ELEC5508 Wireless Engineering Assignment 1 Questions 2025 | UWA

ELEC5508 Assignment Questions

QUESTION 1

(a)Use the velocity-field relations for Si and GaAs shown in Figure 1 to determine the transit time of electrons through a 1 µm distance in these materials for an electric field of (i) 1 kV/cm (ii) 50 kV/cm

(b)Assume that a conduction electron in Si (µn = 1350 cm2/V-s) has a thermal energy kT, related to its mean thermal velocity by Eth= m0vth 2 /2
(i)This electron is placed in an electric field of 100 V/cm. Show that the drift velocity of the electron in this case is small compared to its thermal velocity
(ii)Repeat for a field of 104 V/cm, using the same value of µn . Comment on the actual mobility effects at this higher value of field.

QUESTION 2

(a)Find the maximum percentage of In, that is, the x value for InxG1-x As film grown on GaAs substrate without formation of misfit dislocation, if the final film thickness is 10 nm.

(b)The lattice misfit, ƒ, of a film is defined as 
ƒ≡ [a0s- a0f]/a0f= ∆a0/a 0 

where a0sand a0f are the unstrained lattice constants of the substrate and film, respectively. Find the ƒvalues for InAs-GaAs and Ge-Si systems. (You may assume they are listed as film-substrate.)

QUESTION 3

(a)The built-in voltage for a p-n heterojunction can be given by: 𝑞𝑞𝑉𝑉𝑏𝑏𝑏𝑏=∆𝐸𝐸𝑣𝑣 +𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 �𝑝𝑝 𝑝𝑝0𝑝𝑝𝑛𝑛0∙𝑁𝑁 𝑣𝑣𝑛𝑛 𝑁𝑁 𝑣𝑣𝑝𝑝 � where pp0and pn0 are the thermal equilibrium hole concentrations in the p and n materials, respectively and Nvnand Nvp are the effective density of states functions in the n and p materials, respectively.

1. Using this and the ideal electron affinity rule given in lectures determine ∆Ec, ∆Ev, and V bi for an n-Ge to p-GaAs heterostructure. Use Ge doping of Nd = 1016cm-3 and GaAs doping of Na = 1016cm-3, and T = 300K.
2. Sketch the energy band diagram for this heterostructure, labelling Eo(vacuum level), Ec , Ev, Ef, χn, χp, ∆Ec, ∆Ec, and Vbi . Make sure to explain the choice of heterostructure type (I, II or III) that you have sketched. (b)Repeat question 3(a)(i) for doping of Nd= 1015cm-3and Na= 1015cm-3.