半导体物理与器件-Neamen版

半导体物理与器件(尼曼第四版)答案第一章：半导体材料与晶体1.1 半导体材料的基本特性半导体材料是一种介于导体和绝缘体之间的材料。

它的基本特性包括：1.带隙：半导体材料的价带与导带之间存在一个禁带或带隙，是电子在能量上所能占据的禁止区域。

2.拉伸系统：半导体材料的结构是由原子或分子构成的晶格结构，其中的原子或分子以确定的方式排列。

3.载流子：在半导体中，存在两种载流子，即自由电子和空穴。

自由电子是在导带上的，在外加电场存在的情况下能够自由移动的电子。

空穴是在价带上的，当一个价带上的电子从该位置离开时，会留下一个类似电子的空位，空穴可以看作电子离开后的痕迹。

4.掺杂：为了改变半导体材料的导电性能，通常会对其进行掺杂。

掺杂是将少量元素添加到半导体材料中，以改变载流子浓度和导电性质。

1.2 半导体材料的结构与晶体缺陷半导体材料的结构包括晶体结构和非晶态结构。

晶体结构是指材料具有有序的周期性排列的结构，而非晶态结构是指无序排列的结构。

晶体结构的特点包括：1.晶体结构的基本单位是晶胞，晶胞在三维空间中重复排列。

2.晶格常数是晶胞边长的倍数，用于描述晶格的大小。

3.晶体结构可分为离子晶体、共价晶体和金属晶体等不同类型。

晶体结构中可能存在各种晶体缺陷，包括：1.点缺陷：晶体中原子位置的缺陷，主要包括实际缺陷和自间隙缺陷两种类型。

2.线缺陷：晶体中存在的晶面上或晶内的线状缺陷，主要包括位错和脆性断裂两种类型。

3.面缺陷：晶体中存在的晶面上的缺陷，主要包括晶面位错和穿孔两种类型。

1.3 半导体制备与加工半导体制备与加工是指将半导体材料制备成具有特定电性能的器件的过程。

它包括晶体生长、掺杂、薄膜制备和微电子加工等步骤。

晶体生长是将半导体材料从溶液或气相中生长出来的过程。

常用的晶体生长方法包括液相外延法、分子束外延法和气相外延法等。

掺杂是为了改变半导体材料的导电性能，通常会对其进行掺杂。

常用的掺杂方法包括扩散法、离子注入和分子束外延法等。

半导体物理与器件（教学大纲）Semiconductor Physics and Devices课程编码：12330540学分：课程类别：专业基础课计划学时： 48 其中讲课： 48 实验或实践： 0 上机：0适用专业：IC设计、电信推荐教材：尼曼（Donald H.Neamen）著，赵毅强，姚素英。

解晓东译，《半导体物理与器件》（第3版），电子工业出版社，2010参考书目：D. A. Neamen，《Semiconductor Physics and Devices: Basic Principles》，清华出版社，2003R. T. Pierret著，黄如等译，《半导体器件基础》，电子工业出版社，2004刘恩科、朱秉升、罗晋生等，《半导体物理学》，西安交通大学出版社，2004黄昆、谢希德，《半导体物理学》，科学出版社，1958曾谨言，《量子力学》，科学出版社，1981谢希德、方俊鑫，《固体物理学》,上海科学技术出版社，1961课程的教学目的与任务本课程是集成电路专业的重要选修课之一。

本课程较全面地论述了半导体的一些基本物理概念、现象、物理过程及其规律，并在此基础上选择目前集成电路与系统的核心组成部分，如双极型晶体管（BJT）、金属－半导体场效应晶体管（MESFET）和ＭOS场效应晶体管（MOSFET）等，作为分析讨论的主要对象来介绍半导体器件基础。

学习和掌握这些半导体物理和半导体器件的基本理论和分析方法，为学习诸如《集成电路工艺》、《集成电路设计》等后续课程打下基础，也为将来从事微电子学的研究以及现代VLSI与系统设计和制造工作打下坚实的理论基础。

