hw2_soln - UNIVERSITY OF CALIFORNIA AT BERKELEY College of...

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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 n-type . 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 p-type . b) i) Boron is a Group III element, meaning it is an acceptor. At room temperature, the material will be p-type . 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 don’t, 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 can’t see exactly what the electron and hole mobilities are for N A + N D = 0, but we can 1
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infer that they’re about μ n = 1420 cm 2 / V · s and μ p = 500 cm 2 / V · s. Now we can just plug in these values to the resistivity equation: q = 1 . 602 × 10 - 19 C μ n = 1420 cm 2 / V · s μ p = 500 cm 2 / V · s ρ = 1 q ( n + p ) = 325 . 1 kΩ · cm b) N A = 10 16 cm - 3 N D = 0 cm - 3
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