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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:

enter image description here

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:

enter image description here

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:

enter image description here

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: enter image description hereenter image description here enter image description here

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  • \$\begingroup\$ You might want to recheck the implications of eliminating R2 in your first section. \$\endgroup\$ – WhatRoughBeast Apr 24 '17 at 23:39
  • \$\begingroup\$ TI has a better Filter App for this then you can choose Elliptical, Cauer, Butterworth , exact values or 0.5%' 1% etc. You must learn to define specs first, pass band f, ripple, band reject f and attenuation before starting any design. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 24 '17 at 23:41
  • \$\begingroup\$ @WhatRoughBeast My teacher said that to equate 1 + R2 / R3 to 1 that R2 had to be zero and R3 had to be infinite. I did the calculations according to what he said, but I simulated the circuit and the transfer function gives me zero. I'm new to this, I followed the advice, but I really need help. I've been around this problem for many days now. \$\endgroup\$ – Carmen González Apr 24 '17 at 23:47
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    \$\begingroup\$ @CarmenGonzález - You are confusing the two circuits. What the teacher said only applies to your second circuit. The R2 in the first circuit is not the R2 in the first, so it does not disappear when you set the op amp gain to 1. \$\endgroup\$ – WhatRoughBeast Apr 24 '17 at 23:51
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    \$\begingroup\$ Your title says fourth order filter yet your design is 3rd order. \$\endgroup\$ – Andy aka Apr 25 '17 at 7:28
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This is what one stage of active SallenKey HPF and one passive RC HPF provide

enter image description here

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.

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  • \$\begingroup\$ Sorry I did not realize the technical terms you used. What is UGBW? I do not speak English, sorry. What it means: That higher UGBW is key to a FLAT RESPONSE above the F3dB.? \$\endgroup\$ – Carmen González Apr 25 '17 at 9:04
  • \$\begingroup\$ @CarmenGonzález UGBW: Unity Gain BandWidth \$\endgroup\$ – HatimB Apr 25 '17 at 13:01
  • \$\begingroup\$ @analogsystemsrf What simulator you use? \$\endgroup\$ – Carmen González Apr 28 '17 at 8:42
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Running this through LTSpice AC simulation using two LT1022 single, JFET input, 8.5MHz OpAmps yields: LT1022 Plot -3dB points of ~12kHz and ~6.5MHz.

Running this through LTSpice AC simulation using a LT1208 dual, 45MHz OpAmp yields: LT1208 Plot -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. :)

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