Designing Sallen-Key second-order bandpass filter

I am trying to design a Sallen-Key bandpass filter with a center frequency of 64 Hz, k = 1, Q = 1.8, and H0BP = 1.

How do I approach finding the resistor values that will produce these results? So far I have set both capacitors equal to each other which reduces the problem to figuring out the three resistor values that will produce the right Q, ω0, and H0BP.

Update: sorry I did not post a topology for the BPF. I still have not figured it out but the topology I am working with I believe is called the KRC type. Here it is:

• When you write "Sallen-Key" together with 2nd order and bandpass, you need to write out the exact diagram so that we can validate that it is one of those described by Sallen & Key. I know, off-hand, of one such that has three resistors and two capacitors and has K+1=9Q^2, which does NOT fit your case. This can be broken up into two opamps with different details. So I need to see what you are looking at.
– jonk
Commented Oct 31, 2022 at 5:45
• analog.com/media/en/training-seminars/tutorials/… Commented Oct 31, 2022 at 8:24
• There is a Sallen-Key bandpass (I think it's very similar to Deliyannis) but, it doesn't have a particularly appealing transfer function and, usually, the multiple-feedback topology is a better choice (or any other, really). Otherwise, if the equations you posted are true then you already have three equations with five unknowns. Impose any two of the RC elements and you can solve them for gain, frequency, and quality factor. Commented Oct 31, 2022 at 10:51
• May I ask you - what do you mean with "....it doesn't have a particularly appealing transfer function " ?
– LvW
Commented Oct 31, 2022 at 11:07
• @LvW Personally I just saw others using the @ in front of the names when replying so, when I tried it, as soon as I added the first letter of the desired name, a list appeared. Using <TAB> was just one of the keys I tried and worked. As it turns out, it's not always needed: if @<first_letter> doesn't bring up any names (with or without <TAB>) then it means that @ is not needed (the notification will go automatically). Or when the comments are below your question/answer. Otherwise, the way <TAB> "conjures up" the names is the way to use them: without spaces. Commented Oct 31, 2022 at 19:49

From your expression for wo it is possible to derive the circuit topology you have in mind (one grounded capacitor and a grounded resistor directly at the non-inverting input of the fixed positive-gain amplifier).

However, in order to fully describe your design approach it would be really necessary to write down the corresponding transfer function.

Please note that 3 typical design alternatives do exist (rather simple to design). After inserting the chosen value for K, the equations will look much simpler.

• Equal components (C1=C2 and and R1=R2=R3): In this case, the fixed gain K must be smaller than K=4 (oscillation for K=4). This alternative is NOT recommended. More than that, The DC gain cannot be chosen to be independent on Q.

• Unity gain K=1: Again, two equal capacitors C1=C2 are possible. The value spread for the resistors depends on the Q-value and seems to be acceptable for Q=1.8.

• Gain K=2: The "spread" of the resistor values is much better for the case K=2. This can simply be realized with two equal resistors in the opamps feedback path. Again, both capacitors can be selected to be equal (C1=C2).

bandwidth(-3dB) = f(centre)/Q = 64/1.8 =35.5 Hz

Therefore fcL(-3dB) = 46.2 Hz and fcH(-3dB) = 81.7 Hz

A rule of thumb is that if fcH/fcL > 1.5 then you can cascade 2 standard Sallen-Key filters to obtain the band pass. If fcH/fcL < 1.5 then a different approach is needed.

fcH/fcL = 81.7/46.2 = 1.77 which is > 1.5 and therefore I would say that it is possible to cascade a low-pass S&K followed by high-pass S&K to obtain the required band-pass.

Assuming that you require -12 dB/octave (-40 dB/decade) roll-off, then you need to design a 2-pole, low-pass, unity gain S&K filter with a -3 dB cut-off frequency of 81.7 Hz and a 2-pole, high pass S&K unity gain S&K filter with a -3 dB cut-off frequency of 46.2 Hz.

To have the filters -3dB down at the pole frequency (wn) you require Butterworth filters with their Q of 0.707 (gain of 0.707 at the pole frequency), maximally flat response and roll-off of -12 dB/octave for a 2-pole topology.

This, to me, presents itself as an easier task than trying to deal with those horrendous looking equations which you have posted.

As a design guide to 2-pole Butterworth filters...

• Just a note. Those you show are high pass and low pass filters. Not band pass. Cascading (2) 2nd order doesn't provide a 2nd order band pass. I believe the OP is asking about the case for a 2nd order Sallen-Key band pass. Those are slightly different in topology than you show and, in fact, come in a variety of forms as discussed by Sallen & Key. The problem I have with the OP's question is that the topology isn't provided. Final note: with Q=1.8 the 2nd order band pass filter is NOT a Butterworth. The 2nd order characteristic equation for Butterworth cannot have Q=1.8.
– jonk
Commented Oct 31, 2022 at 6:36
• @Jonk My approach is to create a band-pass using sallen & Key topology making use of a practical approach which simplifies the problem. I have interpreted the Q specification for the band-pass as meaning (centre frequency)/bandwidth rather than how I have used the Q specification for the individual high and low pass filters as being equal to the gain at the pole (and cut-off) frequencies. Obviously the Q (and damping ratio) of a Butterworth filter must be equal to 0.707 as I have clearly stated. Perhaps you could read my answer again.
– user173271
Commented Oct 31, 2022 at 6:51
• @spiderman19 ti.com/lit/an/sboa229/…
– user173271
Commented Oct 31, 2022 at 8:07
• From the task description it is clear the the TO wants to design a second order bandpass. This is the "smallest" order for a bandpass (as a result of applying the lowpass-bandpass transformation for a 1st-order lowpass). A series connection of a 2nd- order lowpass with a 2nd oder highpass will result in a 4th-order bandpass (40dB/dec asymptotoc slope).
– LvW
Commented Oct 31, 2022 at 9:01
• @LvW Sorry for the delay in replying, I've been out all day. Have a look at the ti link which I posted earlier. It seems that you, like the +voters for your comment, are unaware that a band-pass formed from two 2-pole filters can also be referred as a 2nd order filter.
– user173271
Commented Oct 31, 2022 at 20:58

Don't know where your formulas come out ...

Sallen-Key structure with one op-amp.

Here is the result from a note MT222 from Analog Devices.
Shown the pictures with H as parameter, EE&O ... to be "more" checked.

• In principle, there are three differnt S&K bandpass topologies. The TO has chosen another (often used) alternative. The problem with your topology is the sensitivity to the gain-determining resistors R4 and R5.
– LvW
Commented Oct 31, 2022 at 17:34
• Ok. It was just to show the results of this schematic (no sensitivities to components values calculated). What is the schematic used? Commented Oct 31, 2022 at 18:33
• Let me describe the bandpass alternative as follows: The capacitor C2 in your drawing is connected between ground and the node between R1 and C1. Thats all!.
– LvW
Commented Nov 1, 2022 at 8:06

I used your formulas for searching ... value of capacitor.
I made the assumption that all resistors are same value.
I solved HObp for K value. Then use it for solving for ... CC.
I found the values of the capacitors. Note that Q is "weird". Did not calculated it.

Used than all values in my "function" for that circuit.
Found this ... Did not calculated Q.