Active low-pass Sallen-Key Butterworth filter design

kalemaxon89

Joined Oct 12, 2022
235
I have read TI's application note Active Low-Pass Filter Design a couple of times.
Although it is probably explained in the paper, I still don't understand some concepts.
I make a practical case referring again to the document I attached.

Let's assume I need to cut-off my signal at 10kHz and amplify it by a factor of K = 4 .. and let's assume that I have to use (for example) a 2° order Sallen-Key Butterworth filter (I don't care if it's correct to use this filter ... it's just an example where the use of this filter is forced on me!).
The circuit I need to make is this one:

1) Is the K gain tabulated?
I found this table online that "imposes" to choose these fixed gains to ensure that I am using a Butterworth filter ... so I am forced to choose K = 1.586, rigth?
So to achieve an overall K = 4, I will realize an amplifier stage (subsequent to this filter stage) with gain K = 4/1.586 = 2.522

Is everything correct so far?

2) If the answer is yes, it is not very clear to me why these gain values are set, how they were found, and especially what happens if I choose the resistance Rf and (K-1)Rf so that I have a gain K different from 1.586

3) In addition to this K gain contrain, is there any other value/limit/rule that I have to comply with to ensure that I am using a Butterworth filter?
I ask this because online I found this image that seems to impose another limitation:

If so ... I should make sure to use a damping coefficient = 0.707 ... but how?
I know the transfer function of the second order low pass Butterworth filter:

..but in practice I don't know what I should do

Papabravo

Joined Feb 24, 2006
21,225
The theoretical development of the Sallen-Key topology is given in filter design textbooks as well as the original paper.
VanValkenburg, M.E., Analog Filter Design, 1984, pp.171-179
R.P. Sallen and E. L. Key, "A Practical Method for Designing RC Active Filters", IRE Trans. Circuit Theory, vol CT-2, pp 74-85,1955

The topology allows quite a bit of flexibility in choosing component values. Look at Appendix A in the TI paper for successively ways of doing a design. They are:

A.1 Sallen-Key Design Simplifications
Filter design can be simple or tedious, depending on the method used to solve the equations. The following
simplifications are ordered from harder to easier, but note that the easier the design becomes, the more the
design freedom is limited.

A.1.1 Sallen-Key Simplification 1: Set Filter Components as Ratios
...
Start the design by determining the ratios m and n required for the gain and Q of the filter, and then selecting C and
calculating R to set fc .

A.1.2 Sallen-Key Simplification 2: Set Filter Components as Ratios and Gain = 1
...
This sets the gain to 0 dB in the pass band. Start the design by determining the ratios m and n for the required Q of the
filter, and then selecting C and calculating R to set fc.

A.1.3 Sallen-Key Simplification 3: Set Resistors as Ratios and Capacitors Equal
...
The main motivation behind setting the capacitors instead of resistors equal is the limited selection of values in comparison
to resistors. There is interaction between setting fc and Q . Start the design by choosing m and K to set the Q of the circuit,
and then choosing C and calculating R to set fc .

A.1.4 Sallen-Key Simplification 4: Set Filter Components Equal
...
With this simplification, fc and Q are now independent. Q is now determined solely by the gain of the circuit. fc is set by the choice of
RC—choose C and calculate the corresponding R. Since the gain controls the Q of the circuit, further gain or
attenuation is necessary to achieve the desired signal level in the pass band.

Using all four methods described in the TI paper, try each method in turn to see how it works.

