Design second- and third-order Sallen-Keyfilters with one op ampChristopher Paul, Motorola - January 31, 2011

RP Sallen and EL Key of the MassachusettsInstitute of Technologys Lincoln Laboratory in1955 introduced the Sallen-Key analog filtertopology. Engineering literature extensivelydiscusses the second-order section that createstwo filter poles (Figure 1 and references 1 to4). You can also make a third-order filter usingtwo op amps (Figure 2). For filter gains of oneor two, you can make a third-order filter withone op amp (Figure 3). Such a configurationhas been addressed in a limited manner for opamp gains of 1 and 2 (references 5 and 7).Unity-gain filters have low sensitivities tocomponent values, but they can require largeratios of capacitor values. Gain-of-two filtersallow capacitors of similar or identical values,

but generally are much more sensitive.

Using the following design procedure,you can convert sets of two or threepoles into single op-amp filters. Theprocedure does not place unduerestrictions on op-amp gains orcomponent values. You can selectstandard-value capacitors and resistorsand then calculate the remainingresistor values from filterspecifications. The procedure producesdesigns with both low sensitivities and

moderate ranges of user-specified capacitor values. A figure of merit compares filter sensitivities.The tendency of the filters to oscillate can be assessed. The procedure also demonstrates thesuperiority of third-order-filter stopband-leakage characteristics compared with those of second-order filters. You can perform the associated calculations for this procedure in this Excelspreadsheet.

Second-order section design

You first use Equation 1 to determine the transfer function for the second-order section to start thedesign procedure:

You then select standard-value capacitors. If you use capacitor values that are too large, thecapacitors will be expensive or will occupy excessive space. If the values are too small, the PCBs(printed-circuit-boards) and op amps parasitic capacitances will affect the filters response.

In Equation 1, the denominator is (sp2)(sp3), where p2 and p3 are the real-valued and oftencomplex filter poles. By equating the denominator coefficients in s of Equation 1 with those in theexpression containing p2 and p3, you can write an equivalence for a term you define as B and thensolve for AMIN, the minimum op-amp gain for the filter, as equations 2 and 3 show.

You then choose the values of RF and RG that will keep the gain above AMIN, as Equation 4 shows.

You then create equivalences for terms you define as D and E, as equations 5 and 6 show.

This approach lets you conveniently write equations 7 through 10, which define two sets of the tworesistor values.

Resistor values R1A and R2A yield one solution, and R1B and R2B define another. You now choose thenearest standard-value resistors for R1 and R2. The resistor tolerances should be 1% or better, andcapacitor tolerances should be 2% or better. When RFC2>RGC3, the second pair of values will benegative and will constitute an unusable solution.

Third-order section design

Closed-form solutions for third-order filter sections do not exist. However, you can employ numericaltechniques to achieve suitable results. Once again, you start with a derivation of the filter-transferfunction, as Equation 11 shows:

You define the B terms with three equivalences, as equations 12 through 14 show.

You can express the denominator as (sp1)(sp2)(sp3), where p1, p2 and p3 are the poles ofthe filter you are designing. You can then equate appropriate coefficients of powers of s to b0, b1,and b2. By solving Equation 14 for r1 and substituting into equations 12 and 13, you getequations 15 and 16:

The K terms in these two equations are defined in equations 17 through 23.

Note that Equation 15 has a degenerate form when K equals zero. This form allows you to generatefour equivalences based on the value of K by solving for r2 in equations 15 and 16, as equations24 and 25 show:

You can now choose values for the capacitors and for resistors RF and RG and evaluate the K's inequations 17 through 23. Equating equations 24 and 25 and solving for r2 gives sets of values forr2 and r3. You can then easily rearrange Equation 14 to solve for r1. One straightforward way to

find solutions is to iterate r3 from values of 10 to 1 M in a spreadsheet, subtracting expressionsfor r2 from those for r2 in ranges in which both have positive real values. If successive subtractionshave opposing signs, then a solution lies between them.

In searching for solutions, it is sometimes helpful to graph equations 24 and 25. One, many, or nosolutions are possible. If there are no curve intersections, the graphs can show whether a new set ofvalues moves the curves closer to or farther from one. The graph can show cases that find a singlesolution (Figure 4). If you just make arbitrary selections for the capacitors and for resistors RF andRG, you will generally not make a successful design. The sample-filter-design section of this articleprovides guidance for value selections.

Stability of the second-order section

As with any active circuit, oscillation will occur if the zero-phase-shift frequency loop gain exceedsunity. Accordingly, you must calculate that gain. First, break the connection between C2 and the op-amp output of Figure 1. Then, connect a voltage source, VI to C2. You ground the filter input at R2because the source driving the filter must have zero impedance if the filter is to function as youdesign it. Using an op-amp output voltage that you define as VO, you can calculate the transferfunction VO/VI using Equation 26.

