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1、《半導(dǎo)體物理與器件》第四版答案第十章 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions _______

2、_______________________________________________________________________________ Chapter 10 10.1 (a) p-type; inversion (b) p-type; depletion (c) p-type; accumulation (d) n-type; inversion _______________________________________ 10.2 (a) (i) ??? ? ??=i a t fp n N V ln φ

3、 ()??? ? ? ???=10 15 105.1107ln 0259.0 3381.0=V 2 /14?? ????∈=a fp s dT eN x φ ()( ) ()( )( ) 2 /1151914107106.13381.01085.87.114? ? ? ??????=-- 51054.3-?=cm or μ354.0=dT x m (ii) ()??? ? ????=1016105.1103ln 0259.0fp φ 3758.0=V ()(

4、 ) ()( )( ) 2 /1161914103106.13758.01085.87.114? ? ???????=--dT x 51080.1-?=cm or μ180.0=dT x m (b) ()03022.03003500259.0=??? ??=kT V ??? ? ??-=kT E N N n g c i exp 2υ ( )( ) 3 19 19 3003501004.110 8.2?? ? ????= ?? ? ??-?03022.012.1exp 221071.3?= so 111093

5、.1?=i n cm 3- (i)()??? ? ????=11151093.1107ln 03022.0fp φ 3173.0=V ()( ) ()( )( ) 2 /1151914107106.13173.01085.87.114? ? ? ??????=--dT x 51043.3-?=cm or μ343.0=dT x m (ii) ()??? ? ????=11161093.1103ln 03022.0fp φ 3613.0=V ()( ) ()( )( ) 2 /116 1914103106.13613.01085.87.

6、114?????????=--dT x 5 1077.1-?=cm or μ177.0=dT x m _______________________________________ 10.3 (a) ()2 /14m ax ? ? ? ???∈==d fn s d dT d SD eN eN x eN Q φ ()()[] 2 /14fn s d eN φ∈= 1st approximation: Let 30.0=fn φV Then ()281025.1-? ()()()()()()[] 30.01085.87.114106.114

7、 19 --??=d N 141086.7?=?d N cm 3- 2nd approximation: ()2814.0105.11086.7ln 0259.010 14 =??? ? ? ???=fn φV Then ( ) 2 81025.1-? ( )()()()()() [] 2814.01085.87.114106.11419--??=d N 141038.8?=?d N cm 3- (b) ()2831.0105.11038.8ln 0259.01014=??? ? ????=fn φV ()566.02831

8、.022===fn s φφV _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen

9、 Problem Solutions ______________________________________________________________________________________ 10.4 p-type silicon (a) Aluminum gate ???????????? ??++-=fp g m ms e E φχφφ2 We have ???? ??=i a t fp n N V ln φ ()334.0105.1106ln

10、 0259.01015=??? ? ????=V Then ()[]334.056.025.320.3++-=ms φ or 944.0-=ms φV (b) +n polysilicon gate ? ??? ??+-=fp g ms e E φφ2()334.056.0+-= or 894.0-=ms φV (c) +p polysilicon gate ()334.056.02-=???? ??-=fp g ms e E φφ or 226.0+=ms

11、 φV _______________________________________ 10.5 ()3832.0105.1104ln 0259.01016=? ??? ????=fp φV ???? ??++-=fp g m ms e E φχφφ2 ()3832 .056.025.320.3++-= 9932.0-=ms φV _______________________________________ 10.6 (a) 17102??d N cm 3- (b) Not possible - ms φ

12、is always positive. (c) 15102??d N cm 3- _______________________________________ 10.7 From Problem 10.5, 9932.0-=ms φV ox ss ms FB C Q V -=φ (a) ()() 8 14 102001085.89.3--??=∈=ox ox ox t C 710726.1-?=F/cm 2 ()() 719 1010726.1106.11059932.0--???--=FB V

13、 040.1-=V (b) ()() 81410 801085.89.3--??=ox C 710314.4-?=F/cm 2 ()() 7 19 1010314.4106.11059932.0--???--=FB V 012.1-=V _______________________________________ 10.8 (a) 42.0-?ms φV 42.0-==ms FB V φV (b) ()() 78 1410726.1102001085.89.3---?=??=ox C F/cm 2 (

