Talk:Sallen–Key topology

Not very encyclopaedic

This reads more like someone's university EE textbook than an encyclopaedia entry. It would only explain Sallen-Key filters to someone who already knows what they are and how they work. — Preceding unsigned comment added by Gordonjcp (talk • contribs) 06:47, 11 August 2017 (UTC)

I agree, they dive right into formulas without even defining terms! A bunch of gobbleygook, even to many EE students. The formulas are minimally useful if the terms are defined. MacroMyco (talk) 04:54, 10 November 2017 (UTC)

Not very encyclopaedic is inappropriate and should be struck. This article is very well written and succinctly covers the topic. The term Advanced topic might be more appropriate, with a link at the beginning of the article to an introductory wiki on active filters, of which there are several. — Preceding unsigned comment added by 64.201.178.210 (talk) 14:07, 13 November 2018 (UTC)

Figure 1 Node naming

The figure shows the non-inverting input of the amp as a node labeled V-. I would think that this would be labeled V+? Busfault (talk) 16:32, 15 October 2010 (UTC)

The image is trying to show that the operational amplifier in this negative-feedback configuration enforces that the non-inverting input and inverting input match. So the image's author used a $v_{-}$ at the non-inverting input to make the point that the amplifier has negative feedback, which simplifies the analysis. I agree that the image is a little confusing at first (especially if you don't know what to expect). Labeling it with the output voltage may have made more sense. Hopefully the accompanying analysis helps. —TedPavlic (talk/contrib/@) 17:47, 17 May 2011 (UTC)

Relabeling the non-inverting input as Vin would better match the equations in the text. 2600:1700:A3A1:4380:85FC:BAF3:9130:C4BE (talk) 20:05, 7 July 2024 (UTC)

Vin is always the input to the filter, not to the opamp. Constant314 (talk) 20:19, 7 July 2024 (UTC)

Low-pass circuit cutoff attenuation level

There is no set attenuation level for the cutoff of the lowpass circuit. Its 3db frequency is 10222 Hz, not ~15kHz.

I can confirm that. Where it says "the circuit in Figure 3 has an $f_{c}\,$ of $15.9\,{\text{kHz}}\,$ and a $Q\,$ of $0.5\,$", the resistor and capacitor values inserted in the transfer function given yield a -3dB cutoff frequency of ~10.2kHz, not 15.9 kHz as stated. Furthermore, I did start out with $f_{c}$ as 15.9kHz, Q = 0.5, R = 10kHz and C = 1nf, determined m and n from that (which did yield R1 = R2 = 10k and C1 = C2 = 1nf as stated) but still results in a -3dB cutoff at ~10.2 kHz. So something seems to be wrong here (unless I miscalculated somehow). I used m = (1/(2pi*fc*R*C*Q))-1 and n = ( (Q*(m+1))^2 )/m as derived from the formulae for fc and Q. Octave code: ( Neels (talk) 14:21, 6 April 2011 (UTC) )

close all; clear all; f = logspace(log10(1), log10(250e3), 250); w = 2*pi*f; s = j*w; fc = 15.9e3; Q = 0.5; R = 10e3; C = 1e-9; m = 1/(2*pi*fc*R*C*Q) - 1; n = Q^2 * (m + 1)^2 / m; R1 = m*R R2 = R C1 = n*C C2 = C H = 1./( 1 + (R*C*(m+1)).*s + (m*n*R^2*C^2).*(s.^2) ); figure(); hold on; title('freq gain'); H_db = 20*log10(abs(H)); H_ph = mod( arg(H) .* 180 ./ pi , 360) - 180; semilogx(f, H_db, 'g'); line([fc, fc], [min(H_db), max(H_db)]); line([1 250e3], [-3, -3]); grid on; hold off; figure(); hold on; semilogx(f, H_ph, 'g'); title('phase'); grid on;

