《半导体物理与器件》第四版答案第十章-CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIANChapter 1010.1(a) p-type; inversion (b) p-type; depletion (c) p-type; accumulation (d) n-type; inversion_______________________________________ 10.2 (a) (i) ⎪⎪⎭⎫ ⎝⎛=iat fpnN V ln φ ()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1107ln 0259.03381.0=V 2/14⎥⎦⎤⎢⎣⎡∈=a fp s dTeN x φ()()()()()2/1151914107106.13381.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--51054.3-⨯=cmor μ354.0=dT x m (ii) ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1103ln 0259.0fpφ3758.0=V ()()()()()2/1161914103106.13758.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx51080.1-⨯=cmor μ180.0=dT x m (b) ()03022.03003500259.0=⎪⎭⎫⎝⎛=kT V ⎪⎪⎭⎫ ⎝⎛-=kTE N N n gc iexp 2υ ()()319193003501004.1108.2⎪⎭⎫⎝⎛⨯⨯=⎪⎭⎫⎝⎛-⨯03022.012.1exp221071.3⨯=so 111093.1⨯=i n cm 3-(i)()⎪⎪⎭⎫⎝⎛⨯⨯=11151093.1107ln 03022.0fpφ3173.0=V ()()()()()2/1151914107106.13173.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx51043.3-⨯=cmor μ343.0=dT x m (ii) ()⎪⎪⎭⎫⎝⎛⨯⨯=11161093.1103ln 03022.0fpφ3613.0=V ()()()()()2/1161914103106.13613.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx51077.1-⨯=cmor μ177.0=dT x m_______________________________________ 10.3(a) ()2/14max ⎥⎦⎤⎢⎣⎡∈=='d fn s d dT d SDeN eN x eN Q φ()()[]2/14fn s d eN φ∈= 1st approximation: Let 30.0=fn φV Then()281025.1-⨯()()()()()()[]30.01085.87.114106.11419--⨯⨯=d N 141086.7⨯=⇒d N cm 3- 2nd approximation:()2814.0105.11086.7ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fnφVThen()281025.1-⨯()()()()()()[]2814.01085.87.114106.11419--⨯⨯=d N 141038.8⨯=⇒d N cm 3-(b) ()2831.0105.11038.8ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fnφV()566.02831.022===fn s φφV_______________________________________ 10.4p-type silicon(a) Aluminum gate⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫⎝⎛++'-'=fp gmms e E φχφφ2 We have ⎪⎪⎭⎫⎝⎛=i a t fp n N V ln φ ()334.0105.1106ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=VThen()[]334.056.025.320.3++-=ms φ or944.0-=ms φV(b) +n polysilicon gate ⎪⎪⎭⎫⎝⎛+-=fp g mseE φφ2()334.056.0+-= or894.0-=ms φV(c) +p polysilicon gate ()334.056.02-=⎪⎪⎭⎫⎝⎛-=fp gms 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-_______________________________________ 10.7From Problem 10.5, 9932.0-=ms φV oxssms FB C Q V '-=φ (a) ()()814102001085.89.3--⨯⨯=∈=ox ox oxt C710726.1-⨯=F/cm 2 ()()7191010726.1106.11059932.0--⨯⨯⨯--=FB V 040.1-=V(b) ()()81410801085.89.3--⨯⨯=oxC 710314.4-⨯=F/cm 2()()7191010314.4106.11059932.0--⨯⨯⨯--=FB V 012.1-=V_______________________________________ 10.8(a) 42.0-≅ms φV42.0-==ms FB V φV (b)()()781410726.1102001085.89.3---⨯=⨯⨯=oxC F/cm2(i)()()7191010726.1106.1104--⨯⨯⨯-='-=∆ox ss FBC Q V0371.0-=V (ii)()()7191110726.1106.110--⨯⨯-=∆FBV 0927.0-=V (c) 42.0-==ms FB V φV()()781410876.2101201085.89.3---⨯=⨯⨯=ox C F/cm 2(i)()()7191010876.2106.1104--⨯⨯⨯-=∆FBV 0223.0-=V(ii)()()7191110876.2106.110--⨯⨯-=∆FB