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半导体物理与器件(尼曼)第15章答案

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半导体物理与器件(尼曼)第15章答案         课后答案网,用心为你服务!     大学答案 --- 中学答案 --- 考研答案 --- 考试答案   最全最多的课后习题参考答案,尽在课后答案网(www.khdaw.com)! Khdaw团队一直秉承用心为大家服务的宗旨,以关注学生的学习生活为出发点, 旨在为广大学生朋友的自主学习提供一个分享和交流的平台。   爱校园(www.aixiaoyuan.com) 课后答案网(www.khdaw.com) 淘答案(www.taodaan.com)   ...
半导体物理与器件(尼曼)第15章答案
        课后答案网,用心为你服务!     大学答案 --- 中学答案 --- 考研答案 --- 考试答案   最全最多的课后习题参考答案,尽在课后答案网(www.khdaw.com)! Khdaw团队一直秉承用心为大家服务的宗旨,以关注学生的学习生活为出发点, 旨在为广大学生朋友的自主学习提供一个分享和交流的平台。   爱校园(www.aixiaoyuan.com) 课后答案网(www.khdaw.com) 淘答案(www.taodaan.com)   Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 15 Solutions Manual Problem Solutions 225 Chapter 15 Problem Solutions 15.1 The limit of low injection means that n N cmB B0 01 01 10 10 16 15 3( ) ( ) ( )= = = −. . b g Now I AeD n xC B B B = ( )0 = ( ) ( )− − 0 5 1 6 10 20 10 3 10 19 15 4 . . x x b g b g or I AC = 5 33. 15.2 From the junction breakdown curve, for BV VCBO = 1000 , we need the collector doping concentration to be N x cmC ≈ −2 1014 3 Depletion width into the base (neglect Vbi ) x V e N N N Np BC C B C B = ∈ + F HG I KJ F HG I KJ L NM O QP 2 1 1 2/ = ( ) ( )L NM − − 2 11 7 8 85 10 1000 1 6 10 14 19 . . . x x b g × + F HG I KJ FH IK O QP 2 10 5 10 1 5 10 2 10 14 15 15 14 1 2 x x x x / or x mp = 316. µ (Minimum base width) Depletion width into the collector x x xn = ( ) ( )L NM − − 2 11 7 8 85 10 1000 1 6 10 14 19 . . . b g × + F HG I KJ F H I K O QP 5 10 2 10 1 5 10 2 10 15 14 15 14 1 2 x x x x / or x mn = 78 9. µ (Minimum collector width) 15.3 Compute plot 15.4 (a) We have β β β β βeff A B A B= + + Then 180 25 25= + +β βB B or 155 26= β B which yields β B = 5 96. (b) We have β B EA CBi i= or β ββB A A CA CBi i 1+ = F HG I KJ so 5 96 1 25 25 20.( )FH IK + =iCA which yields i ACA = 3 23. 15.5 Sketch 15.6 We want P I V IT CC C rated C rated= ⋅ ⇒ = FH IK 1 2 2 1 2 24 2 20, , which yields I AC rated, .= 3 33 Then R V IL CC C rated = = , . 24 3 33 or RL = 7.2 Ω 15.7 If V VCC = 25 , then I V R A IC CC L C ratedmax . ,( ) = = = < 25 100 0 25 The power P I V I V I RC CE C CC C L= = −a f Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 15 Solutions Manual Problem Solutions 226 To find the maximum power point, set dP dI V I R I C CC C L C= = − = − ( )0 2 25 2 100 which yields I AC = 0 125. So P max . .( ) ( ) ( )( )= −0 125 25 0 125 100 or P W PTmax .( ) = <156 So, maximum VCC is V VCC = 25 15.8 Now R V Ion DS D = Power dissipated in transistor P I V V RD DS DS on = = 2 We have I V D DS = −200 100 so we can write P V V V R DS DS DS on = − ⋅ = FH IK 200 100 2 For T C Ron= =25 2, Ω , Then 200 100 2 2 − ⋅ = F H I K V V VDS DS DS which yields V VDS = 3 92. and P W= − = FH IK( ) 200 3 92 100 3 92 7.69 . . We then have T Ron VDS P 25 50 75 100 2.0 2.33 2.67 3 3.92 4.55 5.20 5.83 7.69 8.89 10.1 11.3 15.9 (a) We have, for three devices in parallel, V V V V 18 2 2.2 5 151 5 . .+ + = ⇒ =( ) or V V= 3 31. Then, I V R = , so that I A1 1839= . I A2 1 655= . I A3 1505= . Now, P IV= , so P W1 6 09= . P W2 5 48= . P W3 4.98= (b) Now V V 1 18 1 3 6 1 2.2 5 1 288 . . .+ + = =FH IK ( ) or V V= 3 88. Then I A P W1 12.16 8 38= =, . I A P W2 21 08 4.19= =. , I A P W3 31 77 6 85= =. , . 15.10 For BV V= 200 , from the junction breakdown curve, we need the drain doping concentration to be N x cmD ≈ −15 1015 3. For the channel length (neglect Vbi ) L V e N N N N D D B D B min / ( ) FHG I KJ F HG I KJ L NM O QP= ∈ + 2 1 1 2a f = ( ) ( )L NM − − 2 117 8 85 10 200 1 6 10 14 19 . . . x x b g × + F HG I KJ F H I K O QP 15 10 10 1 15 10 10 15 16 15 16 1 2 . . / x x or L mmin .