Hints would be appreciated. (And if anyone knows how to indent this box, please go ahead) -- Neels (talk) 14:18, 6 April 2011 (UTC)

You are confusing natural frequency and cut-off frequency (note: if you look at the phase of the transfer function, I think you'll clearly see that it crosses -90 degrees closer to the expected location). The system has an undamped natural frequency that is sometimes confused with the typical definition of cutoff frequency. As the second-order system becomes damped, its initial tendency to oscillate at its natural frequency is still evident, but those oscillates die out quickly and appear to oscillate more slowly. Consequently, the cutoff frequency of the system shifts toward lower frequencies. —TedPavlic (talk/contrib/@) 17:39, 17 May 2011 (UTC)

For more information about natural frequency (damped and undamped) and resonant frequency (which is a frequency of maximum gain), you should see the article on Damping. —TedPavlic (talk/contrib/@) 18:07, 17 May 2011 (UTC)

Transfer function

It'd probably be useful to list the transfer function here. 68.6.101.96 06:46, 19 May 2006 (UTC)

$H(s)={\frac {V_{out}(s)}{V_{in}(s)}}={\frac {R_{1}R_{2}}{R_{1}Z_{1}+Z_{1}Z_{2}+Z_{2}R_{1}+R_{1}R_{2}}}$

where the Z's are the s-domain impedances of the capacitors.

I think thats correct. Fresheneesz 22:57, 19 May 2006 (UTC)

I just added a generic image that includes a generic transfer function (in terms of $Z_{1}$, $Z_{2}$, $Z_{3}$, and $Z_{4}$). --TedPavlic | talk 17:32, 19 November 2007 (UTC)

High-pass configuration transfer function

Any particular reason the transfer function is given in terms of r1, r2, c1, c2 for the lowpass filter and in terms of the q factor, cutoff frequency and stuff for the highpass filter? If no one objects before monday morning utc I'll add the transfer function in terms of the component values.
The current transfer function:
$G={\frac {s^{2}}{s^{2}+2\pi ({\frac {F_{c}}{Q}})s+4\pi ^{2}(F_{c}^{2})}}$
Here's what I get for the component value transfer function:
$G={\frac {s^{2}}{s^{2}+{\frac {s}{\sqrt {C_{1}C_{2}R_{1}R_{2}}}}+{\frac {1}{C_{1}C_{2}R_{1}R_{2}}}}}$ 131.252.212.137 21:34, 24 May 2007 (UTC)

Also, isn't the formula for Q differant depending on if the filter is butterworth, chebyshev, or otherwise?131.252.212.135 17:33, 25 May 2007 (UTC)

The choice of Q is different for the different filter types. For example, a Butterworth has a Q of $1/{\sqrt {2}}$. The form of the filter is always the same -- it's a generic second-order topology. --TedPavlic | talk 17:31, 19 November 2007 (UTC)

First Image Op-Amp

I'm a bit confused by the first image (showing the circuit diagram). I'm accustomed to seeing a resistor along a feedback connection, not just a wire (on the non-inverting side). Is this an omission, is this how the circuit really works, or is the resistor just conventionally implied? Thanks. --joe056

Are you speaking of the feedback loop from the output to the inverting input, or from the output to the junction of R1 and R2? If the former, the usual reason for the inclusion of a resistor when the op-amp is used as a voltage follower (as here) is to control its offset by making sure that the resistance "seen" by each input is identical. In this case, such a resistor would be the sum of R1 and R2, or 20K ohms. If the latter, no resistor is necessary or desirable. This is because the capacitor located between the output and the resistor junction provides just the right amount of frequency-dependent positive feedback to provide the filter's characteristic transition from the passband into the stopband. If a resistor were placed in series or in parallel with the capacitor, the transition characteristic would be compromised.Anoneditor 20:57, 13 July 2007 (UTC)

