I'm trying to figure out the R and C elements for a second-order Bessel filter approximation. Unfortunately, in the below calculations the quality factor is negative, so it is wrong? (I checked by entering the values ​​into this design tool.)

Also, the poles have a positive real part.

In my calculations, I used the tips from this video.

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I often come across the concept that the coefficients in the filter table are for the normalized form of the filter. What does that mean? I read somewhere that they were designated for RC = 1 (or it depends?). In the above example I assumed RC = 0,7861. Is that true?

What does k = fci / fc mean in the table of coefficients? How to take into account the fci/fc parameter in calculations?

How to calculate 3rd-order Bessel low-pass filter?

I am new to this forum. Can anyone tell me how to use equation formatting?


2 Answers 2


the quality factor is negative, so it is wrong?

You have chosen to have a gain greater than three (\$1+\frac{R4}{R3}\$). Sallen Key circuits cannot have a gain equal to or greater than 3. I expect this is why the Okawa Denshi web site calculator got into a quandary. Try making R4 = R3 and see what happens. If you make R3 = 501 Ω (AV = 2.996) then you get a massive Q factor of 250.5 and, if you drop R3 to 500 Ω, Q becomes infinite.

The maximum gain (AV) of 3 comes from this relationship: \$A_V = 3 - \frac{1}{Q}\$

What does k = fci / fc mean in the table of coefficients?

When you have a 3rd order or fourth order filter, there are two distinct filter cut-off frequencies; one for the first stage and one for the second stage.

I am new to this forum. Can anyone tell me how to use equation formatting?

\$\dfrac{\text{top}}{bottom}\$ = \$\dfrac{\text{top}}{bottom}\$

That is just an example but there are many more examples to be found.

I am new to this forum

This isn't a forum; it's not a talking shop; it's not a progressive teaching aid; it's a question and answer site; you ask a question and someone hopefully answers it.

  • \$\begingroup\$ Of course, a stupid mistake. Can you also refer to my question regarding k = fci / fc? How to use this information in calculations, what does it mean? Another question is how this table of coefficients was determined, where do they come from? \$\endgroup\$
    – grandzello
    Sep 10, 2022 at 12:40
  • \$\begingroup\$ I've added the answer to your first question into my answer. Regarding where the table coefficients come from you'll have to raise this as a new question and, I will be as eager as you are to read the answers. It's not like a Butterworth response where that is easy (well, relatively easy) to formulate the pole positions @grandzello \$\endgroup\$
    – Andy aka
    Sep 10, 2022 at 12:52

You ask about a 3rd order Bessel yet you show a 2nd order schematic. First you need to determine the transfer function based on your requirements. And, since this is a Bessel filter then you should know that, usually, these filters are not built for their -3 dB corner frequency but for their flat group delay.

Because you are using that table, which comes from "Opamps for everyone, Ron Mancini" (freely available, last I checked), I'll assume you are looking for a Bessel filter that satisfies a -3 dB @ 225 Hz (as per the poles displayed in your results from the Okawa-Denshi site).

First, start with the 3rd order lowpass prototype Bessel (see this for a table of coefficients):

$$H(s)=\dfrac{15}{s^3+6s^2+15s+15} \tag{1}$$

To find out the -3 dB point you need to solve for \$\omega\$ in:

$$|H(j\omega)|^2=\dfrac12\quad\Rightarrow\quad\omega=2\pi 1.756\;\text{rad}\quad\Rightarrow\quad f_c=1.756\;\text{Hz} \tag{2}$$

Since it's a 3rd order you'll get 6 values, but 4 of them will be complex conjugate, while the 5th will be negative; so, the 6th remains (2). Alternatively, you can use a root-finding algorithm. Since the circuit you're showing is only a 2nd order you need to factor the transfer function into 2nd order stages, with an additional 1st order if the order is odd. For this case it's a 2nd order and a 1st order, and you do that by solving for the roots:

$$\begin{cases} s_{1,2}&=-1.8389\pm j1.7544 \\ s_3&=-2.3222 \end{cases} \tag{3}$$

These poles need to be scaled with the value of \$f_c\$ from (2), after which the transfer function is recreated to have the 2nd and 1st order stages:

$$\begin{align} &\begin{cases} s'_{1,2}&=-1.0472\pm j0.9991 \\ s'_3&=-1.3224 \end{cases} \tag{4} \\ A(s)&=\dfrac{1.3224}{s^2+1.3224} \\ B(s)&=\dfrac{2.0948}{s^2+2.0944s+2.0948} \\ H'(s)&=A(s)B(s) \tag{5} \end{align}$$

The second part is what you can use with the circuit you provided, by first considering that the \$2.0948\$ term in \$B(s)\$ represents \$\omega_{\text{2nd}}^2\$, therefore \$f_{\text{2nd}}=\sqrt{2.0948}\cdot 225=325.65\;\text{Hz}\$. It is this frequency that you need to use for the filter, together with the quality factor calculated as:


The problem comes with the gain for that particular topology (Sallen-Key): it's restricted due to its transfer function, you can see that the term \$k\$ appears in the coefficient for the \$Q\$.

Therefore, you could try a different topology, for example the multiple-feedback (on the same site) which is also more stable. You should know that Sallen-Key has a particularly annoying "feature": at high frequencies, the feedback capacitor has a low impedance and the output appears more and more at the input, causing a high-pass response. I warmly recommend simulating that, because words do not do it justice. I'll leave it to you to discover.

Another reason why I propose the multiple feedback is because now you can keep the whole gain, 4.75 (as it appears from your circuit), which means that for the 1st order stage you don't need another opamp and, instead, you can simply use an RC lowpass. You'll ned to be careful with the values to avoid the loading effect but, it can be done to a certain extent. So, for \$A(s)\$, the frequency needs to be \$\sqrt{1.3224}*225=258.74\;\text{Hz}\$.

Finally, with that much talk out of the way, you can build the filter:

3rd order Bessel, -3 dB @ 225 Hz

I've chosen a multiple-feedback because the gain would have been impossible for Sallen-Key. There are two ways to do it with only one opamp: the unbuffered 1st stage is at the input (V(out1)), or at the output (V(out2)). Compare how they fare against the ideal transfer function (V(test)). Due to the loading effect, out1 is slightly distorted (look at the group delay, it has a slight peak), whereas out2 is better behaved but, only because it has no load. If, however, that load happens to be a high impedance input (such as another opamp) then you're scott-free because out2 and test are very similar. The gain might be slightly different.

  • \$\begingroup\$ Also, here is an excellent resource for MathJax on the SE sites. \$\endgroup\$ Sep 10, 2022 at 15:45

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