半导体器件原理简明教程习题标准答案傅兴华

模拟电⼦技术基础课后练习答案（国防科技⼤学出版社）第⼆章半导体器件习题答案（⼤题）习题：⼀．填空题1. 半导体的导电能⼒与温度、光照强度、掺杂浓度和材料性质有关。

2. 利⽤PN结击穿时的特性可制成稳压⼆极管，利⽤发光材料可制成发光⼆级管，利⽤PN结的光敏性可制成光敏（光电）⼆级管。

3.在本征半导体中加⼊＿＿5价＿＿元素可形成N型半导体，加⼊＿3价＿元素可形成P型半导体。

N型半导体中的多⼦是_⾃由电⼦_______；P型半导体中的多⼦是___空⽳____。

4. PN结外加正向电压时导通外加反向电压时截⽌这种特性称为PN结的单向导电性。

5. 通常情况下硅材料⼆极管的正向导通电压为0.7v ，锗材料⼆极管的正向导通电压为0.2v 。

6..理想⼆极管正向电阻为__0______，反向电阻为_______，这两种状态相当于⼀个___开关____。

7..晶体管的三个⼯作区分别为放⼤区、截⽌区和饱和区。

8.. 稳压⼆极管是利⽤PN结的反向击穿特性特性制作的。

9.. 三极管从结构上看可以分成 PNP 和 NPN 两种类型。

10. 晶体三极管⼯作时有⾃由电⼦和空⽳两种载流⼦参与导电，因此三极管⼜称为双极型晶体管。

11.设晶体管的压降U CE不变，基极电流为20µA时，集电极电流等于2mA，则β=__100__。

12. 场效应管可分为绝缘栅效应管和结型两⼤类，⽬前⼴泛应⽤的绝缘栅效应管是MOS管，按其⼯作⽅式分可分为耗尽型和增强型两⼤类，每⼀类中⼜分为N沟道和P沟道两种。

13. 查阅电⼦器件⼿册，了解下列常⽤三极管的极限参数，并记录填写题表2-1在下表中题表2-1⼆．选择题1.杂质半导体中，多数载流⼦的浓度主要取决于A。

A、杂质浓度B、温度C、输⼊D、电压2.理想⼆极管加正向电压时可视为 B ，加反向电压时可视为＿＿A＿＿。

A.开路B.短路C.不能确定3.稳压管的稳压区是⼆极管⼯作在＿＿D＿＿状态。

A.正向导通B.反向截⽌C.反向导通D.反向击穿4.当温度升⾼时，⼆极管的反向饱和电流将＿＿A＿＿。

第一章半导体器件基础测试题（高三）姓名班次分数一、选择题1、N型半导体是在本征半导体中加入下列物质而形成的。

A、电子；B、空穴；C、三价元素；D、五价元素。

2、在掺杂后的半导体中，其导电能力的大小的说法正确的是。

A、掺杂的工艺；B、杂质的浓度：C、温度；D、晶体的缺陷。

3、晶体三极管用于放大的条件，下列说法正确的是。

A、发射结正偏、集电结反偏；B、发射结正偏、集电结正偏；C、发射结反偏、集电结正偏；D、发射结反偏、集电结反偏；4、晶体三极管的截止条件，下列说法正确的是。

A、发射结正偏、集电结反偏；B、发射结正偏、集电结正偏；C、发射结反偏、集电结正偏；D、发射结反偏、集电结反偏；5、晶体三极管的饱和条件，下列说法正确的是。

A、发射结正偏、集电结反偏；B、发射结正偏、集电结正偏；C、发射结反偏、集电结正偏；D、发射结反偏、集电结反偏；6、理想二极管组成的电路如下图所示，其AB两端的电压是。

A、—12V；B、—6V；C、+6V；D、+12V。

7、要使普通二极管导通，下列说法正确的是。

A、运用它的反向特性；B、锗管使用在反向击穿区；C、硅管使用反向区域，而锗管使用正向区域；D、都使用正向区域。

8、对于用万用表测量二极管时，下列做法正确的是。

A、用万用表的R×100或R×1000的欧姆，黑棒接正极，红棒接负极，指针偏转；B、用万用表的R×10K的欧姆，黑棒接正极，红棒接负极，指针偏转；C、用万用表的R×100或R×1000的欧姆，红棒接正极，黑棒接负极，指针偏转；D、用万用表的R×10，黑棒接正极，红棒接负极，指针偏转；9、电路如下图所示，则A、B两点的电压正确的是。

A、U A=3.5V，U B=3.5V，D截止；B、U A=3.5V，U B=1.0V，D截止；C、U A=1.0V，U B=3.5V，D导通；D、U A=1.0V，U B=1.0V，D截止。