课程的基本要求本课程要求学生掌握半导体物理和半导体器件的基本概念和基本规律，对于基础理论，要求应用简单的模型定性说明，并能作简单的数学处理。

学习过程中，注意提高分析和解决实际问题的能力，并重视理论与实践的结合。

本课程涉及的物理概念和基本原理较多，为了加深对它们的理解，在各章节里都给学生留有一些习题或思考题，这些题目有的还是基本内容的补充。

______________________________________________________________________________________Chapter 1Problem Solutions1.1 (a)fcc: 8 corner atoms 18/1atom6 face atoms32/1atomsTotal of 4 atoms per unit cell (b)bcc: 8 corner atoms 18/1atom1 enclosed atom=1 atom Total of 2 atoms per unit cell(c)Diamond: 8 corner atoms 18/1atom6 faceatoms 32/1atoms4 enclosedatoms= 4 atomsTotal of 8 atoms per unit cell_______________________________________ 1.2 (a)Simple cubic lattice: r a 2Unit cell vol33382rra1 atom per cell, so atom vol 3413r ThenRatio%4.52%10083433rr(b)Face-centered cubic latticerd aa rd22224Unit cell vol 33321622rr a4 atoms per cell, so atom vol3443r ThenRatio%74%10021634433rr (c)Body-centered cubic latticeraa rd3434Unit cell vol 3334ra2 atoms per cell, so atom vol 3423r ThenRatio%68%1003434233r r (d)Diamond lattice Body diagonal raa rd3838Unit cell vol3338r a8 atoms per cell, so atom vol 3483r ThenRatio%34%1003834833rr _______________________________________1.3(a)oA a43.5; From Problem 1.2d,ra38Then oAa r176.18343.583Center of one silicon atom to center ofnearest neighboroAr 35.22______________________________________________________________________________________ (b)Number density22381051043.58cm 3(c)Mass density23221002.609.28105..AN W t At N 33.2grams/cm3_______________________________________1.4(a)4 Ga atoms per unit cell Number density381065.54Density of Ga atoms 221022.2cm34 As atoms per unit cell Density of As atoms 221022.2cm3(b)8 Ge atoms per unit cell Number density381065.58Density of Ge atoms221044.4cm3_______________________________________ 1.5From Figure 1.15 (a)aa d4330.0232oAd 447.265.54330.0(b)aa d7071.022oAd 995.365.57071.0_______________________________________1.674.5423232222sin a a 5.109_______________________________________ 1.7(a) Simple cubic: oAr a 9.32(b)fcc:oAr a515.524(c) bcc:oA r a 503.434(d) diamond:oAra007.9342_______________________________________ 1.8 (a)Br 2035.122035.12oBAr 4287.0(b)oAa 07.2035.12(c)A-atoms: # of atoms1818Density381007.21231013.1cm3B-atoms: # of atoms3216Density381007.23231038.3cm3_______________________________________ 1.9(a)oAr a 5.42# of atoms1818Number density38105.412210097.1cm3______________________________________________________________________________________Mass density AN W t At N ..23221002.65.12100974.1228.0gm/cm3(b)oAr a196.534# of atoms 21818Number density3810196.5222104257.1cm3Mass density23221002.65.12104257.1296.0gm/cm3_______________________________________ 1.10From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is ground,Volume = 0.74 cm3_______________________________________1.11(b)oAa 8.20.18.1(c)Na: Density38108.22/1221028.2cm3Cl: Density221028.2cm3(d)Na: At. Wt. = 22.99 Cl: At. Wt. = 35.45 So, mass per unit cell23231085.41002.645.352199.2221Then mass density21.2108.21085.43823grams/cm3_______________________________________ 1.12(a)oAa 88.122.223Then oA a 62.4Density of A:22381001.11062.41cm3Density of B:22381001.11062.41cm3(b)Same as (a) (c)Same material_______________________________________ 1.13oAa619.438.122.22(a) For 1.12(a), A-atomsSurface density28210619.411a1410687.4cm2For 1.12(b), B-atoms: oAa 619.4Surface density14210687.41acm2For 1.12(a) and (b), Same material(b) For 1.12(a), A-atoms;oAa 619.4Surface density212a1410315.3cm2B-atoms;Surface density______________________________________________________________________________________14210315.321a cm 2For 1.12(b), A-atoms;oAa 619.4Surface density212a1410315.3cm2B-atoms;Surface density14210315.321acm2For 1.12(a) and (b), Same material_______________________________________ 1.14 (a)Vol. Density31oaSurface Density212oa(b)Same as (a)_______________________________________ 1.15 (i)(110) plane(see Figure 1.10(b))(ii) (111) plane(see Figure 1.10(c))(iii) (220) plane,1,1,21,21Same as (110) plane and [110]direction(iv) (321) plane6,3,211,21,31Intercepts of plane at6,3,2sq p [321] direction is perpendicular to(321) plane_______________________________________1.16(a)31311,31,11(b)12141,21,41_______________________________________ 1.17Intercepts: 2, 4, 331,41,21(634) plane_______________________________________ 1.18(a)oAa d 28.5(b)oAa d734.322(c)oAa d048.333_______________________________________ 1.19(a) Simple cubic(i) (100) plane:Surface density2821073.411a141047.4cm 2(ii) (110) plane:Surface density212a141016.3cm 2(iii) (111) plane: Area of planebh21where oAa b 689.62Now2222243222a a a hSooAh793.573.426______________________________________________________________________________________Area of plane881079304.51068923.62116103755.19cm 2Surface density16103755.19613141058.2cm2(b) bcc(i) (100) plane:Surface density 1421047.41acm2(ii) (110) plane: Surface density222a141032.6cm 2(iii) (111) plane:Surface density16103755.19613141058.2cm2(c) fcc(i) (100) plane:Surface density 1421094.82acm2(ii) (110) plane: Surface density222a141032.6cm 2(iii) (111) plane:Surface density16103755.19213613151003.1cm2_______________________________________ 1.20 (a)(100) plane: - similar to a fcc:Surface density281043.52141078.6cm 2(b)(110) plane:Surface density281043.524141059.9cm2(c)(111) plane: Surface density281043.5232141083.7cm2_______________________________________1.21oAr a703.6237.2424(a)#/cm338310703.64216818a2210328.1cm3(b)#/cm222124142a210703.62281410148.3cm2(c)oA a d74.422703.622(d)# of atoms2213613Area of plane: (see Problem 1.19)oAa b4786.92oAa h2099.826Area88102099.8104786.92121bh______________________________________________________________________________________15108909.3cm2#/cm215108909.32=141014.5cm2oAa d87.333703.633_______________________________________ 1.22Density of silicon atoms 22105cm3and4 valence electrons per atom, soDensity of valence electrons 23102cm3_______________________________________ 1.23Density of GaAs atoms22381044.41065.58cm3An average of 4 valence electrons peratom,SoDensity of valence electrons231077.1cm3_______________________________________ 1.24 (a)%10%10010510532217(b)%104%10010510262215_______________________________________ 1.25 (a)Fraction by weight7221610542.106.2810582.10102(b)Fraction by weight5221810208.206.2810598.3010_______________________________________ 1.26Volume density 1631021dcm3So610684.3dcmoAd 4.368We haveoo Aa 43.5Then85.6743.54.368oa d _______________________________________ 1.27Volume density 1531041dcm 3So61030.6dcmoAd630We have oo Aa 43.5Then11643.5630oa d _______________________________________。

