12DB/OCT low-pass filter with the same capacitance parameters

Function of the circuit

The low-pass filter introduced in "12DB/OCT low-pass filter composed of OP amplifier" has inconsistent capacitance of each capacitor to ensure the Q value of 0.707, and it is very difficult to select the capacitor according to the calculated capacitance . In this circuit, the capacitance of C1 and C2 is the same. Once the cut-off frequency is determined, the appropriate value of CO=C1=C2 can be selected, and then the resistance value required for RO can be calculated in reverse.

On the other hand, in order to obtain a certain Q value, the feedback amplifier is made to have a gain, but the gain error will cause the Q value to change.

How the Circuit Works

In this circuit, when A=1, A=0.5, which is no different from a passive filter. When A is greater than 1, Q=1/(3-A), the Q of the 12DB/OCT Butterworth filter is 0.707, and A can be assumed to be: A=3-(I/Q)1.585. According to R4=R3(A-1), when R3=10K, R4 can be 5.85K, and 5.1K and 750 ohms are connected in series.

This method can achieve a general gain of A = 1.585, and then use the circuit connected to the previous or subsequent stage to attenuate it to 1/1.585. The function of feedback capacitor C3 is to suppress the spike generated by the input capacitor C1 of the OP amplifier.

Although there is a certain degree of freedom in setting the cutoff frequency FL, when FL=100KHZ, if R0=10K, then CO≈160PF, the input capacitance C1 or parasitic capacitance of the OP amplifier will have an impact, so CO must be reduced by 5~10PF from the calculated value. In addition, the OP amplifier can be replaced with a single-gain broadband product.