半导体器件物理与工艺+施敏++答案

Solutions Manual to Accompany SEMICONDUCTOR DEVICES

Physics and Technology

2nd Edition

S. M. SZE

UMC Chair Professor

National Chiao Tung University

National Nano Device Laboratories

Hsinchu, Taiwan

John Wiley and Sons, Inc

New York. Chicester / Weinheim / Brisband / Singapore / Toronto

Contents

Ch.1 Introduction--------------------------------------------------------------------- 0 Ch.2 Energy Bands and Carrier Concentration-------------------------------------- 1 Ch.3 Carrier Transport Phenomena-------------------------------------------------- 7 Ch.4p-n Junction--------------------------------------------------------------------16 Ch.5 Bipolar Transistor and Related Devices----------------------------------------32 Ch.6 MOSFET and Related Devices-------------------------------------------------48 Ch.7 MESFET and Related Devices-------------------------------------------------60 Ch.8 Microwave Diode, Quantum-Effect and Hot-Electron Devices---------------68 Ch.9Photonic Devices-------------------------------------------------------------73 Ch.10 Crystal Growth and Epitaxy---------------------------------------------------83 Ch.11 Film Formation----------------------------------------------------------------92 Ch.12 Lithography and Etching------------------------------------------------------99 Ch.13 Impurity Doping---------------------------------------------------------------105 Ch.14 Integrated Devices-------------------------------------------------------------113

CHAPTER 2

1. (a) From Fig. 11a, the atom at the center of the cube is surround by four

equidistant nearest neighbors that lie at the corners of a tetrahedron. Therefore the distance between nearest neighbors in silicon (a = 5.43 Å) is

1/2 [(a /2)2 + (a 2/2)2]1/2 = a 3/4 = 2.35 Å.

(b) For the (100) plane, there are two atoms (one central atom and 4 corner atoms each contributing 1/4 of an atom for a total of two atoms as shown in Fig. 4a)for an area of a 2, therefore we have

2/ a 2 = 2/ (5.43 × 10-8)2 = 6.78 × 1014 atoms / cm 2Similarly we have for (110) plane (Fig. 4a and Fig. 6)(2 + 2 ×1/2 + 4 ×1/4) /a 22 = 9.6 × 1015 atoms / cm 2,and for (111) plane (Fig. 4a and Fig. 6)

(3 × 1/2 + 3 × 1/6) / 1/2(a 2)(a

23) =

2232

a

= 7.83 × 1014 atoms / cm 2.2. The heights at X, Y, and Z point are ,43,41and 43.

3. (a) For the simple cubic, a unit cell contains 1/8 of a sphere at each of the eight corners for a total of one sphere.

4 Maximum fraction of cell filled

= no. of sphere × volume of each sphere / unit cell volume = 1 × 4ð(a /2)3 / a 3 = 52 %

(b) For a face-centered cubic, a unit cell contains 1/8 of a sphere at each of the eight corners for a total of one sphere. The fcc also contains half a sphere at each of the six faces for a total of three spheres. The nearest neighbor distance is 1/2(a 2). Therefore the radius of each sphere is 1/4 (a 2).4 Maximum fraction of cell filled

= (1 + 3) {4ð[(a /2) / 4 ]3 / 3} / a 3 = 74 %.

(c) For a diamond lattice, a unit cell contains 1/8 of a sphere at each of the eight corners for a total of one sphere, 1/2 of a sphere at each of the six faces for a total of three spheres, and 4 spheres inside the cell. The diagonal distance