《半导体集成电路》考试题⽬及参考答案第⼀部分考试试题第0章绪论1.什么叫半导体集成电路？？2.按照半导体集成电路的集成度来分,分为哪些类型,请同时写出它们对应的英⽂缩写？？3.按照器件类型分,半导体集成电路分为哪⼏类？？4.按电路功能或信号类型分,半导体集成电路分为哪⼏类？？5.什么是特征尺⼨？？它对集成电路⼯艺有何影响？？6.名词解释：集成度,wafersize,diesize,摩尔定律？？第1章集成电路的基本制造⼯艺1.四层三结的结构的双极型晶体管中隐埋层的作⽤？？2.在制作晶体管的时候,衬底材料电阻率的选取对器件有何影响？？.3.简单叙述⼀下pn结隔离的NPN晶体管的光刻步骤？？4.简述硅栅p阱CMOS的光刻步骤？？5.以p阱CMOS⼯艺为基础的BiCMOS的有哪些不⾜？？6.以N阱CMOS⼯艺为基础的BiCMOS的有哪些优缺点？？并请提出改进⽅法.7.请画出NPN晶体管的版图,并且标注各层掺杂区域类型.8.请画出CMOS反相器的版图,并标注各层掺杂类型和输⼊输出端⼦.第2章集成电路中的晶体管及其寄⽣效应1.简述集成双极晶体管的有源寄⽣效应在其各⼯作区能否忽略？？.2. 什么是集成双极晶体管的⽆源寄⽣效应？？3.什么是MOS晶体管的有源寄⽣效应？？4.什么是MOS晶体管的闩锁效应,其对晶体管有什么影响?5.消除“Latch-up”效应的⽅法？？6.如何解决MOS器件的场区寄⽣MOSFET效应？？7.如何解决MOS器件中的寄⽣双极晶体管效应？？第3章集成电路中的⽆源元件1.双极性集成电路中最常⽤的电阻器和MOS集成电路中常⽤的电阻都有哪些？？2.集成电路中常⽤的电容有哪些.3.为什么基区薄层电阻需要修正.4.为什么新的⼯艺中要⽤铜布线取代铝布线.5.运⽤基区扩散电阻,设计⼀个⽅块电阻200欧,阻值为1K的电阻,已知耗散功率为20W/c㎡,该电阻上的压降为5V,设计此电阻.第4章TTL电路1.名词解释电压传输特性开门/关门电平逻辑摆幅过渡区宽度输⼊短路电流输⼊漏电流静态功耗瞬态延迟时间瞬态存储时间瞬态上升时间瞬态下降时间瞬时导通时间2.分析四管标准TTL与⾮门（稳态时）各管的⼯作状态？？3.在四管标准与⾮门中,那个管⼦会对瞬态特性影响最⼤,并分析原因以及带来那些困难.4.两管与⾮门有哪些缺点,四管及五管与⾮门的结构相对于两管与⾮门在那些地⽅做了改善,并分析改善部分是如何⼯作的.四管和五管与⾮门对静态和动态有那些⽅⾯的改进.5.相对于五管与⾮门六管与⾮门的结构在那些部分作了改善,分析改进部分是如何⼯作的.6.画出四管和六管单元与⾮门传输特性曲线.并说明为什么有源泄放回路改善了传输特性的矩形性.7.四管与⾮门中,如果⾼电平过低,低电平过⾼,分析其原因,如与改善⽅法,请说出你的想法. 8.为什么TTL 与⾮门不能直接并联？？9.OC 门在结构上作了什么改进,它为什么不会出现TTL 与⾮门并联的问题.第5章MOS 反相器1.请给出NMOS 晶体管的阈值电压公式,并解释各项的物理含义及其对阈值⼤⼩的影响（即各项在不同情况下是提⾼阈值还是降低阈值）.2.什么是器件的亚阈值特性,对器件有什么影响？？3.MOS 晶体管的短沟道效应是指什么,其对晶体管有什么影响？？4.请以PMOS 晶体管为例解释什么是衬偏效应,并解释其对PMOS 晶体管阈值电压和漏源电流的影响.5.什么是沟道长度调制效应,对器件有什么影响？？6.为什么MOS 晶体管会存在饱和区和⾮饱和区之分（不考虑沟道调制效应）？？7.请画出晶体管的D DS I V 特性曲线,指出饱和区和⾮饱和区的⼯作条件及各⾃的电流⽅程（忽略沟道长度调制效应和短沟道效应）.8.给出E/R 反相器的电路结构,分析其⼯作原理及传输特性,并计算VTC 曲线上的临界电压值.9.考虑下⾯的反相器设计问题：给定V DD =5V ,K N `=30uA/V 2,V T0=1V设计⼀个V OL =0.2V 的电阻负载反相器电路,并确定满⾜V OL 条件时的晶体管的宽长⽐(W/L)和负载电阻R L 的阻值.10.考虑⼀个电阻负载反相器电路：V DD =5V ,K N `=20uA/V 2,V T0=0.8V ,R L =200K Ω,W/L=2.计算VTC 曲线上的临界电压值（V OL ,V OH ,V IL ,V IH ）及电路的噪声容限,并评价该直流反相器的设计质量.11.设计⼀个V OL =0.6V 的电阻负载反相器,增强型驱动晶体管V T0=1V ,V DD =5V 1）求V IL 和V IH 2）求噪声容限V NML 和V NMH12.采⽤MOSFET 作为nMOS 反相器的负载器件有哪些优点？？ 13.增强型负载nMOS 反相器有哪两种电路结构？？简述其优缺点.14.以饱和增强型负载反相器为例分析E/E 反相器的⼯作原理及传输特性.15试⽐较将nMOSE/E 反相器的负载管改为耗尽型nMOSFET 后,传输特性有哪些改善？？ 16.耗尽型负载nMOS 反相器相⽐于增强型负载nMOS 反相器有哪些好处？？17有⼀nMOSE/D 反相器,若V TE =2V ,V TD =-2V ,K NE /K ND =25,V DD =2V ,求此反相器的⾼,低输出逻辑电平是多少？？