diodesvoltage
I was trying to understand this circuit under DC conditions, but can someone explain to me why the voltage across the capacitor is 8.1 V? I can't seem to understand. This voltage however does go down to 4.2 V if I increase the simulation speed and wait for couple of seconds but how can it start of it at 8.1 V. Am I missing something?
Here is the circuit I drawn using falstad simulator.
They are simulating a (more or less) real 9V battery. They've modeled the battery as an ideal voltage source of 9V with a series resistance of ~1.5 ohms.
simulate this circuit – Schematic created using CircuitLab
If you work this out, the current is 9V/(11.5 ohms) = 0.783A, so the voltage across the 10 ohm resistor must be 7.83V.
Depending whether the diode is in the conducting or non-conducting state, your circuit could have two different equivalent circuits:
simulate this circuit – Schematic created using CircuitLab
A is the equivalent circuit when the diode is conductiong. B is the equivalent circuit when the diode is not conducting.
You should be able to see that solving Vout is trivial in case A.
In case B you can solve using mesh analysis, or using modified node analysis. Superposition could be applied, but is not needed.
Then you must examine the solution for case B and find any situations where the diode would not be biased correctly to be non-conducting, and replace the solution at those points in time with the solution from case A.
Best Answer
Basically it's your simulator - it can't ever be higher than the 5V supply and 4.2 is pretty normal. Check that your initial conditions don't include some form of charge on the capacitor. Simulators can throw curved-balls sometimes and you did the right thing by asking.
By the way I think @pjc50 may have misread your voltage as being 9V not 5V - or maybe it's me - the picture isn't all that clear.