14、i)()() 7 191010726.1106.1104--???-=-=?ox ss FB C Q V 0371.0-=V (ii)()() 7 19 1110726.1106.110--??-=?FB V 0927.0-=V (c) 42.0-==ms FB V φV ()() 78 1410876.2101201085.89.3---?=??=ox C F/cm 2 (i)()() 7 191010876.2106.1104--???-=?FB V

15、 0223.0-=V (ii)()() 7 19 1110876.2106.110--??-=?FB V 0556.0-=V _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen

16、 Problem Solutions ______________________________________________________________________________________ 10.9 ???? ??++-=fp g m ms e E φχφφ2 where ()365.0105.1102ln 0259.01016=???? ????=fp φV

17、 Then ()365.056.025.320.3++-=ms φ or 975.0-=ms φV Now ox ss ms FB C Q V -=φ or ()ox FB ms ss C V Q -=φ We have ()() 8 14 104501085.89.3--??=∈=ox ox ox t C or 81067.7-?=ox C F/cm 2 So now ()[]() 8 1067.71975.0-??---=ss Q 910

18、92.1-?=C/cm 2 or 10102.1?=e Q ss cm 2- _______________________________________ 10.10 ()3653.0105.1102ln 0259.01016=? ?? ? ????=fp φV ()( ) ()( )( ) 2 /1161914102106.13653.01085.87.114?? ???????=--dT x 510174.2-?=cm ()dT a SD x eN Q =m ax ()()() 5161910

19、174.2102106.1--???= 810958.6-?=C/cm 2 ()() 7 8 14 10301.210 1501085.89.3---?=??=ox C F/cm 2 ()fp ms ox ss SD TN C Q Q V φφ2max ++-= ()() 7 19 10810301.2106.110710958.6---???-?= ()3653.02++ms φ ms φ+=9843.0 (a) n + pol

20、y gate on p-type: 12.1-?ms φV 136.012.19843.0-=-=TN V V (b) p + poly gate on p-type: 28.0+?ms φV 26.128.09843.0+=+=TN V V (c) Al gate on p-type: 95.0-?ms φV 0343.095.09843.0+=-=TN V V _______________________________________ 10.11 ()3161.0105.1103ln 0259.01015=???

21、? ????=fn φV ()( ) ()( )( ) 2/115 1914103106.13161.01085.87.114?? ???????=--dT x 5 10223.5-?=cm ()dT d SD x eN Q =m ax ()()() 5151910223.5103106.1--???= 810507.2-?=C/cm 2 ()() 78 14 10301.210 1501085.89.3---?=??=ox C F/cm 2

22、 ()fn ms ox ss SD TP C Q Q V φφ2m ax -+????????+-= ()() ?????????+?-=---71019810301.2107106.110507.2 ()3161.02-+ms φ ms TP V φ+-=7898.0 (a) n + poly gate on n-type: 41.0-?ms φV 20.141.07898.0-=--=TP V V (b) p + p

23、oly gate on n-type: 0.1+?ms φV 210.00.17898.0+=+-=TP V V (c) Al gate on n-type: 29.0-?ms φV 08.129.07898.0-=--=TP V V _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapte

24、r 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.12 ()3294.0105.1105ln 0259.01015=???? ????=fp φV T

25、he surface potential is ()659.03294.022===fp s φφV We have 90.0-=-=ox ss ms FB C Q V φV Now ()FB s ox SD T V C Q V ++= φmax We obtain 2 /14?? ????∈=a fp s dT eN x φ ()( ) ()( )( ) 2 /115 1914105106.13294.01085.87.114?? ? ??????=-- or 410413

26、.0-?=dT x cm Then ()()()() 4151910413.0105106.1m ax --???=SD Q or ()810304.3m ax -?=SD Q C/cm 2 We also find ()() 8 14 104001085.89.3--??=∈=ox ox ox t C or 810629.8-?=ox C F/cm 2 Then 90.0659.010629.810304.38 8 -+??=--T V or 142.0+=T V V __________________