Thanks. I was talking about the former, and that clarifies. --joe056

Remember that shorting the output of an OpAmp to its inverting input makes the output a "follower" of the non-inverting input. In fact, other followers (like an emitter follower) will work just as well. You may want to look into "complementary transformations" for another spin on the analysis of a SK filter. Alternatively, look into "bootstrapping"--it's a better way to view the "positive feedback" here. --TedPavlic | talk 19:38, 16 November 2007 (UTC)

In my understanding, whether an emitter follower will work as well as an op amp wired as a gain of 1 buffer depends on the circuit in which it is utilized. For example, emitter followers have a DC offset that is temperature dependent. This could be a problem if the follower-powered filter must be DC coupled to the next circuit. Also, the gain of a standard emitter follower is slightly less than one, and it has a much lower input impedance and a much higher output impedance than the gain of 1 op-amp. These factors could compromise its performance in a filter circuit that requires precise shaping of the transition from the passband into the stop band, at least if no adjustment of the standard resistor and capacitor values is made. And, of course, the emitter follower cannot be made to have a gain greater than one, a quality that is useful in creating equal component value Sallen-Key filters.Anoneditor (talk) 06:00, 18 November 2007 (UTC)

All of these are standard issues involving the use of transistor amplifiers regardless of application. That is, they are irrelevant to the *pure* discussion of Sallen-Key filters. Additionally, an SK filter should never be used to drive a heavy load anyway, and so the base-emitter drop of an emitter-follower should be fairly constant (i.e., no thermal runaway). The fact that there is a small DC component is important to the small-signal properties of the filter. It's true that an emitter-follower's gain will be slightly less than one (and non-linear) because of the dependence on collector current, but the input impedance will be fairly constant (and high) and the output impedance will be fairly low; that's the important thing.

The point is that SK filters have nothing to do with operational amplifiers. All they require is some form of amplifier (unity-gain or not) to buffer the output onto the "bottom" of the first divider. All that's important is that the buffer has low output impedance (at relatively all frequencies) and high input impedance (at relatively all frequencies).

A SK filter is nothing more than two cascaded voltage dividers with an output buffer that "bootstraps" its output back onto the first divider in order to improve the corner frequency. --TedPavlic | talk 14:03, 19 November 2007 (UTC)

Ah, you're talking about *purity*. Now I get it. However, the magnitude of the amplifier input and output impedances isn't awfully relevant to the pure SK filter unless you're worried about the shape of the transition into the stop band, which is a practical problem and pretty much the point of using active filters. So, I think it should be said that the amplifier's input and output impedances should be high and low, respectively, when compared to the impedances of the passive components used in the circuit if an easily produced filter characteristic is desired. Anoneditor (talk) 17:14, 19 November 2007 (UTC)

This is the standard assumption when using any amplifier (e.g., operational amplifier, transistor, etc.). The operational amplifiers drawn here are already assumed to be ideal (i.e., no input leakage). Similarly, BJT versions of these designs would also be assumed to be ideal (i.e., infinite current gain). It can confuse the point of the article to introduce these other issues. Yes, they're important, but they're important to EVERY operational amplifier/BJT design and thus implied whenever these active components are used. --TedPavlic | talk 17:29, 19 November 2007 (UTC)

I added an ideal before "operational amplifier" in the discussion of the derivation of the expressions. --TedPavlic | talk 17:38, 19 November 2007 (UTC)

I updated the new section on practical implementation details. Added some images and new sections. Clarified some language. Tried to make format more wikipedia-ish. --TedPavlic | talk 14:48, 21 November 2007 (UTC)

Ted, I am planning to substitute the following "Implementation" section for your "Real filter implementation" section and I thought it proper etiquette to give you a chance to comment before I do. The reason is that I don't think an article on Sallen-Key filters is the appropriate place for a sort of abbreviated filter construction cookbook. My own earlier piece on "Engineering considerations," though much briefer, had some of the same problems and I probably would not have written some of it had I thought it through. I believe it is sufficient simply to note that "real world" departures from the ideal components assumed by the calculations result in approximation of the filter characteristics predicted by the math, and leave it at that.Anoneditor (talk) 03:53, 24 November 2007 (UTC)

As I said before, I would be much happier without ANY of this stuff being talked about here. Electronic filter implementation is a HUGE subject, not a footnote, and generic filter implementation is even bigger.

