This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: UNIVERSITY OF CALIFORNIA AT BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences Homework #2 Solutions Due Tuesday, September 11, 2007 EE 105 Fall 2007 Prof. Liu Problem 1. Doping and Carrier Concentrations a) Ga is a Group III element while Si is a Group IV element, meaning Si has one more valence electron than Ga. Therefore, if the Si atoms exclusively replace Ga atoms, the Si atoms will donate electrons, making the material ntype . Similarly, since As is a Group V element, Si has one fewer valence electron than As. Thus, Si will act as an acceptor when it replaces As atoms and the material will be ptype . b) i) Boron is a Group III element, meaning it is an acceptor. At room temperature, the material will be ptype . The majority carrier concentration is p = N A = 10 17 cm 3 . The minority carrier concentration is n = n 2 i /p = 10 3 cm 3 . ii) The formula from Lecture 1 is as follows: n i = 5 . 2 10 15 T 3 / 2 e E g / 2 kT We need to solve this formula for T when n i = 10 max( n,p ) = 10 18 cm 3 . If you have a graphing calculator, you can use the numerical solver or graphing functions to find that T = 1207 K . If you dont, you can solve by iteration as follows: 10 18 cm 3 = 5 . 2 10 15 T 3 / 2 e E g / 2 kT E g = 1 . 12 eV k = 8 . 62 10 5 eV / K 10 18 5 . 2 10 15 T 3 / 2 = e E g / 2 kT ln bracketleftbigg 10 18 5 . 2 10 15 T 3 / 2 bracketrightbigg = E g / 2 kT T = E g 2 k ln bracketleftBig 10 18 5 . 2 10 15 T 3 / 2 bracketrightBig Starting with an initial guess of T = 500 K, it takes only 7 iterations to converge to T = 1207 K. Problem 2. Resistivity Since mobilities are being obtained from a graph, values within 50 cm 2 / V s are acceptable. a) For undoped silicon, N A = N D = 0 and n = p = n i = 10 10 cm 3 . Using the given chart, we cant see exactly what the electron and hole mobilities are for N A + N D = 0, but we can 1 infer that theyre about n = 1420 cm 2 / V s and p = 500 cm 2 / V s. Now we can just plug in these values to the resistivity equation:...
View Full
Document
 Fall '07
 KingLiu
 Electrical Engineering

Click to edit the document details