半导体物理与器件第四版课后习题答案8

1、页眉内容9.03 1014 cm38.7Chapter 88.1In forward biasThen or(a)1 f1For 10, thenI f 2orVi V2 59.6 mV 60mV(b)1 f 1For 100 , thenI f 2orVi V2 119.3 mV 120 mV2.26 1014cm 38.4一34.5 10 cm4.5 104cm(i) pn Xnor Vap noVa exp 一VtVt lnpn Xnpno0.599 V(ii) n-region - lower doped side8.2_43.214 10 cmn po2ni10 21.5 10Na1

2、5102.8125一 4310 cm337.5 10 cmpno2ni1.5Nd101015-2 101.125510 cm(i) VaVt ln0.1 Nan po0.6165V(a)Va0.45 V,(ii) p-region - lower doped sideornp Xp(b)Va0.55 V,(c)Va0.55 V8.3n po2niNa1.8 106 24 1016pno2ni1.8 106Nd1610(a)Va0.90 V,10.010310 cm3.951210 cm(b) Va1.10V_ 119.88 103cm1.888.11410 cm4.69 1013cm 310

3、5 cm 343.24 10 cm114.0 10 cm8.521.849A/cmIn AJn Xp10 3 1.849 Aor I n 1.85 mA 一，24.521 A/cm一_ 一3I p AJp Xn 104.521 Ap por I p 4.52 mAp(a) I In Ip 1.85 4.52 6.37 mA8.6For an n p silicon diode or15I S 1.8 10 A(a) For Va0.5 V,orI d 4.36 10 7A(b) For Va0.5 V,or15I D I S 1.8 10 A页眉内容8.8Js 1.568 10 4A/cm 2

4、_42.44 10or I 0.244 mA1.568 10Js 5.14510(i) I(ii) I(iii) I1.02910112A/cm14A_ 141.029 103.61101.029 101.721.029 108.16exp14exp510 A14exp410 A0.450.02590.550.02590.650.02598.9We haveor we can write this asso thatIn reverse bias, I is negative, so at0.90, we haveorV 59.6 mV8.10Case 1:IsexpVaVtIs 6.3051

5、015126.305 10 mA3.153 10 8 mA/cm 2Case 2:Va s exp 一Vtor1.093 mA922 10 mA/cmCase 3:AJsVa exp VtSoVt lnAJsVa 0.6502 VThen4I s AJs 108.11Case 4: INd12.73Na710(a) From part (a),Nd0.354Naii10Va exp VtIsmA1.200.72exp 0.02591.014 10 12 mA525.07 10 5cm2Na0.07857orNd电 2.828orNd8.12The cross-sectional area is

6、3A I 10 1042A - 5 10 cmJWe have which yields20_ _ 10 _ ,2Js 2.522 10 A/cmWe can writeWe want or37.071 103 Na7.071 10 a 4.472Ndwhich yields Now We findNd 7.091410andNa 1.0110168.13Plot8.14(a)We have0.101033cm3cmDpDn-and n 2.4noNowpo0.158.4896 10 Asoor(b) Using Einstein's relation, we can write We

7、 haveThen59.710 10 A8.17n e nNd and p e pNa(a) The excess hole concentration is givenAlsoThenbyWe find8.15(a) p-side;orEFi Ef 0.329 eVp no2niNdand8.1610 21.5 1016102.25104 3cm2.828 10 4cm2.828 mAlso on the n-side; orEf EFi 0.407 eV(b) We can findDnDpNoworJ SThenorIsWe findorThenorPn3.81 1014exp1250

8、0.02593204.4264.4260.0259ii1010 156I 1.07 10 A(c) The hole current is orI p 3.278 10 16ThenI sp191.6 105I snVa0.8Vbi232.4 cm /s2 ,8.29 cm /sx42.828 10cm(b) We haveAt x 310 4 cm,2A/cmorJp 30.5966 A/cm1.07 AVd exp Vt(A)_ 141.342 10 A410257V2 10 4.025 10 15 A0.746826V0.8 0.74682610 21.5 10165 100.59746