18.什么是CMOS 电路？？简述CMOS 反相器的⼯作原理及特点. 19.根据CMOS 反相器的传输特性曲线计算V IL 和V IH . 20.求解CMOS 反相器的逻辑阈值,并说明它与哪些因素有关？？ 21.为什么的PMOS 尺⼨通常⽐NMOS 的尺⼨⼤？？22．考虑⼀个具有如下参数的CMOS反相器电路：V DD=3.3VV TN=0.6VV TP=-0.7V K N=200uA/V2K p=80uA/V2计算电路的噪声容限.23.采⽤0.35um⼯艺的CMOS反相器,相关参数如下：V DD=3.3VNMOS：V TN=0.6VµN C OX=60uA/V2(W/L)N=8PMOS：V TP=-0.7Vµp C OX=25uA/V2(W/L)P=12求电路的噪声容限及逻辑阈值.24．设计⼀个CMOS反相器,NMOS：V TN=0.6VµN C OX=60uA/V2PMOS：V TP=-0.7VµP C OX=25uA/V2电源电压为3.3V,L N=L P=0.8um1）求V M=1.4V时的W N/W P.2）此CMOS反相器制作⼯艺允许V TN,V TP的值在标称值有正负15%的变化,假定其他参数仍为标称值,求V M的上下限. 25．举例说明什么是有⽐反相器和⽆⽐反相器.26．以CMOS反相器为例,说明什么是静态功耗和动态功耗.27．在图中标注出上升时间t r,下降时间t f,导通延迟时间,截⽌延迟时间,给出延迟时间t pd的定义.若希望t r=t f,求W N/W P.第6章CMOS静态逻辑门1.画出F=A⊕B的CMOS组合逻辑门电路.2. ⽤CMOS组合逻辑实现全加器电路.3. 计算图⽰或⾮门的驱动能⼒.为保证最坏⼯作条件下,各逻辑门的驱动能⼒与标准反相器的特性相同,N管与P管的尺⼨应如何选取？？4. 画出F=AB+CD的CMOS组合逻辑门电路,并计算该复合逻辑门的驱动能⼒.5．简述CMOS静态逻辑门功耗的构成.6.降低电路的功耗有哪些⽅法？？7. ⽐较当FO=1时,下列两种8输⼊的AND门,那种组合逻辑速度更快？？第7章传输门逻辑⼀,填空1．写出传输门电路主要的三种类型和他们的缺点：（1）,缺点：；（2）,缺点：；（3）,缺点： .2．传输门逻辑电路的振幅会由于减⼩,信号的也较复杂,在多段接续时,⼀般要插⼊ .3.⼀般的说,传输门逻辑电路适合逻辑的电路.⽐如常⽤的和.⼆,解答题1．分析下⾯传输门电路的逻辑功能,并说明⽅块标明的MOS管的作⽤.2.根据下⾯的电路回答问题：分析电路,说明电路的B区域完成的是什么功能,设计该部分电路是为了解决NMOS传输门电路的什么问题？？3．假定反向器在理想的V DD/2时转换,忽略沟道长度调制和寄⽣效应,根据下⾯的传输门电路原理图回答问题.（1）电路的功能是什么？？（2）说明电路的静态功耗是否为零,并解释原因.4.分析⽐较下⾯2种电路结构,说明图1的⼯作原理,介绍它和图2所⽰电路的相同点和不同点.图1图25．根据下⾯的电路回答问题.已知电路B点的输⼊电压为2.5V,C点的输⼊电压为0V.当A点的输⼊电压如图a时,画出X 点和OUT点的波形,并以此说明NMOS和PMOS传输门的特点.A点的输⼊波形6．写出逻辑表达式C=A B的真值表,并根据真值表画出基于传输门的电路原理图.7.相同的电路结构,输⼊信号不同时,构成不同的逻辑功能.以下电路在不同的输⼊下可以完成不同的逻辑功能,写出它们的真值表,判断实现的逻辑功能.图1图28.分析下⾯的电路,根据真值表,判断电路实现的逻辑功能.第8章动态逻辑电路⼀,填空1．对于⼀般的动态逻辑电路,逻辑部分由输出低电平的⽹组成,输出信号与电源之间插⼊了栅控制极为时钟信号的,逻辑⽹与地之间插⼊了栅控制极为时钟信号的 .2.对于⼀个级联的多⽶诺逻辑电路,在评估阶段：对PDN⽹只允许有跳变,对PUN ⽹只允许有跳变,PDN与PDN相连或PUN与PUN相连时中间应接⼊ . ⼆,解答题1.分析电路,已知静态反向器的预充电时间,赋值时间和传输延迟都为T/2.说明当输⼊产⽣⼀个0->1转换时会发⽣什么问题?当1->0转换时会如何?如果这样,描述会发⽣什么并在电路的某处插⼊⼀个反向器修正这个问题.2.从逻辑功能,电路规模,速度3⽅⾯分析下⾯2电路的相同点和不同点.从⽽说明CMOS动态组合逻辑电路的特点.图A图B3.分析下⾯的电路,指出它完成的逻辑功能,说明它和⼀般动态组合逻辑电路的不同,说明其特点.4.分析下⾯的电路,指出它完成的逻辑功能,说明它和⼀般动态组合逻辑电路的不同,分析它的⼯作原理.5.简述动态组合逻辑电路中存在的常见的三种问题,以及他们产⽣的原因和解决的⽅法.6.分析下列电路的⼯作原理,画出输出端OUT的波形.7.结合下⾯电路,说明动态组合逻辑电路的⼯作原理.第9章触发器1. ⽤图说明如何给SR锁存器加时钟控制.2. ⽤图说明如何把SR锁存器连接成D锁存器,并且给出所画D锁存器的真值表3. 画出⽤与⾮门表⽰的SR触发器的MOS管级电路图4. 画出⽤或⾮门表⽰的SR触发器的MOS管级电路图5. 