I designed a 2nd order Butterworth low-pass filter, but it did not work correctly. What is the mistake in this filter?

enter image description here

  • 5
    \$\begingroup\$ Use electronics units. u for microfarad. Did you supply the op-amp? Supply voltage at output? LTC does not like this. \$\endgroup\$
    – Antonio51
    Jun 1 at 13:53
  • 5
    \$\begingroup\$ Missing supply voltages (+ and -) to the opamp. \$\endgroup\$
    – winny
    Jun 1 at 14:02
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    \$\begingroup\$ V2 has no voltage, U1 isn't connected to a power supply. \$\endgroup\$
    – Aaron
    Jun 1 at 14:17
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    \$\begingroup\$ @tryingtobeanengineer I don't know if there is an LTC6752. There is a chopper stabilized LTC7652, though. Is that what you meant? \$\endgroup\$
    – jonk
    Jun 1 at 18:23
  • 2
    \$\begingroup\$ On the device LTC7652 , there is a pin (+) and (-) left unwired which have to be wired to supplies. So, choose a voltage generator as your V2 (disconnect it of its actual place ... it is misplaced) and connect it between the (+) and (-). Choose also its voltage to 5V for example. Wire also the (-) to ground if only one supply is needed. \$\endgroup\$
    – Antonio51
    Jun 1 at 18:30

3 Answers 3



I don't like using applets to solve problems like this. I much prefer going back to the basics fir the various filter equations, so that I know the source of where things come from. It's one thing to let some mysterious process create numbers for you even if that does work. It's another thing to be able to sit down and work out the details for yourself.


Butterworth was a kind of super-genius with mathematics. So I won't dig a lot into how it came to be. You can read the Wiki page on Butterworth for some details.

But the basic visualization for Butterworth is to imagine a unit circle on the \$\left(\sigma,j\omega\right)\$ coordinate system:

enter image description here

(Above is Figure 2.10 from page 2-11 of Wai-Kai Chen's "Passive, Active, and Digital Filters" from the 3rd edition of the series, The Circuits and Filters Handbook.)

The above shows the case for a 2nd order Butterworth. There are four, not two, but four poles. These points can be found by solving the following equation for \$x\$ (where \$N\$ is the order -- in this case \$N=2\$):

$$ 1+\left(-1\right)^N x^{2N}=0 $$

This solves out as \$x=\pm \frac12\sqrt{2}\pm j\frac12\sqrt{2}\$. We don't care about the right-hand side or anything exactly at \$\sigma=0\$. Those parts are just dumped. The resulting poles of interest arrive from whatever remains. In this case, \$x=-\frac12\sqrt{2}\pm j\frac12\sqrt{2}\$ (the left-hand side.)

If you take these two values for the 2nd order case, then you can find the characteristic by \$\left(s-\left(-\frac12\sqrt{2}- j\frac12\sqrt{2}\right)\right)\cdot\left(s-\left(-\frac12\sqrt{2}+ j\frac12\sqrt{2}\right)\right)=s^2 + \sqrt{2}\:s + 1\$

So, the analytic form of the characteristic equation for a 2nd order Butterworth is \$s^2 + \sqrt{2}\:s + 1\$.

(As we are working with the unit circle right now, this is where \$\omega_{_0}=1\$.)

Low-Pass 2nd-order, given \$\omega_{_0}\$

The low-pass transfer function form that includes \$\omega_{_0}\$ is \$A\cdot\frac{\omega_{_0}^{\:2}}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^{\:2}}\$. (In the literature, \$A\$ can also be seen as \$k\$ or \$h\$. It's the voltage gain.) Your diagram shows that the voltage gain you want is \$A=1\$ (\$0\:\text{dB}\$.)

