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nextnano3 - Tutorial

next generation 3D nano device simulator

1D Tutorial

Electron density of states (DOS) of a GaAs quantum well with infinite barriers and
hole density of states (DOS) of a Si hole channel (triangular potential)

Author: Stefan Birner

If you want to obtain the input files that are used within this tutorial, please contact stefan.birner@nextnano.de.
-> 1D_DOS_GaAsQW_10nm.in
-> 1D_DOS_GaAsQW_20nm.in
-> 1D_DOS_GaAsQW_100nm.in
-> 1DSi_triangular_DOS_holes.in

1) Electron density of states (DOS) of a GaAs quantum well with infinite barriers

-> 1D_DOS_GaAsQW_10nm.in
-> 1D_DOS_GaAsQW_20nm.in
-> 1D_DOS_GaAsQW_100nm.in

Here, we calculate the density of states (DOS) for the electrons in a 10 nm, 20 nm and 100 nm GaAs quantum well using infinite barriers.

We are using the 8-band k.p model and calculate the energy dispersion E(k||) = E(kx,ky).
However, for simplicity, and in order to be able to check our k.p results with analytical results of the single-band approximation (i.e. parabolic and isotropic energy dispersion), we modify our 8-band k.p parameters so that the coupling between electrons and holes is switched off. Consequently, we are using an isotropic and parabolic effective mass of 0.067 [m0] for our 8-band k.p model.

The density of states for a single subband in a (GaAs) quantum well system can be calculated to be:
rho2D(E) = m*            / (pi hbar2) =
             = 0.067 [m0] / (pi hbar2) =
             = 0.067 * 2.607275116 * 1037 J-1 m-2 =
             = 0.067 * 4.177314998 * 1018 eV-1 m-2 =
             = 1.746874328 * 1036 J-1 m-2 =
             = 2.798801048 * 1017 eV-1 m-2

[m0] / (pi hbar2) = 2.607275116 * 1037 J-1 m-2 = 4.177314998 * 1018 eV-1 m-2


The following figure shows the electron density of states (DOS) as a function of energy for a GaAs QW with infinite barriers.
The QW width has been assumed to be Lz = 10 nm, Lz = 20 nm and Lz = 100 nm.
In order to compare the four cases
- 10 nm QW
- 20 nm QW
- 100 nm QW
- bulk (analytical formula)
we scale rho2D(E) by dividing by the length of the QW: rho2D ' (E) = rho2D(E) / Lz
where Lz is the length of the QW. Then rho2D ' in units of eV-1 m-3.

For a QW, a sharp steplike density of states is obtained, in contrast to the bulk case where a E1/2 dependence is expected.
(The bulk curve has been calculated analytically.)
Our results are in excellent agreement with analytical results for the two-dimensional DOS (e.g. compare with P. Harrison, Quantum wells, wires and dots or S.L. Chuang, Physics of optoelectronic devices).


The DOS output is contained in the following file: DOS_el_sum_norm.dat
The first column contains the energy in units of [eV], the second column the DOS (sum over all subbands) in units of [eV-1 m-2].
The DOS output for each individual subband is contained in this file: DOS_el_norm.dat
The first column contains the energy in units of [eV], the other columns the DOS of each individual subband in units of [eV-1 m-2].

Note: The nextnano output DOS_el_sum.dat depends on the size of the k|| space region: ~ 1 / k||,max2
         For this output, it seems that the spin is included twice. So these values have to be divided by "2".
         The nextnano output DOS_el_sum_norm.dat  does not depend on the size of the k|| space region. The spin factor is correct.


The following parameters were used:

  •   10 nm QW: 0.25 nm grid resolution; 3000 k|| points in total, i.e. 351 k|| have to be calculated
  •   20 nm QW: 0.50 nm grid resolution; 3000 k|| points in total, i.e. 351 k|| have to be calculated
  • 100 nm QW: 1.00 nm grid resolution; 3000 k|| points in total, i.e. 351 k|| have to be calculated


The following interesting question arises:

  • How does the DOS look like for a QW with finite barriers and an anisotropic and nonparabolic effective mass tensor (8-band k.p)?



Comment: There are different definitions of the density of states.