Chapter 1010.1(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 ()⎪⎪⎭⎫ ⎝⎛⨯⨯=1015105.1107ln 0259.0 3381.0=V 2/14⎥⎦⎤⎢⎣⎡∈=a fp s dTeN x φ()()()()()2/1151914107106.13381.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--51054.3-⨯=cm or μ354.0=dT x m(ii) ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1103ln 0259.0fp φ3758.0=V ()()()()()2/1161914103106.13758.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx51080.1-⨯=cmor μ180.0=dT x m(b) ()03022.03003500259.0=⎪⎭⎫⎝⎛=kT V⎪⎪⎭⎫⎝⎛-=kT E N N n g c i exp 2υ ()()319193003501004.1108.2⎪⎭⎫⎝⎛⨯⨯=⎪⎭⎫⎝⎛-⨯03022.012.1exp221071.3⨯=so 111093.1⨯=i n cm 3-(i)()⎪⎪⎭⎫⎝⎛⨯⨯=11151093.1107ln 03022.0fp φ3173.0=V()()()()()2/1151914107106.13173.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x51043.3-⨯=cm or μ343.0=dT x m(ii) ()⎪⎪⎭⎫⎝⎛⨯⨯=11161093.1103ln 03022.0fp φ3613.0=V()()()()()2/1161914103106.13613.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x51077.1-⨯=cm or μ177.0=dT x m_______________________________________ 10.3(a) ()2/14m ax ⎥⎦⎤⎢⎣⎡∈=='d fn s d dT d SDeN eN x eN Q φ()()[]2/14fns d eN φ∈=1st approximation: Let 30.0=fn φV Then()281025.1-⨯()()()()()()[]30.01085.87.114106.11419--⨯⨯=dN 141086.7⨯=⇒d N cm 3-2nd approximation:()2814.0105.11086.7ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fn φV Then ()281025.1-⨯()()()()()()[]2814.01085.87.114106.11419--⨯⨯=d N 141038.8⨯=⇒d N cm 3-(b) ()2831.0105.11038.8ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fn φV()566.02831.022===fn s φφV _______________________________________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 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 or226.0+=ms 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 is always positive.(c) 15102⨯≅d N cm 3-_______________________________________10.7 From Problem 10.5, 9932.0-=ms φV ox ssms FB C Q V '-=φ (a) ()()814102001085.89.3--⨯⨯=∈=ox ox ox t C 710726.1-⨯=F/cm 2()()7191010726.1106.11059932.0--⨯⨯⨯--=FB V 040.1-=V (b) ()()81410801085.89.3--⨯⨯=ox C 710314.4-⨯=F/cm 2 ()()7191010314.4106.11059932.0--⨯⨯⨯--=FB V012.1-=V _______________________________________10.8 (a) 42.0-≅ms φV 42.0-==ms FB V φV(b) ()()781410726.1102001085.89.3---⨯=⨯⨯=ox C F/cm 2 (i)()()7191010726.1106.1104--⨯⨯⨯-='-=∆ox ss FB C Q V 0371.0-=V (ii)()()7191110726.1106.110--⨯⨯-=∆FB V 0927.0-=V(c) 42.0-==ms FB V φV ()()781410876.2101201085.89.3---⨯=⨯⨯=ox C F/cm 2 (i)()()7191010876.2106.1104--⨯⨯⨯-=∆FB V 0223.0-=V (ii)()()7191110876.2106.110--⨯⨯-=∆FB V0556.0-=V _______________________________________10.9 ⎪⎪⎭⎫ ⎝⎛++'-'=fp g mms e E φχφφ2 where()365.0105.1102ln 0259.01016=⎪⎪⎭⎫ ⎝⎛⨯⨯=fp φV Then ()365.056.025.320.3++-=ms φor975.0-=ms φVNowox ss ms FB C Q V '-=φor ()ox FB ms ss C V Q -='φ We have()()814104501085.89.3--⨯⨯=∈=ox ox ox t C or 81067.7-⨯=ox C F/cm 2 So now ()[]()81067.71975.0-⨯⋅---='ssQ 91092.1-⨯=C/cm 2or10102.1⨯='e Q ss cm 2- _______________________________________10.10 ()3653.0105.1102ln 0259.01016=⎪⎪⎭⎫ ⎝⎛⨯⨯=fp φV ()()()()()2/1161914102106.13653.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x510174.2-⨯=cm()dT a SDx eN Q ='m ax ()()()5161910174.2102106.1--⨯⨯⨯=810958.6-⨯=C/cm 2()()781410301.2101501085.89.3---⨯=⨯⨯=ox C F/cm 2()fp ms ox ss SDTN C Q Q V φφ2max ++'-'= ()()71910810301.2106.110710958.6---⨯⨯⨯-⨯= ()3653.02++ms φ ms φ+=9843.0(a) n + poly 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 φV0343.095.09843.0+=-=TN V V_______________________________________10.11 ()3161.0105.1103ln 0259.01015=⎪⎪⎭⎫ ⎝⎛⨯⨯=fn φV ()()()()()2/1151914103106.13161.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x 510223.5-⨯=cm ()dT d SDx eN Q ='m ax ()()()5151910223.5103106.1--⨯⨯⨯= 810507.2-⨯=C/cm 2 ()()781410301.2101501085.89.3---⨯=⨯⨯=ox C F/cm 2 ()fn ms ox ss SDTP 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 + poly 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 _______________________________________10.12()3294.0105.1105ln 0259.01015=⎪⎪⎭⎫ ⎝⎛⨯⨯=fp φV The surface potential is ()659.03294.022===fp s φφV We have 90.0-='-=oxssms FB C Q V φV Now()FB s oxSDT V C Q V ++'=φmaxWe obtain 2/14⎥⎦⎤⎢⎣⎡∈=a fp s dTeN x φ()()()()()2/1151914105106.13294.