Because the zero-phase-shift frequency occurs when s2=1/(R2R3C2C3), you can solve for theloop gain under that condition using Equation 27.

You then use Equation 28 to convert the expression to negative decibels to obtain the gain margin.

If the result is positive, the circuit will be stable, whereas negative results predict instability. Youshould evaluate Equation 28 with its components at the tolerance extremes that would lead to thehighest possible gains, meaning that you should use the largest possible values for C2, R2, and RF andthe smallest for C3, R3, and RG.

Stability of the third-order section

You use the same procedure to evaluate the stability of a third-order loop. You break the connectionbetween C2 and the op-amp output and then ground the filter input at R1. You then use Equation 29to determine the transfer function between the disconnected side of C2 and the op amps output.

This form of the equation tells you that you should choose R1, R2, C2, and RF at the highest pointswithin their tolerance ranges to maximize gain. Similarly, you should choose C1, R3, C3, and RG at thelowest points. This rule applies to equations 29 through 38. Equation 30 rearranges Equation 29in the standard form to find the gain at the zero-phase-shift frequency, s0.

This equation uses the equivalences you define for the n and d terms in equations 31 through 35.

When the phase shift is zero, the arctangent of the ratio of the imaginary to the real parts of thenumerator of Equation 30 must equal that of the denominator, which means that the ratiosthemselves must be equal:

Imaginary solutions to Equation 36 exist at s0:

You can use Equation 38 to convert to negative decibels to get the gain margin from Equation 29.

As with the second-order stability analysis, positive values indicate stability, and negative valuespredict instability.

Sensitivity

A low-sensitivity-filter design is immune to component variations due to manufacturing tolerances.Filter parameters such as gain and phase shift are sensitive to component tolerances, and soproduction-line filters will have somewhat differing characteristics. You use sensitivity analysis toprevent these differences from becoming unacceptable. You can define the sensitivity of some filterfunction F(x) to a component value, x, using Equation 39.

The partial derivative appears because F(x) is also a function of variables other than x. The otherterms effectively normalize the sensitivity parameter. It is more intuitive to replace the differentialswith small differences and then rewrite the equation (Equation 40).

You can now see that, when you multiply the sensitivity parameter by a small relative change in thevalue of component x, you get an associated relative change in F(x). Rather than directlydifferentiate Equation 39, it is simpler to use the definition of a derivative and evaluate using asmall value of , such as 106 (Equation 41).

One important filter function is the absolute amplitude response, |H(s)|. Monte Carlo evaluationsreveal significant variations of F(x)=|H(sc)| in the vicinity of the filters cutoff frequency sc.Equations defining |H(s)| for third- and second-order sections were derived earlier. Those for thefirst-order sections to form a composite filter with a second-order section are trivial. It is better tocreate a single figure of merit that you can use to compare all filters. One approach might be to takethe root/mean/square of the sensitivities of |H(sc, xi)|for each of a filters i constituent components.You can define parameter S as an aggregate filter sensitivity (Equation 42), taking into accountthat components may have different tolerances.

TOLXI is the tolerance of component xI, so that a component with a 1% tolerance would have a TOL of1, one with a 2% tolerance would have a TOL of 2, and so forth.

Selecting sample-filter designs

It is valuable to design filter sections that implement complex pole pairs j over a range of qualityfactors Q (Equation 43).

A ninth-order, 0.1-dB-ripple Chebyschev lowpass filter offers a good selection of quality factors. Youcan implement second order sections (Figure 1), composite first- and second-order sections (Figure2) and third-order sections (Figure 3) using each of the filters four complex pole pairs. Each third-and first-order section of the composite filters will reflect the single real pole. You can implementsecond-order sections of any Q with equal values of C2 and C3 if the op amp gain is two (Reference

2). You can also do it with a unity gain circuit if C2 is four times the product of Q2 and C3. Thesensitivities of the filters Q and resonance frequencies are much greater if the op amp gain is two,but the ratio of capacitor values for high Q sections can be too high if the gain is one. Byinvestigating filter sections between gains of one and two, you will be able to see that some gainsallow the use of standard-value capacitors whose ratio is considerably less than 4Q2 and yet yield Svalues almost as low as those of the unity-gain filters.

You can use the same approach to get similar results for third-order filter sections. You can findsolutions for op-amp gains of two where C1=C2=C3. You can make unity-gain-filter sections withvalues of C1 equal to C2 and greater than 8Q2C3. Solutions also exist for gains slightly greaterthan unity if C1=C2 and C2/C3RG/RF, where C2/C3 can be much less than 8Q2. In this last case,values of S can approach those of unity gain designs. Try small value variations in ranges that satisfythese conditions to uncover the minimum values of S.