9、V1431.56 10 cm54.1981 10 A41.3997 10 A1.820 10 4A(c) We haveWe can determine that3n po 4.5 10 cmLn 10.72 mThen orJno 0.2615A/cmWe can also findand.一-,2J po 1.724 A/cm pThen at x 3 m,or2J n 31.39A/cm8.18(a) Problem 8.7 ornpVaVt ln n po(b) Problem 8.8or Va VtVtln0.1 Naln 2n2 N0.205 VP no0.623 Vln0.1 N

10、dn： Nd8.19byThe excess electron concentration is givenThe total number of excess electrons isWe may note thatThenWe find that 2 .Dn25 cm /s and Ln 50.0 mAlson po2niNa10 21.5 10158 102.811043cmThen orThen, we find the total number of excess electrons in the p-region to be:一一一_ 一 4(a)Va0.3 V,Np1.5110一

11、_ _5(b)Va0.4 V,Np7.1710Va0.5 V,Np3.401071 kT1(b) Taking the ratioFor Ti300 K,38.61For T2400K,kT1kT20.0259,0.03453,1 kT228.96(i) Germanium:Eg0.66 eVI S2or 1383I S1(ii) Silicon: Eg1.12eVI S2or 1.17105I S18.22PlotSimilarly, the total number of excess holesinLpp nothe n-region is found to be We find tha

12、t2D p 10.0 cm /s and p10.0 mAlso210 2ni1.5 102.25104cm 3Nd1610ThenSo(a)Va0.3 V,Pn2.41103(b)Va0.4 V,Pn1.15510(c)Va0.5 V,Pn5.451068.23First case: or一Va0.50VtIIf ln 2 10 1nlsNow 0.05049T 584.8 KSecond case: 2 or ni 8.2519NowBy trial and error, T 502 KThe reverse-bias current is limiting factor.0.05049

13、VT 0.0259 - 30027108.248.20Thenso or We then haveLp 、Dp po . 10 10 710 3cmor Lp 10 m;WnL p8.21orThenorEg2 0.769 eV(i) pn xnor Va0.1NdVa pno exp Vt0.1Vt ln 2 niN2Va 0.5516V(a) We havewhich can be written in the form(ii) I pAeDp pWnor2niNdexpVaVtor(b)(i) np xp3I p 4.565 10 A6I n 2.26 10 A34.567 10 AI

14、4.567 mAVa0.1 Na npo exp Vtor VaVt ln0.1 NniVa 0.5516VAeDp ni2Va(ii) I p exp WnNdVtI p 4.565 10 5 A p_ _ 4 _I n 2.2597 10 A_42.716 10 Aor I 0.2716mAI d a constant, we haveWe then haveWe haveT 300 K , Vdi 0.60V andkT1kT1 0.0259eV,1 0.0259 VeT 310K ,kT20.02676eV,T22 0.02676V eT 320 K ,kT30.02763eV,T3一

15、 0.02763 V eFor T 310K , which yieldsVd2 0.5827VFor T 320K , which yieldsVd3 0.5653VsinhWnLpThenI gen8.25(a) We can write for the n-region The general solution is of the form The boundary condition at x xngives and the boundary condition atx xn WngivesFrom this equation, we haveThen, from the first

16、boundary condition, weobtainWe then obtainwhich can be written asWe can also findThe solution can now be written as or finally(b)rVa,eDp pno exp 1VtThen 8.26For the temperature range 300 T 320 K,neglect the change in Nc and NThenTaking the ratio of currents, but maintaining8.27(a) We can writewhere

17、C is a constant, independent of temperature.As a first approximation, neglect the variation of Nc and N with temperatureover the range of interest. We can then writewhere Ci is another constant, independent oftemperature. We find or8.28一15 一Is 2.323 10 AWe find0.7665 Vand5 W 6.109 10 cm419105101.6 1

18、01.5 10 6.109 10210 7117.331 10 A8.29(a) Set I s I gen23.0545 102so ni TT 3 Z Z -33.9528 102.50 10_ 1434.734 10 cmThenBy trial and error,T 567 KWe have419_145101.6 104.734 106.109 10210 7Then 6Is I gen 2.314 10 Aor Is I gen 2.314 A(b) From Problem 8.2815 .I s 2.323 10 A_ _ _ 11I gen 7.331 10 AVaI 。c

