# — EDU — Piezo- and Pyroelectric charges in GaN/AlN/GaN wurtzite heterostructure¶

Attention

This tutorial is under construction

## Introduction¶

This tutorial explains how piezo- and pyroelectric polarization constants influence respective charges on interfaces in GaN/AlAn/GaN heterostructure.

You can learn how to introduce a wurtzite structure on nextnano++ in this tutorial. Moreover, this tutorial brings insight into piezoelectricity and pyroelectricity in wurtzite. Section 2.2.3 also helps you to understand piezoelectricity in wurtzite.

The input files simulate a GaN/AlN/GaN wurtzite structure. The structure is grown pseudomorphically on GaN, i.e. the AlN is tensilely strained, whereas the GaN is unstrained. The growth direction [$$0001$$] is along x, the interfaces are in the (y,z) plane. As we initially have [$$0001$$] growth direction, we have Ga-polar GaN (i.e. Ga-face polarity).

## Crystallographic orientation¶

The wurtzite structure belongs to the hexagonal crystal system, therefore you have to be aware of the syntax differences from in a zincblende structure.

15global{}
16    simulate1D{}
17
18    ## This is along [0001] direction: Ga-face polarity
19    crystal_wz{
20        x_hkl = [ 0, 0, 1 ]   # hkil = (0, 0,  0, 1) Miller-Bravais indices
21        y_hkl = [ 1, 0, 0 ]   # hkil = (1, 0, -1, 0) Miller-Bravais indices
22
23    substrate{
24        name = "GaN"
25    }
26}

Although the four-digit Miller-Bravais indices ($$h k i l$$) are usually used in a wurtzite structure, you have to omit $$i$$ in nextnano++ because $$i = h - k$$ holds. x_hkl refers to a plane and perpendicular to the crystal growing direction. Refer to Section 2.6.5 for more information about this topic. As the wurtzite structure lacks c-axis symmetry, the c-plane is polarized. In GaN, + c-plane ($$0001$$) is Ga polar, while the opposite ($$000\overline{1}$$) plane (- c-plane) is N polar. First, we focus on the Ga polarity case, and eventually, we discuss the effect on band offsets in the Na polarity case.

## Strain-induced energy shift¶

### Energy profiles without the strain effects¶

Figure 2.5.1.15 shows the conduction and valence band edges of the heterostructure when no strain is applied. AlN is the barrier for both electrons and holes.

In AlN, the CH (crystal hole) band lies above the HH (heavy hole) and LH (light hole) bands. Meanwhile, the situation is opposite in GaN. This mechanism is explained in [Chuang1996].

Note that heavy and light hole are not degenerate under no-strain condition, unlike in zincblende crystals.

### Including energy shift due to pseudomorphic strain¶

As the substrate in the simulation is GaN, GaN is not unstrained. However, AlN has a smaller lattice constant ($$a_\mathrm{AlN} = 0. 3112\;\mathrm{nm}$$) than the one of GaN ($$a_\mathrm{GaN} = 0. 3189\;\mathrm{nm}$$). Thus, the AlN is strained as following. The biaxial (in-plane) strain is tensile: $$\varepsilon_{\parallel} = (a_\mathrm{substrate} - a) / a = 0.0247429$$ The uniaxial (crystal growth direction) strain is compressive: $$\varepsilon_{\perp} = -2 (c_\mathrm{13} / c_\mathrm{33}) \varepsilon_{\parallel} = -0.0143283$$ The hydrostatic strain is positive which corresponds to an increase in volume for AlN: $$\varepsilon_\mathrm{hy} = Tr(\varepsilon_\mathrm{ij}) = (2 \varepsilon_{\parallel} + \varepsilon_{\perp}) = 0.0351575$$ The strain leads to a shift in the conduction and valence band edges.

The crystal anisotropy leads to two distinct conduction band deformation potentials for the $$\Gamma$$ point in wurtzite. The one is parallel (defpot_absolute_l) and the other one is perpendicular (defpot_absolute_t) to the c axis. You can get those values from database_nnp.in.