27、_____________________ 10.13 ()() 8 14 102201085.89.3--??=∈=ox ox ox t C 710569.1-?=F/cm 2 ()() 1019104106.1??=-ss Q 9104.6-?=C/cm 2 By trial and error, let 16104?=a N cm 3-. Now ()??? ? ????=1016105.1104ln 0259.0fp φ 3832.0=V ()( ) ()( )(

28、 ) 2/116 1914104106.13832.01085.87.114??? ??????=--dT x 510575.1-?=cm ()m ax SD Q ()()() 5161910575.1104106.1--???= 710008.1-?=C/cm 2 94.0-?ms φV Then ()fp ms ox ss SD TN C Q Q V φφ2max ++-= 7 9 710569.1104.610008.1---??-?= ()

29、3832.0294.0+- Then 428.0=TN V V 45.0?V _______________________________________ 10.14 ()() 8 14 101801085.89.3--??=∈=ox ox ox t C 7109175.1-?=F/cm 3- ()() 1019104106.1??=-ss Q 9104.6-?=C/cm 2 By trial and error, let 16105?=d N cm 3- Now ()

30、??? ? ????=1016105.1105ln 0259.0fn φ 3890.0=V ()( ) ()( )( ) 2 /1161914105106.13890.01085.87.114? ? ? ??????=--dT x 510419.1-?=cm ()m ax SD Q ()()() 5161910419.1105106.1--???= 710135.1-?=C/cm 3- 10.1+?ms φV Then Semiconductor Physics and Devic

31、es: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions __________________________________________________________

32、____________________________ ()()fn ms ox ss SD TP C Q Q V φφ2max -++-= () 7 9 7109175.1104.610135.1---??+?-= ()3890.0210.1-+ Then 303.0-=TP V V, which is within the specified value. _______________________________________ 1

33、0.15 We have 710569.1-?=ox C F/cm 2 9104.6-?=ss Q C/cm 2 By trial and error, let 14105?=d N cm 3- Now ()??? ? ????=1014105.1105ln 0259.0fn φ 2697.0=V ()( ) ()( )( ) 2/1141914105106.12697.01085.87.114?? ???????=--dT x 4 10182.

34、1-?=cm ()m ax SD Q ()()() 4 141910182.1105106.1--???= 910456.9-?=C/cm 2 33.0-?ms φV Then ()()fn ms ox ss SD TP C Q Q V φφ2max -++-= ???? ????+?-=---79910569.1104.610456.9 ()2697.0233.0-- 97

35、0.0=V Then 970.0-=TP V V 975.0-? V which meets the specification. _______________________________________ 10.16 (a) 03.1-?ms φV ()() 8 1410 1801085.89.3--??=ox C 7109175.1-?=F/cm 2 Now ox ss ms FB C Q V -=φ ()() 7 1019109175.1106106.103.1--???--=

36、 08.1-=FB V V (b) ()??? ? ???=1015105.110ln 0259.0fp φ 2877.0=V ()( ) ()( )() 2 /115 1914 10106.12877.01085.87.114????????=--dT x 510630.8-?=cm ()m ax SD Q ()()() 5151910630.810106.1--??= 810381.1-?=C/cm 2 Now

37、 ()fp FB ox SD TN V C Q V φ2max ++= ()2877.0208.1109175.110381.17 8+-??=-- or 433.0-=TN V V _______________________________________ 10.17 (a) We have n-type material under the gate, so 2 /14??????∈==d fn s C dT eN t x φ where ()288.0105.110ln 0259.01015=??? ? ???=fn φV The

38、n ()( ) ()( )() 2 /115 1914 10106.1288.01085.87.114????????=--dT x or 410863.0-?==C dT t x cm μ863.0=m (b) ()()fn ms ox ox ss SD T t Q Q V φφ2max -+???? ??∈+-= For an +n polysilicon gate, ()288.056.02--=???? ??--=fn g ms e E φφ or Semiconductor Physic

39、s and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions _______________________________________________

40、_______________________________________ 272.0-=ms φV Now ()()()() 4151910863.010106.1m ax --??=SD Q or ()81038.1m ax -?=SD Q C/cm 2 We have ()() 91019106.110106.1--?=?=ss Q C/cm 2 We now find ( )()() ( ) 814 9 81050010 85.89.3106.11

41、038.1----???+?-=T V ()288.02272.0-- or 07.1-=T V V _______________________________________ 10.18 (b) ???? ??++-=fp g m ms e E φχφφ2 where 20.0-=-χφm V and ()3473.0105.110ln 0259.01016=???? ???=fp φV Then ()3

42、473.056.020.0+--=ms φ or 107.1-=ms φV (c) For 0=ss Q ()fp ms ox ox SD TN t Q V φφ2max ++???? ??∈= We find ()( ) ()( )() 2 /116 191410106.13473.01085.87.114?? ??????=--dT x or 4 1030.0-?=dT x cm μ30.0=m Now ()()()() 416191030.010106.1m ax --?

