Geoffrey,
The voltage applied to the load was
in one case 4.3 V, and in the other 5 V.
Acceptable as in: they meet the boundary
conditions of the solver - as far as I can determine. Of course, they are
not acceptable from a user's point of view, and the simulator should break
out with an error. If users persist they can add the GMIN'ish resistors
themselves
The problem with accepting two parallel
voltage sources is that even the tinyest difference between the two would
lead to numerically unstable results, particularly if there is rounding
involved. Consider
module parent();
electrical a,b;
ground b;
child #(.POT(1.0/(1.0/3.0))) ch1(a,b);
child #(.POT(3.0)) ch2(a,b);
endmodule
If 1.0/(1.0/3.0) gets some rounding
errors (it will!) the results will be quite different.
The safest way around this issue is
to use $mfactor for modellign Kevin's parallel ideal batteries. That also
works for the more complicated battery models. A conditional generate construct
would help if differentiation is needed for the individual cells.
Cheers,
Marq
Marq Kole
Competence Leader Robust Design
Research
NXP Semiconductors
"Geoffrey.Coram" <Geoffrey.Coram@analog.com>
Sent by:
geoffrey.coram@analog.com
29-01-2007 13:11
|
|
To
| Marq Kole <marq.kole@nxp.com>
|
|
cc
| verilog-ams <verilog-ams@eda-stds.org>
|
|
Subject
| Re: Potential Contributions
|
|
Classification
| |
|
Marq Kole wrote:
>
> By the way: when putting two unequal voltage sources in parallel
> both solvers that had been able to solve the above test job would
> still converge (with a warning) but generate huge currents whose
> value relates to the absolute current error of the simulators.
> Only one simulator gave a range warning for the currents. Mind you:
> given the error bounds of the DC solvers, these are acceptable results!
This is acceptable?? What was the voltage applied to the load?
(You may have to deduce this from the current.) I can't see
how any answer is acceptable. The simulators in question must
silently place GMIN-like resistors in series with the voltage
sources -- perhaps then the resulting voltage is the "average"
of the two sourced voltages?
-Geoffrey
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Received on Mon Jan 29 04:49:10 2007