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Could you help me derive this transfer function for this OP-Amp circuit, I keep trying different methods and get no where close!

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    \$\begingroup\$ The voltage at the V+ op amp input is Vo. Use nodal analysis at the z1, z2, z3 node, and the V+ node. \$\endgroup\$ – Chu Mar 20 '18 at 1:07
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    \$\begingroup\$ You'd better showing your work so everyone can see where you go astray. \$\endgroup\$ – anhnha Mar 20 '18 at 1:29
  • \$\begingroup\$ Nodal analysis to work out the voltage at those nodes? Isn't the voltage at z2=Vo, z3=Vo, V+=Vo? What is there to work out at those? \$\endgroup\$ – Jay Makers Mar 20 '18 at 8:20
  • \$\begingroup\$ It's in here \$\endgroup\$ – Andy aka Mar 20 '18 at 10:09
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I just worked this out. Math becomes a little easier if you use admittance instead of impedance. I hope it helps.

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The easiest path is to apply superposition to this op-amp. Split the circuit for \$V_{in}=0\$ and \$V_{out}=0\$. Determine \$\epsilon\$ in both cases and the sum of the determined values is 0 V considering an infinite gain. The first drawing shows the circuit for \$V_{out}=0\$:

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The voltage at the non-inverting pin is simply:

\$\epsilon_1=V_{in}\frac{Z_3}{Z_1+Z_3}\frac{Z_4}{Z_4+Z_2+\frac{Z_1Z_3}{Z_1+Z_3}}\$

Now set \$V_{in}=0\$:

You have: \$\epsilon_2=V_{out}\frac{Z_1}{Z_1+Z_3}\frac{Z_4}{Z_4+Z_2+\frac{Z_1Z_3}{Z_1+Z_3}}-V_{out}\$

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Now solve for \$V_{out}\$ when \$\epsilon_1+\epsilon_2=0\$. You have \$\frac{V_{out}}{V_{in}}=\frac{Z_3Z_4}{Z_1Z_3+Z_2Z_3+Z_3Z_4+Z_1Z_2}\$

If \$Z_3\$ and \$Z_4\$ are 10-nF capacitors, then Mathcad plots the dynamic response of this filter:

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If you now try to develop the expression in which \$Z_3\$ and \$Z_4\$ are capacitance, you'll end up in an ugly result which will need further energy to factor it into a second-order polynomial form. If you apply the FACTs, you'll be there straight away.

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