Electronic band structure of 2DHG in Si inversion layers under arbitrary stress | 1D¶

Introduction¶

This tutorial aims at reproducing figures Fig. 2, Fig. 3., and Fig. 5 of [Wang2004]. These figures are presenting first subband energy contours of 2D hole gas (2DHG) in Si inversion layer with an effective field of 0.5 MV/cm and under several types of stress conditions:

• without any stress applied,

• with an uniaxial 1 GPa stress applied along [110],

• with a biaxial 1.7 GPa compressive stress applied along [100] and [010],

• with a biaxial 1.7 GPa tensile stress applied along [100] and [010].

Coordinate systems¶

As the growth direction [001] is set along the $$z$$-axis in [Wang2004], the electronic band structures are spanned by [100] and [010] corresponding to $$x$$-axis and $$y$$-axis, respectively. Therefore, the wave-vector coordinates for electronic band structures $$k_x$$ and $$k_y$$ correspond to [100] and [010], repectively, as well.

Differently, the growth direction in the simulations presented in this tutorial is always set along the $$x$$-axis with [001] set along it. The remaining directions [100] and [010] are permutated accordingly to align with $$y$$-axis and $$z$$-axis, respectively. Therefore, the wave-vector coordinates for electronic band structures in the simulations $$k_y$$ and $$k_z$$ correspond to [100] and [010], repectively, as well.

As a result, crystallographic directions in the simulations of this tutorial are exactly aligned with the [Wang2004] while the simulation coordinate system is defined differently.

Defining the strain tensor¶

Here, we introduce how to calculate strain and import it to the simulation.

The relationship between the stress tensor ($$\sigma_{ij}$$) and the strain tensor ($$e_{ij}$$) for the crystals with zincblende symmetry is expressed as (2.4.48).

(2.4.48)$\begin{split}\begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{zz} \\ \sigma_{yz} \\ \sigma_{zx} \\ \sigma_{xy} \\ \end{bmatrix} = \begin{bmatrix} C_{11}&C_{12}&C_{12}& & & \\ C_{12}&C_{11}&C_{12}& & & \\ C_{12}&C_{12}&C_{11}& & & \\ & & &C_{44}& & \\ & & & &C_{44}& \\ & & & & &C_{44} \end{bmatrix} \begin{bmatrix} \varepsilon_{xx} \\ \varepsilon_{yy} \\ \varepsilon_{zz} \\ 2\varepsilon_{yz} \\ 2\varepsilon_{zx} \\ 2\varepsilon_{xy} \\ \end{bmatrix}\end{split}$

Hint

See Introduction to strain calculation for further reference.

Uniaxial stress along [110]¶

First, we consider 1 GPa of uniaxial stress along the [110] direction. Uniaxial stress in the orthogonal coordinate system can be calculated using the method shown in uniaxial stress. Then, related stress tensor in GPa units is

$\begin{split}\sigma_{[110]} = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{zx} \\ * & \sigma_{yy} & \sigma_{yz} \\ * & * & \sigma_{zz} \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & -0.5 & -0.5 \\ 0 & -0.5 & -0.5 \\ \end{bmatrix}.\end{split}$

Thus, you can solve the following simultaneous equations to obtain the strain components.

\begin{split}\begin{aligned} \sigma_{xx} & = C_{11}\varepsilon_{xx} + C_{12}\varepsilon_{yy} + C_{12}\varepsilon_{zz} = 165.77 \cdot \varepsilon_{xx} + 63.93 \cdot \varepsilon_{yy} + 63.93 \cdot \varepsilon_{yy} = 0\\ \sigma_{yy} & = C_{12}\varepsilon_{xx} + C_{11}\varepsilon_{yy} + C_{12}\varepsilon_{zz} = 63.93 \cdot \varepsilon_{xx} + 165.77 \cdot \varepsilon_{yy} + 63.93 \cdot \varepsilon_{zz} = -0.5\\ \sigma_{zz} & = C_{12}\varepsilon_{xx} + C_{12}\varepsilon_{yy} + C_{11}\varepsilon_{zz} = 63.93 \cdot \varepsilon_{xx} + 63.93 \cdot \varepsilon_{yy} + 165.77 \cdot \varepsilon_{zz} = -0.5\\ \sigma_{yz} & = 2C_{44}\varepsilon_{yz} = 2 \cdot 79.62 \cdot \varepsilon_{yz} = -0.5\\ \sigma_{zx} & = 2C_{44}\varepsilon_{zx} = 2 \cdot 79.62 \cdot \varepsilon_{zx} = 0\\ \sigma_{xy} & = 2C_{44}\varepsilon_{xy} = 2 \cdot 79.62 \cdot \varepsilon_{xy} = 0\\ \end{aligned}\end{split}

As a result,

\begin{split}\begin{aligned} \varepsilon_{xx} & = 0.00214\\ \varepsilon_{yy} & = \varepsilon_{zz} = -0.00277\\ \varepsilon_{yz} & = -0.00314\\ \varepsilon_{zx} & = \varepsilon_{xy} = 0\\ \end{aligned}\end{split}

This data is contained at 2DHG-strained-bands_Si_Wang_2004_1D_nnp_uniax_strain.dat.

