半導(dǎo)體物理與器件第四版課后習(xí)題答案4

1、Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4By D. A. Neamen Problem Solutions_16Chapter 44.1 where and are the values at 300 K. (a) Silicon(K)(eV)(cm) (b) Germanium (c) GaAs(K)(cm)(cm)_4.2 Plot_4.3(a) By trial and error, K(b) By trial and error, K_4.4 At K, eV At K, eV

2、or or eV Now so cm_4.5 For K, eV For K, eV For K, eV(a) For K, (b) For K, (c) For K, _4.6(a) Let Then To find the maximum value: which yields The maximum value occurs at (b) Let Then To find the maximum value Same as part (a). Maximum occurs at or _4.7 where and Then or _4.8 Plot_4.9 Plot_4.10 Silic

3、on: , eV Germanium: , eV Gallium Arsenide: , eV_4.11 (K)(eV)()(eV)_4.12(a) meV(b) meV_4.13 Let constant Then Let so that We can write so that The integral can then be written as which becomes _4.14 Let for Then Let so that We can write Then or We find that So _4.15 We have For germanium, , Then or T

4、he ionization energy can be written as eV eV_4.16 We have For gallium arsenide, , Then The ionization energy is or eV_4.17(a) eV(b) eV(c) cm(d) Holes(e) eV_4.18(a) eV(b) eV(c) cm(d) eV_4.19(a) eV eV(b) cm(c) p-type_4.20(a) eV cm eV cm(b) eV eV cm_4.21(a) eV cm eV cm(b) eV eV cm_4.22(a) p-type(b) eV

5、cm eV cm_4.23(a) cm cm(b) cm cm_4.24(a) eV(b) eV(c) cm(d) Holes(e) eV_4.25 eV cm cm cm(a) eV(b) eV(c) cm(d) Holes(e) eV_4.26(a) cm eV cm(b) eV cm cm eV eV cm_4.27(a) cm eV cm(b) eV cm cm eV eV cm_4.28 (a) For , Then cm (b) cm_4.29 So We find eV_4.30 (a) Then cm (b) cm_4.31 For the electron concentra

6、tion The Boltzmann approximation applies, so or Define Then To find maximum , set or which yields For the hole concentration Using the Boltzmann approximation or Define Then To find maximum value of , set Using the results from above, we find the maximum at _4.32(a) Silicon: We have We can write For

7、 eV and eV we can write or cm We also have Again, we can write For and eV Then or cm(b) GaAs: assume eVThen or cm Assume eV Then or cm_4.33 Plot_4.34(a) cm cm(b) cm cm(c) cm(d) cm cm cm(e) cm cm cm_4.35(a) cm cm(b) cm cm(c) cm(d) cm cm cm(e) cm cm cm_4.36(a) Ge: cm (i) or cm cm (ii) cm cm(b) GaAs: c

8、m (i) cmcm (ii)cm cm(c) The result implies that there is only one minority carrier in a volume of cm._4.37 (a) For the donor level or (b) We have Now or Then or _4.38(a) p-type(b) Silicon: or cm Then cm Germanium: or cm Then cm Gallium Arsenide: cm and cm_4.39(a) n-type(b) cm cm(c) cm cm_4.40 cm n-t

9、ype_4.41 cm So cm, Then cm so that cm_4.42 Plot_4.43 Plot_4.44 Plot_4.45 so cm cm_4.46(a) p-type Majority carriers are holes cm Minority carriers are electrons cm(b) Boron atoms must be added So cm cm_4.47(a) n-type(b) cm electrons are majority carriers cm holes are minority carriers(c) so cm_4.48 F

10、or Germanium(K) (eV)(cm) and cm(K) (cm) (eV) _4.49(a) For cm, eV cm, eV cm, eV cm, eV (b) For cm, eV cm, eV cm, eV cm, eV_4.50(a) cm so Now By trial and error, K(b) At K, eV At K, eV cm eV then eV(c) Closer to the intrinsic energy level._4.51 At K, eV K, eV K, eV At K, cm At K, cm At K, cm At K and

11、K, cm At K, cm Then, K, eV K, eV K, eV_4.52 (a) For cm, eV cm, eV cm, eV cm, eV (b) For cm, eV cm, eV cm, eV cm, eV_4.53(a) or eV(b) Impurity atoms to be added so eV (i) p-type, so add acceptor atoms (ii) eV Then or cm_4.54 so or cm_4.55(a) Silicon(i) eV(ii)eV cm cm Additional donor atoms(b) GaAs (i) eV (ii)eV cm cm Additional donor atoms_4.56(a) eV(b) eV(c) For part (a); cm cm For part (b): cm cm_4.57 cm Add additional acceptor impurities cm_4.58(a) eV(b) eV(c)(d) eV(e) eV_4.59(a) eV(b) eV