V0556.0-=V_______________________________________ 10.9⎪⎪⎭⎫⎝⎛++'-'=fp gmms e E φχφφ2 where ()365.0105.1102ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fpφVThen()365.056.025.320.3++-=ms φ or975.0-=ms φV Now oxssms FB C Q V '-=φ or()ox FB ms ssC V Q -='φ We have()()814104501085.89.3--⨯⨯=∈=ox ox oxt Cor81067.7-⨯=ox C F/cm 2 So now()[]()81067.71975.0-⨯⋅---='ssQ 91092.1-⨯=C/cm 2 or10102.1⨯='eQ sscm 2- _______________________________________ 10.10 ()3653.0105.1102ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fp φV()()()()()2/1161914102106.13653.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510174.2-⨯=cm()dT a SDx eN Q ='max()()()5161910174.2102106.1--⨯⨯⨯=810958.6-⨯=C/cm 2()()781410301.2101501085.89.3---⨯=⨯⨯=oxC F/cm 2()fp ms oxss 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 φV136.012.19843.0-=-=TN V V (b) p + poly gate on p-type:28.0+≅ms φV26.128.09843.0+=+=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=⎪⎪⎭⎫⎝⎛⨯⨯=fn φV()()()()()2/1151914103106.13161.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510223.5-⨯=cm()dT d SDx eN Q ='max ()()()5151910223.5103106.1--⨯⨯⨯= 810507.2-⨯=C/cm 2()()781410301.2101501085.89.3---⨯=⨯⨯=oxC F/cm 2()fn ms ox ss SDTPC Q Q V φφ2max -+⎥⎥⎦⎤⎢⎢⎣⎡'+'-=()()⎥⎦⎤⎢⎣⎡⨯⨯⨯+⨯-=---71019810301.2107106.110507.2()3161.02-+ms φms TP V φ+-=7898.0(a) n + poly gate on n-type:41.0-≅ms φV20.141.07898.0-=--=TP V V (b) p + poly gate on n-type:0.1+≅ms φV210.00.17898.0+=+-=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=⎪⎪⎭⎫⎝⎛⨯⨯=fpφVThe 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()()()()4151910413.0105106.1max --⨯⨯⨯='SDQ or()810304.3max -⨯='SDQ C/cm 2 We also find ()()814104001085.89.3--⨯⨯=∈=ox ox oxt Cor810629.8-⨯=ox C F/cm 2 Then90.0659.010629.810304.388-+⨯⨯=--T Vor142.0+=T V V_______________________________________10.13 ()()814102201085.89.3--⨯⨯=∈=ox ox oxt C710569.1-⨯=F/cm 2 ()()1019104106.1⨯⨯='-ssQ9104.6-⨯=C/cm 2 By trial and error, let 16104⨯=a N cm 3-.Now ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1104ln 0259.0fpφ3832.0=V ()()()()()2/1161914104106.13832.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510575.1-⨯=cm()max SDQ ' ()()()5161910575.1104106.1--⨯⨯⨯= 710008.1-⨯=C/cm 2 94.0-≅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 oxt C7109175.1-⨯=F/cm 3-()()1019104106.1⨯⨯='-ssQ 9104.6-⨯=C/cm 2 By trial and error, let 16105⨯=d N cm 3- Now ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1105ln 0259.0fnφ3890.0=V()()()()()2/1161914105106.13890.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510419.1-⨯=cm()max SDQ ' ()()()5161910419.1105106.1--⨯⨯⨯= 710135.1-⨯=C/cm 3- 10.1+≅ms φV Then()()fn ms oxssSDTP C Q Q V φφ2max -+'+'-=()797109175.1104.610135.1---⨯⨯+⨯-=()3890.0210.1-+Then 303.0-=TP V V, which is within the specified value._______________________________________10.15We have 710569.1-⨯=ox C F/cm 29104.6-⨯='ssQ C/cm 2 By trial and error, let 14105⨯=d N cm 3- Now ()⎪⎪⎭⎫⎝⎛⨯⨯=1014105.1105ln 0259.0fnφ2697.0=V ()()()()()2/1141914105106.12697.