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I am trying to figure out how to derive this second-order op-amp circuit.

I understand that I should be trying to find Vo/Vi, but I’m endup calculating a function that leaves me with a bandwidth much higher than the listed specification.

my working of deriving the op-amp circuit

*I do realize that I can divide through with a factor of R2, but that still would not change the bandwidth.

I created a bode plot that gave me a natural frequency of 662Hz, which using \$B=\frac{w_n}{2\zeta}\$ leaves me with a bandwidth of 172Hz, while the active filter on the circuit board is listed at 100Hz.

Bode plot created through matlab "bode" function:

Bode plot created through matlab "bode" function

is there somewhere where I am going wrong with the derivation, or is the method of how I approach calculating bandwidth wrong?

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  • \$\begingroup\$ I believe I asked a very similar question here electronics.stackexchange.com/questions/561883/…. It seems the formula \$B = \frac{\omega_n}{2\zeta} \$ is intended for bandpass filters. \$\endgroup\$
    – Carl
    May 16, 2021 at 10:16
  • \$\begingroup\$ The magnitude of the transfer function looks similar to a Butterworth response (maximally flat). Therefore, the 3dB- bandwidth is equal to the systems pole frequency wp=1/R1R2C1C2. This frequency can be found where the phase shift is -90deg. \$\endgroup\$
    – LvW
    May 16, 2021 at 10:21
  • \$\begingroup\$ @LvW Yes, whereas the listed specification of the filter is a bandwidth of 100Hz. so by these calculations, you are saying the bandwidth is around that 600 mark? \$\endgroup\$
    – Phillip Kjær
    May 16, 2021 at 10:28
  • \$\begingroup\$ You probably left out the 2\$\pi\$ factor. Also, the plotting must have considered \$\omega\$ as f, not \$2\pi f\$. \$\endgroup\$
    – a concerned citizen
    May 16, 2021 at 10:54
  • \$\begingroup\$ @PhillipKjær Your ammendment to your question has now made it erroneous, as the formula for bandwidth of bandpass filter is \$B=\frac{\omega_n}{2\zeta} \$, and not \$B=\frac{\omega_n}{\zeta/2} \$ as you have written. \$\endgroup\$
    – Carl
    May 16, 2021 at 10:57

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is there somewhere where I am going wrong with the derivation, or is the method of how I approach calculating bandwidth wrong?

When you calculated 662 Hz you actually calculated 662 radians per second. To convert to hertz you divide 662.134 radians per second (the precise value using your component values) by 2π to get 105.3819 Hz.

Cut-off frequency for a low pass Sallen Key filter is: -

$$F_C = \dfrac{1}{2\pi\sqrt{R_1R_2C_1C_2}}$$

Which is precisely what your algebra is telling you.

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