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1、-半导体物理与器件第四版课后习题答案7-第 7 页Chapter 7 (a) (i) V (ii) V (iii) V (b) (i) V (ii) V (iii) V_ 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_(a) Silicon (K) For cm; V ; V ; V ; V(b) GaAs (K) For cm; V ; V ; V ; V(c) Silicon (400 K)
2、, cm For cm; V ; V ; V ; V GaAs(400 K), cm For cm; V ; V ; V ; V_(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_(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_ (b) or cm or cm (c) V_ 200 K; ; cm 300 K; ; cm 400 K; ; cm Fo
3、r 200 K; V For 300 K; V For 400 K; V_ So or which yields cm cm cm or m cm or mNow V/cm(a) From part (a), we can write which yields cm cm cm or m cm or m V/cm_ or V (b) or cmm Now or cmm (c) or V/cm_ (a) V (b) increases as temperature decreases AtK, we can write At K, eV So Then V We find_ Using the
4、procedure from Problem 7.10, we can write, for K, At K, V For V, K At K, eV Also Then V V_(a) For cm, or eV For cm or eV Then or V_ or V (b) or cm (c) or cm (d) or V/cm_ 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)_ We find (a) (i) For
5、,;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(b) increases as the doping increases, and the electric field extends further into the low-doped side of the pn junction
6、._ V (i) For , cm or m (ii) For V, cm or m (i)For , V/cm (ii)For V, V/cm_ V(b) cm or m cm or m cm or m Also m(c) V/cm F or pF_ We find cm cm(b) cm or m cm or m(c) V/cm(d) F/cm_ V So (c) For a larger doping, the space charge width narrows which results in a larger capacitance._ or V Now or or so that
7、 V which yields V or V We have so that V which yields V or V We have so that V which yields V_ (a) or We find V V We find or (b) or (c) or_ (a) We have or For V, we find or V (b) Then Now so We can then write which yields cm and cm_ V So which yields V_ V (i) For , pF (ii) For V, pF V (i) For , pF(i
8、i) For V, pF_ V F HmH(b) (i) For V, pF HzMHz (ii) For V, pF HzMHz_ Let V cm cm_ By trial and error, cm, cm, V(a) From part (a), By trial and error, cm, cm, V_ or V (b) or cm Also or cm(c) For m, we have which becomes We find V_ An junction with cm,(a) A one-sided junction and assume . Then or which
9、yields V (b) so cmm (c) or V/cm_ V (i) For V, F (ii) For V, F (iii) For V, F_ cm cm V_ Plot_ (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_ (a) For m, So At m, So Then At , , so or V/cm (c) Magnitude o
10、f 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_ or Then cm Or cm_ cm_(a) For cm, from Fig
11、ure 7.15, V(b) For cm, V_(a) From Equation (7.36), Set and V Then V So V(b) V Then So V_ For a silicon junction with cm and V, then, neglecting we have or cmm_ We find V Now so which yields cm Now Then which yields V or V_ Assume silicon: For an junction Assume (a) For m which yields V (b) For m whi
12、ch yields V Note: From Figure 7.15, the breakdown voltage is approximately 300 V. So, in each case, breakdown is reached first._ Impurity gradien cm From Figure 7.15, V_ (a) For the linearly graded junction Then Now At and , So Then (b) Set at , then Then_ We have that Then which yields cm_ Let cm Then V Now cm (i) For V, pF (ii) For , pF_