That being said, if that information is not mentioned elsewhere, perhaps it could be offloaded into a different article. Or just trash it? Thoughts? --TedPavlic | talk 12:41, 28 November 2007 (UTC)

Implementation

The calculations above assume that all components used are ideal. This means that each amplifier shown has infinite input impedance, no output impedance and no phase shift from input to output at any frequency of interest. It also means that all resistors and capacitors are exactly the values stated, with no tolerance for error. It also means that the filter is driven by a signal source that has no impedance at any frequency of interest. Most of these assumptions are invalid in the actual implementation of these circuits because these ideal components do not exist. Therefore, the response of an actual filter will only approximate the theoretical response indicated by these calculations. How close this approximation is depends on how close the components utilized approximate the ideal.

I think even this snippet is too much. Once you start listing some of the assumptions, you run the risk of not listing them all. Adding a "For example" might help. However, most texts that I use that refer to Sallen Key filters say something like "Using ideal operational amplifier theory" before calculations and nothing else. I think something that simple would probably be best. I think it's wrong to imply that Sallen Key filters have anything to do with operational amplifiers. I like Horowitz and Hill's (The Art of Electronics) treatment of SK filters, as they simply use a generic "x1" buffer and (if I remember correctly) they mention that the buffer gain is arbitrary.

So, why don't we just take the snippet, add your "For example" qualifier, refer to "any kind of amplifier" to eliminate the inference that it has to be the op-amps shown in the article, and make it look like this?

------------

Implementation

The calculations above assume that all components used are ideal. This means, for example, that any kind of amplifier used in the implementation of the filter has infinite input impedance, no output impedance and no phase shift from input to output at any frequency of interest. It also means that all resistors and capacitors are exactly the values stated, with no tolerance for error. And it also means that the filter is driven by a signal source that has no impedance at any frequency of interest. Most of these assumptions are invalid in any actual implementation of these circuits because these ideal components do not exist. Therefore, the response of an actual filter will only approximate the theoretical response indicated by these calculations. How close this approximation is depends on how close the components utilized approximate the ideal.

------------

That way, we get something in the article that directly points out the issue and gives most, if not all, of the parameters in question. Anoneditor 22:31, 1 December 2007 (UTC)

I don't think it's necessary to say anything about phase shift. Because the buffer is simply a real gain (i.e., no imaginary part), it's implied that there is no INTERNAL phase shift at any frequency. Adding that there is zero output impedance (at all frequencies) reinforces that there is no phase shift. Otherwise, such a caveat sounds fine. --TedPavlic | talk 14:06, 3 December 2007 (UTC)

I've substituted the snippet for your earlier work, with the parameter of phase shift removed. Though I don't agree with you if you're saying that real-world buffers have no phase shift from input to output, leaving it out doesn't make much difference because the factors listed are merely examples, rather than all parameters. Anoneditor 23:39, 3 December 2007 (UTC)

Perhaps there needs to be an "operational amplifier examples" section that includes the four (generic/LPF/HPF/BPF) examples on the page. At the top of THAT section could be something saying that the operational amplifiers are assumed to be ideal. That would satisfy me because it doesn't imply that Sallen Key filters must be implemented with OA's, plus it acknowledges the idealization assumptions. --TedPavlic | talk 12:41, 28 November 2007 (UTC)

Perhaps, but that will require a significant rewriting of the article and, maybe, a change in the illustrations. Don't you think it would be enough to mention specifically that operational amplifiers are shown in the schematics because that is usually the way these filters are implemented but that other types of amplifiers can be used as well? Anoneditor 22:31, 1 December 2007 (UTC)

I don't think it would be a significant rewriting. I certainly wasn't talking about generating new images. All I was saying is that the existing article's presentation of the OpAmp material could be changed so that it was more explicit that OpAmps are being used AS AN EXAMPLE.

