Design Article
Comment
anthonyandriano
This spreadsheet is very confusing. It's pretty obnoxious to require poles in ...
anonymous user
Hi Jack, I've just downloaded from the website and don't have the problem, nor ...
Design second- and third-order Sallen-Key filters with one op amp
Christopher Paul, Motorola
1/31/2011 6:15 PM EST
RP Sallen and EL Key of the Massachusetts Institute of Technology’s Lincoln Laboratory in 1955 introduced the Sallen-Key analog filter topology. Engineering literature extensively discusses the second-order section that creates two filter poles (Figure 1 and references 1 to 4). You can also make a third-order filter using two op amps (Figure 2). For filter gains of one or two, you can make a third-order filter with one op amp (Figure 3). Such a configuration has been addressed in a limited manner for op amp gains of 1 and 2 (references 5 and 7). Unity-gain filters have low sensitivities to component values, but they can require large ratios of capacitor values. Gain-of-two filters allow capacitors of similar or identical values, but generally are much more sensitive.
Using the following design procedure, you can convert sets of two or three poles into single op-amp filters. The procedure does not place undue restrictions on op-amp gains or component values. You can select standard-value capacitors and resistors and then calculate the remaining resistor values from filter specifications. The procedure produces designs 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 the superiority 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 Excel spreadsheet.
Second-order section design
You first use Equation 1 to determine the transfer function for the second-order section to start the design procedure:

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



Third-order section design
Closed-form solutions for third-order filter sections do not exist. However, you can employ numerical techniques to achieve suitable results. Once again, you start with a derivation of the filter-transfer function, as Equation 11 shows:






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

As with any active circuit, oscillation will occur if the zero-phase-shift frequency loop gain exceeds unity. 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 R2 because the source driving the filter must have zero impedance if the filter is to function as you design it. Using an op-amp output voltage that you define as VO, you can calculate the transfer function VO/VI using Equation 26.



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



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



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 so production-line filters will have somewhat differing characteristics. You use sensitivity analysis to prevent these differences from becoming unacceptable. You can define the sensitivity of some filter function F(x) to a component value, x, using Equation 39.


Selecting sample-filter designs
It is valuable to design filter sections that implement complex pole pairs α±jβ over a range of quality factors Q (Equation 43).

You can use the same approach to get similar results for third-order filter sections. You can find solutions for op-amp gains of two where C1=C2=C3. You can make unity-gain-filter sections with values of C1 equal to C2 and greater than 8×Q2×C3. Solutions also exist for gains slightly greater than unity if C1=C2 and C2/C3≈RG/RF, where C2/C3 can be much less than 8×Q2. In this last case, values of S can approach those of unity gain designs. Try small value variations in ranges that satisfy these conditions to uncover the minimum values of S.
Comparing sample-filter designs
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 these designs have op-amp gains of two and relatively high quality factors. You should be cautious when using 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 composite filters, those having similar or the same op-amp gains and implementing the same poles. If you ignore the unstable designs, there is little difference in the aggregate sensitivities. A composite design, which requires an additional op amp, has about the same sensitivity as a third-order single section, meaning that you can convert a third-order composite design into a third-order single section by removing an op amp and adjusting component values. This result exhibits little or no penalty 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 section’s cutoff frequency to get additional 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 the design by ensuring that other design procedures give component values identical to these. Because few 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 in Table 1 were built and tested using high-gain-bandwidth op amps. The measured dc gains were normalized to unity. Spot checks of each filter were made against the continuous graph of their theoretical responses (Figure 5).

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 filter’s input to the op amp’s output (Reference 8). Because the output impedance of an op amp rises with falling open-loop gain at higher frequencies, it and the current from the input combine to yield stopband leakage, an unexpected 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 a third-order single section decreases to (VIN/(R1+R2))/(1+sC1R1R2/(R1+R2)). Figure 6 shows the measured 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-gain-bandwidth op amp and with components having one-tenth of the impedances the table lists. The stopband 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 to determine at what point leakage appeared in the third-order section. The inputs of both filters were driven at 10V rms at frequencies greater than 1500 Hz where filter attenuation ensured that output saturation of the ±12V-powered op amp was not a concern.

| References |
|
Navigate to related information


anonymous user
4/13/2011 10:28 AM EDT
Well described and potentially useful, but the Excel file as distributed is locked and password protected for all cells when opened.
Sign in to Reply
Chris Paul
4/22/2011 5:10 PM EDT
The boxes highlit in yellow are unlocked. They are available for the user to specify design requirements. The other boxes are locked and the spreadsheet is password protected.
Sign in to Reply
anonymous user
7/21/2011 11:01 AM EDT
Chis,
Sorry, but in my version of excel apparently even the yellow boxes are locked. The message that the cell or chart is locked and that the sheet must be unlocked is all I can ever get.
Sign in to Reply
anonymous user
8/10/2011 2:41 PM EDT
Hi Jack, I've just downloaded from the website and don't have the problem, nor do some others I know who have tried it. If you want to publish an e-mail address, I can try sending you this 7Meg file.
Sign in to Reply
anthonyandriano
1/4/2013 10:02 AM EST
This spreadsheet is very confusing. It's pretty obnoxious to require poles in radians instead of hertz. I've never seen a filter design require input from the user regarding real and imaginary pole locations. Not useful.
Sign in to Reply