仔细观察下⾯RS触发器的版图,判断它是或⾮门实现还是与⾮门实现6. 仔细观察下⾯RS触发器的版图,判断它是或⾮门实现还是与⾮门实现7. 下图给出的是⼀个最简单的动态锁存器,判断它是否有阈值损失现象,若有,说明阈值损失的种类,给出两种解决⽅案并且阐述两种⽅案的优缺点,若没有,写出真值表.8. 下图给出的是⼀个最简单的动态锁存器,判断它是否有阈值损失现象,若有,说明阈值损失的种类,给出两种解决⽅案并且阐述两种⽅案的优缺点,若没有,写出真值表.9. 下图给出的是⼀个最简单的动态锁存器,判断它是否有阈值损失现象,若有,说明阈值损失的种类,给出两种解决⽅案并且阐述两种⽅案的优缺点,若没有,写出真值表.10. 解释下⾯的电路的⼯作过程画出真值表.（提⽰注意图中的两个反相器尺⼨是不同的）11. 解释下⾯的电路的⼯作过程画出真值表.12. 解释静态存储和动态存储的区别和优缺点⽐较.13. 阐述静态存储和动态存储的不同的的存储⽅法.14. 观察下⾯的图,说明这个存储单元的存储⽅式,存储的机理.15. 观察下⾯的图,说明这个存储单元的存储⽅式,存储的机理.16. 说明锁存器和触发器的区别并画图说明17. 说明电平灵敏和边沿触发的区别,并画图说明18. 建⽴时间19. 维持时间20. 延迟时间21. 连接下⾯两个锁存器使它们构成主从触发器,并画出所连的主从触发器的输⼊输出波形图22. 简述下时钟重叠的起因所在23. 下图所⽰的是两相时钟发⽣器,根据时钟信号把下⾯四点的的波形图画出24. 反相器的阈值⼀般可以通过什么进⾏调节25. 施密特触发器的特点26. 说明下⾯电路的⼯作原理,解释它怎么实现的施密特触发.27. 画出下⾯施密特触发器的⽰意版图.28. 同宽长⽐的PMOS和NMOS谁的阈值要⼤⼀些第10章逻辑功能部件1, 根据多路开关真值表画出其组合逻辑结构的CMOS电路图.2, 根据多路开关真值表画出其传输门结构的CMOS电路图.3,计算下列多路开关中P管和N管尺⼨的⽐例关系.4,根据下列电路图写出SUM和C0的逻辑关系式,并根据输⼊波形画出其SUM和C0的输出波形.ABCiK1K0Y1 1 D01 0 D10 1 D20 0 D3K1K0Y1 1 D01 0 D10 1 D20 0 D35,计算下列逐位进位加法器的延迟,并指出如何减⼩加法器的延迟.6,画出传输门结构全加器的电路图,已知下图中的P=A⊕B.7,试分析下列桶型移位器各种sh输⼊下的输出情况.8,试分析下列对数移位器各种sh输⼊下的输出情况.第11章存储器⼀,填空1．可以把⼀个4Mb的SRAM设计成[Hirose90]由32块组成的结构,每⼀块含有128Kb,由1024⾏和列的阵列构成.⾏地址（X）,列地址（Y）,和块地址（Z）分别为,,位宽.2．对⼀个512×512的NOR MOS,假设平均有50%的输出是低电平,有⼀已设计电路的静态电流⼤约et.于0.21mA(输出电压为1.5V时),则总静态功耗为,就从计算的到的功耗看,这个电路设计的（“好”或“差”）.3.⼀般的,存储器由,和三部分组成. 4．半导体存储器按功能可分为：和；⾮挥发存储器有, 和；⼆,解答题1．确定图1中ROM中存放地址0,1,2和3处和数据值.并以字线WL[0]为例,说明原理.图1⼀个4×4的ORROM2．画⼀个2×2的MOSOR型ROM单元阵列,要求地址0,1中存储的数据值分别为01和00.并简述⼯作原理.3.确定图2中ROM中存放地址0,1,2和3处的数据值.并简述⼯作原理.图2⼀个4×4的NORROM4．画⼀个2×2的MOSNOR型ROM单元阵列,要求地址0,1中存储的数据值分别为01和01.并简述⼯作原理.5．如图3为⼀个4×4的NORROM,假设此电路采⽤标准的0.25µmCMOS⼯艺实现,确定PMOS上拉器件尺⼨使最坏的情况下V OL值不会⾼于1.5V(电源电压为2.5V).这相当于字线摆为1V.NMOS尺⼨取(W/L)=4/2.图3⼀个4×4的NORROM6.确定图4中ROM中存放地址0,1,2和3处和数据值.并简述⼯作原理.图4⼀个4×4的NANDROM7．画⼀个2×2的MOSNAND型ROM单元阵列,要求地址0,1中存储的数据值分别为10和10.并简述⼯作原理.8.预充电虽然在NORROM中⼯作的很好,但它应⽤到NANDROM时却会出现某些严重的问题.请解释这是为什么？？9.sram,flashmemory,及dram的区别？？10.给出单管DRAM的原理图.并按图中已给出的波形画出X波形和BL波形,并⼤致标出电压值.11．试问单管DRAM单元的读出是不是破坏性的？？怎样补充这⼀不⾜？？（选作）有什么办法提⾼refreshtime？？12.给出三管DRAM的原理图.并按图中已给出的波形画出X和BL1波形,并⼤致标出电压值.（选作）试问有什么办法提⾼refreshtime？？13．对1TDRAM,假设位线电容为1pF,位线预充电电压为1.25V.在存储数据为1和0时单元电容Cs（50fF）上的电压分别et.于1.9V和0V.这相当于电荷传递速率为4.8%.求读操作期间位线上的电压摆幅.14.给出⼀管单元DRAM的原理图,并给出版图.。

