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1、1Solutions Manual to AccompanySEMICONDUCTOR DEVICES Physics and Technology 2nd EditionS. M. SZE UMC Chair Professor National Chiao Tung University National Nano Device Laboratories Hsinchu, TaiwanJohn Wiley and Sons, Inc New York. Chicester / Weinheim / Brisband / Singapore / Toronto0Contents Ch.1 I
2、ntroduction- 0 Ch.2 Energy Bands and Carrier Concentration- 1 Ch.3 Carrier Transport Phenomena - 7 Ch.4 p-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 Devi
3、ces - 68 Ch.9 Photonic 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- 1131CHAPTER 21. (a) From Fig. 11a, the atom at the center of the cube is surround by fourequidistant nearest neighbo
4、rs that lie at the corners of a tetrahedron. Thereforethe distance between nearest neighbors in silicon (a = 5.43 ) is1/2 (a/2)2 + (a2/2)21/2 = a3/4 = 2.35 .(b) For the (100) plane, there are two atoms (one central atom and 4 corner atomseach contributing 1/4 of an atom for a total of two atoms as s
5、hown in Fig. 4a)for an area of a2, therefore we have2/ a2 = 2/ (5.43 10-8)2 = 6.78 1014 atoms / cm2Similarly we have for (110) plane (Fig. 4a and Fig. 6)(2 + 2 1/2 + 4 1/4) /a22 = 9.6 1015 atoms / cm2,and for (111) plane (Fig. 4a and Fig. 6)(3 1/2 + 3 1/6) / 1/2(a2)(a 23) = 2 232a = 7.83 1014 atoms
6、/ cm2.2. The heights at X, Y, and Z point are , 43, 41and43.3. (a) For the simple cubic, a unit cell contains 1/8 of a sphere at each of the eightcorners for a total of one sphere.4 Maximum fraction of cell filled= no. of sphere volume of each sphere / unit cell volume= 1 4?(a/2)3 / a3 = 52 %(b) For
7、 a face-centered cubic, a unit cell contains 1/8 of a sphere at each of theeight corners for a total of one sphere. The fcc also contains half a sphere ateach of the six faces for a total of three spheres. The nearest neighbor distanceis 1/2(a2). Therefore the radius of each sphere is 1/4 (a2).4 Max
8、imum fraction of cell filled= (1 + 3) 4?(a/2) / 4 3 / 3 / a3 = 74 %.(c) For a diamond lattice, a unit cell contains 1/8 of a sphere at each of the eightcorners for a total of one sphere, 1/2 of a sphere at each of the six faces for atotal of three spheres, and 4 spheres inside the cell. The diagonal
9、 distance2between (1/2, 0, 0) and (1/4, 1/4, 1/4) shown in Fig. 9a isD = 21222222+ + aaa= 34aThe radius of the sphere is D/2 = 38a4 Maximum fraction of cell filled= (1 + 3 + 4)3 3834 a/ a3 = ?3/ 16 = 34 %.This is a relatively low percentage compared to other lattice structures.4. 1d = 2d = 3d = 4d =
10、 d1d+2d+3d+4d= 01d (1d+2d+3d+4d) = 1d 0 = 02 1d+1d2d+1d3d+ 1d4d= 0 4d2+ d2 cos?12 + d2cos?13 ? d2cos?14 ? d2 +3 d2 cos? 04 cos? = 31?= cos-1 (31) ? 109.47? ?5. Taking the reciprocals of these intercepts we get 1/2, 1/3 and 1/4. The smallestthree integers having the same ratio are 6, 4, and 3. The pl
11、ane is referred to as(643) plane.6. (a) The lattice constant for GaAs is 5.65 , and the atomic weights of Ga and Asare 69.72 and 74.92 g/mole, respectively. There are four gallium atoms andfour arsenic atoms per unit cell, therefore4/a3 = 4/ (5.65 10-8)3 = 2.22 1022 Ga or As atoms/cm2,Density = (no.
12、 of atoms/cm3 atomic weight) / Avogadro constant= 2.22 1022(69.72 + 74.92) / 6.02 1023 = 5.33 g / cm3.(b) If GaAs is doped with Sn and Sn atoms displace Ga atoms, donors areformed, because Sn has four valence electrons while Ga has only three. Theresulting semiconductor is n-type.7. (a) The melting
13、temperature for Si is 1412 C, and for SiO2 is 1600 C. Therefore,SiO2 has higher melting temperature. It is more difficult to break the Si-Obond than the Si-Si bond.(b) The seed crystal is used to initiated the growth of the ingot with the correctcrystal orientation.(c) The crystal orientation determines the semiconductors chemical and electrical3properties, such as the etch rate, trap density, breakage plane etc.(d) The temperating of the crusible and the pull rate.8. Eg (T) = 1.17 636)(4.73x1024+TTfor Si Eg ( 100 K) = 1.163
