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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: enter image description here

And its transfer function in s-domain will be equal to: enter image description here

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.

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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!

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  • \$\begingroup\$ 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? \$\endgroup\$ – Sean87 Oct 5 '11 at 13:36
  • \$\begingroup\$ 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). \$\endgroup\$ – MikeJ-UK Oct 5 '11 at 15:22

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