Putting so much emphasis on "operational amplifiers" gives the young reader the impression that an analog designer picks a 741 off the shelf whenever an active filter is needed. Technically, any difference amplifier with near infinite gain (with frequency compensation as needed for feedback) will be an adequate operational amplifier. Because the operational amplifier will be connected as a buffer (possibly with some gain), the "OA" design can be simplified significantly on silicon. Thus, for modern designers, it's more important that a unity gain buffer amplifier is used than an operational amplifier.

That being said, because OA's come in such well-known and easy-to-use-in-the-lab packages, the OA provides a good teaching tool for explaining active filter implementation.

So, I think a good way to go is to introduce the SK filter in terms of buffers and then present the content that makes up most of the current page as "operational amplifier examples." No additional content is needed. Someday, if someone could replicate the buffer graphic from Horowitz and Hill, that might be useful. However, in the meanwhile just changing the language will suffice. --TedPavlic | talk 14:06, 3 December 2007 (UTC)

I don't think I have any argument with such a change. Have at it! Anoneditor 23:39, 3 December 2007 (UTC)

Q Expressions Wrong

The expressions for Q seem wrong for the filters here. In particular, it looks like "1/Q" is what's meant here, and I think that (at least for the low-pass case) the C2 needs to be swapped for C1. Thoughts? --TedPavlic | talk 19:43, 16 November 2007 (UTC)

Everything is fixed now (and improved?). --TedPavlic | talk 17:24, 19 November 2007 (UTC)

High pass example possibly misleading

Figures 2,3,4 renumber the Z3/Z4 components from Figure 1 in reversed order. For the low-pass calculations there is a compensation Z3=1/sC2, Z4=1/sC1 (swaps C1 and C2). This leads to compensation in the calculation for 'n' which makes up for it. For the high-pass filter it is however not trivial that 'm' needs to be calculated differently from what is described in the low-pass section. For high-pass the following holds: R1 = R, R2 = mR, C1 = C, C2 = nC. As I'm not all that familiar with filters, I haven't applied this on the page yet and want your opinion on it first. Would it maybe be easier to modify the images, so both high/low pass can use the same 'm' and 'n' formulas? —Preceding unsigned comment added by 78.27.11.131 (talk) 16:32, 7 March 2009 (UTC)

I also find the index nr swapping irritating. Neels (talk) 01:40, 7 April 2011 (UTC)

The indexes on the first image can be changed easily. It's an SVG file, and so they can be changed with a text editor. The changes would then just have to be pushed down through everything else in the page. —TedPavlic (talk/contrib/@) 16:43, 17 May 2011 (UTC)

Changes have been made. Ordering should be consistent throughout. Hopefully I caught all of the text fixes. —TedPavlic (talk/contrib/@) 17:02, 17 May 2011 (UTC)

Band-pass filter

1. Is there an expression for the quality factor for the band-pass filter? How do you reduce the bandwidth of the response and narrow the spike? ICE77 (talk) 21:41, 14 May 2011 (UTC)

I'm a little confused. There are several links (probably too many) to Q factor throughout the article. The bandwidth is set by $\omega _{0}$, and the peak response is set by the Q factor (or damping ratio). Just as in the low-pass and high-pass examples, there are several parameter choices that can be used for each desired bandwidth and peak shape. It is up to the designer to choose one set of parameter values. Having said that, the resistor and capacitor component choices all depend on each other. In general, a change of two components (for example, to change the gain of the VCVS) will require a change in many other components to maintain the prior properties of the device. —TedPavlic (talk/contrib/@) 23:09, 14 May 2011 (UTC)

Your comment does not answer my question. Beside, I only see two expression for Q and not even one link. The choice of values for C and R will certainly change the location of the center frequency. Still, no expression for Q is given for the band-pass circuit.