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=-- or410413.0-⨯=dT x cm Then()()()()4151910413.0105106.1m ax --⨯⨯⨯='SDQ or()810304.3m ax -⨯='SDQ C/cm 2 We also find()()814104001085.89.3--⨯⨯=∈=ox ox ox t C or810629.8-⨯=ox C F/cm 2 Then90.0659.010629.810304.388-+⨯⨯=--T Vor142.0+=T V V_______________________________________10.13()()814102201085.89.3--⨯⨯=∈=ox ox ox t C 710569.1-⨯=F/cm 2()()1019104106.1⨯⨯='-ssQ 9104.6-⨯=C/cm 2By trial and error, let 16104⨯=a N cm 3-.Now ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1104ln 0259.0fp φ3832.0=V()()()()()2/1161914104106.13832.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x 510575.1-⨯=cm ()m ax SDQ ' ()()()5161910575.1104106.1--⨯⨯⨯= 710008.1-⨯=C/cm 294.0-≅ms φV Then()fp ms oxss SDTN C Q Q V φφ2max ++'-'=79710569.1104.610008.1---⨯⨯-⨯=()3832.0294.0+- Then 428.0=TN V V 45.0≅V_______________________________________10.14()()814101801085.89.3--⨯⨯=∈=ox ox ox t C 7109175.1-⨯=F/cm 3- ()()1019104106.1⨯⨯='-ssQ 9104.6-⨯=C/cm 2By trial and error, let 16105⨯=d N cm 3- Now()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1105ln 0259.0fn φ3890.0=V()()()()()2/1161914105106.13890.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x510419.1-⨯=cm()m ax SDQ ' ()()()5161910419.1105106.1--⨯⨯⨯= 710135.1-⨯=C/cm 3-10.1+≅ms φV Then()()fn ms ox ss SDTP C Q Q V φφ2max -+'+'-= ()797109175.1104.610135.1---⨯⨯+⨯-= ()3890.0210.1-+Then 303.0-=TP V V, which is within thespecified value. _______________________________________ 10.15 We have 710569.1-⨯=ox C F/cm 2 9104.6-⨯='ssQ 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 410182.1-⨯=cm ()m ax SDQ ' ()()()4141910182.1105106.1--⨯⨯⨯= 910456.9-⨯=C/cm 233.0-≅ms φVThen ()()fn ms oxss SDTP C Q Q V φφ2max -+'+'-= ⎪⎪⎭⎫⎝⎛⨯⨯+⨯-=---79910569.1104.610456.9 ()2697.0233.0--970.0=V Then 970.0-=TP V V 975.0-≅ V which meets the specification._______________________________________ 10.16(a) 03.1-≅ms φV()()814101801085.89.3--⨯⨯=ox C 7109175.1-⨯=F/cm 2Now oxss ms FB C Q V '-=φ()()71019109175.1106106.103.1--⨯⨯⨯--= 08.1-=FB V V(b) ()⎪⎪⎭⎫ ⎝⎛⨯=1015105.110ln 0259.0fp φ 2877.0=V ()()()()()2/115191410106.12877.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dT x 510630.8-⨯=cm()m ax SDQ ' ()()()5151910630.810106.1--⨯⨯= 810381.1-⨯=C/cm 2 Now ()fp FB oxSDTN V C Q V φ2max ++'=()2877.0208.1109175.110381.178+-⨯⨯=-- or 433.0-=TN V V_______________________________________10.17 (a) We have n-type material under the gate, so2/14⎥⎦⎤⎢⎣⎡∈==d fn s C dT eN t x φ where()288.0105.110ln 0259.01015=⎪⎪⎭⎫ ⎝⎛⨯=fn φVThen()()()()()2/115191410106.1288.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--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 or272.0-=ms φV Now ()()()()4151910863.010106.1m ax --⨯⨯='SD Q or ()81038.1m ax -⨯='SDQ C/cm 2 We have()()91019106.110106.1--⨯=⨯='ssQ C/cm 2 We now find ()()()()81498105001085.89.3106.11038.1----⨯⨯⨯+⨯-=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()3473.056.020.0+--=ms φ or 107.1-=ms φV (c) For 0='ss Q ()fp ms ox ox SDTN t Q V φφ2max ++⎪⎪⎭⎫⎝⎛∈'= We find()()()()()2/116191410106.13473.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dT xor 41030.0-⨯=dT x cm μ30.0=m Now()()()()416191030.010106.1m ax --⨯⨯='SDQ or ()810797.4m ax -⨯='SDQ C/cm 2Then()()()()14881085.89.31030010797.4---⨯⨯⨯=T V()3473.02107.1+- or00455.0+=T V V 0≅V _______________________________________ 10.19Plot _______________________________________ 10.20 Plot_______________________________________ 10.21 Plot _______________________________________10.22 Plot_______________________________________10.23 (a) For 1=f Hz (low freq), ()()814101201085.89.3--⨯⨯=∈=ox ox ox t C 710876.2-⨯=F/cm 2a st s ox ox oxFB eNV t C ∈⎪⎪⎭⎫ ⎝⎛∈∈+∈=' ()()()()()()()16191481410106.11085.87.110259.07.119.3101201085.89.3----⨯⨯⎪⎭⎫ ⎝⎛+⨯⨯= 710346.1-⨯='FB C F/cm 2 dTs ox ox oxx t C ⋅⎪⎪⎭⎫ ⎝⎛∈∈+∈='minNow ()3473.0105.110ln 0259.01016=⎪⎪⎭⎫ ⎝⎛⨯=fp φV ()()()()()2/116191410106.13473.