Returning to the Butterworth

From the analytic expression for the Butterworth, \$s^2 + \sqrt{2}\:s + 1\$, we can now see that \$2\zeta=\sqrt{2}\$ and therefore the unitless damping factor for a 2nd order Butterworth is \$\zeta=\frac{\sqrt{2}}{2}\approx 0.7071\$ and therefore also \$Q=\frac{\sqrt{2}}{2}\approx 0.7071\$.

(A problem with Butterworth is that it is pure mathematics and doesn't deal with non-linear minimax optimizations, which can cope better with part variations/errors. This is where Chebyshev finds its place. But that's for another time.)

Your Sallen-Key Low-Pass Topology

The transfer function for your topology tells me that \$\omega_{_0}=\frac1{\sqrt{R_1\, R_2\, C_1\, C_2}}\$ and \$\zeta=\frac12\left(R_1+R_2\right)C_2\,\omega_{_0}\$. I see you chose \$R_1=R_2\$ (doesn't matter the value for now), so this implies (you can work out the details given that \$\zeta=\frac{\sqrt{2}}{2}\$ and the aforementioned \$\omega_{_0}\$) that \$C_1=2\, C_2\$.

So we have \$R=R_1=R_2\$ and \$C_1=2\, C_2\$. From that, we can find that \$C_1=\frac{\sqrt{2}}{R\,\omega_{_0}}\$ and that \$C_2=\frac1{\sqrt{2}\, R\, \omega_{_0}}\$.

Plugging in your existing resistor and capacitor values (have to take it on faith, for now), I find \$f_{_0}=768.228253\:\text{Hz}\$ or \$\omega_{_0}=4826.92047\:\frac{\text{rad}}{\text{s}}\$. Assuming you keep \$R=R_1=R_2=1\:\text{k}\Omega\$, then find \$C_2\approx 146.5\:\text{nF}\$ and \$C_1\approx 293.0\:\text{nF}\$.

Note that this means your capacitor values were swapped, just as Andy said they were.

But the wonderful thing now is that you can see exactly how that is true, rather than seeing it from some applet that uses formulas but fails to explain how they were developed or even what they are, in particular.

The above allows you to fish. The applet gives you a fish. I prefer knowing how to fish.


It also appears that you don't know how to provide power supplies to Spice simulations. You'll need to work that out. It's important. But that's a separate question, really.


Finally, winding up with \$C_1=2\, C_2\$ can be inconvenient. As it turns out, for this topology it must be the case that \$C_1\$ is at least twice as large as \$C_2\$ (this can be proven from the above.) But as is often the case, you don't find capacitor values that are exactly double another's value. Instead, for example, you might have \$22\:\text{nF}\$ and \$10\:\text{nF}\$. So, instead of 2 shown here, you'd use 2.2 as the factor. And then use this factor to work out the resistor values, from there. Resistors come in a much greater array of value choices, so focusing first on the capacitors is probably better.

If you label the capacitor ratio as \$\gamma=\frac{C_2}{C_1}\$, where \$0\lt \gamma \lt \frac12\$, then the ratio of the two resistors, \$\rho=\frac{R_1}{R_2}\$, is also established for this Butterworth filter. (\$\rho=\frac{\gamma^2-4\gamma+2+2\left(1-\gamma\right)\sqrt{1-2\gamma}}{\gamma^2}\$.) In general, you want to keep \$\gamma\$ as close to \$\frac12\$ as possible, though. So this constrains the capacitor choices.

For example, if I select \$C_1=22\:\text{nF}\$ and \$C_2=10\:\text{nF}\$ then \$\gamma=\frac5{11}\$ and \$\rho=\frac{12\sqrt{11}+47}{25}\approx 3.472\$. With \$f_{_0}=1\:\text{kHz}\$, I'd find the values of \$19.994\:\text{k}\Omega\$ and \$5.759\:\text{k}\Omega\$ for the two resistors (either one can be either value, as it turns out, though there will be other issues based upon the realistic opamps that might be applied here.)

