nextnano^{3}  Tutorial
next generation 3D nano device simulator
1D Tutorial
Resistance of a bulk ntype doped silicon sample
Author:
Stefan Birner
If you want to obtain the input file that is used within this tutorial, please contact stefan.birner@nextnano.de.
> bulk_n_Si_current_1D_simba_nn3.in
 input file for the nextnano^{3}
software
> bulk_n_Si_current_1D_nn3.in /
*_nnp.in  input file for the nextnano^{3}
and nextnano++ software
Resistance of a bulk ntype doped silicon sample
> bulk_n_Si_current_1D_simba_nn3.in
Experiment
 Si sample of 1 cm x 1 cm x 1 µm
 ohmic resitance R = 5 kOhm at 300 K
 applied voltage 1 V (along d = 1 cm)
 ntype phosphorous doping with a concentration of 1 * 10^{16}
cm^{3}
Simulation
We consider a onedimensional ntype doped Si sample of length d = 1 cm at room
temperature (300 K).
The Si sample is ntype doped with phosphorous (P) donors with a doping
concentration of N_{D} = 1 x 10^{16} cm^{3}.
At both ends of the device there are ohmic contacts.
We vary the applied voltage in steps of 0.1 V from 0 V to 1 V (i.e. 10 voltage
sweeps:
$voltagesweep ).
Electron mobility
a) bulk_n_Si_current_1D_simba_nn3.in  SIMBA mobility model
b) bulk_n_Si_current_1D_nn3.in  constant mobility model
a) For the mobility which depends on the concentration of ionized impurities we
assume
mobilitymodelsimba0 and use the following parameters:
$mobilitymodelsimba
!
!
materialname =
Si !
taken from the
SIMBA
manual
!
nalphadoping =
0.73 ! []
nNrefdoping =
1.072e17 ! [1/cm^{3}]
nmumin
= 55.2 ! [cm^{2}/Vs]
nmudoping =
1374.0 ! [cm^{2}/Vs]
This leads to an electron mobilty of (for details, see
$mobilitymodelsimba )
µ_{e} = 55.2 cm^{2}/Vs +
1374 cm^{2}/Vs / [ 1 + ( 1*10^{16} cm^{3}
/ 1.072*10^{17} cm^{3})^{0.73} ] =
1222.58 cm^{2}/Vs
The mobility output can be found in this file:
current/mobility_V010.dat (V010 corresponds to an
applied voltage of 1 V in our example.)
The second column contains the electron mobility, the third column the hole
mobility (for each grid point) in units of [cm^{2}/Vs].
Comparison:
 InSb has mobilities of 4 * 10^{5} cm^{2}/Vs.
 Twodimensional electron gases (2DEGs) in AlGaAs/GaAs
heterostructures have mobilities of the order ~10^{7} cm^{2}/Vs.
b) bulk_n_Si_current_1D_nn3.in  constant mobility model
In this input file, a constant mobility is used.
µ_{e} =
1417 cm^{2}/Vs
Mean drift velocity
The mean drift velocity of the electrons at an applied electric field F of F
= U / d = 1 V / 1 cm = 1 V/cm is given as follows:
v_{d,e} = µ * F = µ * U / d = 1222.58 cm^{2}/Vs *
1 V / 1cm = 1222.58 cm/s = 12.23 m/s
The drift velocity output can be found in this file:
current/drift_velocity_V010.dat (V010 corresponds to
an applied voltage of 1 V in our example.)
The second column contains the electron drift velocity, the third column the
hole drift velocity (for each grid point) in units of [cm/s].
Comparison:
 InSb has mean drift velocities of 4 * 10^{5} cm/s = 4
km/s (at a field of 1 V/cm).
 Twodimensional electron gases (2DEGs) in AlGaAs/GaAs
heterostructures have mean drift velocities of the order ~100 km/s (at a field
of 1 V/cm).
Scattering time
The effective scattering time of the electrons t_{eff,e} can be calculated as follows:
t_{eff,e} = µ * m_{e,cond} / e = 1222.58 cm^{2}/Vs *
0.258 m_{0} / e = 1.79 * 10^{13} s = 0.18
ps
where the conductivity electron mass is given by m_{e,cond }= 3 / (1/0.916^{ }+ 2/0.19) m_{0 }=
0.258 m_{0}.
Comparison:
 InSb (m_{e} = 0.0135 m_{0}) has an effective
scattering time of 3.1 ps.
 Twodimensional electron gases (2DEGs) in AlGaAs/GaAs
heterostructures (m_{e} = 0.2 m_{0}) have an effective
scattering time of the order 1.1 ns.
Mean free path
The mean free path is the distance traveled between two collisions.
Assuming that the mean free path is given by l_{mfp} = v_{d,e}
* t_{eff,e} we obtain:
 Si:
l_{mfp} = v_{d,e} * t_{eff,e} = 0.0022 nm
 InSb:
l_{mfp} = 12.4 nm
 AlGaAs/GaAs 2DEG: l_{mfp} = 110 µm
Resistance / Conductivity
The calculated current density j (in units of [A/m^{2}] for a 1D
simulation) can be found in this file:
current/IV_characteristics.dat
The calculated value for an applied voltage of 1 V is j = 19507 A/m^{2}
= 1.9507 A/cm^{2}.
Taking into account the dimensions of the Si sample, this corresponds to a total
current I of
I = 19507 A/m^{2} * 1 cm * 1
µm = 1.9507 * 10^{4} A = 0.2 mA
The ohmic resistance is thus given by R = U / I = 1 V / 1.9507 * 10^{4}
A = 5105.2 Ohm = 5.1 kOhm
The conductivity sigma is given by sigma = j / F = µ_{e} n e = 19507 A/m^{2}
/ 1 V/cm = 195 Ohm^{1}m^{1}
and can be found in this file:
current/conductivity_V010.dat
The conductivity is related to the resistance as follows: sigma = j / F = (I
/ A) / (U / d) = 1 / ( w R )
where w is the thickness of the sample. (Here, w = 1 µm.)
 Please help us to improve our tutorial. Send comments to
support
[at] nextnano.com .
