%path = "physics/circuits/thevenin" %kind = kinda["examples"] %level = 13

We will derive the gain of a band-stop filter using Thevenin’s method.

Our starting circuit is the following from Op Amps for EveryOne (5-10).

The input voltage is against the ground. We redraw the circuit to reflect this:

We want to find \(G=\frac{V_o}{V_i}\). \(V_o\) is the voltage at the rightmost R. We will find the Thevenin equivalent there.

Next we find the currents using Kirchhoff’s current law (KCL) and voltage law (KVL). There are two loops where current flows. There is no current at the R at which we have made an open circuit

The resulting equations are

\[\begin{split}\begin{array}{l l l} V_i - I_2 R - \frac{I_2}{iwC} - I_1 R & = 0\\ V_i - \frac{I_1 - I_2}{iwC} - I_1 R & = 0 \end{array}\end{split}\]

We solve for \(I_1\) and \(I_2\):

\[\begin{split}\begin{array}{l l} I_1 &= \frac{\omega C V_i (-2 i+C R \omega)}{-1-3 C R i \omega+C^2 R^2 \omega^2}\\ I_2 &= -\frac{i \omega C V_i}{-1-3 C R i \omega+C^2 R^2 \omega^2} \end{array}\end{split}\]

The small loop at \(V_{th}\) with the now known currents can be used to calculate

\[V_{th}=\frac{I_2}{iwC} + I_1 R\]

Next we will need the Thevenin impedance equivalent. For this we remove the source \(V_i\) and calculate the impedance as seen from \(V_{th}\).

We redraw to see a little better, what is parallel and what is serial

With this we get

\[Z_{th}=\left(\frac{1}{i \omega C}+\frac{R \frac{1}{i \omega C}}{R+\frac{1}{i \omega C}}\right) || R = \frac{R (1+2 i \omega C R)}{1+3 i \omega C R - C^2 R^2 \omega^2}\]

and

\[V_o = \frac{R}{R + Z_{th}}\]

Then we finally have the gain:

\[G = \frac{V_o}{V_i} = \frac{(-i+C R \omega)^2}{-2-5 i \omega C R+C^2 R^2 \omega^2} = \frac{(1+i \omega C R)^2}{2+5 i \omega C R-C^2 R^2 \omega^2} = \frac{(1+ s\tau)^2}{2+5 s\tau+(s\tau)^2}\]

We have replaced \(\tau=R C\) and \(s=i \omega\), as is custom for filters. The denominator can be solved for \(s\tau\) (-4.56,-0.44) and the product of the solutions is 2. Therefore

\[G = \frac{(1+ s\tau)^2}{2(1+\frac{s\tau}{0.44})(1+\frac{s\tau}{4.56})}\]