second order is -40db. Third order is -60db, etc. But the Butterworth filter does not have as sharp a cut-off slope as the other filter types. The Butterworth filter has a more linear phase response in its pass band.

The transfer function for the Butterworth filter can be written in terms of its poles of its characteristic equation (the denominator of the function.) There are no zeroes in the function which accounts for its monotonic characteristic.

H(S) = Ho /(II(k=1,n) (S-Sk)/Wc)

Ho = DC gain
II(k=1,n) means the product of the factors to follow
S=complex radian frequency = j*2*Pi*f
Sk’s = roots or pole locations in terms or radian frequency
Wc = j*2*Pi*Fo , the radian frequency for the cut-off frequency Fo
k,n is an index range for the product of factors to indicate the order of the filter from 1 to order n

The Butterworth polynomials have been calculated for normalized Butterworth filters (by setting the cut-off radian frequency Wc = 1.) We will list the first 3 Butterworth polynomials here. Higher orders can be looked up in standard tables found in most filter texts, or on the Internet by searching on “Butterworth filters.”

n=1 s+1

n=2 s^2 + sqrt( 2) * s + 1

n=2 (s + 1) * (s^2 + s + 1)

The root of the above polynomials are all negative and lie on the unit circle surrounding 0,0 on the complex plane.