3376binary_wz{
3377    name = AlN
3378
3379    conduction_bands{
3380        Gamma{
3381            defpot_absolute_l = -20.5       # Vurgaftman2 (a1) along c axis
3382            defpot_absolute_t = -3.9        # Vurgaftman2 (a2) perpendicular to c axis
3383        }
3384    }
3385}

We replace defpot_absolute_l to $$a_\mathrm{c,c axis}$$ and defpot-absolute_t to $$a_\mathrm{c,a axis}$$ for convenience. Thus, the conduction band edge energy including the hydrostatic energy shift is given by:

\begin{split}\begin{aligned} E_\mathrm{c}' &= E_\mathrm{c} + a_\mathrm{c,c axis}\varepsilon_{\perp} + 2a_\mathrm{c,a axis}\varepsilon_{\parallel} \\ &= 4.712 + (-20.5 \times (-0.0143283)) + 2 (-3.9) \times 0.0247429 \\ &= 4.712 + 0.10073553 \\ &= 4.81274\;\mathrm{eV} \end{aligned}\end{split}

So in this particular example, the barrier for electrons is increased.

Note

Data for uniaxial deformation potentials of other minima than Gamma are not available yet. At the Gamma point, the uniaxial deformation potential is zero.

The six valence band deformation potentials (d1, d2, d3, d4, d5, and d6) arise from a full treatment of the effect of strain on the six-band Hamiltonian. These values are also specified in database_nnp.in.

3376binary_wz{
3377    name = AlN
3378
3379    valence_bands{
3380        defpotentials = [ -17.1, 7.9, 8.8, -3.9, -3.4, -3.4 ] # d1, d2, d3, d4, d5, d6, respectively, Vurgaftman2
3381    }
3382}

In contrast to zincblende, an absolute deformation potential for the valence band is not needed. The shifts of the valence bands are obtained by diagonalizing the Bir-Pikus strain Hamiltonian. This is a general approach as it gives the correct shifts for arbitrary orientations. Note that this holds only for valence bands.

In our example, the tensile strain in AlN shifts all holes upwards, - the heavy hole by $$0.32847\;\mathrm{eV}$$, - the light hole by $$0.32877\;\mathrm{eV}$$ and - the crystal field split-off hole by $$0.64726\;\mathrm{eV}$$, thus reducing the barrier for the holes.

## Polarization Effects¶

### Pyroelectric polarization (spontaneous polarization)¶

The wurtzite material GaN, AlN, and InN are pyroelectric materials and thus show a spontaneous polarization. The pyroelectric field $$P^\mathrm{py}$$ is along the hexagonal x direction (along c-axis) and it is always negative. GaN: $$P^\mathrm{py} = -0.034\;\mathrm{C/m}^2$$, AlN: $$P^\mathrm{py} = -0.090\;\mathrm{C/m}^2$$ At the interfaces, we have a discontinuity of $$P^\mathrm{py}(x)$$. The pyroelectric polarizations at the interfaces are determined as follows:

1st interface at

$100\;\mathrm{nm} (\mathrm{GaN/AlN}): P^\mathrm{py}(\mathrm{left point}) - P^\mathrm{py}(\mathrm{right point}) = -0.034 + 0.090 = 0.056\;\mathrm{C/m^2}$

2nd interface at

$117\;\mathrm{nm} (\mathrm{AlN/GaN}): P^\mathrm{py}(\mathrm{left point}) - P^\mathrm{py}(\mathrm{right point}) = -0.090 + 0.034 = -0.056\;\mathrm{C/m^2}$

The interface charge of $$-0.056\;\mathrm{C/m}^2$$ corresponds to $$34.952 \times 10^{12}\;\mathrm{electrons/cm}^2$$. Once having determined the pyroelectric polarization $$P^\mathrm{py}(x)$$, you can compute the pyroelectric charge density.

$\rho_\mathrm{py}(\boldsymbol{x}) = - div P^\mathrm{py}(\boldsymbol{x})$

If the c-axis is oriented along the x-axis as in our example, this equation reduces to

$\rho_\mathrm{py}(x) = - d/dx P^\mathrm{py}(x).$

The pyroelectric field and charge density (per $$\mathrm{cm}^3$$) are respectively given in the output files polarization_vector_pyroelectric_simulation.dat and density_pyroelectric_charge.dat.

### Piezoelectric polarization¶

If AlN is strained, piezoelectric fields arise. In GaN, the piezoelectric polarization is zero as there is no strain. The piezoelectric constants are specified in database_nnp.in.

3376binary_wz{
3377    name = AlN
3378
3379    piezoelectric_consts{
3380        e31 = -0.50  e33 = 1.79   # Vurgaftman1 (Vurgaftman2 lists d_ij (/= e_ij !) parameters.)
3381        e15 = -0.48               # [Tsubouchi1985] (experiment) and [Momida2016] and O. Ambacher
3382    }
3383}

Be aware that e15 is not relevant for [$$0001$$] growth direction.

