Solution of the Poisson equation for different charge density profiles¶

Input file:
• 1D_Poisson_dipole_nnpp.in

• 1DPoisson_linear_nnp.in

• 1D_Poisson_delta_nnpp.in

Note

If you want to obtain the input files that are used within this tutorial, please check if you can find them in the installation directory. If you cannot find them, please submit a Support Ticket.

Scope:

In this tutorial we show solution of Poisson equation for constant, linear and delta-function like charge density profile of positive and negative charges.

Output files:
• bias_00000\density_electron.dat, bias_00000\density_hole.dat

• bias_00000\electric_field.dat

• bias_00000\potential.dat

1) Dipole: Constant charge density profile of positive and negative charge¶

Input file: 1D_Poisson_dipole_nnpp.in

The following figures (Figure 2.5.2.58 and Figure 2.5.2.59) show a dipole charge density distribution where

• the left region (from x = 0 nm to x = 10 nm) carries a constant positive charge density (resulting from ionized donors $$N_D^+$$) and

• the right region (from x = 10 nm to x = 20 nm) carries a constant negative charge density (resulting from ionized acceptors $$N_A^-$$).

We have to solve the Poisson equation:

$\frac{d^2 \phi}{d x^2} = - \frac{\rho}{\epsilon_r \epsilon_0}$

Figure 2.5.2.60 shows the corresponding electric field distribution and Figure 2.5.2.61 shows the electrostatic potential profile

The electric field is given by

$E(x) = - \frac{d \phi}{d x}$

and has a linear dependence (~ -$$x$$) because the electrostatic potential has a quadratic dependence (~ $$x^2$$). The maximum value of the electric field is given by:

$E_{\mathrm{max}} = \frac{\rho}{\epsilon_r \epsilon_0}\cdot x_0 = \frac{e \cdot 1\cdot10^{18} \mathrm{cm}^{-3}}{ 12.93 \cdot 8.8542\cdot10^{-12} \mathrm{As/Vm} } \cdot 10 \mathrm{nm} = 139.95 \mathrm{kV/cm}$

where $$x_0$$ is the width of the positive (or negative) charge density region, and $$\epsilon_r$$ = 12.93 is the static dielectric constant of $$GaAs$$.

The drop of the electrostatic potential between 0 nm and 20 nm is simply given by the area that is below the graph of the electric field:

$\Delta\phi = \frac{1}{2} E_{\mathrm{max}} \cdot 20 \mathrm{nm} = 139.95 \mathrm{mV}$

2) Linear charge density profile of positive and negative charge¶

Input file: 1D_Poisson_linear_nnpp.in

The following figures (Figure 2.5.2.62 and Figure 2.5.2.63) show a linearly varying charge density distribution where

• the left region (from x = 0 nm to x = 10 nm) carries a linearly decreasing positive charge density (resulting from ionized donors $$N_D^+$$) and

• the right region (from x = 10 nm to x = 20 nm) carries a linearly increasing negative charge density (resulting from ionized acceptors $$N_A^-$$).

Figure 2.5.2.64 shows the corresponding electric field distribution and Figure 2.5.2.65 shows the electrostatic potential profile

The electric field shows a quadratic dependence (~ $$-x^2$$) whereas the electrostatic potential shows a cubic dependence (~ $$x^3$$).

3) Delta-function like charge density profile of positive and negative charges¶

Input file: 1D_Poisson_delta_nnpp.in

The following figures (Figure 2.5.2.66 and Figure 2.5.2.67) show a delta-function like charge density distribution where

• in the middle of the structure (x = 0 nm) there is a constant positive charge density of width 1 nm (resulting from ionized donors $$N_D^+$$) and

• at the boundaries of the structure there are constant negative charge densities of width 1 nm each (resulting from ionized acceptors $$N_A^-$$).

Figure 2.5.2.68 shows the corresponding electric field distribution and Figure 2.5.2.69 shows the electrostatic potential profile