# Calculating symbolic input impedance of Multiple Feedback Low-Pass filters

I am trying to design a filter and I am Following this guide. The second step is determining the input impedance of the circuit.

The filter is a MFB Low-Pass type: And its transfer function in s-domain will be equal to: For calculating the symbolic input impedance, I recon that I have to use Thevenin Equialent of the circuit as seen by the input (voltage source). Will this be the calculating the resistor in prallel and series and then adding the series resistance of capacitors to the result? what will happen to the OpAmp, will it have effect on the input impedance calculation?

Hints and tips are pretty much welcome!

Thanks.

Take it in stages. The input impedance is $\frac{V_{IN}}{I_{IN}}$

The input current is given by $I_{IN}=\frac{V_{IN}-V_X}{R_1}$
... where $V_X$ is the voltage at the junction of R1, R3, R4, C2

Using KCL, Vx is given by $\frac{V_{IN}-V_X}{R_1}+\frac{V_{OUT}-V_X}{R_4}+\frac{0-V_X}{\frac{1}{sC_2}}+\frac{0-V_X}{R_3}=0$

Use your transfer function to express Vout in terms of Vin and you're there ... eventually ... after some substitution!

• Thanks. Can you explain what happend to the current which comes back from OpAmp to R4 (feedback resistor). Also the OpAmp should be ignored in this calculations? Is it only about the junction which connects R1, R3, R4 and C2? Oct 5 '11 at 13:36
• No it isn't being ignored. In the calculation for Vx, the term (Vout-Vx)/R4 takes care of the op-amp (the current through R4). Oct 5 '11 at 15:22

Simpler explanation for the passband:

• C2 and C5 behave like open circuits, effectively not there.
• Op-amp input impedance is infinite, so no current flows through R3, and it is effectively not there.
• The op-amp feedback maintains a voltage of 0 V at the inverting input, which is also effectively the junction between R1 and R4.

So the junction between R1 and R4 is a virtual ground and the input impedance = R1.