nextnano^{3}  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)
Note: This tutorial's copyright is owned by Stefan Birner,
www.nextnano.com.
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 8band k.p model and calculate the energy dispersion E(k_{})
= E(k_{x},k_{y}).
However, for simplicity, and in order to be able to check our k.p results
with analytical results of the singleband approximation (i.e. parabolic and
isotropic energy dispersion), we modify our 8band 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 [m_{0}] for our 8band
k.p model.
The density of states for a single subband in a (GaAs) quantum well system can be calculated
to be:
rho^{2D}(E) = m*
/ (pi h_{bar}^{2}) =
= 0.067
[m_{0}] / (pi h_{bar}^{2}) =
= 0.067
* 2.607275116 * 10^{37} J^{1} m^{2} =
= 0.067
* 4.177314998 * 10^{18} eV^{1} m^{2} =
=
1.746874328 * 10^{36} J^{1} m^{2} =^{
} =
2.798801048 * 10^{17} eV^{1} m^{2}
[m_{0}] / (pi h_{bar}^{2}) = 2.607275116 * 10^{37} J^{1} m^{2}
= 4.177314998 * 10^{18} 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 L_{z} = 10 nm, L_{z} = 20 nm
and L_{z} = 100 nm.
In order to compare the four cases
 10 nm QW
 20 nm QW
 100 nm QW
 bulk (analytical formula)
we scale
rho^{2D}(E) by dividing by the length of the QW:
rho^{2D ' }(E) =
rho^{2D}(E) / L_{z}
where L_{z} is the length of the QW. Then rho^{2D '} 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 E^{1/2} dependence is expected.
(The bulk curve has been calculated analytically.)
Our results are in excellent agreement with analytical results for the
twodimensional 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 nextnano³ output DOS_el_sum.dat depends on the size of the k_{}
space region: ~ 1 / k_{,max}^{2
} For this output, it seems
that the spin is included twice. So these values have to be divided by "2".
The nextnano³ 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 (8band k.p)?
Comment: There are different definitions of the density of states.
 P. Harrison: rho^{2D}(E) = m*
/ (pi h_{bar}^{2})
 G.
Bastard: rho^{2D}(E) = m*
/ (pi h_{bar}^{2}) * L_{x} L_{y}
 S.L.
Chuang: rho^{2D}(E) = m*
/ (pi h_{bar}^{2}) / L_{z}
Other results (benchmark):
 Paul Harrison for a 20 nm GaAs QW with infinite barriers:
1.746847 * 10^{36} J^{1} m^{2} =
2.798757 * 10^{17} eV^{1} m^{2} for an electron
mass of m* = 0.067 [m_{0}].
This corresponds to
rho^{2D ' }(E) =
rho^{2D}(E) / L_{z} = 1.399 * 10^{25} eV^{1} m^{3}
f
 S.L. Chuang for a 10 nm GaAs QW (p. 90) at the first step (E_{1}
= 56.5 meV): 2.78 * 10^{25} eV^{1} m^{3}
 S.L. Chuang for bulk GaAs (p. 90): 100 meV above E_{c}: 3.69 * 10^{25} 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 SiO_{2} oxide layer is included in the simulations.
 The Si/SiO_{2} valence band offset (VBO) was assumed to be 4.3
eV.
 The grid spacing in real space was 0.1 nm.
 The following 6band k.p parameters are employed for both the Si and
the SiO_{2} layer:
6x6kpparameters = 6.53d0 4.64d0
8.32d0 ! L, M, N
0.044d0
! Delta_{splitoff} [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(k_{x},k_{y}) is discretized on a
square up to k_{x,max} = k_{y,max} = 0.7 [1/Angstrom]
krange =
0.7d0 ! [1/Angstrom]
numkpparallel = 4761 ! 4761
= 595 x 8 + 1
using a total of 4761
points in the twodimensional (k_{x},k_{y}) space.
Due to symmetry, we only have to discretize ~1/8^{th} of the total 2D
k_{} space, i.e. the 6band 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 pchannel 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 splitoff
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 splitoff hole
are also shown.
At k_{} = 0, the energies and the squares of the spinup and
spindown wave functions are identical.
This is not the case for k_{} /= 0 because the quantum well
potential is unsymmetric.
The following figure shows the twodimensional 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 spinup and spindown states are
not equal any more leading to different DOS for spinup and spindown.
Here, we want to emphasize the different DOS for heavy hole,
light hole and
splitoff hole (thus we added the spinup and spindown DOS for each
subband).
One can nicely see that the hole DOS differs dramatically from the steplike DOS
of the analytical isotropic and parabolic effectivemass 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 topmost heavy hole state is at E_{hh1}
= 0.672 eV
 the topmost light
hole state is at E_{lh1} = 0.696 eV
 the topmost splitoff hole state
is at E_{so1} = 0.784 eV .
Obviously, between 0 eV and the topmost 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 splitoff hole DOS differs.
Namely, it is not clear why Krishnan's splitoff 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(k_{x},k_{y})
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 .
