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
121        radiative = 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++.

../../../_images/1D_GaN_n_doped_IV.svg

Figure 2.5.1.22 IV curve of an n-doped GaN single layer.

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