Talk:Sallen Key filter

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

Contents

[edit] 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 Z1, Z2, Z3, and Z4). --TedPavlic | talk 17:32, 19 November 2007 (UTC)

[edit] 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_1C_2R_1R_2}}+\frac{1}{C_1C_2R_1R_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)

[edit] 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)

[edit] 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)