Solutions Manual to Accompany SEMICONDUCTOR DEVICES Physics and Technology nd 2 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 1 Contents Ch.1 Introduction--------------------------------------------------------------------- 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 Devices ---------------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-------------------------------------------------------------113 0 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 + ( 2a /2)2]1/2 = 3a /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 a2, therefore we have 2/ a2 = 2/ (5.43 × 10-8)2 = 6.78 × 1014 atoms / cm2 Similarly we have for (110) plane (Fig. 4a and Fig. 6) (2 + 2 ×1/2 + 4 ×1/4) / 2a 2 = 9.6 × 1015 atoms / cm2, and for (111) plane (Fig. 4a and Fig. 6) (cid:230) 3(cid:246) 2 (3 × 1/2 + 3 × 1/6) / 1/2( 2a )( (cid:231) (cid:247) a ) = = 7.83 × 1014 atoms / cm2. Ł 2ł (cid:231)(cid:230) 3(cid:247)(cid:246) (cid:231) (cid:247) a2 Ł 2 ł 2. The heights at X, Y, and Z point are 3 1 and 3 . 4, 4, 4 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 × 4(cid:240)(a/2)3 / a3 = 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) {4(cid:240)[(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 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 1 between (1/2, 0, 0) and (1/4, 1/4, 1/4) shown in Fig. 9a is 1 (cid:230) a(cid:246) 2 (cid:230) a(cid:246) 2 (cid:230) a(cid:246) 2 a D = (cid:231) (cid:247) +(cid:231) (cid:247) +(cid:231) (cid:247) = 3 2 Ł 2ł Ł 2ł Ł 2ł 4 a The radius of the sphere is D/2 = 3 8 4 Maximum fraction of cell filled Ø 4p (cid:230) a (cid:246) ø 3 = (1 + 3 + 4)Œ (cid:231) 3(cid:247) œ / a3 = (cid:240) 3/ 16 = 34 %. º 3 Ł 8 ł ß This is a relatively low percentage compared to other lattice structures. 4. d = d = d = d = d 1 2 3 4 d +d +d +d = 0 1 2 3 4 d • (d +d +d +d ) = d • 0 = 0 1 1 2 3 4 1 d 2+d •d +d •d + d •d = 0 1 1 2 1 3 1 4 4d2+ d2 cosŁ + d2cosŁ + d2cosŁ = d2 +3 d2 cosŁ= 0 12 13 14 - 1 4 cosŁ = 3 - 1 Ł= cos-1 ( ) = 109.470 . 3 5. Taking the reciprocals of these intercepts we get 1/2, 1/3 and 1/4. The smallest three integers having the same ratio are 6, 4, and 3. The plane is referred to as (643) plane. 6. (a) The lattice constant for GaAs is 5.65 Å, and the atomic weights of Ga and As are 69.72 and 74.92 g/mole, respectively. There are four gallium atoms and four arsenic atoms per unit cell, therefore 4/a3 = 4/ (5.65 × 10-8)3 = 2.22 × 1022 Ga or As atoms/cm2, Density = (no. 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 are formed, because Sn has four valence electrons while Ga has only three. The resulting semiconductor is n-type. 7. (a) The melting temperature for Si is 1412 ºC, and for SiO is 1600 ºC. Therefore, 2 SiO has higher melting temperature. It is more difficult to break the Si-O 2 bond than the Si-Si bond. (b) The seed crystal is used to initiated the growth of the ingot with the correct crystal orientation. (c) The crystal orientation determines the semiconductor’s chemical and electrical 2 properties, such as the etch rate, trap density, breakage plane etc. (d) The temperating of the crusible and the pull rate. 4.73x10- 4T2 8. E (T) = 1.17 – for Si g (T +636) \ E ( 100 K) = 1.163 eV , and