In filter design we use fc and Q as parameters for a second order transfer function. In mechanics we use the damping ratio "ζ". Both represent the same thing, except for a scale factor.

$$\zeta\;=\;\cfrac{1}{2Q}$$

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Ian0

Joined Aug 7, 2020
9,792
There are two version of Sallen & Key filters for which values are commonly tabulated:
Equal Component Value: where the two Rs and the two Cs are the same, and the gain varies.
Unity Gain: where the gain is always unity, but the two Rs and Cs can vary.
It is possible to make a filter with a given fc and Q with any positive gain greater than unity.
(MFB filters can do it with any negative gain)
I use this utility to calculate the values
http://sim.okawa-denshi.jp/en/Fkeisan.htm

kalemaxon89

Joined Oct 12, 2022
235
The theoretical development of the Sallen-Key topology is given in filter design textbooks as well as the original paper.
VanValkenburg, M.E., Analog Filter Design, 1984, pp.171-179
R.P. Sallen and E. L. Key, "A Practical Method for Designing RC Active Filters", IRE Trans. Circuit Theory, vol CT-2, pp 74-85,1955

The topology allows quite a bit of flexibility in choosing component values. Look at Appendix A in the TI paper for successively ways of doing a design. They are:

A.1 Sallen-Key Design Simplifications
Filter design can be simple or tedious, depending on the method used to solve the equations. The following
simplifications are ordered from harder to easier, but note that the easier the design becomes, the more the
design freedom is limited.

A.1.1 Sallen-Key Simplification 1: Set Filter Components as Ratios
...
Start the design by determining the ratios m and n required for the gain and Q of the filter, and then selecting C and
calculating R to set fc .

A.1.2 Sallen-Key Simplification 2: Set Filter Components as Ratios and Gain = 1
...
This sets the gain to 0 dB in the pass band. Start the design by determining the ratios m and n for the required Q of the
filter, and then selecting C and calculating R to set fc.

A.1.3 Sallen-Key Simplification 3: Set Resistors as Ratios and Capacitors Equal
...
The main motivation behind setting the capacitors instead of resistors equal is the limited selection of values in comparison
to resistors. There is interaction between setting fc and Q . Start the design by choosing m and K to set the Q of the circuit,
and then choosing C and calculating R to set fc .

A.1.4 Sallen-Key Simplification 4: Set Filter Components Equal
...
With this simplification, fc and Q are now independent. Q is now determined solely by the gain of the circuit. fc is set by the choice of
RC—choose C and calculate the corresponding R. Since the gain controls the Q of the circuit, further gain or
attenuation is necessary to achieve the desired signal level in the pass band.

Using all four methods described in the TI paper, try each method in turn to see how it works.

In filter design we use fc and Q as parameters for a second order transfer function. In mechanics we use the damping ratio "ζ". Both represent the same thing, except for a scale factor.

$$\zeta\;=\;\cfrac{1}{2Q}$$
There are two version of Sallen & Key filters for which values are commonly tabulated:
Equal Component Value: where the two Rs and the two Cs are the same, and the gain varies.
Unity Gain: where the gain is always unity, but the two Rs and Cs can vary.
It is possible to make a filter with a given fc and Q with any positive gain greater than unity.
(MFB filters can do it with any negative gain)
I use this utility to calculate the values
http://sim.okawa-denshi.jp/en/Fkeisan.htm
Thank you both.

So if I follow the guidelines on page 19 of the attached document:

and assuming: m = 0.3 n = 3 K = 2 (gain) C1 = 10nF e R1 = 10k
I get this result:

The cutoff frequency is correct, but Q calculated is 2.37 ... while in the simulator it is 0.24
There is something wrong :/

Ian0

Joined Aug 7, 2020
9,792
Thank you both.

So if I follow the guidelines on page 19 of the attached document:
View attachment 311340

and assuming: m = 0.3 n = 3 K = 2 (gain) C1 = 10nF e R1 = 10k
I get this result:
View attachment 311341

The cutoff frequency is correct, but Q calculated is 2.37 ... while in the simulator it is 0.24
There is something wrong :/
You are using the unity-gain version.

kalemaxon89

Joined Oct 12, 2022
235
You are using the unity-gain version.
Hold on, I'm following point A.1.1 not the 1.2 so it's not unity-gain version.