ICE77 (talk) 18:42, 15 May 2011 (UTC)

Q factor is discussed several times throughout the page above the bandpass filter section, and the treatment of the Q factor is essentially identical. The bandpass filter section is still a second-order filter, and so it still has a second-order characteristic equation in the denominator of its rational transfer function. You can read the Q factor straight out of the linear coefficient (i.e., the coefficient of the pure $s$ term in the middle of the characteristic equation). Consequently, the $H(s)$ equation already given in the bandpass section had an underbrace pointing out that that coefficient was $\omega _{0}/Q$. So if you know $\omega _{0}$ (from the constant part of the characteristic equation, which is $\omega _{0}^{2}$), and you have a desired $Q$, you can choose the other components to make the linear coefficient give you your $Q$. I've added one line under the $f_{0}$ description breaking out the Q factor as well. Is that more like what you were looking for? —TedPavlic (talk/contrib/@) 05:41, 16 May 2011 (UTC)

2. I'm not sure if this is a comment or if this should be part of the text of the article: "Caution, The Rb and Ra in the figure above have been swapped around, this causes a gain differant to expected when using this circuit design." If it should be part of the text, I think the sentence should start with "Caution: the Rb ...". ICE77 (talk) 21:41, 14 May 2011 (UTC)

Well, certainly something should be changed. If the text doesn't match the figure, then the text should be changed. On first glance though, I don't understand the "Caution". The $R_{a}--R_{b}$ divider has $R_{a}$ going to ground, and so the $(1+R_{b}/R_{a})$ gain makes sense. That is, $V_{-}=V_{out}R_{a}/(R_{a}+R_{b})$ which means $V_{out}=(1+R_{b}/R_{a})V_{-}$. So I'm not sure what the "Caution" text is complaining about. Do you? Either way, something should be changed. At a minimum, the "Caution" stuff should be removed. —TedPavlic (talk/contrib/@) 23:09, 14 May 2011 (UTC)

For example, if resistor A is made to be an open circuit (as in the unity-gain case), then $R_{a}\approx \infty$. Consequently, the gain $(1+R_{b}/R_{a})\approx 1$ is correct. —TedPavlic (talk/contrib/@) 23:12, 14 May 2011 (UTC)

Regarding the "caution" sentence, I noticed you removed it. I don't know why it was there. Either way, if the image is consistent with the definition of G, it really doesn't matter. The caution was probably there to warn about a mismatch but I really don't see it anyway. I made some simulations with PSpice and I confirm that the equation is consistent to the image. I also agree on your last statement. When the value of R_A is very large, the op-amp is in a buffer mode and the VCVS circuit becomes a SK circuit. In fact I see that the frequency response drops by 12bB/octave at the center frequency with Q=0.5 because of two poles (just like for low and high pass circuits).

ICE77 (talk) 18:42, 15 May 2011 (UTC)

You can verify the $(1+B/A)$ gain with a simple node-voltage analysis as well (similar to what was done in the unity-gain case at the top of the page, but without the unity gain). —TedPavlic (talk/contrib/@) 05:41, 16 May 2011 (UTC)

Passband Non-Unity Gain

This article would be a bit more complete were it to point out that passband gain can be altered by simply overlaying the gain resistors from a Non-Inverting Amplifier [[1]]. — Preceding unsigned comment added by 69.140.49.200 (talk) 13:19, 3 June 2011 (UTC)

Please read the lead (i.e., the first paragraph) of the article. Then read the section with the title "VCVS Example: Bandpass configuration". Did you miss these sections? Perhaps your point is that these things could be highlighted more to make it more obvious? —TedPavlic (talk/contrib/@) 16:51, 3 June 2011 (UTC)