Let's look at this schematic for the 2nd order low-pass Butterworth (as documented by Sallen-Key):

enter image description here

Here's the response:

enter image description here


My use of \$\gamma\$ and \$\rho\$ corresponds directly to the use in Sallen & Key's TR-50 paper from 1954:

enter image description here

Here is the appropriate low-pass filter topology they list in their catalog:

enter image description here

As you can see, they offer four different pairings of design parameters for the same topology which are linked to their design formulas:

enter image description here enter image description here

Note that their use of the term, dissipation factor (\$d\$), is now antiquated. Modern teaching uses the damping factor, \$\zeta\$. For conversion purposes, \$d=2\zeta\$.

  • 2
    \$\begingroup\$ I understand what you say about applets because mostly they are unreliable. But in all fairness to the Okawa Denshi site I linked, I've been using it for years so, on this occasion it's one with a good history of being right. \$\endgroup\$
    – Andy aka
    Jun 1 at 20:13
  • \$\begingroup\$ Thank you so much for the advices, I changed the capacitor values but it must be C1=2C2 because it is a final project and teacher gave us these values. I am trying to improve my knowledge in LTSpice, again, thank you so much! \$\endgroup\$ Jun 1 at 20:15
  • \$\begingroup\$ @Andyaka I use applets, too. :) However, I also try to separately require from myself the ability to solve problems even if stranded on a desert isle, using nothing more than my fingers and sand. I very much dislike depending upon tools that may, or may not, be around when I need them. I always have my brain and my hands (until I'm dead.) Also, there's much to be said for understanding something than not understanding the same something. One is a tool deigner. The other is a tool user. Very different things. \$\endgroup\$
    – jonk
    Jun 1 at 20:15
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    \$\begingroup\$ Yeah, I wouldn't use a tool without knowing theory @jonk hence why I sometimes write my own !! \$\endgroup\$
    – Andy aka
    Jun 1 at 20:48
  • 1
    \$\begingroup\$ @tryingtobeanengineer If you want to double the value of a capacitor then just put two of the same value in parallel. \$\endgroup\$
    – James
    Jun 2 at 7:03

I designed a 2nd order Butterworth low-pass filter below the image but it did not work correctly, what is the mistake in this filter?

You have the two capacitor values swapped. If you had C1 = 290 nF and C2 = 148 nF then you'll get a Q-factor of very nearly 0.7. Not quite a Butterworth response but nearly right: -

enter image description here

Image/tool from this website.

Try C1 = 295 nF to get a closer-to-Butterworth true response. Also, make sure your device is correctly powered.

Also note that the LTC6752 is a comparator and not an op-amp and may produce some unusual effects as a linear filter: -

2.45V to 5.25V Input Supply and 1.71V to 3.5V Output Supply (Separate Supply Option)

I'm sure you might want to think about choosing a more suitable device.

You also have a voltage source (V2) connected to your device output pin and that is wrong - you need to let that output node be "free" and not have it connected to a voltage source.

  • \$\begingroup\$ Thank you so much for the advices! I changed capacitor values but I cannot change the LTC6752 because teacher wants to be this. I updated my circuit, thank you so much! \$\endgroup\$ Jun 1 at 20:17

I don't think you can use a comma for a decimal in LTspice. It doesn't work in my installation. Perhaps locale affects it but I wouldn't count on it, you should use a decimal point.

Also, if you want to use scientific notation for values you can just use an E like this

4.7E3 = 4.7K
4.7E-6 = 4.7u

For capacitors you should use u or uF for microFarads, n or nF for nanoFarads, and p or pF for picoFarads.

I use the scientific notation for large numbers for example if I was doing an AC analysis to 10 Gigahertz I would use 1E9.

Megohms should use Meg not M as LTspice is not case sensitive and M will be the same as m which is milli, or 1/1000th.

In LTspice:
1Meg = 1 million
1M = 1 milli

  • \$\begingroup\$ Thank you for information :) ! \$\endgroup\$ Jun 1 at 20:12

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