43、?=SD Q or ()8 10797.4m ax -?=SD Q C/cm 2 Then ( )( ) ()( ) 14 8 81085.89.31030010797.4---???=T V ()3473 .02107.1+- or 00455.0+=T V V 0?V _______________________________________ 10.19 Plot _______________________________________ 10.20 Plot _____________

44、__________________________ 10.21 Plot _______________________________________ 10.22 Plot _______________________________________ 10.23 (a) For 1=f Hz (low freq), ()() 8 14 101201085.89.3--??=∈=ox ox ox t C 710876.2-?=F/cm 2 a s t s ox ox ox FB eN V t

45、 C ∈??? ? ??∈∈+∈= ()() ()()( ) ( )() 161914 8 1410106.11085.87.110259.07.119.3101201085.89.3----????? ??+??= 710346.1-?=FB C F/cm 2 dT s ox ox ox x t C ????? ??∈∈+∈=min Now ()3473.0105.110ln 0259.01016=???? ???=fp φV ()( ) ()( )() 2 /116191410106.134

46、73.01085.87.114? ? ??????=--dT x 51000.3-?=cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen

47、 Problem Solutions ______________________________________________________________________________________ Then ()() () 5 814min 1000.37.119.3101201085.89.3---??? ? ??+??=C 810083.3-?=F/cm 2 C (inv)710876.2-?==ox C F/cm 2 (b) 1=f MHz (high freq),

48、 7 10876.2-?=ox C F/cm 2 (unchanged) 710346.1-?=FB C F/cm 2 (unchanged) 8min 10083.3-?=C F/cm 2 (unchanged) C (inv)8min 10083.3-?==C F/cm 2 (c) 10.1-?==ms FB V φV ()fp FB ox SD TN V C Q V φ2max ++= Now ()dT a SD x eN Q =m ax ( )()()5 1

49、6 19 1000.31010 6.1--??= 81080.4-?=C/cm 2 ()3473.0210.110 876.21080.47 8 +-??=--TN V 2385.0-=TN V V _______________________________________ 10.24 (a) 1=f Hz (low freq), ()() 8 14101201085.89.3--??=∈=ox ox ox t C 710876.2-?=F/cm 2 a s t s ox ox ox FB eN

50、 V t C ∈??? ? ??∈∈+∈= ()()()()()()() 14 19 14 8 14105106.11085.87.110259.07.119.3101201085.89.3????? ? ??+ ??= ---- 810726.4-?=FB C F/cm 2 dT s ox ox ox x t C ???? ? ??∈∈+∈=min Now ()2697.0105.1105ln 0259.010 14 =??? ? ? ???=fn φV ()( ) ()( )( ) 2/114

51、 1914 105106.12697 .01085.87.114?? ???????=--dT x 410182.1-?=cm Then ()() () 4 814 min 10182.17.119.3101201085.89.3---??? ? ??+??=C 910504.8-?=F/cm 2 C (inv)710876.2-?==ox C F/cm 2 (b) 1=f MHz (high freq), 710876.2-?=ox C F/cm 2 (unchanged) 810726.4-

52、?=FB C F/cm 2 (unchanged) 9min 10504.8-?=C F/cm 2 (unchanged) C (inv)9min 10504.8-?==C F/cm 2 (c) 95.0?=ms FB V φV ()fn FB ox SD TP V C Q V φ2max -+- = Now ()dT d SD x eN Q =m ax ()()() 4141910182.1105106.1--???= 910456.9-?=C/cm 2 Then ()2697.0

