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How do I calculate the cutoff frequency and the gain of a second order sallen key low pass filter? For example if I want to have a cutoff 35KHz what would be the values of the two resistances and the two capacitors

enter image description here

enter image description here Using R1=R2=4700 as it was suggested

As you can see with those values , I have succeeded in creating the low pass filter I want , but the cutoff frequency is very high (I needed 35kHz, the graph is voltage to frequency) . What should I change?

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  • \$\begingroup\$ What gain do you want, exactly? \$\endgroup\$ – jonk Dec 29 '17 at 18:54
  • \$\begingroup\$ @jonk I edited the pictures , By mistake I uploaded a schematic without values. The gain is not the problem at the moment , it is something I will need in the future . My problem now is the cutoff frequency being very high \$\endgroup\$ – maverick98 Dec 29 '17 at 18:56
  • \$\begingroup\$ Sallen Key topology is different if you want gain, or don't. And if you want equal valued components, or not. I suppose it also helps to know the damping you want, as well. One could just toss you a fish, rather than teach you how to fish. But that would be just doing a design for you. Wouldn't it? \$\endgroup\$ – jonk Dec 29 '17 at 19:00
  • \$\begingroup\$ @jonk I am just asking for help , you can clearly see I put in the work and I am not asking others to do my work for me \$\endgroup\$ – maverick98 Dec 29 '17 at 19:02
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    \$\begingroup\$ @Maverick98 If you just want to build a filter, I suggest using using a filter calculator. Check out: analog.com/designtools/en/filterwizard (for example). TI and Linear also have them. \$\endgroup\$ – pgvoorhees Dec 29 '17 at 19:04
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Here's an example run from LTSpice for a very simple unity gain, equal component value Sallen Key filter.

enter image description here

The values used are common values rather than exact values to hit your frequency precisely. But they get close.

Looks about right to me.


Here's the chart done differently:

enter image description here

That may also help you.


The reason why the -6 dB point is selected as the "cross-over" has a lot of context. I'm neither competent to explain it fully nor do I have the time to try. I know a few things, is all.

But I can summarize the basics:

  1. It's convention and people will understand you better if you use terms they know in the way they know them.
  2. If you look at the first graphic I posted here (I suppose, in a way I'll explain later on, the 2nd one also shows this), you can see that the output (the solid line) stays "flat" for a while. Then it goes through a transition period. Then it seems to follow a fairly straight line downward. It would be nice to find a way to select a point in the transition period that helps delineate between the flat spot before and the sloped part after. It turns out that the "equidistant" midpoint is the -6 dB point for voltage in a low pass filter.

The filter leaves the input alone (is flat) up until some point. In the first chart shown above, it is pretty flat until it nears \$20\:\textrm{kHz}\$. Then it starts to turn. The turn is finished by the time you get to about \$60\:\textrm{kHz}\$. Once you are there, it's a straight line down at a rate of -40 dB per decade of frequency (for a 2nd order filter.)

The half-voltage point, or -6 dB voltage, is the center of the transition period. And people share this meaning when they speak of filters like this.

I like the 2nd chart I added above because it makes this point in mathematical fashion. Look at the shape of that curve. It is downward curving (2nd derivative is negative) until it reaches some frequency. Then, although continuing to decline, it is upward curving (2nd derivative is positive.) The -6 dB point is exactly where the 2nd derivative transitions from negative to positive -- and hits zero. This is the mathematical reason why this point was chosen.

So this special corner point has mathematical reasoning, visual reasoning, and convention to support its use.

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  • \$\begingroup\$ But from the graph of Vout , how can we see the cutoff frequency \$\endgroup\$ – maverick98 Dec 29 '17 at 19:27
  • \$\begingroup\$ @Maverick98 Since this is voltage out, you look for the -6 dB point (and/or where you see -90 degrees.) \$\endgroup\$ – jonk Dec 29 '17 at 19:28
  • \$\begingroup\$ oh I see , compared to my graphs what has changed, is it possible to replace the db on the left with voltage so I can compare? \$\endgroup\$ – maverick98 Dec 29 '17 at 19:30
  • \$\begingroup\$ @Maverick98 \$10^{\frac{-6}{20}}\approx 0.5\$ so you are looking for where the voltage drops to half. \$\endgroup\$ – jonk Dec 29 '17 at 19:31
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    \$\begingroup\$ @Maverick98 There are lots of good reasons. It would take a chapter or two to put them into context. I'll add some notes to my answer, though. Not much, just a little. But perhaps enough to help out on this question. \$\endgroup\$ – jonk Dec 29 '17 at 19:37
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The cutoff frequency is calculated using the following equation.

\begin{equation} f_c = \frac{1}{2 \pi \sqrt{R_1 R_2 C_1 C_2}} \end{equation}

Using your current values, you have a cutoff frequency of 338627.5 Hz, which is about ten times higher than desired. Sticking with \$ C_1 = C_2 = 1nF \$ and \$ R_1 = R_2 \$, you should use \$ R_1 = R_2 = 4.7k \Omega \$ to (approximately) get your desired cutoff frequency.

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  • \$\begingroup\$ I have actually tried that , the thing is the graph is worse than what I posted , see my edit \$\endgroup\$ – maverick98 Dec 29 '17 at 19:05
  • \$\begingroup\$ Actually, it's not worse at all ! Your cutoff frequency is about 34kHz on your graph. \$\endgroup\$ – altai Dec 29 '17 at 20:02
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Lets assume you are trying to design a Butterworth filter which is what's commonly used for audio work to get a maximally flat pass band and a roll off of 6dB/octave/pole (20dB/decade/pole).

Your current design doesn't fall into any standard filter category, neither Butterworth nor Bessel nor Chebychev. It has a Q of 0.5 which is even lower than a Bessel filter giving a gentle corner rolloff. If you want to design an equal value Butterworth filter (R1=R2, C1=C2) then you have to give the op amp a non-inverting gain of 1.586 which raises the Q to a Butterworth specification of 0.707. In this situation the cut-off frequency = 1/(2*PIRC).

If you want to have unity gain (as in your design) then for a low pass Butterworth response (Q = 0.707) you must set:- R1 = R2 and C2 must equal 2*C1 (using your label designations)

and the cut-off frequency now equals 1/(2*PI*Sqrt(R1*R2*C1*C2))

If you require a low pass unity gain filter with a cut-off frequency of 35KHz then I would suggest :- C1 = 1nF C2 = 2nF (2 1nFs in parallel) and R1 = R2 = 3K3

These values give an fc of about 34kHz where fc is the -3dB point (frequency where the output amplitude is 0.707 times the input amplitude.

The 0.707 is coincidentally the same value as the Q of a Butterworth filter.

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