| Author | Message | ||
     Jianhong Yang (yangjh) New member Username: yangjh Post Number: 1 Registered: 08-2005 |
I try to use Flexpde solving the set of equations like this in 2D: V: div(k*grad(V))+(p-n+C)=0 n: div(-un*n*grad(V)+Dn*grad(n)-R=0 p: -div(-up*p*grad(V)-Dp*grad(p))-R=0 These are essentially Poisson's equation plus the continuity equations,in which V,n,and p stand for electrostatic potential,electron concentration,and hole concentration, respectively, in a semiconductor. The problems are: (1) the natural(n or p)=0 and jump(n or p) boundary conditions don't allow a converged solution. In my problem, one or more regions having zero (n,p) are enclosed in the nonzero (n,p) region, but I still have to solve V in that zero (n,p) region(e.g.,MOS structure) (2) Because the prolem is of large quantities for very small dimensions,I cann't get a converged solution; it always appears floating-point overflow during the running (even for equilibrium state). Is there anyone who has encountered similar problems? Could experts show me the handling techniques? (showing me with a MOS structure as an example is the best).Any help is much appriciated. Thanks. | ||
| Jianhong Yang (yangjh) New member Username: yangjh Post Number: 2 Registered: 08-2005 |
The above problems seem disappear,but a "new" problem exists: the particle densities are negative in some area in the domain. The densities should be physically positive. What can I do then? Thanks in advance. | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 432 Registered: 06-2003 |
You have to send me the script, I can't guess. FlexPDE allows an enormous breadth of problem description, and many of the things you can say do not constitute well-posed PDE problems. You can post the script here, or send it to me directly. | ||
| Jianhong Yang (yangjh) Junior Member Username: yangjh Post Number: 3 Registered: 08-2005 |
Thank you for your reply, Mr.R.G.Nelson. After many trivials, the run of the equations as well as the solutions look like what I expect. Thank you again. |