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTx51000.3-⨯=cmThen ()()()5814min 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), 710876.2-⨯=ox C F/cm 2 (unchanged) 710346.1-⨯='FBC F/cm 2 (unchanged) 8min10083.3-⨯='C F/cm 2 (unchanged) C '(inv)8min10083.3-⨯='=C F/cm 2 (c) 10.1-≅==ms FB V φV()fp FB oxSDTN V C Q V φ2max ++'=Now()dT a SDx eN Q ='m ax ()()()516191000.310106.1--⨯⨯=81080.4-⨯=C/cm 2 ()3473.0210.110876.21080.478+-⨯⨯=--TN V 2385.0-=TN V V_______________________________________10.24(a) 1=f Hz (low freq), ()()814101201085.89.3--⨯⨯=∈=ox ox oxt C 710876.2-⨯=F/cm 2a st s ox ox oxFB eNV t C ∈⎪⎪⎭⎫ ⎝⎛∈∈+∈='()()()()()()()141914814105106.11085.87.110259.07.119.3101201085.89.3⨯⨯⨯⎪⎭⎫ ⎝⎛+⨯⨯=---- 810726.4-⨯='FBC F/cm 2 dTs ox ox oxx t C ⋅⎪⎪⎭⎫⎝⎛∈∈+∈='minNow()2697.0105.1105ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fn φV()()()()()2/1141914105106.12697.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x 410182.1-⨯=cmThen()()()4814min 10182.17.119.3101201085.89.3---⨯⎪⎭⎫ ⎝⎛+⨯⨯='C 910504.8-⨯=F/cm 2C '(inv)710876.2-⨯==ox C F/cm 2(b) 1=f MHz (high freq),710876.2-⨯=ox C F/cm 2 (unchanged)810726.4-⨯='FBC F/cm 2 (unchanged) 9min10504.8-⨯='C F/cm 2 (unchanged) C '(inv)9min10504.8-⨯='=C F/cm 2 (c) 95.0≅=ms FB V φV()fn FB oxSDTP V C Q V φ2max -+'-=Now()dT d SDx eN Q ='m ax ()()()4141910182.1105106.1--⨯⨯⨯= 910456.9-⨯=C/cm 2Then()2697.0295.010876.210456.979-+⨯⨯-=--TP V378.0+=TP V V_______________________________________10.25The amount of fixed oxide charge at x is ()x x ∆ρ C/cm 2By lever action, the effect of this oxide charge on the flatband voltage is()x x t x C V ox ox FB ∆⎪⎪⎭⎫⎝⎛-=∆ρ1 If we add the effect at each point, we must integrate so that ()dx t x x C V oxt oxoxFB⎰-=∆01ρ _______________________________________10.26 (a) We have ρx Q t SS ()='∆ Then∆V C x x t dx FB ox ox ox t=-()z 10ρ ≈-'F H G I K J F H I K-z 1C t t Q t dx ox ox oxox oxSSt t t ∆∆b g =-'--=-'F H I K 1C Q t t t t Q C ox SS ox ox SSox ∆∆a for ∆V Q t FB SS ox ox=-'∈F H G I K J =-⨯⨯⨯⨯---()16108102001039885101910814...b g b g b gb gor∆V FB =-00742.V(b) We have ρx Q t SS ox()='=⨯⨯⨯--16108102001019108.b g b g =⨯=-64103.ρONow ∆V C x x t dx C t xdx FB oxox oxOox oxoxt t =-=-()zz10ρρor ∆V t FB O oxox=-∈ρ22=-⨯⨯⨯---()6410200102398851038214...bg b g b gor∆V FB =-00371.V (c) ρρx x t O ox()F H G I KJ =We find12216108102001019108t Q ox O SS O ρρ='⇒=⨯⨯⨯--.b gb g or ρO =⨯-128102. Now ∆V C t x x t dx FB ox ox O ox t ox =-⋅⋅F H G I KJ z110ρ =-⋅z122C t x dx ox O oxox t ρaf which becomes ∆V t t x t FB ox oxO oxox O oxox t =-∈⋅⋅=-∈F H G I KJ 1332302ρρaf Then∆V FB =-⨯⨯⨯---()12810200103398851028214...b g b g b gor 0494.0-=∆FB V V_______________________________________10.27 Sketch_______________________________________10.28 Sketch_______________________________________10.29 (b)⎪⎪⎭⎫⎝⎛-=-=2ln i d a t bi FB n N N V V V ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯-=2101616105.11010ln 0259.0or695.0-=FB V V(c) Apply 3-=G V V, 3≅ox V VFor 3+=G V V,sdx d ∈-=Eρ n-side: d eN =ρ1C x eN eN dx d sd s d +∈-=E ⇒∈-=E0=E at n x x -=, then snd x eN C ∈-=1 so()n s dx x eN +∈-=E for 0≤≤-x x n In the oxide, 0=ρ, so=E ⇒=E 0dxd constant. From the boundary conditions, in the oxidesn d x eN ∈-=E In the p-region,2C x eN eN dx d sa sa s+∈=E ⇒∈+=∈-=Eρ 0=E at ()p ox x t x +=, then ()[]x x teN p oxsa-+∈-=EAt ox t x =, snd sp a x eN x eN ∈-=∈-=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 φFor zero bias, we can write bi p ox n V V V V =++where p ox n V V V ,, are the voltage drops acrossthe n-region, the oxide, and the p-region, respectively. For the oxide:soxn d ox ox t x eN t V ∈=⋅E =For the n-region:()C x x x eN x V n s d n '+⎪⎪⎭⎫ ⎝⎛⋅+∈=22Arbitrarily, set 0=n V at n x x -=, thensnd x eN C ∈='22so that()()22n sdn x x eN x V +∈=At 0=x , snd n x eN V ∈=22which is