Third order filter - I need to verify calculations

Question

I need to construct a Butterworth high pass filter from two sections: one of second order and one of first order with a cut-off frequency of \$12\, KHz\$. The second order section is as follows:

The transfer function of this section I calculated:

\$ H(s)=\frac{s^2}{\frac{1}{R_2R_4C_1C_3}+\frac{1}{R_4C_1}s+\frac{1}{R_4C_3}s+s^2}\$

The first order section is as follows:

The transfer function of this section calculated by me is:

\$H(s)=\frac{1+\frac{R_2}{R_3}}{1+\frac{1}{R_1C_1s}}=\frac{\big(1+\frac{R_2}{R_3}\big)s}{s+\frac{1}{R_1C_1}}\$

Through the Butterworth polynomials:

\$H_{LP}(S)=\frac{1}{S^2+S+1}\frac{1}{S+1}\$

To transform into a high pass, I have to make the transformation:

\$ S\rightarrow \frac{\omega_C}{s}\$

\$H_{HP}(s)=\frac{1}{\frac{{\omega_C}^2}{s^2}+\frac{\omega_C}{s}+1}\frac{1}{\frac{\omega_C}{s}+1}=\frac{s^2}{{\omega_C}^2+\omega_Cs+s^2}\frac{s}{\omega_C+s}\$

The coefficients are matched and a system of equations is made:

System of equations for the first order section: \$1+\frac{R_2}{R_3}=1 \rightarrow R_2=0; R_3=\infty\$ (Buffer)

\$\frac{1}{R_1C_1}=\omega_C\leftrightarrow \frac{1}{R_1C_1}=75398,22\$

I considered that \$C_1=100\,nF\$. So:

\$R_1=132,63\,\Omega\$

System of equations for the second order section:

\$\frac{1}{R_2R_4C_1C_3}={\omega_C}^2\$

\$\big(\frac{1}{R_4C_1}+\frac{1}{R_4C_3}\big)=\omega_C\$

I considered that: \$C_3=100\,nF\$

So:

\$\frac{1}{R_2R_4C_3}=568,49 \rightarrow R_2=66,31\,\Omega\$

\$\big(\frac{1}{R_4\times 10^{-7}}+\frac{1}{R_4\times 10^{-7}}\big)=75398,22\rightarrow R_4=265,26\Omega\$

The final circuit which is supposed to have a cutoff frequency \$f0=12\,KHz\$ is as follows:

I would like if you could check the calculations and if you could tell me if this circuit I built has the correct cut-off frequency.

Error of Qucs simulator:

Answer 1

This is what one stage of active SallenKey HPF and one passive RC HPF provide

Instinctively, I used the default opamp model, but edited the UGBW to be 10MHz instead of default 1MHz. That higher UGBW is key to a FLAT RESPONSE above the F3dB.

Answer 2

Running this through LTSpice AC simulation using two LT1022 single, JFET input, 8.5MHz OpAmps yields: -3dB points of ~12kHz and ~6.5MHz.

Running this through LTSpice AC simulation using a LT1208 dual, 45MHz OpAmp yields: -3dB points of ~12kHz and ~40MHz.

Seems the upper range is heavily dependent on the unity gain bandwidth (UGBW) of the particular op-amp chosen. Otherwise the math looks good to me.

When picking the capacitor value, try to fit resistors into the kilo-Ohm range. This is to prevent drive current from influencing stages. Some opamps can only drive a few mA, and will struggle with low resistances and large capacitances.

Note that if you wanted to build this circuit, a solderless breadboard may cause trouble due to parasitic capacitance and inductance. Even a purpose-built PCB would likely require some tuning to match calculations. So calcs and sims are only good for ball-park estimates to real performance. Some tweaking is required. :)