Solved Homework 1 Questions - Microelectronic Circuits | ECE 3040, Assignments of Electrical and Electronics Engineering

Download Solved Homework 1 Questions - Microelectronic Circuits | ECE 3040 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040B: Homework #1, Solutions January 18, 2009 Page 1 of 4 GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING ECE 3040B: Microelectronic Circuits Spring Semester 2009, Homework #1 SOLUTIONS 1. Diamond Lattice (20 points) (a) The lattice constant of Ge at room temperature is a = 5.65 Å. Determine the number of Ge atoms per cm3. (b) Calculate the density of crystalline silicon at room temperature using the lattice constant a = 5.43Å, the atomic weight m = 28.09 g/mol, and the Avogadro’s number 6.023·1023 mol-1. Each diamond type crystal (e.g. Si and Ge) has 8 atoms per unit cell: 8 atoms at the corners each of them shared between 8 unit cells 6 atoms on the faces each shared between 2 cells 4 atoms at “inside” locations, e.g. (¼, ¼, ¼) and similar. 8* (1/8)+ 6*(1/2) + 4 = 8. (a) Number of Ge atoms per cm3 = Number of atoms per unit cell divided by unit cell volume a3: #of Geatoms = 8 a3 = 4.4 ⋅1022 cm−3 (b) Density of silicon = # of atoms per cm-3 (= 5 1022 cm-3 from class, or calculated similar to part (a)) * weight of 1 mol (28.09g/mol) / # atoms per mol (6 1023 mol-1) ρ = 5 ⋅10 22 m Navogadro = 8 ⋅28.09 6.023 ⋅1023 g cm3 = 2.33 g cm3 ECE 3040B: Homework #1, Solutions January 18, 2009 Page 2 of 4 2. Cubic Crystal Structures (20 points) In terms of the lattice constant a, what is the distance between nearest neighbor atoms in (a) a bcc lattice? (b) an fcc lattice? 3. Pierret, Problem 1.5, parts (a)–(d) (20 points)