# Bandwidth calculation of second order Sallen-Key low pass filter

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

• 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.
– Carl
May 16 at 10:16
• 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.
– LvW
May 16 at 10:21
• @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? May 16 at 10:28
• You probably left out the 2$\pi$ factor. Also, the plotting must have considered $\omega$ as f, not $2\pi f$. May 16 at 10:54
• @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.
– Carl
May 16 at 10:57

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