The attached file is with images from the simulations I did. The first circuit is the circuit I need to calculate something for. The green signal is the input and the blue signal is the output. I need to calculate the time it takes for the output signal to reach its maximum value and the time it takes for it to reach its minimum value. So I know that the output does not go bellow 1.35V because of the diode and the voltage source in series (not 2V because of the diode forward voltage). This is what I did. The rise and fall of the output voltage is because of the capacitor. So I need to calculate the time for the capacitor to charge and discharge. To do it faster I used this site: http://mustcalculate.com/electronics/capacitorchargeanddischarge.php. When charging the time constant should be determined by the resistors in parallel (10k and 1k in parallel make 0.909k) and the capacitor capacitance. On the site I filled that the capacitor will charge with voltage source of 5V (tried with 4.35V too), from 1.35 to 4.58 (from the simulation). Compared it to the simulation and the simulation showed a lot slower charging time. When discharging the simulation showed even slower time compared to the calculated values. Only considered the 10k resistor when discharging.
I was also wondering why in that second circuit (3rd image) the output voltage is limited closer to 2. Why the forward voltage on the diode fall?
I will be very thankful if someone can explain this to me.

 

I think your first mistake is 10k and 1k resistors. You claim that they are in parallel. The picture shows that they ARE NOT in parallel. So. Either you are wrong or picture is wrong. Which is it?

 

I think your first mistake is 10k and 1k resistors. You claim that they are in parallel. The picture shows that they ARE NOT in parallel. So. Either you are wrong or picture is wrong. Which is it?

I am pretty sure they are in parallel if you consider the diode for ideal (and even if you don't it shouldn't matter so much). And if you think I am wrong then what resistance should I take when calculating the time constant for charging the capacitor.

 

I am pretty sure they are in parallel if you consider the diode for ideal (and even if you don't it shouldn't matter so much). And if you think I am wrong then what resistance should I take when calculating the time constant for charging the capacitor.

Ok.
Remind me the setup for ideal diode.

I can tell of hand that when diode is not conducting, it act as open circuit. Which means there is no current flowing through the 1k resistor. Which means that you can throw 1k resistor away and forget about it. Which means that only 10k matters. So your tau=R*C=10k*100pF.

Basically, I think the problem is a bit more complicated that you think. You have a situation where you have two cases:
Case 1: V_diode < V_forward of diode: V_diode < 2 volts, no current flow though 1k resistor, all current is going through 10k resistor.

Case 2: V_diode = or > V_forward of diode, in this case the diode is conducting, and like you pointed out it acts as a short, so 1k and 10k are in parallel.

 

Ok.
Remind me the setup for ideal diode.

I can tell of hand that when diode is not conducting, it act as open circuit. Which means there is no current flowing through the 1k resistor. Which means that you can throw 1k resistor away and forget about it. Which means that only 10k matters. So your tau=R*C=10k*100pF.

Basically, I think the problem is a bit more complicated that you think. You have a situation where you have two cases:
Case 1: V_diode < V_forward of diode: V_diode < 2 volts, no current flow though 1k resistor, all current is going through 10k resistor.

Case 2: V_diode = or > V_forward of diode, in this case the diode is conducting, and like you pointed out it acts as a short, so 1k and 10k are in parallel.

You are right about that. That's the way I see it too. But my problem is that the simulation results and the math don't seem to fit and I don't know why.

 

You are right about that. That's the way I see it too. But my problem is that the simulation results and the math don't seem to fit and I don't know why.

What that second power supply, V2, doing there? Am I reading it right that it is dc, 2 volts? Are you accounting for it? Also is D2 same as D1? D2 forward voltage is also 2 volts?

 

What that second power supply, V2, doing there? Am I reading it right that it is dc, 2 volts? Are you accounting for it? Also is D2 same as D1? D2 forward voltage is also 2 volts?

The diodes are the same. That second power supply and the diode does not let the output voltage be less than 2V (a bit less because the diode is not ideal). I am not considering it in the calculations because I don't think it matters (not 100% sure).

 

No need for more replies guys. Figured it out not long ago. It seems that I have been doing everything right except a very stupid mistake when calculating. Wasted a lot of time on that stupid mistake.