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schematic

simulate this circuit – Schematic created using CircuitLab

I want to calculate the time constant τ of the filter. Can anybody help me?

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  • \$\begingroup\$ If you are familiar with using the Laplace transform to represent impedances, you can calculate the transfer function by dividing R2 by the parallel equivalent of R1 and 1/sC1. You will realize that the gain will be relatively flat till a certain cutoff frequency (I'll let you determine it) and then it will start growing at 20dB/dec after that. \$\endgroup\$ – joribama Apr 21 at 6:55
  • \$\begingroup\$ Nodal analysis at the inverting input will give the TF. \$\endgroup\$ – Chu Apr 21 at 7:23
  • \$\begingroup\$ I am familiar with the laplace transformation, but it seems like the transferfunction has no zeros. I end up with the following: R2 * ( C * s * R1 + 1)/R1 \$\endgroup\$ – Electro by night Apr 21 at 7:41
  • \$\begingroup\$ If you want the time constant of this circuit, reduce the input voltage to 0 V (replace the source by a short circuit) then disconnect capacitor \$C_1\$ and determine the resistance "seen" from its connections. That resistance multiplied by \$C_1\$ is your time constant while the inverse gives the pole. More details in this APEC seminar: cbasso.pagesperso-orange.fr/Downloads/PPTs/… \$\endgroup\$ – Verbal Kint Apr 21 at 7:47
  • \$\begingroup\$ Verbal Klint-I think, in your answer it would be helpful to mention if the opamp is to be considered as active or not. \$\endgroup\$ – LvW Apr 21 at 10:03
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Looking at this circuit by considering its time constant is the right way to go. This is actually the basis for the fast analytical techniques or FACTs. In the considered circuit, set \$s=0\$ (open the capacitor) and determine the gain linking \$V_{in}\$ to \$V_{out}\$ as illustrated in sketch (b) below. Considering a perfect op-amp, the dc gain is immediate and equal to:

\$G_0=-\frac{R_2}{R_1}\$

To determine the time constant of this circuit, simply reduce the excitation to 0 and "look" at the resistance offered by \$C_1\$'s connecting terminals as shown in sketch (c). In this drawing, zeroing the excitation to 0 V is similar to replacing the source by a short circuit. In this mode, again considering a perfect op-amp, the left connection of \$C_1\$ is grounded while the right-one, considering the virtual ground, is also 0 V. The time constant is thus:

\$\tau=0\times C_1 = 0\$ implying that there is no pole in this circuit. Actually, there is one and it is coming from the op-amp but it is not set by the external components.

For the zero, what condition would imply that the stimulus would not propagate through the circuit to form a response at \$s=s_z\$? Well, if the parallel combination of \$C_1\$ and \$R_1\$ offers an infinite impedance, there is no response. An impedance becomes infinite when its denominator \$D(s)\$ equals zero. In other words, what is the pole of \$Z_1\$? It is the resistance \$R_1\$. The zero is located at:

\$\omega_z=\frac{1}{R_1C_1}\$ and our transfer function is:

\$G(s)=G_0(1+\frac{s}{\omega_z})\$

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The Mathcad file is there:

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