I tried to create a simple simulation in Orcad PSPICE. For some reasons, the measured AC voltage across the capacitor C is shifted -20 Vdc. Its value should vary in the range [-20V, +20V] as far as I'm concerned. Very appreciate if anyone can give me an idea of what's going on here. Thank you very much.

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  • \$\begingroup\$ You either haven't simulated for long enough - or RC is too large. Try simulating for 10 * RC (you may want to reduce R to less than 1G :-) Once you understand that, there may be fixes involving different initial conditions... \$\endgroup\$ – Brian Drummond Dec 21 '17 at 13:30
  • \$\begingroup\$ Thank Brian. The initial condition is Vc(t=0s)=0V. I did try to run over 10 * time constant but the result didn't change. \$\endgroup\$ – dejoker Dec 21 '17 at 13:59
  • \$\begingroup\$ Then are you sure that is a current source? \$\endgroup\$ – Brian Drummond Dec 21 '17 at 14:13
  • \$\begingroup\$ Yes, it is. I'm modeling a practical AC sinusoidal current source (1 ideal current source // big resistor) \$\endgroup\$ – dejoker Dec 21 '17 at 14:14

The initial charge of the capacitor is 0. You then start sucking out current through the current source (it starts with the negative cycle). The voltage can only decrease. Eventually the current source goes positive, the capacitor will get some charge again, and its voltage will increase back to the initial condition, 0 volt. Then the cycle repeats.

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  • \$\begingroup\$ Thank pipe. Intuitively, your explanation really makes sense. What I don't understand here is if I do the calculation, Vc(t) has no DC component. Am I missing something here? Thank you very much. \$\endgroup\$ – dejoker Dec 21 '17 at 13:58
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    \$\begingroup\$ @dejoker Ok, I haven't done this in a long time, but is it possible that you calculate this using an integral, and the constant of integration is in play? You just assume it's zero, when it can be any number. \$\endgroup\$ – pipe Dec 21 '17 at 14:33
  • \$\begingroup\$ @dejoker the issue is the reference point. BEcause the current source is tied to ground on one side it is not really an AC source the way you are thinking of one. The reference point you are calculating is based on a reference in the middle of the current source. Try using two identical 2A sources in series and put ground in between them. \$\endgroup\$ – Trevor_G Dec 21 '17 at 14:35
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    \$\begingroup\$ @pipe maybe you didn't do this in a long time, but you certainly did not forget how it works ;) I second your opinion. \$\endgroup\$ – Vladimir Cravero Dec 21 '17 at 14:43
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    \$\begingroup\$ @dejoker ya I tried that yesterday.. for some reason spice did not like having two current sources in series like that. But that aside, your issue here is your math. This question is pretty much done. You should really start a new question and show your math so someone can see, and explain to you, where your math goes sideways. \$\endgroup\$ – Trevor_G Dec 22 '17 at 14:06

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