19、 VaSo I I s exp I gen exp Vt2Vt.一 一 22 .Is1.50 10 A0.5366 V192 101.6 101.8 10 4.201 10I gen6.04910 14a(ii)I recI ro expVa(iii) Irec141010 86.43610 9 A0.8exp2 0.02593.058 10 7A141.0(iv) I rec 6 10 exp 2 0.025951.453 10 5A8.31Using results from Problem 8.30, we findVa0.4 V, Id 7.64 10 16 A,10I rec 1.3

20、5 10 A,10It 1.35 10 10A12Va0.6 V, Id 1.73 10 A-一一 9 _I rec6.44 10 A,i ec9It 6.44 10 A9Va0.8 V, 1d 3.90 10 AI rec 3.06 10 7 A,It 3.10 10 7 AVa1.0V, I d 8.80 10 6A-5 .I rec 1.45 10 A,一4 5 -It2.33 10 A2Va1.2V. I d 1.99 10 A4 44 4 -I rec6.90 10 A,1 ec.一一 一 2 -It2.06 10 A8.32Plot 8.33Plot 8.34We have tha

21、tLet pO nO O and n pniWe can writeandWe also haveso thatThenDefineaeVa IkTandEFn ErkTThen the recombination rate can be written as orTo find the maximum recombination rate, setorwhich simplifies toThe denominator is not zero, so we have orThen the maximum recombination rate becomes or which can be w

22、ritten asIf VakT e , then we can neglect the(-1)term in the numerator and the (+1) term in thedenominator, so we finally have 8.35We have1931In this case, G g 4 10 cm sand isa constant through the space charge region.ThenWe findorVbi0.659 VAlsoor4W 2.35 10 cmThenor3 .2J gen 1.5 10 A/cm8.36or112Js 1.

23、638 10 A/cmNowWe want or which can be written as We find orVd0.548V8.3781.16 10 For Cd 11.6nF91.16 10 For Cd 1.16nF8.38Q (a) Cd,For Id 1.2mAV 105.79 10 8.40Reverse bias0.790V5.1078 10.Vbi VrVr(V)Cj(pF) 1.55552.12332.624 C(b) For I d 0.12mA115.79 10 13.81805.7470.206.6500.408.179Forward biasFor Na Nd

24、 I po Ino C8.39For a p n diodegdDQ,CdVtDQ pO2VtNowgd10 3Q QO210 S3.860.0259and10 3 10 7Cd4 QQ9f2 0.0259We havewhere2 fWe obtainf10 kHz ,Z25.9J0.0814f100kHz,Z25.9 J0.814f1MHz ,Z23.6J7.41f10 MHz ,Z2.38J7.49Then14V aI po 1.006 10 exp AVa(V) Cd(F)+ Cj(F)=CTotal (F)8.41For a p n diode, I pO InO, then Now

25、pO62.5 10 F/A 2VtThen or pO 1.3 107 sAt 1 mA, or 9 Cd 2.5 10 Fn-region: soorThe total resistance is or(b)which yieldsI 1.38 mA8.42(i) Cd8.44 or We can write (a) (i) For I d 1 mA, or V 0.567 V (ii) For I d 10 mA, or V 1.98V (b) Set R 0 (i) For I D 1 mA, or V 0.417 V (ii) For ID 10 mA, or V 0.477 Vpo

26、p02VorI poorI po(ii) Ipo(iii)(b)2Vt Cd2 0.0259 10一 7p01045.18 10 A0.518 mA2 niexpNdVaVtrd0.618VM 0.0259Id 0.518 10502Vt Cd 2 0.0259 0.25 10(i) I po7p0101.295 10 4Aor Ipo0.1295 mApv(ii) Va30.1295 100.0259 ln -2.25 100.5821V(iii)d0.025930.1295 102008.43(a) p-region: soor8.454or I d 8.09375 10 AVa 0.4896 VVt0.02594(a) I d - 4.3167 10 Ard 600.4733V8.461 dI d (a)rddVaor which yields (b) which yield