Simulation of a nonlinear capacitor

While it is true that LTSpice supports it, i believe that it is generic to all spice v.3

Haven't a clue if that is over OP's head or not, but now i know who to call when the math is getting tough for me.

I get the same result, but have a problem that v is imaginary for times greater than 0.9 sec. I guess the way to interpret this is that this "capacitor" really discharges to 0 V in a finite time, and simply remains 0 thereafter.

Mark

Hello Mark,

When the capacitor is discharged by the resistor, its energy is changed to heat in the resistor. That means that all the energy is lost.

E = 1/2*C*V^2

After V has become 0, E will be 0:

-> V remains 0V after the discharge.

Best regards, Helmut

"JosephKK" schrieb im Newsbeitrag news: snipped-for-privacy@4ax.com...

Hello,

A nice wish regarding SPICE3, but it's not available there! I have the feeling there isn't much activity around SPICE3 since a decade.

Helmut

The current base version from Berkeley is v.3f5. The initial development of v.3 included all kinds of added non-linear components changes and a complete set of linear or non-linear dependant sources.

"JosephKK" schrieb im Newsbeitrag news: snipped-for-privacy@4ax.com...

Hello Joseph,

SPICE 3f5 is from 1997.

Do you have any link about a manual-page with an arbitrary capacitor model? I couldn't find one.

Best regards, Helmut

It does not seem to be listed on the Berkeley SPICE models page:

but with a pair of non-linear dependant sources is should be pretty easy to make a macro for one. In fact they provide an annotated subcircuit for a non-linear capacitor in the non-linear source description.

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Hello Helmut,

Much of what you say is also true of standard linear capacitors, which of course do not go to zero volts in a finite time.

After reviewing my derivation, I found that as V --> 0, the current becomes infinitely large, thereby draining all the charge. At V=0, the differential equation I had come up with is no longer valid, so I guess it makes sense in a way that the solution is not valid for times beyond when V reaches 0.

Regards, Mark