半导体物理与器件第四版课后习题答案7 内容参考

Chapter 7材料供借鉴#7.1 (a) (i) V (ii) V (iii) V (b) (i) V (ii) V (iii) V_7.2 Si: cm Ge: cm GaAs: cm and V (a)cm, cm Then Si: V Ge: V GaAs: V (b)cm, cm Si: V Ge: V GaAs: V (c)cm, cm Si: V Ge: V GaAs: V_7.3(a) Silicon (K) For cm; V ; V ; V ; V(b) GaAs (K) For cm; V ; V ; V ; V(c) Silicon (400 K), cm For cm; V ; V ; V ; V GaAs(400 K), cm For cm; V ; V ; V ; V_7.4(a) n-side or eV p-side or eV (b) or V (c) or V (d) or cmm Now or cmm We have or V/cm_7.5(a) n-side or eV p-side or eV (b) or V (c) or V (d) or cmm By symmetry cmm Now or V/cm_7.6 (b) or cm or cm (c) V_7.7 200 K; ; cm 300 K; ; cm 400 K; ; cm For 200 K; V For 300 K; V For 400 K; V_7.8 So (a) or which yields cm cm cm or m cm or mNow V/cm(b) From part (a), we can write which yields cm cm cm or m cm or m V/cm_7.9(a) or V (b) or cmm Now or cmm (c) or V/cm_7.10 (a) V (b) increases as temperature decreases AtK, we can write At K, eV So Then V We find _7.11 Using the procedure from Problem 7.10, we can write, for K, At K, V For V, K At K, eV Also Then V V_7.12(b) For cm, or eV For cm or eV Then or V_7.13(a) or V (b) or cm (c) or cm (d) or V/cm_7.14 Assume silicon, so or (a)cm, m (b)cm, m (c)cm, m Now (a)V (b)V (c)V Also Then (a)m (b)m (c)m Now (a) (b) (c)_7.15 We find (a) (i) For ,;V (ii) ; V (iii) ; V (iv) ; V (i) For, ; V/cm (ii) ; V/cm (iii) ; V/cm (iv) ; V/cm (b) (i) For ,;V (ii) ; V (iii) ; V (iv) ; V (i) For, ;V/cm (ii) ; V/cm (iii) ; V/cm (iv) ; V/cm(c) increases as the doping increases, and the electric field extends further into the low-doped side of the pn junction._7.16(a) V(b) (i) For , cm or m (ii) For V, cm or m(c) (i)For , V/cm (ii)For V, V/cm_7.17(a) V(b) cm or m cm or m cm or m Also m(c) V/cm(d) F or pF_7.18(a) We find cm cm(b) cm or m cm or m(c) V/cm(d) F/cm_7.19(a) V(b) So (c) For a larger doping, the space charge width narrows which results in a larger capacitance._7.20(a) or V Now or or so that V which yields V(b) or V We have so that V which yields V(c) or V We have so that V which yields V_7.21 (a) or We find V V We find or (b) or (c) or _7.22 (a) We have or For V, we find or V (b) Then Now so We can then write which yields cm and cm_7.23 V So which yields V_7.24(a) V (i) For , pF (ii) For V, pF(b) V (i) For , pF(ii) For V, pF_7.25 V(a) F HmH(b) (i) For V, pF HzMHz (ii) For V, pF HzMHz_7.26 Let V(a) cm(b) cm_7.27 (a) By trial and error, cm, cm, V(b) From part (a), By trial and error, cm, cm, V_7.28(a) or V (b) or cm Also or cm(c) For m, we have which becomes We find V_7.29 An junction with cm,(a) A one-sided junction and assume . Then or which yields V (b) so cmm (c) or V/cm_7.30(a) V(b) (i) For V, F (ii) For V, F (iii) For V, F_7.31(a) cm(b) cm(c) V_7.32 Plot_7.33(a) (c) p-region or We have at Then for n-region, or n-region, or We have at so that for , we have We also have at Then which gives Then for we have _7.34 (a) For m, So At m, So Then At , , so or V/cm (c) Magnitude of potential difference is Let at , then Then we can write At m or V Potential difference across the intrinsic region or V By symmetry, the potential difference across the p-region space-charge region is also 3.863 V. The total reverse-bias voltage is then V_7.35(a) or Then cm(b) Or cm_7.36 cm_7.37(a) For cm, from Figure 7.15, V(b) For cm, V_7.38(a) From Equation (7.36), Set and V Then V So V(b) V Then So V_7.39 For a silicon junction with cm and V, then, neglecting we have or cmm_7.40 We find V Now so which yields cm Now Then which yields V or V_7.41 Assume silicon: For an junction Assume (a) For m which yields V (b) For m which yields V Note: From Figure 7.15, the breakdown voltage is approximately 300 V. So, in each case, breakdown is reached first._7.42 Impurity gradien cm From Figure 7.15,