( ) = 184 µ For the drift region W x x min . . . ( ) ( ) ( )L NM = − − 2 11 7 8 85 10 200 1 6 10 14 19 b g × + F HG I KJ FH IK O QP 10 15 10 1 15 10 10 16 15 15 16 1 2 . . / x x or W mmin( ) = 12.3 µ Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 15 Solutions Manual Problem Solutions 227 15.11 (b) In saturation region, I K V V VD n GS T GS= − = −a f a f2 20 25 4. and V V I R IDS DD D D= − = − ( )40 10 We find V V I A V V satGS D DS DS= = = > ( )5 0 25 37.5, . , V V I A V V satGS D DS DS= = = > ( )6 1 30, , V V I A V V satGS D DS DS= = = > ( )7 2.25 17.5, , For V VGS = 8 and V VGS = 9 , transistor is biased in the nonsaturation region. For V VGS = 8 . I V V VD DS DS DS= − = − −( )40 10 0 25 2 8 4 2. We find V V I ADS D= =2.92 3 71, . For V VGS = 9 , I V V VD DS DS DS= − = − −( )40 10 0 25 2 9 4 2. and we find V V I ADS D= =188 3 81. , . Power dissipated in the transistor is P I VT D DS= . We find V V P WGS T= =5 9.375, V V P WGS T= =6 30, V V P WGS T= =7 39.4, V V P WGS T= =8 10 8, . V V P WGS T= =9 7.16, 15.12 T T Pdev amb dev case case ambD− = −− −θ θa f which can be written as θ θdev case dev amb case amb T T PD − − = − − = 175 25 10 6 9 − − = °C W/ Now P T T D rated j amb dev case , ,max = − = − − θ 175 25 9 or P WD rated, .= 16 7 15.13 P T T D rated j amb dev case , ,max = − − θ or θ dev case j amb D rated T T P− = −,max , = − = ° 150 25 50 2.5 C W/ Then T T Pdev amb D dev case case amb− = +− −θ θa f so 150 25 2.5− = + − PD case ambθa f or 125 2.5= + − PD case ambθa f 15.14 We have P I V WD D DS= ⋅ = =( )( )4 5 20 Now T T Pdev amb D dev case case snk snk amb− = + +− − −θ θ θa f or Tdev − = + + =( )25 20 1 75 0 8 3 111. . which yields T Cdev = °136 Also T T Pdev case dev caseD− = ⋅ = =− ( )( )θ 20 1 75 35. so T Tcase dev= − = −35 136 35 or T Ccase = °101 And T T P Ccase snk D case snk− = ⋅ = = °− ( )( )θ 20 0 8 16. so T Tsnk case= − = −16 101 16 or T Csnk = °85 15.15 We have T T Pdev amb D dev case case amb− = +− −θ θa f so 200 25 25 3− = + − θ case amba f or θ case amb C W− = °4 / Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 15 Solutions Manual Problem Solutions 228 15.16 We have θ dev case j amb D rated T T P C W − = − = − = ° ,max , / 175 25 15 10 Now P T T D j amb dev case case snk snk amb = − + + − − − ,max θ θ θ = − + + 175 25 10 1 4 or P WD = 10 15.17 We have α α1 2 1+ = . Now α β β1 1 11 = + and α β β2 2 21 = + so α α β β β β1 2 1 1 2 21 1 1+ = + + + = which can be written as 1 1 1 1 1 1 2 2 1 1 2 = + + + + + β β β β β β a f a f a fa f or 1 1 1 11 2 1 2 2 1+ + = + + +β β β β β βa fa f a f a f Expanding, we find 1 1 2 1 2+ + +β β β β = + + +β β β β β β1 1 2 2 1 2 which yields β β1 2 1= 15.18 The reverse-biased p-well to substrate junction corresponds to the J2 junction in an SCR. The photocurrent generated in this junction will be similar to the avalanche generated current in an SCR, which can trigger the device. 15.19 Case 1: Terminal 1(+), terminal 2(-), and IG negative. This triggering was discussed in the text. Case 2: Terminal 1(+), terminal 2(-), and IG positive. Gate current enters the P2 region directly so that J3 becomes forward biased. Electrons are injected from N2 and diffuse into N1, lowering the potential of N1. The junction J2 becomes more forward biased, and the increased current triggers the SCR so that P2N1P1N4 turns on. Case 3: Terminal 1(-), terminal 2(+), and IG positive. Gate current enters the P2 region directly so that the J3 junction becomes more forward biased. More electrons are injected from N2 into N1 so that J1 also becomes more forward biased. The increased current triggers the P1N1P2N2 device into its conducting state. Case 4: Terminal 1(-), terminal 2(+), and IG negative. In this case, the J4 junction becomes forward biased. Electrons are injected from N3 and diffuse into N1. The potential of N1 is lowered which increases the forward biased potential of J1. This increased current then triggers the P1N1P2N2 device into its conducting state. Problem Solutions
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