the voltagedrop across the n-region. Because ofsymmetry, p n V V =. Then for zero bias, wehavebi ox n V V V =+2 which can be written as bi sox n d s n d V t x eN x eN =∈+∈2or 02=∈-+ds bi ox n n eN V t x x Solving for n x , we obtain dbis ox ox n eN V t t x ∈+⎪⎪⎭⎫ ⎝⎛+-=222 If we apply a voltage G V , then replace bi V by G bi V V +, so ()dG bi s ox ox p n eN V V t t x x +∈+⎪⎪⎭⎫ ⎝⎛+-==222 We find2105008-⨯-==p n x x()()()()()1619142810106.1695.31085.87.11210500---⨯⨯+⎪⎪⎭⎫ ⎝⎛⨯+ which yields510646.4-⨯==p n x x cmNow soxn d ox t x eN V ∈=()()()()()()148516191085.87.111050010646.410106.1----⨯⨯⨯⨯=or359.0=ox V V We also findsnd p n x eN V V ∈==22()()()()()142516191085.87.11210646.410106.1---⨯⨯⨯=or67.1==p n V V V_______________________________________10.30(a) n-type (b) We have731210110210200---⨯=⨯⨯=ox C F/cm 2Also ()()7141011085.89.3--⨯⨯=∈=⇒∈=ox ox ox ox ox ox C t t C or 61045.3-⨯=ox t cm 5.34=nm o A 345= (c)oxssms FB C Q V '-=φ or 71050.080.0-'--=-ssQwhich yields8103-⨯='ssQ C/cm 21110875.1⨯=cm 2- (d) ⎪⎪⎭⎫ ⎝⎛∈⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛∈∈+∈='d s s ox ox ox FB eN e kT t C()()[][6141045.31085.89.3--⨯÷⨯= ()()()()()⎥⎥⎦⎤⨯⨯⨯⎪⎭⎫ ⎝⎛+--161914102106.11085.87.110259.07.119.3 which yields81082.7-⨯='FBC F/cm 2 or156=FB C pF_______________________________________10.31 (a) Point 1: Inversion 2: Threshold3: Depletion4: Flat-band5: Accumulation_______________________________________10.32 We have ()()[]fp ms x GS ox nV V C Q φφ2+---=' ()()max SD ssQ Q '+'- Now let DS x V V =, so ()⎩⎨⎧--='DS GS ox n V V C Q ()()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡+-'+'+fp ms ox ss SD C Q Q φφ2m ax For a p-type substrate, ()max SDQ ' is a negative value, so we can write()⎩⎨⎧--='DS GS ox n V V C Q()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡++'-'-fp ms ox ss SD C Q Q φφ2m ax Using the definition of threshold voltage T V ,we have()[]T DS GS ox nV V V C Q ---=' At saturation()T GS DS DS V V sat V V -== which then makes nQ 'equal to zero at the drain terminal._______________________________________10.33(a) ()[]222DS DS T GS n D V V V V L W k I --⋅'= ()()()()[]22.02.04.08.028218.0--⎪⎭⎫ ⎝⎛= 0864.0=mA (b) ()22T GS n D V V LW k I -⋅'= ()()24.08.08218.0-⎪⎭⎫ ⎝⎛= 1152.0=mA(c) Same as (b), 1152.0=D I mA(d) ()22T GS n D V V L W k I -⋅'=()()24.02.18218.0-⎪⎭⎫ ⎝⎛= 4608.0=mA _______________________________________ 10.34 (a) ()[]222SDSD T SG p D V V V V LW k I -+⋅'= ()()()()[]225.025.04.08.0215210.0--⎪⎭⎫ ⎝⎛= 103.0=D I mA(b) ()22T SG p D V V LW k I +⋅'= ()()24.08.015210.0-⎪⎭⎫ ⎝⎛= 12.0=mA(c) ()22T SG p D V V L W k I +⋅'=()()24.02.115210.0-⎪⎭⎫ ⎝⎛=48.0=mA(d) Same as (c), 48.0=D I mA_______________________________________10.35(a) ()22T GS n D V V LW k I -⋅'=()28.04.126.00.1-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=L W26.9=⇒LW(b) ()()28.085.126.926.0-⎪⎭⎫ ⎝⎛=D I06.3=mA(c) ()[]222DSDS T GS n D V V V V L W k I --⋅'= ()()()()[]215.015.08.02.1226.926.0--⎪⎭⎫ ⎝⎛=271.0=mA_______________________________________10.36(a) Assume biased in saturation region()22T SG p D V V L W k I +⋅'=()()2020212.010.0T V +⎪⎭⎫ ⎝⎛=289.0+=⇒T V VNote: 0.1=SD V V 289.00+=+>T SG V V V So the transistor is biased in the saturation region.(b) ()()2289.04.020212.0+⎪⎭⎫ ⎝⎛=D I570.0=mA(c) ()()[()15.0289.06.0220212.0+⎪⎭⎫⎝⎛=D I()]215.0-or293.0=D I mA_______________________________________10.37 ()()781410138.3101101085.89.3---⨯=⨯⨯=ox C F/cm 2 ()()()()2.122010138.342527-⨯==L W C K ox n n μ310111.1-⨯=A/V 2=1.111 mA/V 2(a) 0=GS V , 0=D I 6.0=GS V V, ()15.0=sat V DS V, ()()()245.06.0111.1-=sat I D 025.0=mA2.1=GS V V, ()75.0=sat V DS V, ()()()245.02.1111.1-=sat I D 625.0=mA8.1=GS V V, ()35.1=sat V DS V,()()()245.08.1111.1-=sat I D 025.2=mA4.2=GS V V, ()95.1=sat V DS V,()()()245.04.2111.1-=sat I D 225.4=mA (c)0=D I for 45.0≤GS V V 6.0=GS V V,()()()()[]21.01.045.06.02111.1--=D I 0222.0=mA 2.1=GS V V,()()()()[]21.01.045.02.12111.1--=D I 156.0=mA 8.1=GS V V,()()()()[]21.01.045.08.12111.1--=D