Risposte dell'homework di Electron Devices and Components

Answers to Electron Devices and Components Homework

1. The wavelength of electromagnetic wave propagating in Silicon and GaAs depends on their

refractive index n and n , which are respectively

Si GaAs = 3.9, = 3.927

Remembering that the formula to calculate the wavelength is

= ∙

and calculating the wavelengths at the given frequencies, the following results have been obtained:

Frequency Silicon [Si] Gallium Arsenide [GaAs]

50 GHz 1538,46 µm 1527,88 µm

250 GHz 307,69 µm 305,58 µm

1 THz 76,92 µm 76,39 µm

From the table, it's evident that the wavelengths of the electromagnetic waves in both Silicon and

GaAs at the given frequencies are much larger than the resolution of extreme UV lithography (13.5 nm).

This highlights the significant difference in scale between typical RF/microwave frequencies and the

nanometer-scale precision required in advanced lithographic techniques.

2. The calculation of the electric field invokes the following formula

=

which expresses the dependance only on the voltage applied and on the distance between the plates

of the capacitor. The dielectric in between, though, increases the total capacitance of the capacitor,

as the following formula shows:

=

0

in which ε is the dielectric permittivity of the air, ε depends on the relative permittivity of the material,

0 r

A stands for the area of the parallel plates and d the distance in between.

Doing the calculations, in the first case (SiO , ε = 3.9), I get

2 r

= 6.916

and in the second one (Hafnia, ε = 20), I get

r = 35.416

3. Looking for the lattice constant of Ge crystal, NanoHub expresses a lattice constant of 5.65791

Å, while Joffe website states that is equal to 5.431 Å. The distribution of these electrons in the orbitals

follows the Aufbau principle, Hund's rule, and the Pauli exclusion principle. The electron configuration

for Germanium is:

Ge: [Ar] 3d 4s 4p

10 2 2

A complete and detailed description of the orbital distribution can be done. In particular, it’s important

to remember that the electrons in the outer shell for this atom are the ones present in 4s and 4p

orbitals, which are 2 for each one, for a total of 4 valence atoms. Due to this fact, it’s possible to see

Ge as element of fourth group in the periodic table.

4. Graph of the Fermi-Dirac and Maxwell-Boltzmann distribution function. Designed in MATLAB

Looking at the formulas for the Fermi-Dirac and Maxwell-Boltzmann statistics, they are respectively

1

() = (− )/

1+