Default value for capacitor instance initial condition

[mildspice]

Hello QSpice forum! I am a beginner in Spice circuit simulation.

The documentation for the capacitor shows the default value for IC as 0.
I tried to simulate a simple RC circuit and got these curves.
It looks like the actual default IC value for a capacitor is 1. Is this a bug or a typo in the documentation?

Thank you!

[Olin]

I’m new too, but isn’t IC the initial current at start of simulation? That’s what your plot of voltage over time also shows.

[jay314159265]

Try setting v1 to 5v and see what happens

[KSKelvin]

In SPICE (not only Qspice), transient simulation (.tran) starts (i.e. at t=0s) with certain initial condition. In default, initial voltage/current conditions are calculated from bias point / DC operating point simulation (.op) [Operation point values not corresponding to plots (using PWL controlled DC supplies)]. In short, SPICE performed a .op simulation before .tran begin, as to take the initial condition value for .tran to be used.

For your circuit, DC operating point is to assume C1 fully charged over time, where V(X)=V1.
I think Qspice Help should describe IC as “not set” to prevent confusion for SPICE beginner.

[mildspice]

I am almost sure “IC” stands for “Initial Condition”. For a capacitor the IC is a certain user-specified voltage across it. There is the IC for an inductor as well. It is a certain user-specified current through it.

I wonder if there are other basic devices that have the IC defined and what its meaning is in those cases

[mildspice]

@jay314159265

Try setting v1 to 5v and see what happens

Thank you jπ for the suggestion. I know see the flaw in my thinking. I’ve got to wrap my head around how to think about the initial conditions in terms of simulations.

[Olin]

Oops, I meant to say initial condition. Initial current doesn’t make any sense for a capacitor. I just tried, but apparently you can’t edit posts in this forum to fix errors like that.

[mildspice]

Thank you @KSKelvin for the explanation.

Interestingly, for the inductor the default value of IC is “none” (matching your suggestion) while for the capacitor it is “0.0”.

Also, I found this reply from Mike Engelhardt from July 2023 talking about the use of the UIC option of the .tran command

Initial conditions not working

UIC has different meanings to different simulators. To Berkeley SPICE, it means don’t do the .op, i.e., it’s like the PSpice skipbp. However, in Berkeley SPICE, that is when instance initial conditions are used. I might change the QSPICE behavior to be more like the PSpice behavior because that is more common. Anyway, as of today, just remove “uic”: [image] If you want to use UIC, you have to put the condition on the instance. But then the .ic V(…)=… statements are ignored: [image]

If the UIC option is used, what ICs will be used for the components without explicitly defined ICs? Will the defaults be used - “none” for inductors and “0.0” for capacitors?

[KSKelvin]

Refer to this link : UIC
Node voltage and current are considered zero at t=0 if not specified with instance parameter IC or .ic directive when UIC is utilized. Qspice will not store the t=0 data point if UIC is set.