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?
Best Answer
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.