I have designed this Sallen-Key to give a 2nd order low pass circuit with a cut-off frequency of 150Hz. Another design specification is that R1=R2 and C1=C2. The issue is when simulated on LT-Spice this sallen-key returns a low pass bode plot with a cut off frequency of around 100Hz. Have I calculated the resistor and capacitor values incorrectly? What could be causing this issue? enter image description here

  • \$\begingroup\$ Why don't you talk a little about your calculations rather than your results? How did you arrive at your values? For example,, here's my example of designing a band-pass filter using a low-pass and a high-pass combination. Can you show us your work, like that? (And yes, I can see about 159 Hz.) Note that voltage is different than power. What do you read when you find -6 db? If half power is -3 db power, what is the voltage down by? \$\endgroup\$ – jonk May 24 '20 at 2:43
  • \$\begingroup\$ Sorry can you explain why I would go to -6dB I am a bit confused. Also I am working on getting my methodology for you. \$\endgroup\$ – madge2111 May 24 '20 at 2:55
  • \$\begingroup\$ Decibels has always been and always will be a power specification. Not a voltage one. It can be adapted to voltage, knowing the concept that power is proportional to voltage-squared. But the "squared" part of this, applied to logarithms, means "times 2." Aren't you aware of this? But there is an alternative way to look at this, that's actually better, though. If you look at the phase plot, the inflection point also carries similar meaning. Where do you see the phase inflection taking place? (2nd derivative equals zero.) \$\endgroup\$ – jonk May 24 '20 at 3:05
  • \$\begingroup\$ (By the way, \$2.2 \: \text{k} \Omega\$ and \$470 \: \text{nF}\$ might be a little closer.) \$\endgroup\$ – jonk May 24 '20 at 3:09
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    \$\begingroup\$ What filter alignment are you aiming for? If you chose an overdamped filter. the response will be well below -3dB at the pole frequency (while a sufficiently underdamped filter will actually peak) \$\endgroup\$ – user_1818839 May 24 '20 at 14:36

With equal values for both resistors and capacitors, the Q of the circuit is 0.5. At the natural resonant frequency of 2nd order filters, the Q value is the magnitude of the transfer function. So, at 159.15 Hz, the magnitude of the transfer function is half or down 6 dB. I expect that at around 100 Hz, the magnitude of the transfer function is down 3 dB.

Another design specification is that R1=R2 and C1=C2.

This will always mean a Q factor of 0.5 (critical damping = 1) and so, you would have to reduce both capacitors to produce a higher natural resonant frequency in order to get a 3 dB point at 150 Hz. It's the same for an RLC low pass filter too: -

enter image description here

In the design I've chosen R = 200 ohms, L = 100 mH and C = 10 uF to achieve a natural resonant frequency of 159.15 Hz and a Q of 0.5. In the upper graph I've position the cursor at 159 Hz and you can see that the magnitude of the transfer function is -6 dB.

In the lower graph I've moved the cursor to around 100 Hz and revealed that the magnitude is more like -3 dB.

Pictures from this interactive tool.

  • \$\begingroup\$ .....for my opinion, the term "natural resonant frequency of 2nd order filters" can lead to misunderstandings (because...in particular, for low-Q circuits there is no "resonance" effect). I think, we should use the term "pole frequency" instead which also appears in the classical transfer functions. But it is true and it is good that you have mentioned the fact that at w=wp the magnitude of a unity-gain lowpass is identical to the Qp value (in this caase Qp=0.5). This also explains the -6dB drop at w=wp. \$\endgroup\$ – LvW May 24 '20 at 11:04
  • \$\begingroup\$ @LvW I understand your point but I still prefer to use \$\omega_n\$. \$\endgroup\$ – Andy aka May 24 '20 at 11:10
  • \$\begingroup\$ Andy aka.....the denominator of a second-order transfer function is [1+s/(wpQp)+(s/wp)²] and for s=jwp this expression reduces to 1/Qp.. Hence, the magnitude of the transfer function at w=wp is Ao*Qp. The term wp is the so called POLE frequency (which is the magnitude of the pointer to the pole). And the term wn is the imaginary part of this pointer (wp=sigma+jwn). The term wn gives the frequency of damped oscillation in case of large Qp values (natural frequency). Therefore, the magnitude of the transfer function gives the Qp value at the frequency w=wp (and not at w=wn). \$\endgroup\$ – LvW May 24 '20 at 17:04
  • \$\begingroup\$ I disagree. The natural resonant frequency is a circle in the pole zero diagram and not fixed to the jw axis. \$\endgroup\$ – Andy aka May 24 '20 at 17:49
  • \$\begingroup\$ Andy aka...so I do not know what do you mean with "natural resonant frequency"...what is the definition? I only know the following two: (1) Pole frequency wp (magnitude of the pointer to the pole p) and the terms for the real and imag. parts of this pole: p=(sigma+jwn). (2) This defines the "natural frequency" wn=Imag. part of the pole p. Example (passive RLC lowpas): wp=SQRT(1/LC) and wn=SQRT[1/LC - (R/2L)²] . The frequency wn gives the frequency of damped oscillations. The pole is p=-sigma (+-) jwn with sigma=-R/2L. What is your definition? \$\endgroup\$ – LvW May 25 '20 at 9:39

If you're trying to achieve a Butterworth response then you can't have same value components with unity gain.

Either make your C1 twice the value of your C2 and use the formula below or give the op amp a gain of 1.586.

Unity gain B'worth filter

B'worth filter

  • \$\begingroup\$ Does he speak about a Butterworth filter? \$\endgroup\$ – LvW May 24 '20 at 9:29

For the shown filter stage with equal components and a unity gain amplifier the Qp-value (pole quality factor) is Qp=0.5 and the POLE FREQUENCY is simply wp=1/RC=1000 rad/s or fp=159.2 Hz. This is the frequency where the phase shift is excatly -90 deg .

However, the pole frequency is identical to the 3dB cut-off frequency fo for Qp=0.7071 only (Butterworth response).

For Qp-values below 0.7071 (as in the present case) the 3dB cut-off fo is always lower than the pole frequency fp.

You should use a BODE diagram with better resolution (down to -10....-20 dB only). Then you will see that the display will be in accordance with these calculations.

Remark: In your case with Qp=0.5 there is a double real pole on the neg.real axis of the s-plane. For a SINGLE pole we have a 3dB cut-off at wp=wo. Hence, for a double pole we have -6dB at wp=1000 rad/s (fp=159. Hz)


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