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx410182.1-⨯=cm()max SDQ ' ()()()4141910182.1105106.1--⨯⨯⨯= 910456.9-⨯=C/cm 2 33.0-≅ms φV Then()()fn ms oxssSDTP C Q Q V φφ2max -+'+'-=⎪⎪⎭⎫⎝⎛⨯⨯+⨯-=---79910569.1104.610456.9 ()2697.0233.0--970.0=VThen 970.0-=TP V V 975.0-≅ V which meets the specification._______________________________________10.16(a) 03.1-≅ms φV ()()814101801085.89.3--⨯⨯=oxC 7109175.1-⨯=F/cm 2Now oxssms FB C Q V '-=φ ()()71019109175.1106106.103.1--⨯⨯⨯--=08.1-=FB V V (b) ()⎪⎪⎭⎫⎝⎛⨯=1015105.110ln 0259.0fpφ2877.0=V ()()()()()2/115191410106.12877.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTx510630.8-⨯=cm()max 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, so 2/14⎥⎦⎤⎢⎣⎡∈==d fn s C dTeN t x φwhere ()288.0105.110ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯=fnφVThen ()()()()()2/115191410106.1288.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTxor410863.0-⨯==C dT t x cm μ863.0=m (b)()()fn ms oxoxss SDT t Q Q V φφ2max -+⎪⎪⎭⎫⎝⎛∈'+'-= For an +n polysilicon gate,()288.056.02--=⎪⎪⎭⎫⎝⎛--=fn gms e E φφ or272.0-=ms φV Now()()()()4151910863.010106.1max --⨯⨯='SDQor()81038.1max -⨯='SDQ C/cm 2 We have()()91019106.110106.1--⨯=⨯='ssQ C/cm 2We now find()()()()81498105001085.89.3106.11038.1----⨯⨯⨯+⨯-=T V()288.02272.0-- or07.1-=T V V_______________________________________10.18(b) ⎪⎪⎭⎫⎝⎛++'-'=fp gmms e E φχφφ2 where20.0-='-'χφmV and ()3473.0105.110ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯=fpφVThen()3473.056.020.0+--=ms φ or107.1-=ms φV(c) For 0='ssQ ()fp ms oxox SDTN t Q V φφ2max ++⎪⎪⎭⎫⎝⎛∈'= We find ()()()()()2/116191410106.13473.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTxor41030.0-⨯=dT x cm μ30.0=m Now()()()()416191030.010106.1max --⨯⨯='SDQ or()810797.4max -⨯='SDQ C/cm 2 Then ()()()()14881085.89.31030010797.4---⨯⨯⨯=TV()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 oxt C710876.2-⨯=F/cm 2a st s ox ox oxFBeNV t C ∈⎪⎪⎭⎫ ⎝⎛∈∈+∈='()()()()()()()16191481410106.11085.87.110259.07.119.3101201085.89.3----⨯⨯⎪⎭⎫ ⎝⎛+⨯⨯= 710346.1-⨯='FBC F/cm 2 dT soxox oxx t C ⋅⎪⎪⎭⎫ ⎝⎛∈∈+∈='minNow()3473.0105.110ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯=fpφV()()()()()2/116191410106.13473.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTx51000.3-⨯=cmThen()()()5814min1000.37.119.3101201085.89.3---⨯⎪⎭⎫ ⎝⎛+⨯⨯='C810083.3-⨯=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)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 ='max ()()()516191000.310106.1--⨯⨯= 81080.4-⨯=C/cm 2 ()3473.0210.110876.21080.478+-⨯⨯=--TNV 2385.0-=TNV V_______________________________________10.24(a) 1=f Hz (low freq), ()()814101201085.89.3--⨯⨯=∈=ox ox oxt C710876.2-⨯=F/cm 2a st s ox ox oxFBeNV t C ∈⎪⎪⎭⎫ ⎝⎛∈∈+∈='()()()()()()()141914814105106.11085.87.110259.07.119.3101201085.89.3⨯⨯⨯⎪⎭⎫⎝⎛+⨯⨯=---- 810726.4-⨯='FBC F/cm 2 dT soxox oxx t C ⋅⎪⎪⎭⎫ ⎝⎛∈∈+∈='minNow()2697.0105.1105ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fnφV()()()()()2/1141914105106.12697.