第8章 半导体二极管、三极管1004 N 型 半 导 体 的 多 数 载 流 子 是 电 子，因 此 它 应（ ）。

(a) 带 负 电 (b) 带 正 电 (c) 不 带 电2002当 温 度 升 高 时，半 导 体 的 导 电 能 力 将（ ）。

(a) 增 强 (b) 减 弱 (c) 不 变3007在 PN 结 中 形 成 空 间 电 荷 区 的 正、负 离 子 都 带 电，所 以 空 间 电 荷 区 的 电 阻 率（ ）。

(a) 很 高 (b) 很 低(c)等 于N 型 或 P 型 半 导 体 的 电 阻 率 4010半 导 体 二 极 管 的 主 要 特 点 是 具 有（ ）。

(a) 电 流 放 大 作 用(b) 单 向 导 电 性 (c) 电 压 放 大 作 用 5011理 想 二 极 管 的 正 向 电 阻 为（ ）。

(a) 零 (b) 无 穷 大 (c) 约 几 千 欧 6014二 极 管 接 在 电 路 中， 若 测 得 a 、b 两 端 电 位 如 图 所 示，则 二 极 管 工 作 状 态 为（ ）。

(a) 导 通 (b) 截 止 (c) 击 穿7015如 果 把 一 个 小 功 率 二 极 管 直 接 同 一 个 电 源 电 压 为1.5V 、内 阻 为 零 的 电 池 实 行 正 向 连 接，电 路 如 图 所 示，则 后 果 是 该 管（ ）。

(a) 击 穿 (b) 电 流 为 零 (c) 电 流 正 常 (d) 电 流 过 大 使 管 子 烧 坏-6.3VD8016 电 路 如 图 所 示 ，二 极 管 D 为 理 想 元 件，U S =5 V ，则 电 压u O =（ ）。

(a) U s (b) U S / 2 (c) 零Du O9018电 路 如 图 所 示, 所 有 二 极 管 均 为 理 想 元 件，则 D 1、D 2、D 3的 工 作 状 态 为（ ）。

(a) D 1导 通，D 2、D 3 截 止 (b) D 1、D 2截 止 ， D 3 导 通 (c) D 1、D 3截 止， D 2导 通 (d) D 1、D 2、D 3均 截 止0V+6V10019电 路 如 图 所 示，二 极 管 D 1、D 2 为 理 想 元 件，判 断D 1、D 2的 工 作 状 态 为（ ）。

