Open PDE toolbox by typing ‘pdetool’ in the command prompt Resize the working area from Options -> Axis Limits..

Potential distribution of charged dielectric circular disk

Draw a rectangular region (20x20 or any other shape) where you want to solve the PDE. Name the geometry, will help defining complicated geometry. Draw a small circle (0.2 diameter with 0,0 center) that we would treat as Point charge source.

Select PDE -> PDE Mode Enable Show Subdomain Labels Click on one of the Subdomain (1 or 2) Select Type of PDE (Elliptic for solving Poission’s/Laplace Eq) Small circle (region 2) is considered as a charged source (ρ=1) with dielectric constant 1 (Poission region).

Define rest of the solution region (region 1), with dielectric Constant of 1 and no charge source (ρ=0, i.e, f=0, Laplace region). Switch to Boundary mode.

Select all the edges (Shift+ click) of solution region (region 1), click and choose Dirichlet boundary (red). Dirichlet boundary enforces specific value of solution at selected Boundary edge. Set r=0, which will imply V=0 at these Edges. Dirichlet is the default boundary condition in PDE Toolbox. Initiate simulation mesh.

Refine simulation mesh for more accurate numerical result. Solve PDE

Open Plot Selection Click plot Select Height (3-D) plot

Potential of charged metal circular disk

Draw circular metal disk inside rectangular Solution region. Change Set formula as subtraction of charged.metal from solution.region (- instead of +) Switch to PDE Mode and define solution region (1) With dielectric constant of 1, no charge (ρ=0, i.e, f=0, Laplace region).

Switch to Boundary Mode Select Dirichlet boundary at the metal edge. Set r=10. This enforces V=10 volt at metal edge.

Enforce V=0 volt at the edges of solution region. Mesh and solve

Parallel plate capacitor

Draw metal plates of parallel plate capacitor (1x20 rectangle or other size). Subtract metallic rectangles from solution region (to enforce Diriclet boundary).

Switch to PDE mode and make sure there is no charge (ρ=0, i.e, f=0, Laplace region) in solution region

10V 0V Set 10V and 0V at two capacitor plates (Dirichlet condition). Set Neumann boundary condition At the edges of simulation region. Neumann boundary (blue) sets the slope of solution (rate of change). Set q=0 to enforce continuous boundary at simulation edge. # In previous two cases, we could not apply Neumann boundary, because we needed to define two voltage values in order to get non-trivial solution.

Mesh and solve Add two rectangular (2x20 or other size) dielectric region between parallel plates

From PDE Mode, define dielectric constant of dielectric1 region as 5. And dielectric constant of dielectric2 region As 0.1. These dielectric regions will overwrite the Background solution region. Make sure there is no charge in dielectric1, dielectric2 or solution region (ρ=0, i.e, f=0, Laplace region).

Set boundary condition (10V and 0V at capacitor plates, Neumann condition at the edges of solution region). Dielectric1 and dielctric2 are part of solution region. We do not need to apply any boundary condition at their edges. Mesh, and solve.

Change color map from cool to jet for better visualization.

pn junction diode electrostatic metal contact

Draw the representative pn junction structure (relative dimensions and charge density are important From this problem) Rectangular simulation region (60x20) Space charged region of p-side (10x20) Space charged region of n-side (10x20)

Switch to PDE Mode Specify the charge density and dielectric Constant of semiconductor (11.68 for Si). P-side (region 3) ρ=-0.01 (negative charge in p side, Poission region), ε=11.68 N-side (region 4) ρ=0.01 (positive charge in N side, Poission region), ε=11.68 Rest of the simulation region (1 and two), ρ=0 and ε=11.68

pn junction comes with two metal contacts, one in the p side and other in the n side. (We can consider other sides as semiconductor or continuous.) Apply Neumann boundary to every edge except one of the metal contact on either p side or n side. You may apply 0V or (any other voltage) on p-side Metal contact.

Mesh and solve Is the diode floating/ forward bias/ reverse bias? What will happen if you apply voltage on the other metal also?