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Given the following circuit, I would like to calculate the time constant \$ \tau \$.

I know that in an RC circuit \$\tau=R_{eq}\cdot C \$.

enter image description here

I've also found out that \$\frac{d v_{c}}{d t}=-\frac{2-\alpha}{R C} v_{c}(t)+\frac{E(1-\alpha)}{R C}\$, where \$E=e(t)\$. Let's define \$\lambda=\frac{1}{\tau}\$. My textbook considers \$\lambda=-\frac{2-\alpha}{RC}\$, so that \$\tau=-\frac{RC}{2-\alpha}\$.

Instinctively I would simplify the resistors so that I can find \$R_{eq}\$ in order to find \$\tau\$, but I don't know if that would be appropriate. Also why \$\lambda=-\frac{2-\alpha}{RC}\$?

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2 Answers 2

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It is easier to re-draw the circuit in a simpler representation to see if things are clearer:

enter image description here

Then, to determine the time constant, you turn the excitation off and replace the source by a short circuit. Redraw the circuit again in this mode:

enter image description here

To determine the resistance seen from the capacitor's terminals, connect a test generator \$I_T\$ which develops a voltage \$V_T\$ across its terminal. Determine \$\frac{V_T}{I_T}\$ and you're done:

enter image description here

You can see that \$R_2\$ is in parallel with the resistance we want. Temporarily disconnect \$R_2\$ to bring it back later across the intermediate result. The equation is \$V_T=I_TR_1+kV_T\$. Factor and reorganize to find that \$R=(\frac{R_1}{1-k})||R_2\$. A quick Mathcad sheet with a SPICE sim and the operating point confirms the result:

enter image description here

So, unless I misinterpret the controlled source symbol (a controlled voltage source I believe), the time constant is \$\tau=C((\frac{R_1}{1-k})||R_2\$). If both resistances are equal to \$R\$, the expression simplifies to \$\tau=\frac{CR}{2-k}\$

When you determine the natural time constants of a circuit, you always turn the excitation off and look through the energy-storing element's terminal what resistance \$R\$ do you see. That \$R\$ multiplied by the capacitor in this case forms the time constant. This is the basic approach I described in the book I wrote on fast analytical circuits techniques or FACTs.

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  • \$\begingroup\$ Why are we allowed to take out R2 and then consider it again? Thanks! \$\endgroup\$
    – Kevin
    Jan 21, 2020 at 22:50
  • \$\begingroup\$ When you want to determine a resistance across two terminals and you see that other resistances are directly connected in parallel with these terminals - and provided the current in these resistors is not driving any source or used elsewhere in the circuit - then you can temporarily disconnect these resistors and bring them back in parallel with the intermediate result. No magic here, simple inspection which lets you gain a lot of time. \$\endgroup\$ Jan 22, 2020 at 7:35
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Well, we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Using KCL, we can write:

$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$

Using KVL, we can write:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_3}{\text{R}_2}\\ \\ \text{I}_2=\frac{\text{V}_2-\text{V}_3}{\text{R}_1}\\ \\ \text{I}_3=\frac{\text{V}_x-\text{V}_3}{\text{R}}\\ \\ \text{V}_1=\text{V}_2-\text{V}_x \end{cases}\tag2 $$

Now, using the fact that \$\text{R}\$ is not a resistor but a capacitor and assuming that the initial condition is equal to zero. We can set:

$$\text{R}\space\space\space\to\space\space\space\frac{1}{\text{sC}}\tag3$$

Besides that we also know that:

$$\text{V}_1=-\alpha\text{V}_3\tag4$$

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