《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。

Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V xt x m ,,2222ψ⋅+∂ψ∂- ()tt x j ∂ψ∂=, Assume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m exp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j exp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu exp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u kSetting ()()x u x u 1= for region I, the equation becomes:()()()()021221212=--+x u k dx x du jk dxx u d α where222mE=α Q.E.D.In Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu exp This equation can be written as:()()()2222x x u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV OSetting ()()x u x u 2= for region II, this equation becomes()()dx x du jk dxx u d 22222+ ()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O α where again222mE=α Q.E.D._______________________________________3.3We have()()()()021221212=--+x u k dx x du jk dxx u d α Assume the solution is of the form: ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp The first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexp and the second derivative becomes()()[]()[]x k j A k j dxx u d --=ααexp 2212 ()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differential equation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k ()[]0exp =+-⨯x k j B α We find that00= Q.E.D. For the differential equation in ()x u 2 and the proposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2()[]x k j D +-+βexp for 0<<-x b .The first boundary condition is ()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dx dudx du which yields()()()C k B k A k --+--βαα()0=++D k β The third boundary condition is ()()b u a u -=21 which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp()()[]b k j D -+-+βexp and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp()[]0exp =+-b k j D β The fourth boundary condition isbx a x dx dudx du -===21 which yields()()[]a k j A k j --ααexp()()[]a k j B k j +-+-ααexp ()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexp and can be written as ()()[]a k j A k --ααexp()()[]a k j B k +-+-ααexp()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________ 3.5(b) (i) First point: πα=aSecond point: By trial and error, πα729.1=a (ii) First point: πα2=aSecond point: By trial and error, πα617.2=a_______________________________________3.6(b) (i) First point: πα=aSecond point: By trial and error, πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a_______________________________________ 3.7ka a aaP cos cos sin =+'αααLet y ka =, x a =α Theny x x xP cos cos sin =+'Consider dy dof this function.()[]{}y x x x P dy d sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydxx sin sin -=- Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-'For πn ka y ==, ...,2,1,0=n 0sin =⇒y So that, in general,()()dk d ka d a d dy dxαα===0 And 22 mE=α Sodk dEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221 α This implies thatdk dE dk d ==0α for an k π= _______________________________________ 3.8(a) πα=a 1π=⋅a E m o 212()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J 34E E E -=∆1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________3.9(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o ()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πoE19105172.2-⨯=J o E E E -=∆11919105172.2104114.3--⨯-⨯= 2010942.8-⨯=Jor 559.0106.110942.81920=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka . From Problem 3.5, πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV_______________________________________3.10(a) πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 12E E E -=∆1919104114.310830.7--⨯-⨯= 19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV (b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J 34E E E -=∆1818103646.1109242.1--⨯-⨯= 1910597.5-⨯=Jor 50.3106.110597.51919=⨯⨯=∆--E eV_____________________________________3.11(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα727.0=a oπ727.022=⋅a E m o o()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=Jo E E E -=∆11919108030.1104114.3--⨯-⨯= 19106084.1-⨯=Jor 005.1106.1106084.11919=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6,πα515.12=aπ515.1222=⋅a E m o()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J23E E E -=∆191810830.7103646.1--⨯-⨯= 1910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV_______________________________________3.12For 100=T K, ()()⇒+⨯-=-1006361001073.4170.124gE164.1=g E eV200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV 600=T K, 032.1=g E eV_______________________________________3.13The effective mass is given by1222*1-⎪⎪⎭⎫⎝⎛⋅=dk E d mWe have()()B curve dkE d A curve dk E d 2222> so that ()()B curve m A curve m **<_______________________________________ 3.14The effective mass for a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p We have that()()B curve dkEd A curve dk E d 2222> so that ()()B curve m A curve m p p **<_______________________________________ 3.15Points A,B: ⇒<0dk dEvelocity in -x directionPoints C,D: ⇒>0dk dEvelocity in +x directionPoints A,D: ⇒<022dk Ednegative effective massPoints B,C: ⇒>022dkEd positive effective mass _______________________________________3.16For A: 2k C E i =At 101008.0+⨯=k m 1-, 05.0=E eV Or ()()2119108106.105.0--⨯=⨯=E J So ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4o m m 488.0=* For B: 2k C E i =At 101008.0+⨯=k m 1-, 5.0=E eV Or ()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C 3711025.1-⨯=⇒CNow ()()37234121025.1210054.12--*⨯⨯==C m 321044.4-⨯=kg or o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_______________________________________ 3.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C 3921025.6-⨯=⇒C()()39234221025.6210054.12--*⨯⨯-=-=C m31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C 382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________ 3.18(a) (i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν1410429.3⨯=Hz(ii) 141010429.3103⨯⨯===νλc E hc 51075.8-⨯=cm 875=nm(b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm_______________________________________ 3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0=k , and is negativearound 2π±=k . _______________________________________ 3.20()[]O O k k E E E --=αcos 1 Then()()()[]O k k E dkdE ---=ααsin 1()[]O k k E -+=ααsin 1 and()[]O k k E dk E d -=ααcos 2122Then221222*11 αE dk Ed m o k k =⋅== or212*αE m =_______________________________________ 3.21(a) ()[]3/123/24lt dn m m m =*()()[]3/123/264.1082.04oom m =o dn m m 56.0=*(b)o o l t cnm m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cn m m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lh hh dp m m m +=*()()[]3/22/32/3082.045.0o om m +=[]o m ⋅+=3/202348.030187.0o dp m m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=*()()()()om ⋅++=2/12/12/32/3082.045.0082.045.0 o cp m m 34.0=*_______________________________________ 3.23For the 3-dimensional infinite potential well, ()0=x V when a x <<0, a y <<0, and a z <<0. In this region, the wave equation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψ Use separation of variables technique, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=⋅+XYZ mEDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ mEz Z Z y Y Y x X XLet01222222=+∂∂⇒-=∂∂⋅X k x X k x X X xx The solution is of the form: ()x k B x k A x X x x cos sin +=Since ()0,,=z y x ψ at 0=x , then ()00=X so that 0=B .Also, ()0,,=z y x ψ at a x =, so that ()0=a X . Then πx x n a k = where ...,3,2,1=x n Similarly, we have2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅From the boundary conditions, we find πy y n a k = and πz z n a k = where...,3,2,1=y n and ...,3,2,1=z n From the wave equation, we can write022222=+---mE k k k z y xThe energy can be written as()222222⎪⎭⎫ ⎝⎛++==a n n n m E E z y x n n n z y x π _______________________________________ 3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by()332a dk k dk k g T ⋅=ππ where222 mEk =We can then writemEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121 Substituting these expressions into the density of states function, we have()dE E mmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233 ππ Noting thatπ2h=this density of states function can be simplified and written as()()dE E m h a dE E g T ⋅⋅=2/33324π Dividing by 3a will yield the density of states so that()()E h m E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k a n E m n ==*π Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n *⋅=21dE Em dk n⋅⋅⋅=*2211 Then()dE Em a dE E g n T ⋅⋅⋅=*2212 π Divide by the "volume" a , so ()Em E g n *⋅=21πSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o n m m 08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E h m 22/332/33224+*-⋅⋅=π()()2/332/323224kT hm n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV()()19106.10259.0-⨯= 2110144.4-⨯=J Then ()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3- or 19100.6⨯=c g cm 3-(ii) At 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯= 21105253.5-⨯=J Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3- or 191024.9⨯=c g cm 3-(b) GaAs, o nm m 067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π ()()2/3542102288.1kT ⨯=(i) At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m 3- or 171027.9⨯=c g cm 3-(ii) At 400=T K, 21105253.5-⨯=kT J ()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3-181043.1⨯=c g cm 3-_______________________________________ 3.27(a) Silicon, o p m m 56.0=* ()()E E h mE g p-=*υυπ32/324()dE E E h mg E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E hm 32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT hmp-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.4310969.2-⨯⨯=υg2510116.4⨯=m3-or 191012.4⨯=υg cm 3- (ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m3-or 191034.6⨯=υg cm 3- (b) GaAs, o p m m 48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3- or 191027.3⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υg cm 3-_______________________________________ 3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1 For c E E =; 0=c g1.0+=c E E eV; 4610509.1⨯=c g m 3-J 1-2.0+=c E E eV; 4610134.2⨯=m 3-J 1-3.0+=c E E eV; 4610614.2⨯=m 3-J 1- 4.0+=c E E eV; 4610018.3⨯=m 3-J 1- (b) ()E E h m g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4 For υE E =; 0=υg1.0-=υE E eV; 4510634.5⨯=υg m 3-J 1-2.0-=υE E eV; 4510968.7⨯=m 3-J 1-3.0-=υE E eV; 4510758.9⨯=m 3-J 1-4.0-=υE E eV; 4610127.1⨯=m 3-J 1-_______________________________________ 3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pncmm g g υ_______________________________________3.30 Plot_______________________________________ 3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66=(ii) ()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________ 3.32()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F exp 11(a) kT E E F =-, ()()⇒+=1exp 11E f()269.0=E f (b) kT E E F 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f(c) kT E E F 10=-, ()()⇒+=10exp 11E f ()51054.4-⨯=E f_______________________________________ 3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F exp 1111or()⎪⎪⎭⎫ ⎝⎛-+=-kT E E E f F exp 111(a) kT E E F =-, ()269.01=-E f (b) kT E E F 5=-, ()31069.61-⨯=-E f(c) kT E E F 10=-, ()51054.41-⨯=-E f_______________________________________3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp F f 61066.5-⨯=kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +; ()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+; ()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =; ⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kTE -υ; ()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯= kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=_______________________________________3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1 ()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So ()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp ()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c +-=-+υOr midgap c F E E E E =+=2υ_______________________________________ 3.3622222ma n E n π =For 6=n , Filled state()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eV For 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eV Therefore 8.1240.9<<F E eV_______________________________________ 3.37(a) For a 3-D infinite potential well()222222⎪⎭⎫ ⎝⎛++=a n n n mE z y x π For 5 electrons, the 5th electron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x π()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--E eV For the next quantum state, which is empty, the quantum state is 2,2,1===z y x n n n . This quantum state is at the same energy, so 35.2=F E eV(b) For 13 electrons, the 13th electronoccupies the quantum state 3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE 1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eVThe 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is at the same energy, so 746.5=F E eV_______________________________________ 3.38The probability of a state at E E E F ∆+=1 being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111or()⎪⎭⎫ ⎝⎛∆+=-kT E E f exp 11122so ()()22111E f E f -= Q.E.D. _______________________________________3.39(a) At energy 1E , we want01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F FThis expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F F or()⎪⎪⎭⎫⎝⎛-=kT E E F 1exp 01.01Then()100ln 1kT E E F += orkT E E F 6.41+= (b)At kT E E F 6.4+=, ()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F which yields()01.000990.01≅=E f_______________________________________ 3.40 (a)()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kT which yields 740=T K_______________________________________ 3.41 (a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b) At 1000=T K, 08633.0=kT eV Then()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫ ⎝⎛-+=E for 99.7% (d)At F E E =, ()21=E f for all temperatures_______________________________________ 3.42(a) For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =, 82.030.012.12=-=-E E F eV Then()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b) For 4.02=-E E F eV,72.01=-F E E eVAt 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E f_______________________________________ 3.43(a) At 1E E =()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f F or()61032.9-⨯=E fAt 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.012.1expor()191066.11-⨯=-E f (b) For 4.02=-E E F ,02.11=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________ 3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛⨯kT E E kT F exp 1or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kT E E kT E E kT dE E df F F (a) At 0=T K, For()00exp =⇒=∞-⇒<dE dfE E F()0exp =⇒+∞=∞+⇒>dEdfE E FAt -∞=⇒=dEdfE E F(b) At 300=T K, 0259.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()65.91110259.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1-(c) At 500=T K, 04317.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________ 3.45(a) At midgap E E =,()⎪⎪⎭⎫⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kT E kTE E E f g F2exp 11exp 11Si: 12.1=g E eV, ()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for()101007.4-⨯=E fGe: 66.0=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E for()61093.2-⨯=E f GaAs: 42.1=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0242.1exp 11E for()121024.1-⨯=E f(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._______________________________________3.46(a) ()⎥⎦⎤⎢⎣⎡--=kT E E f F F exp ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 108or ()810ln 60.0+=kT()032572.010ln 60.08==kT eV ()⎪⎭⎫⎝⎛=3000259.0032572.0Tso 377=T K(b) ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 106()610ln 60.0+=kT()043429.010ln 60.06==kT ()⎪⎭⎫⎝⎛=3000259.0043429.0Tor 503=T K_______________________________________ 3.47(a) At 200=T K,()017267.03002000259.0=⎪⎭⎫⎝⎛=kT eV⎪⎪⎭⎫ ⎝⎛-+==kT E E f F F exp 1105.019105.01exp =-=⎪⎪⎭⎫ ⎝⎛-kT E E F()()()19ln 017267.019ln ==-kT E E F 05084.0=eV By symmetry, for 95.0=F f , 05084.0-=-F E E eVThen ()1017.005084.02==∆E eV (b) 400=T K, 034533.0=kT eV For 05.0=F f , from part (a),()()()19ln 034533.019ln ==-kT E E F 10168.0=eVThen ()2034.010168.02==∆E eV _______________________________________。