Hint

For guidance on importing strain to simulation follow — NEW — Importing files.

Biaxial compressive stress along [100] and [010]¶

Next, we consider 1.7 GPa of biaxial compressive stress along [100] and [010]. Related stress tensor in GPa units is

$\begin{split}\sigma_{[110]} = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{zx} \\ * & \sigma_{yy} & \sigma_{yz} \\ * & * & \sigma_{zz} \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & -1.7 & 0 \\ 0 & 0 & -1.7 \\ \end{bmatrix}.\end{split}$

Thus, you can solve the following simultaneous equations to obtain the strain components as well as in the case of uniaxial stress.

\begin{split}\begin{aligned} \sigma_{xx} & = C_{11}\varepsilon_{xx} + C_{12}\varepsilon_{yy} + C_{12}\varepsilon_{zz} = 165.77 \cdot \varepsilon_{xx} + 63.93 \cdot \varepsilon_{yy} + 63.93 \cdot \varepsilon_{yy} = 0\\ \sigma_{yy} & = C_{12}\varepsilon_{xx} + C_{11}\varepsilon_{yy} + C_{12}\varepsilon_{zz} = 63.93 \cdot \varepsilon_{xx} + 165.77 \cdot \varepsilon_{yy} + 63.93 \cdot \varepsilon_{zz} = -1.7\\ \sigma_{zz} & = C_{12}\varepsilon_{xx} + C_{12}\varepsilon_{yy} + C_{11}\varepsilon_{zz} = 63.93 \cdot \varepsilon_{xx} + 63.93 \cdot \varepsilon_{yy} + 165.77 \cdot \varepsilon_{zz} = -1.7\\ \sigma_{yz} & = 2C_{44}\varepsilon_{yz} = 2 \cdot 79.62 \cdot \varepsilon_{yz} = 0\\ \sigma_{zx} & = 2C_{44}\varepsilon_{zx} = 2 \cdot 79.62 \cdot \varepsilon_{zx} = 0\\ \sigma_{xy} & = 2C_{44}\varepsilon_{xy} = 2 \cdot 79.62 \cdot \varepsilon_{xy} = 0\\ \end{aligned}\end{split}

As a result,

\begin{split}\begin{aligned} \varepsilon_{xx} & = 0.00727\\ \varepsilon_{yy} & = \varepsilon_{zz} = -0.00277\\ \varepsilon_{yz} & = \varepsilon_{zx} = \varepsilon_{xy} = 0\\ \end{aligned}\end{split}

This data is contained at band-structure-2DHG_Si_Wang_2004_1D_nnp_biax_comp_strain.dat.

Biaxial tensile stress along [100] and [010]¶

Next, we consider 1.7 GPa of biaxial tensile stress along [100] and [010]. You just need to change the signs of the strain components in the previous section.

Therefore,

\begin{split}\begin{aligned} \varepsilon_{xx} & = -0.00727\\ \varepsilon_{yy} & = \varepsilon_{zz} = 0.00277\\ \varepsilon_{yz} & = \varepsilon_{zx} = \varepsilon_{xy} = 0\\ \end{aligned}\end{split}

This data is contained at band-structure-2DHG_Si_Wang_2004_1D_nnp_biax_tens_strain.dat.

Simulation results¶

No stress applied¶

First, Figure 2.4.278 (a) shows the energy dispersion under no strain. This corresponds to Fig.2 in [Wang2004]. The electric field is applied to model a triangular well potential, which causes the inversion layer. The magnitude is $$0.5\;\mathrm{MV/cm}$$ along the crystal growth direction.

The energy dispersion is in Dispersions\dispersion_quantum_region_kp6_XXXX.fld.

Next, Figure 2.4.278 (b) shows the energy dispersion under uniaxial compressive strain. This is equivalent to Fig.3 in [Wang2004]. Note that the uniaxial stress is $$1.0\;\mathrm{GPa}$$ and the direction is [110]. Furthermore, the same magnitude of the electric field is applied as well as in under no strain.

Next, Figure 2.4.278 (c), (d) shows the energy dispersion under compressive / tensile biaxial strain, respectively. This corresponds to Fig.5 in [Wang2004]. Note that the biaxial stress is $$1.7\;\mathrm{GPa}$$ and the direction is in-plane. The same magnitude of the electric field is applied as well as in under no strain. Here, the lowest subband is composed by heavy hole in (a), whereas light hole composes the lowest subband in (b).

Overall, our simulation results match very well with the results in [Wang2004].

Last update: 07/03/2024