# OP-Amp active filter analysis

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

• The voltage at the V+ op amp input is Vo. Use nodal analysis at the z1, z2, z3 node, and the V+ node. – Chu Mar 20 '18 at 1:07
• You'd better showing your work so everyone can see where you go astray. – anhnha Mar 20 '18 at 1:29
• 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? – Jay Makers Mar 20 '18 at 8:20
• It's in here – Andy aka Mar 20 '18 at 10:09

I just worked this out. Math becomes a little easier if you use admittance instead of impedance. I hope it helps.

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 $V_{out}=0$:

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

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:

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.