E (600 K) = 1.032 eV g g 5.405x10- 4T 2 E (T) = 1.519 – for GaAs g (T +204) \ E ( 100 K) = 1.501 eV, and E (600 K) = 1.277 eV . g g 9. The density of holes in the valence band is given by integrating the product N(E)[1-F(E)]dE from top of the valence band (E taken to be E = 0) to the V bottom of the valence band E : bottom p = (cid:242) Ebottom N(E)[1 – F(E)]dE (1) { [ ( ) ]} 0 [ ] where 1 –F(E) = 1- 1/ 1 + e E- EF /kT = 1+e(E- EF)/kT - 1 If E – E >> kT then F [ ( ) ] 1 – F(E) ~ exp - E - E kT (2) F Then from Appendix H and , Eqs. 1 and 2 we obtain p = 4(cid:240)[2m / h2]3/2(cid:242) Ebottom E1/2 exp [-(E – E) / kT ]dE (3) p F 0 Let x a E / kT , and let E = - ¥ , Eq. 3 becomes bottom p = 4(cid:240)(2m / h2)3/2 (kT)3/2 exp [-(E / kT)](cid:242) - ¥ x1/2exdx p F 0 where the integral on the right is of the standard form and equals p / 2. 4 p = 2[2(cid:240)m kT / h2]3/2 exp [-(E / kT)] p F By referring to the top of the valence band as E instead of E = 0 we have, V p = 2(2(cid:240)m kT / h2)3/2 exp [-(E – E ) / kT ] p F V or p = N exp [-(E –E ) / kT ] V F V where N = 2 (2(cid:240)m kT / h2)3 . V p 10. From Eq. 18 N = 2(2(cid:240)m kT / h2)3/2 V p The effective mass of holes in Si is m = (N / 2) 2/3 ( h2 / 2(cid:240)kT ) p V (cid:231)(cid:230) 2.66· 1019 · 106m- 3 (cid:247)(cid:246) 23 (6.625· 10- 34)2 = (cid:231)Ł 2 (cid:247)ł 2p(1.38· 10- 23)(300) = 9.4 × 10-31 kg = 1.03 m . 0 Similarly, we have for GaAs m = 3.9 × 10-31 kg = 0.43 m . p 0 11. Using Eq. 19 3 ( ) ( ) E =(E + E ) 2+ kT ln N N i C V 2 V C Ø 2 ø = (E + E )/ 2 + (3kT / 4) ln (m m )(6) 3 (1) C V Œº p n œß At 77 K E = (1.16/2) + (3 × 1.38 × 10-23T) / (4 × 1.6 × 10-19) ln(1.0/0.62) i = 0.58 + 3.29 × 10-5 T = 0.58 + 2.54 × 10-3 = 0.583 eV. At 300 K E = (1.12/2) + (3.29 × 10-5)(300) = 0.56 + 0.009 = 0.569 eV. i At 373 K E = (1.09/2) + (3.29 × 10-5)(373) = 0.545 + 0.012 = 0.557 eV. i Because the second term on the right-hand side of the Eq.1 is much smaller compared to the first term, over the above temperature range, it is reasonable to assume that E is in the center of the forbidden gap. i (cid:242) Etop(E - E ) E - E e- (E- EF)/kTdE C C 12. KE = EC (cid:242) Etop E - E e- (E- EF)/kTdE x” (E- EC) C EC (cid:230) 5(cid:246) (cid:242) ¥ x32e- xdx G (cid:231)Ł 2(cid:247)ł 1.5· 0.5· p = kT 0 = kT = kT (cid:242) ¥ x12e- xdx G (cid:231)(cid:230) 3(cid:247)(cid:246) 0.5 p 0 Ł 2ł 3 = kT. 2 13. (a) p = mv = 9.109 × 10-31 ×105 = 9.109 × 10-26 kg–m/s h 6.626· 10- 34 l = = = 7.27 × 10-9 m = 72.7 Å p 9.109· 10- 26 m 1 (b) l = 0 l = × 72.7 = 1154 Å . n m 0.063 p 14. From Fig. 22 when n = 1015 cm-3, the corresponding temperature is 1000 / T = 1.8. i So that T = 1000/1.8 = 555 K or 282 . 15. From E – E = kT ln [N / (N – N )] c F C D A which can be rewritten as N – N = N exp [–(E – E ) / kT ] D A C C F Then N – N = 2.86 × 1019 exp(–0.20 / 0.0259) = 1.26 × 1016 cm-3 D A or N = 1.26 × 1016 + N = 2.26 × 1016 cm-3 D A A compensated semiconductor can be fabricated to provide a specific Fermi energy level. 