I have such circuit:


simulate this circuit – Schematic created using CircuitLab

Vin is either equal to Vcc or to 0.

How do I go about determining Vout?


closed as too broad by Bimpelrekkie, laptop2d, PeterJ, RoyC, Finbarr Apr 8 at 0:10

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ You can start by ignoring R1, does that help? However, note that "open" is not the same as 0V. \$\endgroup\$ – Spehro Pefhany Apr 5 at 12:43
  • 4
    \$\begingroup\$ How can I determine the output in this circuit? By learning about circuit analysis. Hundreds of books have been written on the subject. I see no reason why anyone would need to explain that here as well. If this is asked of you then surely you must have had some education on the subject. If not then get studying. \$\endgroup\$ – Bimpelrekkie Apr 5 at 12:47

The first thing you need to know is that the capacitor charges fully to the Vcc at steady state i.e the capacitor maintains 3.3V with respect to ground.And once the capacitor is fully charged, it acts like a open switch which stops the flow of current. Since the Vout is directly connected to the capacitor, the output voltage is equal to the voltage across capacitor that is 3.3V.

Now, when the Vin is equal to 3.3V, nothing happens to the circuit.


Please ignore the values of the components I have in my circuit. Also ignore the names in your circuit and look at mine now. I tried editing the circuit but the 'edit' button has disappeared in the preview section. Sorry for the inconvenience.

In this case. node 1 is at 3.3V. This means that there is no any potential difference across the resistors R1 and R2. So no any current flows through any of the resistors. Also the capacitor doesn't discharge due to the fact that there are no any nodes at lower potential with respect to the capacitor's potential.

When the Vin goes to zero, something happens. Since the node 1 is at zero volt, the capacitor finds 'somewhere' to dump its potential. So the capacitor stats to discharge through the resistor 1(in my figure). Due to the discharge of the capacitor, the voltage across it starts to decrease as well. Once the capacitor is fully discharged, the output voltage is zero. For the instantaneous voltage across the capacitor, I suggest you reading this topic. Once you get the instantaneous voltage across the capacitor, you get the instantaneous Vout

  • \$\begingroup\$ Thank you for the answer. It is what I was looking for. The main issue for me is that I was used to seeing closed loop circuits rather than such schematics with open ends and therefore couldn't analyse it properly. Thanks! \$\endgroup\$ – Aivaras Kazakevičius Apr 5 at 13:44

I'm guessing you need to know Vout when the circuit has reached it's stationary state.

When that happens the capacitor charges to a voltage so that no current flow through it. So you can "remove" it from your circuit.

Then you need to evaluate both cases, when Vin = 0V and when Vin = 3.3V.


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