Some feedback on article

Good work, Ted. I just worked on a Sallen–Key (SK) filter wizard at work and had to go from knowing zero to being able to automate SK design. This is the first site I used, and it got me started. Anyway, here's three places that this article could have been more helpful to me. First, the low pass and high pass examples do not discuss the gain resistors, which of course I had to support. It is in the band-pass discussion, so I don't understand why it's not in low-pass and high-pass. Second, the benefits of SK filters are somewhat buried. Another site says SK filters are relatively insensitive to component values and that component values are fairly close to each other. Pole control is low, making high $Q$ difficult to control. Third, and this is a complaint I have with all SK sites I found, the simplification described by letting $R_{1}=mR_{2}$ and $C_{1}=nC_{2}$ is not nearly as good as letting $R_{1}=mR$, $R_{2}=R/m$, $C_{1}=nC$ and $C_{2}=C/n$. The formulas for $Q$ and $f_{c}$ become:

$f_{c}={\frac {1}{2\pi RC}}$

$Q={\frac {mn}{m^{2}+1+(mn)^{2}(1-K)}}$

where $K$ is the gain when we add the negative feedback resistors. This eliminates the dependence between $f_{c}$ and $Q$ calculations. In this case $R={\sqrt {R_{1}R_{2}}}$ and $C={\sqrt {C_{1}C_{2}}}$. Further simplification comes when I assume $R_{1}=R_{2}$ (i.e., $m=1$), although in my case I care mostly about the area on an IC occupied by the passives. $R_{1}=R_{2}$ comes very close to the best possible minimum for gains up to $20\times$. Board-level users may care more about minimizing the cost of the capacitors, and might prefer to use $C_{1}=C_{2}$. WaywardGeek (talk) 11:09, 11 October 2011 (UTC)

Thanks for your comments, WaywardGeek. If I recall correctly, much of the basic features of the web page (e.g., the examples picked) were present before I touched the page. I just cleaned up some of the math and (hopefully) some of the presentation. The gain is non-unity only for the band-pass case because a Sallen–Key (SK) filter is unity gain by definition. Once you add non-unity gain, the filter becomes a more general VCVS filter. This is explained briefly in the lede of the article. Certainly the introductory information on SK filters could (and probably should) be expanded to make their relevance more clear; however, this encyclopedia article should not turn itself into a guide on SK design. Finally, regarding the design choice of letting $R_{1}=mR_{2}$ (etc) instead of setting $R_{1}=m^{2}R_{2}$ and introducing a fictitious $R$ (and $C$), I think many references avoid this approach so that the effect of the actual components are clear. Again, because the encyclopedia article is not meant to be a design guide, different choices are made. The clarity of the content might be compromised by introducing symbols for components that only exist in the abstract. —TedPavlic (talk/contrib/@) 16:59, 11 October 2011 (UTC)

Super-Unity-Gain

This is the most useless set of links. Does this phrase just mean 'a gain greater than 1'? If you search for "super unity gain" in google you just get lots of identical sentences about sallen-key topology filters! Can someone who understands this article adjust the phrasing :) thank you! — Preceding unsigned comment added by 94.116.245.123 (talk) 09:29, 12 December 2012 (UTC)

Bandpass filter simulates as a bandstop filter

The section claims to detail a bandpass filter. However, the schematic given in Figure 5 simulates as a notch/stopband filter. Can someone confirm or possibly correct this?

Edit: Never mind, capacitor was in Farads, not micro or nano Farads.

Design Choices

Although it is questionable whether this section really belongs to an encyclopedic article, if it is present, even when just intended as an example, it would be wise to introduce a procedure that is feasible in practice. I experienced that undergraduate students, who tried to follow the previous version as a design guideline, got stuck with a few equations from which they were unable to obtain the component values. By the relative simple modification that I made to this section, the dimensioning of $f_{0}$ becomes independent from $Q$.WimdeValk (talk) 20:55, 12 October 2017 (UTC)