I've the following circuit:

enter image description here

Question: What are the time constants of the voltage at node \$\text{Y}_2\$ when the up part of the puls comes and when the down part of the puls comes?


The voltage at node \$\text{Y}_2\$ can by using the following analysis:

  1. The input voltage (starts at \$t=0\$ at \$8\$ V and is high for \$\frac{1}{20}\$ seconds and turn than back to \$0\$) is given by (in the s-domain using the Lapalce transform): $$\text{u}_\text{in}\left(\text{s}\right)=\int_0^\frac{1}{20}8e^{-\text{s}t}\space\text{d}t=8\cdot\frac{1-\exp\left(-\frac{\text{s}}{20}\right)}{\text{s}}\tag1$$
  2. The relation between the voltage at node \$\text{Y}_2\$ and the input voltage is given by: $$\frac{\text{u}_{\text{Y}_2}\left(\text{s}\right)}{\text{u}_\text{in}\left(\text{s}\right)}=\frac{600+\frac{1}{10^{-6}\text{s}}}{600+600+\frac{1}{10^{-6}\text{s}}}=\frac{5000+3\text{s}}{5000+6\text{s}}\tag2$$

So, the voltage at node \$\text{Y}_2\$ is given by:


Plotting the function (in the time domain gives):

enter image description here

So, for finding the time constant (\$\approx63\text{%}\$ of the maximum value) we need to solve:

$$8\cdot\left(1-\frac{1}{e}\right)=\mathcal{L}_\text{s}^{-1}\left[8\cdot\frac{1-\exp\left(-\frac{\text{s}}{20}\right)}{\text{s}}\cdot\frac{5000+3\text{s}}{5000+6\text{s}}\right]_{\left(t\right)}\space\Longleftrightarrow\space$$ $$t\approx0.368223\space\text{ms}\tag4$$

Am I correct in my analysis or not? Because I also thought about using \$\tau=\text{R}\cdot\text{C}\$ so then we get \$\tau=(600+600)\cdot10^{-6}=1.2\$ miliseconds, is that right or is my previous analysis right? Because I got two different answers

  • \$\begingroup\$ Just a note. The two R values aren't summed for your cross-check \$\tau\$. \$\endgroup\$ – jonk Nov 29 '18 at 19:25
  • \$\begingroup\$ @jonk Okay, but what is my mistake there? \$\endgroup\$ – Klopjas Nov 29 '18 at 19:26
  • \$\begingroup\$ What does \$Y_2\$ see looking back in? (Assuming you weren't doing all your math and were just trying to take a first-guess at it.) \$\endgroup\$ – jonk Nov 29 '18 at 19:27
  • \$\begingroup\$ @jonk just one resistor of \$600\space\Omega\$, I think :) \$\endgroup\$ – Klopjas Nov 29 '18 at 19:28
  • \$\begingroup\$ What about your ideal voltage source (pulse) as well as your capacitor? Would \$Y_2\$ only see one of those? Or would it see both? I haven't checked your math analysis, but I think you may have a reasonable answer given my own guess at about where it should be close to without doing the math. (Of course, I could be wrong. I'm just pointing out that my "off the cuff" computation came close to your result.) \$\endgroup\$ – jonk Nov 29 '18 at 19:31

I did not check your calculations, but I can tell you that this question can be solved much easier:

The time constant at Y1 is the time constant of a well known and simple RC circuit, which (google tells us) is T=RC with R=Ri+R1 and thus T is 1.2ms.

As the voltage potential at Y2 is always exactly half way between V1 and Y1, the resulting voltage at Y2 is always just a (Y-axis) scaled version of the voltage at Y1. So Y2 must have the same time constant as Y1 which is T=1.2ms.


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