Ian0

Joined Aug 7, 2020
9,792
Hold on, I'm following point A.1.1 not the 1.2 so it's not unity-gain version.
But the screenshot you sent of Okawa IS the unity gain version.

kalemaxon89

Joined Oct 12, 2022
235
But the screenshot you sent of Okawa IS the unity gain version.
You are right! I should have scrolled down ... now I'll try the simulation again
Sorry for my oversight

kalemaxon89

Joined Oct 12, 2022
235
But the screenshot you sent of Okawa IS the unity gain version.

The choice of Q is arbitrary, if I want a Butterworth filter I will set Q = 0.707 .. for Bessel Q = 0.577 etc.
However, I don't understand why the simulator has that range limit for capacitor selection That causes the error in the screenshot

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LvW

Joined Jun 13, 2013
1,753
The choice of Q is arbitrary, if I want a Butterworth filter I will set Q = 0.707 .. for Bessel Q = 0.577 etc.
However, I don't understand why the simulator has that range limit for capacitor selection That causes the error in the screenshot
In theory, you can choose capacitor and/or resistor ratios as you like - as long as they are able to realize the desired filter function.
However, in practice there are some restrictions with respect to realization (tolerances and parasitic properties).
Do you think it would be desirable to have two capacitors in the filter circuit having a ratio of 100 or even 1000 (1nF and 1µF in one filter circuit) ?

In this context, the so-called "component spread" comes into play - and this number is one of the quality criteria which is used to compare several filter topologies and/or several strageties for the the dimensioning.

Example:
* S&K topology with unity gain: Very simple and exact gain realization (100% feedback), but relatively high component spread.
* S&K topology with equal components: Best component ratios (of unity) - however, two resistors are required for gain realization. And the gain (and with it the Q value) is extremely sensitive to the resistor ratio (which must be between 1 and 4). For a gain of 4 the amplifier oscillates.
* S&K topology with a gain of "2": Good tradeoff between gain realization (two equal resistors) and componenet spread (which is much better than for unity gain).

___________________________________________________
Similar considerations can be made to compare two different filter topolgies which each other.
In theory, for ideal amplifiers and ideal passive componenets all available structure alternatives give the same filter function.
However, in practice all the realization alternatives have different properties - in particular it is the sensitivity to tolerances.

Examples:
* S&K topology: Relatively good (small) "active" sensitivity to opamp non-idealities and tolerances and - at the same time - relatively bad (large) passive sensitivities to parts tolerances.
* Multi-feedback-topology (MFB): Relatively large active sensitivity figures and relatively small passive sensitivities.

The mentioned considerations are two of several aspects that are taken into account when selecting a certain filter topology for a specific application.

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kalemaxon89

Joined Oct 12, 2022
235
In theory, you can choose capacitor and/or resistor ratios as you like - as long as they are able to realize the desired filter function.
However, in practice there are some restrictions with respect to realization (tolerances and parasitic properties).
Do you think it would be desirable to have two capacitors in the filter circuit having a ratio of 100 or even 1000 ? (1nF and 1µF).

In this context, the so-called "component spread" comes into the game - and this number is one of the quality criteria which is used to compare several filter topologies and/or several strageties for the the dimensioning.

Example:
* S&K topology with unity gain: Very simple and exact gain realization (100% feedback), but relatively high component spread.
* S&K topology with equal components: Best component ratios (of unity) - however, two resistors are required for gain realization. And the gain is extremely sensitive to the resistor ratio (which must be between 1 and 4). For a gain of 4 the amplifier oscillates.
* S&K topology with a gain of "2": Good tradeoff between gain realization (two equal resistors) and componenet spread (which is much better than for unity gain).

Similar considerations can be nmade to compare two different filter topolgies which each other.
In theory, for ideal amplifiers and ideal passive componenets both are identical.
However, in practice both realizaton alternatives have different properties - in particular it is the sensitivity to tolerances:
* S&K topology: Relatively good (small) "active" sensitivity to opamp non-idealities and tolerances and - at the same time - relatively bad (large) passive sensitivities to parts tolerances.
* Multi-feedback-topology (MFB): Relatively large active sensitivity figures and relatively small passive sensitivities.

