Problem with initial conditions for large Capacitor value

This works fine (I doubt this would be dependent upon the version of spectre - I'm using a recent MMSIM15.1 version though):

// test of ic on big capacitor
parameters del=10 cap=2m
V1 (n1 0) vsource type=pulse rise=1n fall=1n delay=del val0=0 val1=1
R1 (n1 n2) resistor r=1k
C1 (n2 0) capacitor c=cap ic=0.5

tran tran stop=del*5

I get this:

Regards,

Andrew.

Hi Andrew,

Thanks a lot. 

I use spectre  version 11.1.0 64bit, sub-version  11.1.0.341.isr6.

I would like to clearly explain the scenario of my problem.

With Time constant, RC = 0.05 s.

This is my circuit:

CASE 1A  :

R10 (in out) resistor r=50K
V0 (in 0) vsource type=sine ampl=1 freq=250
V11 (net04 net05) vsource type=sine ampl=1 freq=250
C0 (out 0) capacitor c=1u

CASE 1B  :

R10 (in out) resistor r=50K
V0 (in 0) vsource type=sine ampl=1 freq=250
V11 (net04 net05) vsource type=sine ampl=1 freq=250
C0 (out 0) capacitor c=1u ic=10m

Comments : Initial condition satisfied

CASE 2A:

R10 (in out) resistor r=50
V0 (in 0) vsource type=sine ampl=1 freq=250
V11 (net04 net05) vsource type=sine ampl=1 freq=250
C0 (out 0) capacitor c=1m

CASE 2B:

R10 (in out) resistor r=50
V0 (in 0) vsource type=sine ampl=1 freq=250
V11 (net04 net05) vsource type=sine ampl=1 freq=250
C0 (out 0) capacitor c=1m ic=10m

Comments : Initial condition satisfied

CASE 3A:

R10 (in out) resistor r=5
V0 (in 0) vsource type=sine ampl=1 freq=250
V11 (net04 net05) vsource type=sine ampl=1 freq=250
C0 (out 0) capacitor c=10m

CASE 3B:

R10 (in out) resistor r=5
V0 (in 0) vsource type=sine ampl=1 freq=250
V11 (net04 net05) vsource type=sine ampl=1 freq=250
C0 (out 0) capacitor c=10m ic=10m

Comments : Initial condition not satisfied

CASE 4A:

R10 (in out) resistor r=50m
V0 (in 0) vsource type=sine ampl=1 freq=250
V11 (net04 net05) vsource type=sine ampl=1 freq=250
C0 (out 0) capacitor c=1

CASE 4B:

R10 (in out) resistor r=50m
V0 (in 0) vsource type=sine ampl=1 freq=250
V11 (net04 net05) vsource type=sine ampl=1 freq=250
C0 (out 0) capacitor c=1 ic=10m

Comments : Initial condition not satisfied.

So, if you observe all the cases, it is seen that cases 3 and 4 do not correspond to the initial conditions setting. I wonder why that happens in Spectre. I would really appreciate if this is clarified.

Thanks in advance.

Regards,
Pallavi

Your problem is not the large capacitor value but the small resistor value. Quoting from the Spectre User Guide (support.cadence.com/.../cos;src=pubs;releaseName=;DocumentType=manuals;q=/spectreuser/spectreuser15.1/chap17.html#297523):

The Spectre simulator sets initial conditions on a node by attaching a voltage source through a resistor. The default value of this resistor is 1, but you can control the value through the options parameter rforce. You can lower the value of rforce to bring the voltage values into agreement in one of the following ways:

• Through the Analog Options window in the Analog Design Environment (Simulation -> Options -> Analog -> Algorithm)
• Inserting an options statement in the netlist. An example is given below:

myOptions options rforce=1m

Hi Pallavi,

The issue is actually caused by the resistors being so small, not the capacitors being large. In fact in the last case, spectre gives you a warning telling you about the problem (probably does in case 3 too - I didn't go back and re-check):

Notice from spectre during IC analysis, during transient analysis `tran'.
   C0: Initial condition computed for node out is in error by 9.52381 mV
     (95.2381 %).
     Decrease `rforce' to reduce error in computed initial conditions.
     However, setting rforce too small may result in convergence
     difficulties or in the matrix becoming singular.

Initial conditions are met by applying a voltage source in series with a resistor, rforce which defaults to 1ohm. So with a 5 ohm resistor, you'd expect to get the initial condition of approx 5/6*10mV - which is 8.333mV which is pretty much what you see. With a 50mOhm resistor, you're going to get a tiny proportion of the initial condition - hence the message above. If I set the option rforce=1m then you get closer to the initial condition (I would be wary of setting it too much lower than this though - generally speaking tiny resistors are bad for convergence). This rforce can be set on Simulation->Options->Analog or by adding:

myopts options rforce=1m 

in the netlist.

Of course, your circuit is presumably rather artificial - 1F capacitors are not very common on IC circuits... and spectre has been optimised to work with real-life circuits rather than artificial ones like this (so be wary of using excessively large or small component values).

Regards,

Andrew.