I 289.0=mA 4.2=GS V V,()()()()[]21.01.045.04.22111.1--=D I 422.0=mA_______________________________________10.38()()814101101085.89.3--⨯⨯=∈=ox ox ox t C 710138.3-⨯=F/cm 2L WC K ox p p 2μ=()()()()2.123510138.32107-⨯=41061.9-⨯=A/V 2=0.961 mA/V 2(a) 0=SG V , 0=D I6.0=SG V V, ()25.0=sat V SD V()()()235.06.0961.0-=sat I D 060.0=mA2.1=SG V V, ()85.0=sat V SD V()()()235.02.1961.0-=sat I D 694.0=mA 8.1=SG V V, ()45.1=sat V SD V()()()235.08.1961.0-=sat I D02.2=mA4.2=SG V V, ()05.2=sat V SD V()()()235.04.2961.0-=sat I D04.4=mA (c)0=D I for 35.0≤SG V V6.0=SG V V()()()()[]21.01.035.06.02961.0--=D I 0384.0=mA 2.1=SG V V ()()()()[]21.01.035.02.12961.0--=D I154.0=mA8.1=SG V V ()()()()[]21.01.035.08.12961.0--=D I 269.0=mA 4.2=SG V V()()()()[]21.01.035.04.22961.0--=D I 384.0=mA_______________________________________10.39(a) From Problem 10.37,111.1=n K mA/V 2 For 8.0-=GS V V, 0=D I0=GS V , ()8.0=sat V DS V()()()28.00111.1+=sat I D 711.0=mA8.0+=GS V V, ()6.1=sat V DS V()()()28.08.0111.1+=sat I D 84.2=mA6.1=GS V V, ()4.2=sat V DS V()()()28.06.1111.1+=sat I D 40.6=mA_______________________________________10.40 Sketch _______________________________________10.41 Sketch _______________________________________ 10.42We have ()T DS T GS DS V V V V sat V -=-=so that()T DS DS V sat V V +=Since ()sat V V DS DS >, the transistor is always biased in the saturation region. Then()2T GS n D V V K I -=where, from Problem 10.37,111.1=n K mA/V 2and 45.0=T V V10.43From Problem 10.38, 961.0=p K mA/V 2()()[]22SD SD T SG p D V V V V K I -+=()T SG p V SDDd V V K V I g SD +=∂∂=→20For 35.0≤SG V V, 0=d g For 35.0>SG V V,()()35.0961.02-=SG d V g For 4.2=SG V V,()()35.04.2961.02-=d g 94.3=mA/V_______________________________________10.44(a) GS D m V I g ∂∂=()()[]{}22DS DS T GS n GSV V V V K V --∂∂=()DS n V K 2=()()05.0225.1n K =5.12=⇒n K mA/V 2(b) ()()()[()]205.005.03.08.025.12--=D I 594.0=mA(c) ()()23.08.05.12-=D I125.3=mA_______________________________________10.45We find that 2.0≅T V V Now ()()T GS oxn D V V LC W sat I -⋅=2μ where ()()814104251085.89.3--⨯⨯=∈=ox ox oxt C or81012.8-⨯=ox C F/cm 2We are given 10=L W . From the graph, for 3=GS V V, we have ()033.0≅sat I D , then ()2.032033.0-⋅=LC W oxn μ or310139.02-⨯=LC W oxn μor()()3810139.01012.81021--⨯=⨯n μwhich yields342=n μcm 2/V-s_______________________________________10.46 (a)()T GS DS V V sat V -= or8.48.04=⇒-=GS GS V V V(b) ()()()sat V K V V K sat I DS n T GS n D 22=-= so()244102n K =⨯- which yields μ5.12=n K A/V 2 (c) ()2.18.02=-=-=T GS DS V V sat V Vso ()sat V V DS DS > ()()()258.021025.1-⨯=-sat I Dor ()μ18=sat I D A(d)()sat V V DS DS <()[]22DS DS T GS n D V V V V K I --= ()()()()[]25118.0321025.1--⨯=-orμ5.42=D I A_______________________________________10.47(a) ()()814101801085.89.3--⨯⨯=ox C 7109175.1-⨯=F/cm 2(i)()()7109175.1450-⨯=='ox n nC k μ 510629.8-⨯=A/V 2 or μ29.86='nk A/V 2 (ii)()()22T GS nD V V L W k sat I -⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛'= ()24.02208629.08.0-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=L W24.7=⇒L W(b) (i) ()()7109175.1210-⨯=='ox p p C k μ 510027.4-⨯=A/V 2or μ27.40='p k A/V 2(ii) ()()22T SG p D V V L W k sat I +⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛'= ()24.02204027.08.0-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=L W5.15=⇒LW_______________________________________ 10.48 From Problem 10.37, 111.1=n K mA/V 2(a) ()()[]{}22DS DS T GS n GS mL V V V V K V g --∂∂= ()()()()1.02111.12==DS n V K so 222.0=mL g mA/V (b) (){}2T GS n GS ms V V K V g -∂∂=()()()45.05.1111.122-=-=T GS n V V K so 33.2=ms g mA/V _______________________________________10.49From Problem 10.38, 961.0=p K mA/V 2(a) ()()[]{}22SD SD T SG p SGmL V V V V K Vg -+∂∂= ()()()()1.02961.02==SD p V K or 192.0=mL g mA/V (b) ()[]2T SG p SGms V V K V g +∂∂=()()()35.05.1961.022-=+=T SG p V V K or 21.2=ms g mA/V_______________________________________10.50 (a) oxa s C N e ∈=2γNow ()()814101501085.89.3--⨯⨯=oxC 710301.2-⨯=F/cm 2 Then()()()()716141910301.21051085.87.11106.12---⨯⨯⨯⨯=γ 5594.0=γV 2/1 (b) ()3890.0105.1105ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fpφV (i)()()()()()2/1161914105106.13890.