The piezoelectric polarization in AlN is:

$P^\mathrm{pz} = e33 \varepsilon_{\perp} + e31(\varepsilon_{\parallel} + \varepsilon_{\parallel}) = 1.79 \cdot (-0.0143283) - 0.50 \cdot 2 \cdot (0.0247429) = -0.050390\;\mathrm{C/m}^2$

GaN: $$P^\mathrm{pz} = 0$$ (no strain), AlN: $$P^\mathrm{pz} = -0.050390\;\mathrm{C/m}^2$$ At the interfaces, we have a discontinuity of $$P^\mathrm{pz}(x)$$. The piezoelectric polarizations at the interfaces are determined as follows:

1st interface at

$100\;\mathrm{nm} (\mathrm{GaN/AlN}): P^\mathrm{pz}(\mathrm{left point}) - P^\mathrm{pz}(\mathrm{right point}) = 0 + 0.050390 = 0.050390\;\mathrm{C/m^2}$

2nd interface at

$117\;\mathrm{nm} (\mathrm{AlN/GaN}): P^\mathrm{pz}(\mathrm{left point}) - P^\mathrm{pz}(\mathrm{right point}) = -0.050390 - 0 = -0.050390\;\mathrm{C/m^2}$

The interface charge of $$-0.050390\;\mathrm{C/m}^2$$ corresponds to $$31.451 \times 10^{12}\;\mathrm{electrons/cm}^2$$. Once having determined the piezoelectric polarization $$P^\mathrm{pz}(x)$$ you can compute the piezoelectric charge density.

$\rho_\mathrm{pz}(\boldsymbol{x}) = - div P^\mathrm{pz}(\boldsymbol{x})$

In our 1D example, the symmetry reduces this equation to

$\rho_\mathrm{pz}(x) = - d/dx P^\mathrm{pz}(x)$

The piezoelectric field and charge density (per $$\mathrm{cm}^3$$) are respectively given in the output files polarization_vector_piezoelectric_simulation.dat and density_piezoelectric_charge.dat.

### Electrostatic potential of piezo- and pyroelectric charges¶

The electrostatic potential is the solution of the nonlinear Poisson equation.

$-div(\varepsilon(\boldsymbol{r}) \cdot grad\phi(\boldsymbol{r})) = \rho(\boldsymbol{r})$

The charge density $$\rho$$ contains the (static) piezo and pyroelectric charge densities as well as the electron and hole charge densities and ionized donors and acceptors. The ionized impurity atoms depend on the electrostatic potential $$\phi$$, however, the piezo and pyroelectric charge densities do not.

Here we plot the electrostatic potential in units of $$\mathrm{eV}$$ for three different cases:

1. include only pyroelectric charges (dark blue line)

2. include only piezoelectric charges (brown line)

3. including both pyro- and piezoelectric charges (purple line)

In our example, the pyro and piezoelectric contributions are of equal magnitude. The Fermi level is set to zero. The electrostatic potential in units of $$\mathrm{eV}$$ has to be subtracted from the conduction and valence edges. The resulting band structure is plotted in Figure 2.5.1.17 (b) (The pyro and piezoelectric charges were considered in this calculation). Note that the conduction band is pulled below the Fermi level (around $$98\;\mathrm{nm}$$) and the valence band above the Fermi level (around $$118\;\mathrm{nm}$$)

### N-face polarity versus Ga-face polarity¶

We simulate GaN/AlN/GaN wurtizte structure in case of the N-face polarity. The only difference from the previous sections is the polarity: N-face instead of Ga-face. We have to switch from [$$0001$$] growth direction to [$$000\overline{1}$$] growth direction (x $$\rightarrow$$ -x). In order to keep a right-handed coordinate axes system, we also invert the y direction (y $$\rightarrow$$ -y). The third direction (z direction) is calculated automatically.

15global{}
16    simulate1D{}
17
18    ## This is along [000-1] direction: N-face polarity
19    crystal_wz{
20        x_hkl = [0, 0, -1]   # hkil = (0,  0, 0, -1) Miller-Bravais indices
21        y_hkl = [-1, 0, 0]   # hkil = (-1, 0, 1,  0) Miller-Bravais indices
22
23    substrate{
24        name = "GaN"
25    }
26}

Figure 2.5.1.18 shows the Fermi level (at $$0\;\mathrm{eV}$$), the electrostatic potential and the conduction and valence band edges (heavy and light hole) of the N-face heterostructure.

Note the difference in comparison to Figure 2.5.1.17 above for the Ga-face case. The location of the 2DEGs and 2DHGs is reversed.

An alternative way of switching from Ga-face to N-face would be to change the sign of the pyroelectric constants.

Last update: 08/03/2024