16. From Fig. 28a we can draw the following energy-band diagrams: 4 17. (a) The ionization energy for boron in Si is 0.045 eV. At 300 K, all boron impurities are ionized. Thus p = N = 1015 cm-3 p A n = n2 / n = (9.65 × 109)2 / 1015 = 9.3 × 104 cm-3. p i A The Fermi level measured from the top of the valence band is given by: E – E = kT ln(N /N ) = 0.0259 ln (2.66 × 1019 / 1015) = 0.26 eV F V V D (b) The boron atoms compensate the arsenic atoms; we have p = N – N = 3 × 1016 – 2.9 × 1016 = 1015 cm-3 p A D Since p is the same as given in (a), the values for n and E are the same as p p F in (a). However, the mobilities and resistivities for these two samples are different. 18. Since N >> n, we can approximate n = N and D i 0 D p = n2 / n = 9.3 ×1019 / 1017 = 9.3 × 102 cm-3 0 i 0 (cid:230) E - E (cid:246) From n = n exp(cid:231) F i (cid:247) , 0 i Ł kT ł we have E – E = kT ln (n / n) = 0.0259 ln (1017 / 9.65 × 109) = 0.42 eV F i 0 i The resulting flat band diagram is : 5 19. Assuming complete ionization, the Fermi level measured from the intrinsic Fermi level is 0.35 eV for 1015 cm-3, 0.45 eV for 1017 cm-3, and 0.54 eV for 1019 cm-3. The number of electrons that are ionized is given by n @ ND[1 – F(ED)] = ND / [1 + e- (ED- EF)/kT ] Using the Fermi levels given above, we obtain the number of ionized donors as n = 1015 cm-3 for N = 1015 cm-3 D n = 0.93 × 1017 cm-3 for N = 1017 cm-3 D n = 0.27 × 1019 cm-3 for N = 1019 cm-3 D Therefore, the assumption of complete ionization is valid only for the case of 1015 cm-3. 1016 1016 20. N + = = D 1+ e- (ED- EF)/kT 1+e- 0.135 1016 = = 5.33 × 1015 cm-3 1 1+ 1.145 The neutral donor = 1016 – 5.33 ×1015 cm-3 = 4.67 × 1015 cm-3 NO 4.76 4 The ratio of D = = 0.876 . N+ 5.33 D 6 CHAPTER 3 1. (a) For intrinsic Si, m = 1450, m = 505, and n = p = n = 9.65· 109 n p i 1 1 We have r= = = 3.31· 105 W -cm qnm + qpm qn (m + m ) n p i n p (b) Similarly for GaAs, m = 9200, m = 320, and n = p = n = 2.25· 106 n p i 1 1 We have r= = = 2.92· 108 W -cm. qnm + qpm qn (m + m ) n p i n p 2. For lattice scattering, m (cid:181) T-3/2 n 200- 3/2 T = 200 K, m = 1300· = 2388 cm2/V-s n 300- 3/2 400- 3/2 T = 400 K, m = 1300· = 844 cm2/V-s. n 300- 3/2 1 1 1 3. Since = + m m m 1 2 1 1 1 \ = + m = 167 cm2/V-s. m 250 500 4. (a) p = 5· 1015 cm-3, n = n2/p = (9.65· 109)2/5· 1015 = 1.86· 104 cm-3 i m = 410 cm2/V-s, m = 1300 cm2/V-s p n 1 1 r = » = 3 W -cm qm n+qm p qm p n p p (b) p = N – N = 2· 1016 – 1.5· 1016 = 5· 1015 cm-3, n = 1.86· 104 cm-3 A D m = m (N + N ) = m (3.5· 1016) = 290 cm2/V-s, p p A D p m = m (N + N ) = 1000 cm2/V-s n n A D 1 1 r = » = 4.3 W -cm qm n+qm p qm p n p p 7 (c) p = N (Boron) – N + N (Gallium) = 5· 1015 cm-3, n = 1.86· 104 cm-3 A D A m = m (N + N + N ) = m (2.05· 1017) = 150 cm2/V-s, p p A D A p m = m (N + N + N ) = 520 cm2/V-s n n A D A r = 8.3 W -cm. 5. Assume N - N >> n, the conductivity is given by D A i s » qnm = qm(N - N ) n n D A We have that 16 = (1.6· 10-19)m(N - 1017) n D Since mobility is a function of the ionized impurity concentration, we can use Fig. 3 along with trial and error to determine m and N For example, if we n D . choose N = 2· 1017, then N = N + + N - = 3· 1017, so that m » 510 cm2/V-s D I D A n which gives s = 8.16. Further trial and error yields N » 3.5· 1017 cm-3 D and m » 400 cm2/V-s n which gives s » 16 (W -cm)-1. 6. s = q(m n+m p) = qm (bn+n2 /n) n p p i From the condition ds/dn = 0, we obtain n = n / b i Therefore 8