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x510419.1-⨯=cm()m ax SDQ ' ()()()5161910419.1105106.1--⨯⨯⨯=710135.1-⨯=C/cm 2 ()fp FB oxSDTO V C Q V φ2max ++'= ()3890.025.010301.210135.177+-⨯⨯=-- 7713.0=VL WC K ox n n 2μ=()()()()2.12810301.24507-⨯=410452.3-⨯=A/V 2 or 3452.0=n K mA/V 2 For 0=D I , 7713.0==TO GS V V V For 5.0=D I ()()27713.03452.0-=GS V 975.1=⇒GS V V (c) (i) For 0=SB V , 7713.0==TO T V V V (ii) 1=SB V V,()()[1389.025594.0+=∆T V()]389.02-2525.0=V024.12525.07713.0=+=T V V (iii) 2=SB V V,()()[2389.025594.0+=∆T V ()]389.02-4390.0=V210.14390.07713.0=+=T V V (iv) 4=SB V V,()()[4389.025594.0+=∆T V()]389.02-7294.0=V501.17294.07713.0=+=T V V _______________________________________10.51()3473.0105.110ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯=fp φ V[]fpSBfpT V V φφγ22-+=∆()()[5.23473.0212.0+=()]3473.02- or114.0=∆T V VNow T TO T V V V ∆+= 114.05.0+=TO V 386.0=⇒TO V V _______________________________________ 10.52 (a) ()()814102001085.89.3--⨯⨯=ox C710726.1-⨯=F/cm 2oxds C N e ∈=2γ ()()()()715141910726.11051085.87.11106.12---⨯⨯⨯⨯= 2358.0=γV 2/1 (b) ()3294.0105.1105ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=fnφV []fn BS fnT V V φφγ22-+-=∆()()[BS V +-=-3294.022358.022.0()]3294.02- 39.2=⇒BS V V_______________________________________10.53(a) +n poly-to-p-type 0.1-=⇒ms φV ()288.0105.110ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯=fp φValso 2/14⎥⎦⎤⎢⎣⎡∈=a fp s dTeN x φ()()()()()2/115191410106.1288.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=-- or410863.0-⨯=dT x cm Now()()()()4151910863.010106.1m ax --⨯⨯='SDQ or()81038.1m ax -⨯='SDQ C/cm 2 Also()()814104001085.89.3--⨯⨯=∈=ox ox ox t C or81063.8-⨯=ox C F/cm 2 We find ()()91019108105106.1--⨯=⨯⨯='ss Q C/cm 2 Then ()fp ms oxss SD T C Q Q V φφ2m ax ++'-'=()288.020.11063.81081038.1898+-⎪⎪⎭⎫ ⎝⎛⨯⨯-⨯=--- or 357.0-=T V V(b) For NMOS, apply SB V and T V shifts in apositive direction, so for 0=T V , we want 357.0+=∆T V V. So[]fp SB fpoxa s T V C N e V φφ222-+∈=∆or()()()()81514191063.8101085.87.11106.12357.0---⨯⨯⨯=+ ()()[]288.02288.02-+⨯SB V or[]576.0576.0211.0357.0-+=SB V which yields 43.5=SB V V_______________________________________10.54 Plot_______________________________________10.55 (a)()T GS oxn m V V L C W g -=μ()T GS oxoxn V V t L W -∈=μ ()()()()()65.0510*******.89.340010814-⨯⨯=--or26.1=m g mS Nowsm m m s m m m r g g g r g g g +=='⇒+='118.01which yields⎪⎭⎫⎝⎛-=⎪⎭⎫ ⎝⎛-=18.0126.1118.011m s g r or 198.0=s r k Ω (b) For 3=GS V V, 683.0=m g mS Then ()()602.0198.0683.01683.0=+='m g mS or 88.0683.0602.0=='m m g g which is a 12% reduction._______________________________________10.56 (a) The ideal cutoff frequency for no overlap capacitance is,()222L V V C g f T GS n gs m T πμπ-==()()()24102275.04400-⨯-=π or 17.5=T f GHz (b) Now ()M gsT m T C C g f +=π2 where ()L m gdT M R g C C +=1 We find()()4410201075.0--⨯⨯=ox gdT C C()()814105001085.89.3--⨯⨯= ()()4410201075.0--⨯⨯⨯ or1410035.1-⨯=gdT C F Also ()T GS oxn m V V LC W g -=μ()()()()()()84144105001021085.89.34001020----⨯⨯⨯⨯= ()75.04-⨯or3108974.0-⨯=m g SThen ()1410035.1-⨯=M C ()()[]331010108974.01⨯⨯+⨯- or 1310032.1-⨯=M C F Now()()W L C C ox gsT 41075.0-⨯+= ()()814105001085.89.3--⨯⨯= ()()44410201075.0102---⨯⨯+⨯⨯ or1410797.3-⨯=gsT C F We now find ()M gsT mTC C g f +=π2 ()1314310032.110797.32108974.0---⨯+⨯⨯=π or 01.1=T f GHz _______________________________________10.57 (a) For the ideal case()4610221042-⨯⨯==ππυL f ds Tor 18.3=T f GHz(b) With overlap capacitance (using the values from Problem 10.56), ()MgdT mT C C g f +=π2 We findds ox m W C g υ= ()()()()86144105001041085.89.31020---⨯⨯⨯⨯= or3105522.0-⨯=m g S We have()L m gdT M R g C C +=1 ()1410035.1-⨯=()()[]331010105522.01⨯⨯+⨯- or 1410750.6-⨯=M C F。