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I have recently been trying to solve this transfer function an my professor keeps saying I have been doing wrong and I finally need help. I know my Gain is \$1+ \frac{R_A}{R_B}\$. but I keep getting extra stuff in my numerator. also I need to design it to 1 kHz cut off frequency. I test the transfer function out by using Matlab and seeing where the highest dB it can reach.

  • 1
    \$\begingroup\$ This is a Sallen-Key high pass filter. \$\endgroup\$
    – Chu
    Commented Aug 3, 2019 at 23:30
  • \$\begingroup\$ The highest gain an opamp can reach is not determined by the filter. \$\endgroup\$
    – Voltage Spike
    Commented Aug 4, 2019 at 5:20
  • \$\begingroup\$ Kahled, can you help us help you? There are a lot of things being asked for in your problem statement. All the way to doing a PCB design, in fact. What do you know about transfer functions, generally? (By the way, the standard form us \$\frac{K\,\omega_{_0}^2}{s^2+2\,\zeta\,\omega_{_0}\,s+\omega_{_0}^2}\$ where \$K\$ is the gain and not as the form shown in the problem.) What transfer function did you arrive at, and how? Do you understand the idea of a damping factor, \$\zeta\$, or the idea of \$Q\$? Have you read Sallen & Key's paper(s)? \$\endgroup\$
    – jonk
    Commented Aug 4, 2019 at 17:05

1 Answer 1


Regarding to derive the transfer function of this second order high pass active filter consider, for example, to apply the KCL to nodes A and B on the figure below (assuming an ideal op. amp.):

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$$ \left\{\begin{matrix} \frac{V_A-V_i}{\frac{1}{sC_1}}+ \frac{V_A-V_O}{R_2}+\frac{V_A-V_B}{\frac{1}{sC_2}}=0 & (1)\\ \frac{V_B-V_A}{\frac{1}{sC_2}}+ \frac{V_B}{R_1}+0= 0 & (2) \end{matrix}\right. $$


$$ V_B=\frac{V_oR_B}{R_A+R_B} $$

in (1) and (2):

$$\left\{\begin{matrix} (V_A-V_i)sC_1+ \frac{V_A-V_O}{R_2}+(V_A-\frac{V_oR_B}{R_A+R_B})sC_2=0 & (3)\\ (\frac{V_oR_B}{R_A+R_B}-V_A)sC_2+\frac{V_oR_B}{(R_A+R_B)R_1}=0 & (4) \end{matrix}\right.$$

From (4): $$ V_A=\frac{V_oR_B}{R_A+R_B}\left ( 1+\frac{1}{sR_1C_2} \right ) $$

Returning with this expression on (3) and, after a long algebraic manipulation, it's possible to obtain \$\frac{V_o}{V_i}\$ as:

$$ \frac{V_o(s)}{V_i(s)}=\frac{\left( 1+\frac{R_A}{R_B}\right)R_1R_2C_1C_2s^2}{R_1R_2C_1C_2s^2+\left[ R_2C_1+R_2C_2 - \frac{R_A}{R_B}R_1C_2 \right]s+1} $$

The cut-off frequency (in Hz) is given by:

$$ f_c=\frac{1}{2\pi\sqrt{R_1R_2C_1C_2}} $$


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