  • P. Harrison:   rho2D(E) = m* / (pi hbar2)
  • G. Bastard:   rho2D(E) = m* / (pi hbar2)  * Lx Ly
  • S.L. Chuang: rho2D(E) = m* / (pi hbar2)  / Lz


Other results (benchmark):

  • Paul Harrison for a 20 nm GaAs QW with infinite barriers:
    1.746847 * 1036 J-1 m-2 = 2.798757 * 1017 eV-1 m-2 for an electron mass of m* = 0.067 [m0].
    This corresponds to rho2D ' (E) = rho2D(E) / Lz = 1.399 * 1025 eV-1 m-3 f
  • S.L. Chuang for a 10 nm GaAs QW (p. 90) at the first step (E1 = 56.5 meV): 2.78 * 1025 eV-1 m-3
  • S.L. Chuang for bulk GaAs (p. 90): 100 meV above Ec: 3.69 * 1025 eV-1 m-3




2) Hole density of states (DOS) of a Si hole channel (triangular potential)

-> 1DSi_triangular_DOS_holes.in

Here, we calculate the density of states (DOS) for the holes in a 10 nm wide Si channel that is modeled by a triangular potential, corresponding to an applied electric field of 5 MV/cm.

  • A 1 nm SiO2 oxide layer is included in the simulations.
  • The Si/SiO2 valence band offset (VBO) was assumed to be 4.3 eV.
  • The grid spacing in real space was 0.1 nm.
  • The following 6-band k.p parameters are employed for both the Si and the SiO2 layer:
      6x6kp-parameters = -6.53d0  -4.64d0  -8.32d0 ! L, M, N
                          0.044d0                  !
    Deltasplit-off [eV]
    Note that our L, M, N parameters are the same as in
    M.V. Fischetti, JAP 80, 2234 (1996)
    but Fischetti's parameters (L',M',N) have to be converted because his parameters are defined differently.
      ==> L = L' - 1, M = M' - 1, N = N'
  • The energy dispersion E(kx,ky) is discretized on a square up to kx,max = ky,max = 0.7 [1/Angstrom]
     k-range         = 0.7d0 ! [1/Angstrom]
     num-kp-parallel = 4761  ! 4761 = 595 x 8 + 1
    using a total of 4761 points in the two-dimensional (kx,ky) space.
    Due to symmetry, we only have to discretize ~1/8th of the total 2D k|| space, i.e. the 6-band k.p Hamiltonian (dimension of matrix: 600 x 600) has to be diagonalized 595 times.

This tutorial is similar to the following paper:

Hole transport in p-channel Si MOSFETs
S. Krishnan, D. Vasileska, M.V. Fischetti
Microelectronics Journal 36, 323 (2005)


The following figure shows the valence band edges of heavy, light and split-off hole at an electric field of 5 MV/cm.
The heavy and light hole valence band edges are degenerate because the silicon layer is assumed to be unstrained.
The square of the wave functions of the heavy hole, light hole and split-off hole are also shown.
At k|| = 0, the energies and the squares of the spin-up and spin-down wave functions are identical.
This is not the case for k|| /= 0 because the quantum well potential is unsymmetric.


The following figure shows the two-dimensional density of states (DOS) for the holes in the 10 nm triangular Si channel at an electric field of 5 MV/cm.
At k|| /= 0 the energies for spin-up and spin-down states are not equal any more leading to different DOS for spin-up and spin-down.
Here, we want to emphasize the different DOS for heavy hole, light hole and split-off hole (thus we added the spin-up and spin-down DOS for each subband).
One can nicely see that the hole DOS differs dramatically from the step-like DOS of the analytical isotropic and parabolic effective-mass approximation. Only the heavy hole DOS is close to this simple approximation.

With respect to this energy scale, the valence band maximum is located at 0 eV.
At k|| = 0,
- the top-most heavy   hole state is at Ehh1 = -0.672 eV
- the top-most light     hole state is at Elh1 = -0.696 eV
the top-most split-off hole state is at Eso1 = -0.784 eV.
Obviously, between 0 eV and the top-most state, the DOS is zero for this subband.

Our heavy and light hole DOS is in good agreement with the paper of Krishnan et al. However, the split-off hole DOS differs.
Namely, it is not clear why Krishnan's split-off hole DOS seems to start at an energy which is equal to the heavy hole energy.


See also this tutorial for more information on the energy dispersion E(kx,ky) of a triangular Si channel:
k|| energy dispersion of holes in unstrained and strained silicon inversion layers


  • Please help us to improve our tutorial. Send comments to support [at] nextnano.com.