# I–V characteristics of n-doped GaN single layer¶

## Introduction¶

This tutorial shows the accuracy of drifft-diffusion model implemented in nextnano++ on a simple example: a single layer of an n-doped GaN. We compare the I–V characteristics obtained by nextnano++ with analytical solutions.

## IV characteristics of an n-doped GaN single layer¶

The conductivity $$\sigma$$ and the resistivity $$\rho$$ of an n-type doped GaN sample can be calculated analytically, following formulas:

\begin{align}\begin{aligned}\sigma = q\mu_{n}n,\\\rho = d/\sigma,\end{aligned}\end{align}

where $$q$$ is electron charge, $$n$$ is concentration of electron carriers, $$\mu_{n}$$ is mobility of electrons, and $$d$$ is thickness of the material.

This is a good check for the results obtained with nextnano++ simulations. The thickness of the GaN layer is $$d = 100\;\mathrm{nm}$$.

The structure we are dealing with consists of bulk GaN that is sandwiched between two contacts. The whole structure has the following dimensions:

material

width ($$\mathrm{nm}$$)

doping

contact

$$10$$

n-GaN

$$100$$

$$1\times10^{18}\;\mathrm{cm^{-3}}$$

contact

$$10$$

As you see, the GaN is n-type doped with a donor concentration of $$N_{D} = 1\times10^{18}\;\mathrm{cm^{-3}}$$. The energy level is chosen to be $$0.01507\;\mathrm{eV}$$ below the conduction band edge.

70impurities{
71    donor{ name = "Si_donor" degeneracy = 2 energy = 0.01507 }
72}


This leads to the electron density of $$5.2846\times 10^{17}\;\mathrm{cm^{-3}}$$. This is also equivalent to the concentration of the ionized donors. The result obtained by another commercial software is $$5.355\times 10^{17}\;\mathrm{cm^{-3}}$$.

61contacts{
62    ohmic{ name = "left_contact" bias = 0.0 }
63    ohmic{
64        name = "right_contact"
65        !WHEN $biassweep bias = [$biasstart, $biasend ] 66 !WHEN$biassweep steps = $biassteps 67 !WHEN$nosweep   bias = $biasstart 68 } 69}  If $biassweep = 1, sweeping bias takes place. Otherwise, if $biassweep = 0 and $nosweep (= 1 - $biassweep) = 1, sweeping bias is not applied. Since the bias is swept from $$0.00\;\mathrm{V}$$ to $$0.10\;\mathrm{V}$$, $biasstart is set to 0.0 and $biasend is set to 0.1. In addition, $biassteps is equal to 10.

We take the GaN mobility to be constant: $$\mu_{n} = 100\;\mathrm{cm^2/Vs}$$. The mobility model that is applied is called constant and described as below.

116currents{
117    mobility_model = constant
118    recombination_model{
119        SRH = no
120        Auger = no
122    }
123    output_currents{}
124}


We sweep the voltage at the right contact and calculate the current density for $$0.00\;\mathrm{V}$$, $$0.01\;\mathrm{V}$$, $$0.02\;\mathrm{V}$$, …, $$0.10\;\mathrm{V}$$ (10 steps).

## Results¶

### 1D¶

The current-voltage (IV) characteristic can be found in the following file: IV_characteristics.dat. Figure 2.5.1.22 shows the IV curve obtained by nextnano++.

The figure shows that the GaN layer is an ohmic resistor. From Figure 2.5.1.22, you can obtain a resistivity of the n-GaN layer of $$1.1819\times 10^{-6}\;\mathrm{\Omega cm^2}$$. Another commercial software results in $$1.43\times 10^{-6}\;\mathrm{\Omega cm^2}$$.

A good check is the analytic formula given above. From this, you can obtain:

$\sigma_{n} = e\mu_{n}n = 1.6022 \times 10^{-19}\;\mathrm{As} \times 100\;\mathrm{cm^{2}/Vs} \times 5.2846\times 10^{17}\;\mathrm{cm^{-3}} = 8.4670\;\mathrm{A/Vcm}$
$\rho = d/\sigma = 100\;\mathrm{nm} / (8.46700\;\mathrm{A/Vcm}) = 1.1811 \times 10^{-6}\;\mathrm{\Omega cm^2}$

Another analytical result with the other commercial software is $$1.168\times 10^{-6}\;\mathrm{\Omega cm^2}$$.

Thus, you can see that the nextnano++ result agrees better with the analytical result than the result by the other commercial software.

### 2D¶

Now, we try the same structure in a 2D nextnano++ simulation to check if the 2D result agrees with the 1D one. The input file IV_GaN_n_doped_2D_nnp.in is used for this section. The width of the sample along the y direction is $$200\;\mathrm{nm}$$. The x direction is the same as in 1D.

Note that the unit for the current in a 2D simulation is $$\mathrm{[A/cm]}$$. Dividing this two-dimensional current value by the width of the device (in our case $$200\;\mathrm{nm}$$), we obtain the current density in units of $$\mathrm{[A/cm^2]}$$ which is the usual unit of a 1D simulation. As our simple 2D example structure is basically equivalent to a 1D structure, we can easily compare our 2D results with the 1D results to check for consistency.

voltage

current ($$\mathrm{A/cm}$$) (nextnano++ 2D)

current density ($$\mathrm{A/cm^2}$$) (nextnano++ 2D*)

current density ($$\mathrm{A/cm^2}$$) (nextnano++ 1D)

$$0$$

$$0$$

$$0$$

$$0$$

$$0.02$$

$$0.33845$$

$$16922.4$$

$$16922.4$$

$$0.04$$

$$0.67689$$

$$33844.7$$

$$33844.7$$

$$0.06$$

$$1.0153$$

$$50767.0$$

$$50767.0$$

$$0.08$$

$$1.3538$$

$$67689.2$$

$$67689.2$$

$$0.10$$

$$1.6922$$

$$84611.2$$

$$84611.3$$

$$*$$ Here, the current density of the 2D simulation is obtained by dividing the current $$\mathrm{[A/cm]}$$ by the width $$200\;\mathrm{nm}$$.

From the IV characteristics obtained from the 2D simulation, you can obtain a resistivity of the n-GaN layer of $$1.1819\times 10^{-6}\;\mathrm{\Omega cm^{2}}$$ which agrees very well with the 1D result (1D: $$1.1819\times 10^{-6}\;\mathrm{\Omega cm^2}$$).

### 3D¶

Of course, it is also possible to simulate this structure in 3D. In this case, the unit of the current is $$\mathrm{[A]}$$ and have to be divided by the area of the device perpendicular to the current flow direction to obtain the units of $$\mathrm{[A/cm^{2}]}$$.

Last update: nn/nn/nnnn