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx410182.1-⨯=cmThen()()()4814min10182.17.119.3101201085.89.3---⨯⎪⎭⎫ ⎝⎛+⨯⨯='C910504.8-⨯=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 ='max ()()()4141910182.1105106.1--⨯⨯⨯= 910456.9-⨯=C/cm 2 Then ()2697.0295.010876.210456.979-+⨯⨯-=--TPV 378.0+=TPV V_______________________________________10.25The amount of fixed oxide charge at x is()x x ∆ρ C/cm 2By lever action, the effect of this oxide chargeon the flatband voltage is ()x x t x C V oxoxFB ∆⎪⎪⎭⎫⎝⎛-=∆ρ1 If we add the effect at each point, wemustintegrate so that ()dx t x x C V oxt oxoxFB ⎰-=∆01ρ _______________________________________10.26(a) We have ρx Q tSS ()='∆Then ∆V C x x t dx FB oxoxoxt =-()z10ρ ≈-'F H G I K J F H I K -z 1C t t Q t dx oxox oxox oxSSt tt ∆∆bg =-'--=-'F H I K 1C Q t t t t Q CoxSS ox oxSS ox∆∆a f or∆V Q t FB SSox ox=-'∈F H G I KJ =-⨯⨯⨯⨯---()16108102001039885101910814...b gb gb g b gor∆V FB =-00742.V(b)We haveρx Q t SS ox()='=⨯⨯⨯--16108102001019108.b gb g =⨯=-64103.ρONow∆V C x x t dx C t xdx FB oxox ox O ox oxoxt t =-=-()zz10ρρor ∆V t FB O oxox=-∈ρ22=-⨯⨯⨯---()6410200102398851038214...bgbg bgor∆V FB =-00371.V (c) ρρx x t O ox()F H G I KJ =We find12216108102001019108t Q ox O SSO ρρ='⇒=⨯⨯⨯--.bgb gor ρO =⨯-128102. Now ∆V C t x x t dx FB oxoxOoxt ox=-⋅⋅F H G I KJ z110ρ=-⋅z122C tx dx oxOoxoxt ρa fwhich becomes ∆V t t xt FB oxoxOoxoxO ox oxt =-∈⋅⋅=-∈F H G I KJ 133232ρρa fThen∆V FB =-⨯⨯⨯---()12810200103398851028214...bgb 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 0259.0 or695.0-=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 sn d xeN C ∈-=1so ()n sdx x eN +∈-=E for 0≤≤-x x n In the oxide, 0=ρ, so=E ⇒=E0dxd constant. From the boundary conditions, in the oxidesnd x eN ∈-=E In the p-region,2C x eN eN dx d sa s a s +∈=E ⇒∈+=∈-=Eρ 0=E at ()p ox x t x +=, then ()[]x x t eN p ox sa-+∈-=E At ox t x =, sn d s p a xeN x eN ∈-=∈-=ESo 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 eNx V n sdn '+⎪⎪⎭⎫ ⎝⎛⋅+∈=22 Arbitrarily, 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 thevoltagedrop across the n-region. Because of symmetry, p n V V =. Then for zero bias, we havebi ox n V V V =+2which can be written asbi soxn d s n d V t x eN x eN =∈+∈2or02=∈-+dsbi 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 byG 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 cm Now 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 have 731210110210200---⨯=⨯⨯=ox C F/cm 2 Also()()7141011085.89.3--⨯⨯=∈=⇒∈=ox ox ox ox ox oxC t t Cor61045.3-⨯=ox t cm 5.34=nm oA 345= (c) oxss ms FB C Q V '-=φ or71050.080.0-'--=-ss Qwhich yields8103-⨯='ssQ C/cm 21110875.1⨯=cm 2- (d)⎪⎪⎭⎫ ⎝⎛∈⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛∈∈+∈='dss ox ox oxFBeN e kT t