53、295.010876.210456.97 9 -+??-=--TP V 378.0+=TP V V _______________________________________ 10.25 The amount of fixed oxide charge at x is ()x x ?ρ C/cm 2 By lever action, the effect of this oxide charge on the flatband voltage is ()x x t x C V ox ox FB ???? ? ??-=?ρ1 If

54、we add the effect at each point, we must integrate so that ()dx t x x C V ox t ox ox FB ? - =?0 1ρ _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neam

55、en Problem Solutions ______________________________________________________________________________________ 10.26 (a) We have ρx Q t SS ()=? Then ?V C x x t dx FB ox ox ox

56、t =-()z 10ρ ≈-F H G I K J F H I K -z 1C t t Q t dx ox ox ox ox ox SS t t t ??b g =---=- F H I K 1C Q t t t t Q C ox SS ox ox SS ox ??a f or ?V Q t FB SS ox ox =-∈F H G I K J =-????---()161081020010398851019108 14 ...b g b g b g b g

57、or ?V FB =-00742.V (b) We have ρx Q t SS ox ()= = ???--161081020010 19 10 8 .b g b g =?=-64103 .ρO Now ?V C x x t dx C t xdx FB ox ox ox O ox ox ox t t =-=-()z z 10 ρρ or ?V t FB O ox ox =- ∈ρ2 2 = -???

58、---()6410 20010239885103 8 2 14 ...b g b g b g or ?V FB =-00371.V (c) ρρx x t O ox ()F H G I K J = We find 1 22161081020010 19108 t Q ox O SS O ρρ=?=???--.b g b g or ρO =?-128102 . Now ?V C t x x t dx FB ox ox O ox t ox =- ??F H G I K J

59、z 110ρ =-?z 12 2 C t x dx ox O ox ox t ρa(bǔ)f which becomes ?V t t x t FB ox ox O ox ox O ox ox t =-∈??=-∈F H G I K J 1332 302 ρρa(bǔ)f Then ?V FB =-???---()128102001033988510 28214...b g b g b g or 0494.0-=?FB V V ______________

60、_________________________ 10.27 Sketch _______________________________________ 10.28 Sketch _______________________________________ 10.29 (b) ??? ? ??-=-=2ln i d a t bi FB n N N V V V ()()()() ??? ? ?????-=2 101616105.11010ln 0259.0 or 695.0-=FB V V

61、 (c) Apply 3-=G V V, 3?ox V V For 3+=G V V, s dx d ∈-=E ρ n-side: d eN =ρ 1C x eN eN dx d s d s d +∈-=E ?∈-=E Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen

62、 Problem Solutions ______________________________________________________________________________________ 0=E at n x x -=, then s n d x eN C ∈-=1 so ()n s d x x eN +∈- =E for 0≤≤-x x n In the oxide, 0

63、=ρ, so =E ?=E 0dx d constant. From th e boundary conditions, in the oxide s n d x eN ∈-=E In the p-region, 2C x eN eN dx d s a s a s +∈ =E ?∈+=∈-=E ρ 0=E at () p ox x t x +=, then ()[] x x t eN p ox s a -+∈- =E At ox t x =, s n d s p a x eN x eN ∈- =∈

64、- =E So that n d p a x N x N = Since d a N N =, then p n x x = The potential is ? E -=dx φ For zero bias, we can write bi p ox n V V V V =++ where p ox n V V V ,, are the voltage drops across the n-region, the oxide, and the p-region, respectively. For the

65、 oxide: s ox n d ox ox t x eN t V ∈=?E = For the n-region: ()C x x x eN x V n s d n +??? ? ???+∈=22 Arbitrarily, set 0=n V at n x x -=, then s n d x eN C ∈=22 so that ()()22n s d n x x eN x V +∈= At 0=x , s n d n x eN V ∈=22 which is the voltage drop across the n-region. Because

66、 of symmetry, p n V V =. Then for zero bias, we have bi ox n V V V =+2 which can be written as bi s ox n d s n d V t x eN x eN =∈+∈2 or 02=∈-+d s bi ox n n eN V t x x Solving for n x , we obtain d bi s ox ox n eN V t t x ∈+???? ??+-=2 22 If we apply a voltage G V , then replace bi V by G bi V V +, so ()d G bi s ox ox p n eN V V t t x x +∈+???? ??+-