The mentioned considerations are two of several aspects that are taken into account when selecting a certain filter topology for a specific application.
Thanks for the information, but I still don't understand why the simulator has that range limit for capacitor selection.
In the example considered in post #4 I did the calculations and I don't think there are any problems inherent component ratios, in fact I found C1 = 10nF and C2 = 30nF (similar thing with resistors).
So I don't understand why the online simulator gives me that error (see post #9).

Maybe the problem is the quality factor Q = 2.37 I chose (randomly) in the example .. if I set Q = 0.7 everything works.
So I repeat .. why do the hand calculations look correct to me while this site does not?

LvW

Joined Jun 13, 2013
1,753
Maybe the problem is the quality factor Q = 2.37 I chose (randomly) in the example .. if I set Q = 0.7 everything works.
So I repeat .. why do the hand calculations look correct to me while this site does not?
How did you arrive at a value as large as Q=2.37?

kalemaxon89

Joined Oct 12, 2022
235
How did you arrive at a value as large as Q=2.37?
It is a randomly chosen number just because I wanted to test the simulator online; while hand calculations provide feasible capacitor resistor values, the simulator gives me error .. and having never implemented a filter I was wondering why.

LvW

Joined Jun 13, 2013
1,753
It is a randomly chosen number just because I wanted to test the simulator online; while hand calculations provide feasible capacitor resistor values, the simulator gives me error .. and having never implemented a filter I was wondering why.
When you want an helpful answer you should show us your "hand calculations" and the selected componenet values you have used to start the design.

Ian0

Joined Aug 7, 2020
9,792
It is a randomly chosen number just because I wanted to test the simulator online; while hand calculations provide feasible capacitor resistor values, the simulator gives me error .. and having never implemented a filter I was wondering why.
It needs two slightly different capacitor values that are quite close together. It's telling you to use E24 series capacitors.
You're lucky - I always get the error messages in Japanese.
It's also telling you (though not is so many words) that you need close tolerance capacitors to achieve a Q that high.

kalemaxon89

Joined Oct 12, 2022
235
It needs two slightly different capacitor values that are quite close together. It's telling you to use E24 series capacitors.
You're lucky - I always get the error messages in Japanese.
It's also telling you (though not is so many words) that you need close tolerance capacitors to achieve a Q that high.
Thank you!

Ian0

Joined Aug 7, 2020
9,792
Thank you!
I just fiddled about with the component values until it gave me a result!

kalemaxon89

Joined Oct 12, 2022
235
It needs two slightly different capacitor values that are quite close together. It's telling you to use E24 series capacitors.
You're lucky - I always get the error messages in Japanese.
It's also telling you (though not is so many words) that you need close tolerance capacitors to achieve a Q that high.
In fact, the simulator (in this specific case) must satisfy: G-1=1 < C1/C2 < 1.044=(1/2Q)^2 --> C1 = 1.044*C2

Papabravo

Joined Feb 24, 2006
21,225
If you look at a pole location diagram for a 2nd order Butterworth filter there is one and only one Q value that gives the desired response. The larger the value of Q, the closer the poles get to the imaginary axis. A Q of 2.37 locates the poles at an angle to the negative real axis of ±77.92°.
The response in this case will be a long way from maximally flat. This results in a great deal of peaking and very little damping.

If you want to play theoretical games with filters, I highly recommend using a simulator to simulate transfer functions in their standard form.

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Audioguru again

Joined Oct 21, 2019
6,688
The high-Q filter has a frequency with a peak level and it rings like a bell for a fairly long time because it is almost oscillating.
A Butterworth Q produces a sharp cutoff corner but rings a little.
A Bessel low-Q produces no ringing but the cutoff corner is droopy with a dull corner.