C()()[][6141045.31085.89.3--⨯÷⨯=()()()()()⎥⎥⎦⎤⨯⨯⨯⎪⎭⎫ ⎝⎛+--161914102106.11085.87.110259.07.119.3which yields81082.7-⨯='FBC F/cm 2 or156=FB C pF_______________________________________10.31(a) Point 1: Inversion 2: Threshold 3: Depletion 4: Flat-band 5: Accumulation_______________________________________10.32We 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 nV V C Q()()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡+-'+'+fpms ox ss SDC Q Q φφ2maxFor a p-type substrate, ()max SD Q ' is a negative value, so we can write()⎩⎨⎧--='DS GS ox nV V C Q()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡++'-'-fp ms ox ss SD C Q Q φφ2maxUsing the definition of thresholdvoltage 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 thedrain terminal._______________________________________10.33 (a) ()[]222DS DS T GS n D V V V V LW k I --⋅'=()()()()[]22.02.04.08.028218.0--⎪⎭⎫ ⎝⎛=0864.0=mA(b) ()22T GS n D V V L W 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 L Wk 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-⎪⎭⎫ ⎝⎛=DI 06.3=mA(c) ()[]222DS DS T GS n D V V V V LW 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 Wk 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---⨯=⨯⨯=oxC F/cm 2()()()()2.122010138.342527-⨯==L W C K ox n n μ 310111.1-⨯=A/V 2=1.111 mA/V 2 (a) 0=GS V , 0=D I6.0=GS V V, ()15.0=sat V DS V, ()()()245.06.0111.1-=sat I D025.0=mA2.1=GS V V, ()75.0=sat V DS V, ()()()245.02.1111.1-=sat I D 625.0=mA 8.1=GS V V, ()35.1=sat V DS V, ()()()245.08.1111.1-=sat I D 025.2=mA 4.2=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 oxt C710138.3-⨯=F/cm 2 L 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 D060.0=mA2.1=SG V V, ()85.0=sat V SD V ()()()235.02.1961.0-=sat I D 694.0=mA8.1=SG V V, ()45.1=sat V SD V ()()()235.08.1961.0-=sat I D 02.2=mA4.2=SG V V, ()05.2=sat V SD V ()()()235.04.2961.0-=sat I D 04.4=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 I 154.0=mA 8.1=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 Problem10.37,111.1=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 alwaysbiased 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 V______10.43From 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 gFor 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) GSDm V I g ∂∂=()()[]{}22DSDS 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 I 125.3=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 Cor81012.8-⨯=ox C F/cm 2We are given 10=L W . From the graph, for3=GS V V, we have ()033.0≅sat I D , then ()2.032033.0-⋅=LC W oxn μ or310139.02-⨯=L C W oxn μ or()()3810139.01012.81021--⨯=⨯n μ which 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 DSn T GS n D 22=-= so()244102n K =⨯- which yieldsμ5.12=n K A/V 2 (c)()2.18.02=-=-=T GS DS V V sat V V so ()sat V V DS DS >()()()258.021025.1-⨯=-sat I D or()μ18=sat I D A (d)()sat V V DS DS <()[]22DSDS T GS n D V V V V K