Comparing sample-filter designs

You can generate a set of sample filter designs using the Excel spreadsheet to do the mathematics(Click here to view Table 1). The first column assigns a trial number to each design. Thesubsequent two columns yield pole values that are multiplied by a 21000 frequency-scale factor toachieve a 1-kHz cutoff frequency. The next two columns give the values of Q and 4Q2 for eachcomplex pole pair. The next column describes the orders of the implemented filter sections. Columnswith headings of C1, C2, C3, RG, and RF contain the values you have selected, whereas those withheadings of R1, R2, and R3 list calculated values that the nearest standard 1%-tolerance valuesapproximate. The subsequent column lists gain margin, using worst-case values within tolerances of1% for resistors and 2% for capacitors. The following column presents the aggregate sensitivity, S,of the third-order composite (Figure 2) or third-order single section (Figure 3). The next columnshows S only for the second-order section in the composite filter. The final column indicates whetherother design procedures have confirmed the filter-component values in the table.

Perhaps the most notable finding is that the gain margins of trials 3, 6, and 9 (in red) are negative,meaning that these designs will be unstable at component-value-tolerance extremes. All of thesedesigns have op-amp gains of two and relatively high quality factors. You should be cautious whenusing such designs. Among the composite designs, virtually all the sensitivity lies in the second-order sections, and little exists in the first.

You can also compare the S values for the third-order single sections and their companion compositefilters, those having similar or the same op-amp gains and implementing the same poles. If youignore the unstable designs, there is little difference in the aggregate sensitivities. A compositedesign, which requires an additional op amp, has about the same sensitivity as a third-order singlesection, meaning that you can convert a third-order composite design into a third-order singlesection by removing an op amp and adjusting component values. This result exhibits little or nopenalty in component-tolerance sensitivity. If your design requires a specific second-order response,you can add a real pole high enough above the second-order sections cutoff frequency to getadditional stopband attenuation.

Building and measuring filters

It is a good idea to validate the solutions to these equations with actual physical filters. Alternatively,for composite and single-section third-order filters with gains of two, you can gain confidence in thedesign by ensuring that other design procedures give component values identical to these. Becausefew if any alternative procedures exist for third-order single-section designs with gains other than

one or two, you must evaluate those designs by building the circuit. The designs of trials 8 and 11 inTable 1 were built and tested using high-gain-bandwidth op amps. The measured dc gains werenormalized to unity. Spot checks of each filter were made against the continuous graph of theirtheoretical responses (Figure 5).

Stopband leakage

There is one additional benefit to a third-order design. The second-order designs suffer from a high-frequency leakage current through R2 and C2 from the filters input to the op amps output(Reference 8). Because the output impedance of an op amp rises with falling open-loop gain athigher frequencies, it and the current from the input combine to yield stopband leakage, anunexpected signal at the output. In a third-order filter, C1 shunts much of this current to ground.Although the high-frequency current in a second-order section is simply VIN/R2, the current for athird-order single section decreases to (VIN/(R1+R2))/(1+sC1R1R2/(R1+R2)). Figure 6 shows themeasured and theoretical results of the Table 1 filter designs of trials 11 (second-order section only)and 8 (third-order single section). To exacerbate stopband leakage, they both use a low-gai--bandwidth op amp and with components having one-tenth of the impedances the table lists. Thestopband leakage of the second-order Trial 11 section is evident, whereas it is absent from the third-order Trial 8 section. Because the measurement noise floor was about 70 dBV, it is not possible todetermine at what point leakage appeared in the third-order section. The inputs of both filters weredriven at 10V rms at frequencies greater than 1500 Hz where filter attenuation ensured that outputsaturation of the 12V-powered op amp was not a concern.

References1. Williams, Arthur B, and Fred Taylor, Electronic Filter Design Handbook, McGraw-Hill, 1981.2. Van Valkenburg, ME, Analog Filter Design, Van Valkenburg, M.E., Holt, Reinhart and Winston,1982.3. Cano, Martin, A new set of Sallen-Key filter equations, EDN, Oct 1, 2009.4. Texas Instruments FilterPro, 2001-2006, Texas Instruments Corp.5. Williams, Arthur B, Active Filter Design, Artech House Inc, 1975.6. Parker, Glenn A, Reducing Active Filter Costs Using Symbolic Three-Pole Synthesis, RF Design,March 1996.7. Beis, Uwe, Design and Dimensioning of Active Filters."8. Cano, Martin, Eliminate Sallen-Key stopband leakage with a voltage follower, EDN, May 14,2009, pg 17.