半导体物理与器件第四版课后习题答案8.doc

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8By D. A. Neamen Problem Solutions_Chapter 8198.1 In forward bias Then or (a) For , then or mVmV (b) For , then or mVmV_8.2 cm cm (a) V, cm or cm(b) V, cm cm(c) V _8.3 cm cm(a) V, cm cm(b) V cm cm(c) _8.4(a) cm cm (i) or V (ii) n-region - lower doped side(b) cm cm (i) V (ii) p-region - lower doped side_8.5(a) A/cm A or mA(b) A/cm A or mA(c) mA_8.6 For an silicon diode or A (a) For V, or A (b) For V, or A_8.7 A/cm(a) A or mA(b) A_8.8 (a) A/cm A(b) (i) A (ii) A (iii) A_8.9 We have or we can write this as so that In reverse bias, is negative, so at , we have or mV_8.10 Case 1: AmA mA/cm Case 2: or mA mA/cm Case 3: So V Then mA Case 4: mA cm_8.11(a) or (b) From part (a), or _8.12 The cross-sectional area is cm We have which yields A/cm We can write We want or = which yields Now We find cm and cm_8.13 Plot_8.14 (a) We have and so or (b) Using Einsteins relation, we can write We have and Also Then _8.15 (a) p-side; or eV Also on the n-side; or eV (b) We can find cm/s cm/s Now or A/cm Then or A We find or AA (c) The hole current is or (A) Then _8.16(a) A(b) A(c) V V cm(d) A(e) A A Now A Then A_8.17 (a) The excess hole concentration is given by We find cm and cm m Then or cm(b) We have At cm, or A/cm (c) We have We can determine that cm and m Then or A/cm We can also find A/cm Then at m, or A/cm_8.18(a) Problem 8.7 or V(b) Problem 8.8 or V_8.19 The excess electron concentration is given by The total number of excess electrons is We may note that Then We find that cm/s and m Also cm Then or Then, we find the total number of excess electrons in the p-region to be: (a)V, (b)V, (c)V, Similarly, the total number of excess holes in the n-region is found to be We find that cm/s and m Also cm Then So (a)V, (b)V, (c)V, _8.20 Then so or We then have or Then or eV_8.21 (a) We have which can be written in the form or (b) Taking the ratio For K, , For K, , (i) Germanium: eV or (ii) Silicon: eV or _8.22 Plot_ 8.23 First case: or V Now K Second case: or Now By trial and error, K The reverse-bias current is limiting factor._8.24 cm or m; (a) (i) or V (ii) A A A or mA (b) (i) or V (ii) A A A or mA_8.25 (a) We can write for the n-region The general solution is of the form The boundary condition at gives and the boundary condition at gives From this equation, we have Then, from the first boundary condition, we obtain We then obtain which can be written as We can also find The solution can now be written as or finally (b) = Then _8.26 For the temperature range K, neglect the change in and . Then Taking the ratio of currents, but maintaining a constant, we have We then have We have K , V and eV, V K , eV, V K , eV, V For K , which yields V For K , which yields V_8.27(a) We can write where C is a constant, independent of temperature. As a first approximation, neglect the variation of and with temperature over the range of interest. We can then write where is another constant, independent of temperature. We find or _8.28(a) A(b) We find V and cmThen A(c)_8.29(a) Set , so cmThen By trial and error, K We have Then A or A(b) From Problem 8.28 A A So V_8.30 cm/s cm/s (a)(i) A (ii) A (iii) A (iv) A(b) V cm(i)Then A (ii) A(iii) A (iv) A_8.31 Using results from Problem 8.30, we find V, A, A, A V, A A, A V, A A, A V, A A, A V. A A, A_8.32 Plot_8.33 Plot_8.34 We have that Let and We can write and We also have so that Then Define and Then the recombination rate can be written as or To find the maximum recombination rate, set or which simplifies to The denominator is not zero, so we have or Then the maximum recombination rate becomes or which can be written as If , then we can neglect the (-1) term in the numerator and the (+1) term in the denominator, so we finally have Q.E.D._8.35 We have In this case, cms and is a constant through the space charge region. Then We find or V Also or cm Then or A/cm_8.36 or A/cm Now We want or which can be written as We find or V_8.37(a) F or nF(b) F or nF_8.38(a) , For mA C(b) For mA C_8.39 For a diode , Now S and F We have where We obtain kHz , kHz , MHz , MHz , _8.40 Reverse bias V F (V) (pF)10 1.5555 2.1233 2.6241 3.8180 5.747 6.650 8.179 Forward bias For Then A (V) (F) + (F) = (F)0.20 0.40 0.60 _8.41 For a diode, , then Now F/A Then or s At 1 mA, or F_8.42(a) (i) or A or mA(ii) V (iii) (b) (i) A or mA (ii) V (iii) _8.43 (a) p-region: so or n-region: so or The total resistance is or (b) which yields mA_8.44 or We can write (a) (i) For mA, or V (ii) For mA, or V (b) Set (i) For mA, or V (ii) For mA, or V_8.45(a) or A V(b) A V_8.46 (a) or which yields (b) which yields _8.47 (a) If Then we have or We find (b) If , then which yields _8.48(a) erf erf=