I --= ()()()()[]25118.0321025.1--⨯=- orμ5.42=D I A_______________________________________10.47 (a) ()()814101801085.89.3--⨯⨯=oxC7109175.1-⨯=F/cm 2(i)()()7109175.1450-⨯=='ox n nC k μ 510629.8-⨯=A/V 2or μ29.86='nk A/V 2 (ii)()()22T GS n D V V L W k sat I -⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛'= ()24.02208629.08.0-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=L W24.7=⇒LW(b) (i) ()()7109175.1210-⨯=='ox p p C k μ510027.4-⨯=A/V 2 or 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_______________________________________10.48From Problem 10.37, 111.1=n K mA/V 2(a) ()()[]{}22DSDS T GS n GSmL V V V V K V g --∂∂=()()()()1.02111.12==DS n V Kso 222.0=mL g mA/V (b) (){}2T GS n GSms V V K V g -∂∂=()()()45.05.1111.122-=-=T GS n V V Kso 33.2=ms g mA/V_______________________________________10.49From Problem 10.38, 961.0=p K mA/V 2(a) ()()[]{}22SDSD T SG p SGmL V V V V K V g -+∂∂=()()()()1.02961.02==SD p V Kor 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 Kor 21.2=ms g mA/V_______________________________________10.50 (a) oxa s C N e ∈=2γNow ()()814101501085.89.3--⨯⨯=ox C 710301.2-⨯=F/cm 2Then()()()()716141910301.21051085.87.11106.12---⨯⨯⨯⨯=γ 5594.0=γV 2/1 (b) ()3890.0105.1105ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fp φV(i)()()()()()2/1161914105106.13890.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510419.1-⨯=cm()max SDQ ' ()()()5161910419.1105106.1--⨯⨯⨯= 710135.1-⨯=C/cm 2()fp FB oxSDTO V C Q V φ2max ++'=()3890.025.010301.210135.177+-⨯⨯=--7713.0=VLWC K ox n n 2μ=()()()()2.12810301.24507-⨯=410452.3-⨯=A/V 2 or 3452.0=n K mA/V 2For 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=⎪⎪⎭⎫⎝⎛⨯=fpφ V[]fp SB fp T 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--⨯⨯=oxC 710726.1-⨯=F/cm 2oxd s C Ne ∈=2γ()()()()715141910726.11051085.87.11106.12---⨯⨯⨯⨯=2358.0=γV 2/1 (b) ()3294.0105.1105ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=fnφV[]fn BS fn T 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()()()()4151910863.010106.1max --⨯⨯='SDQor()81038.1max -⨯='SDQ C/cm 2 Also()()814104001085.89.3--⨯⨯=∈=ox ox oxt Cor81063.8-⨯=ox C F/cm 2 We find()()91019108105106.1--⨯=⨯⨯='ssQ C/cm 2 Then ()fp ms oxss SDT C Q Q V φφ2max ++'-'=()288.020.11063.81081038.1898+-⎪⎪⎭⎫⎝⎛⨯⨯-⨯=--- or357.0-=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 []fpSB 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 oxox n V V t L W -∈=μ()()()()()65.0510*******.89.340010814-⨯⨯=-- or26.1=m g mS Nowsm m m s m m mr g g g r g g g +=='⇒+='118.01 which yields⎪⎭⎫⎝⎛-=⎪⎭⎫ ⎝⎛-=18.0126.1118.011ms g r or198.0=s r k Ω(b) For 3=GS V V, 683.0=m g mS Then()()602.0198.0683.01683.0=+='mg mSor88.0683.0602.0=='m m g g which is a 12% reduction._______________________________________10.56(a) The ideal cutoff frequency for no overlapcapacitance is, ()222LV V C g f T GS n gs m T πμπ-==()()()24102275.04400-⨯-=πor17.5=T f GHz (b) Now。