西安邮电大学微电子学系商世广半导体器件试题库常用单位：在室温（ T = 300K ）时，硅本征载流子的浓度为n i = 1.510×10/cm3电荷的电量 q= 1.6 ×10-19Cn2/V sp2/V s μ=1350 cmμ=500 cmε0×10-12F/m=8.854一、半导体物理基础部分（一）名词解释题杂质补偿：半导体内同时含有施主杂质和受主杂质时，施主和受主在导电性能上有互相抵消的作用，通常称为杂质的补偿作用。

非平衡载流子：半导体处于非平衡态时，附加的产生率使载流子浓度超过热平衡载流子浓度，额外产生的这部分载流子就是非平衡载流子。

迁移率：载流子在单位外电场作用下运动能力的强弱标志，即单位电场下的漂移速度。

晶向：晶面：（二）填空题1．根据半导体材料内部原子排列的有序程度，可将固体材料分为、多晶和三种。

2．根据杂质原子在半导体晶格中所处位置，可分为杂质和杂质两种。

3．点缺陷主要分为、和反肖特基缺陷。

4．线缺陷，也称位错，包括、两种。

5．根据能带理论，当半导体获得电子时，能带向弯曲，获得空穴时，能带向弯曲。

6．能向半导体基体提供电子的杂质称为杂质；能向半导体基体提供空穴的杂质称为杂质。

7．对于 N 型半导体，根据导带低E C和 E F的相对位置，半导体可分为、弱简并和三种。

8．载流子产生定向运动形成电流的两大动力是、。

9．在 Si-SiO 2系统中，存在、固定电荷、和辐射电离缺陷 4 种基本形式的电荷或能态。

10．对于N 型半导体，当掺杂浓度提高时，费米能级分别向移动；对于P 型半导体，当温度升高时，费米能级向移动。

（三）简答题1．什么是有效质量，引入有效质量的意义何在？有效质量与惯性质量的区别是什么？2．说明元素半导体Si 、 Ge中主要掺杂杂质及其作用？3．说明费米分布函数和玻耳兹曼分布函数的实用范围？4．什么是杂质的补偿，补偿的意义是什么？（四）问答题1．说明为什么不同的半导体材料制成的半导体器件或集成电路其最高工作温度各不相同？要获得在较高温度下能够正常工作的半导体器件的主要途径是什么？（五）计算题1．金刚石结构晶胞的晶格常数为a，计算晶面（ 100）、（ 110）的面间距和原子面密度。

______________________________________________________________________________________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 77.1⎪⎪⎭⎫⎝⎛=2ln i d a t bi n N N V V (a)(i)()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101515105.1102102ln 0259.0bi V611.0=V(ii)()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101615105.1102102ln 0259.0bi V671.0=V(iii)()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101715105.1102102ln 0259.0bi V731.0=V (b)(i)()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101517105.1102102ln 0259.0bi V731.0=V(ii)()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101617105.1102102ln 0259.0bi V790.0=V(iii)()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101717105.1102102ln 0259.0bi V850.0=V_______________________________________ 7.2Si: 10105.1⨯=i n cm 3-Ge: 13104.2⨯=i n cm 3- GaAs: 6108.1⨯=i n cm 3- ⎪⎪⎭⎫⎝⎛=2ln i d a t bi n N N V V and 0259.0=t V V (a)1410=d N cm 3-, 1710=a N cm 3-' Then Si: 635.0=bi V V Ge: 253.0=bi V V GaAs: 10.1=bi V V(b)16105⨯=d N cm 3-, 16105⨯=a N cm 3- Si: 778.0=bi V V Ge: 396.0=bi V V GaAs: 25.1=bi V V (c)1710=d N cm 3-, 1710=a N cm 3- Si: 814.0=bi V V Ge: 432.0=bi V VGaAs: 28.1=bi V V_______________________________________ 7.3(a) Silicon (300=T K)()()⎥⎥⎦⎤⎢⎢⎣⎡⨯=210105.1ln 0259.0d a bi N N VFor 1410==d a N N cm 3-; 4561.0=bi V V1510= ; 5754.0=V1610= ; 6946.0=V1710= ; 8139.0=V(b) GaAs (300=T K)()()⎥⎥⎦⎤⎢⎢⎣⎡⨯=26108.1ln 0259.0d a bi N N VFor 1410==d a N N cm 3-;9237.0=bi V V1510= ; 043.1=V1610= ; 162.1=V1710= ; 282.1=V(c) Silicon (400 K), 034533.0=kT 121038.2⨯=i n cm 3-For 1410==d a N N cm 3-;2582.0=bi V V1510= ; 4172.0=V1610= ; 5762.0=V1710= ; 7353.0=VGaAs(400 K), 91029.3⨯=i n cm 3- For 1410==d a N N cm 3-;7129.0=bi V V1510= ; 8719.0=V1610= ; 031.1=V1710= ; 190.1=V_______________________________________ 7.4(a) n-side⎪⎪⎭⎫⎝⎛=-i d Fi F n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1105ln 0259.0or3294.0=-Fi F E E eV p-side⎪⎪⎭⎫⎝⎛=-i a F Fi n N kT E E ln ()⎪⎪⎭⎫⎝⎛⨯=1017105.110ln 0259.0or4070.0=-F Fi E E eV (b)4070.03294.0+=bi V or7364.0=bi V V (c)⎪⎪⎭⎫⎝⎛=2ln i d a t bi n N N V V ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯=2101517105.110510ln 0259.0or7363.0=bi V V (d)2/112⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛∈=d adabis n N N N N e Vx()()()⎢⎣⎡⨯⨯=--1914106.1736.01085.87.1122/11517151710510110510⎥⎥⎦⎤⎪⎭⎫⎝⎛⨯+⎪⎪⎭⎫ ⎝⎛⨯⨯ or410426.0-⨯=n x cm μ426.0=m Now()()()⎢⎣⎡⨯⨯=--1914106.1736.01085.87.112p x2/11517171510510110105⎥⎥⎦⎤⎪⎭⎫ ⎝⎛⨯+⎪⎪⎭⎫ ⎝⎛⨯⨯or4100213.0-⨯=p x cm μ0213.0=mWe havesnd x eN ∈=E max()()()()()14415191085.87.1110426.0105106.1---⨯⨯⨯⨯=or4max 1029.3⨯=E V/cm_______________________________________ 7.5(a) n-side⎪⎪⎭⎫⎝⎛=-i d Fi F n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1102ln 0259.0or3653.0=-Fi F E E eV p-side⎪⎪⎭⎫⎝⎛=-i a F Fi n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1102ln 0259.0or3653.0=-F Fi E E eV (b)3653.03653.0+=bi V or7306.0=bi V V (c)⎪⎪⎭⎫⎝⎛=2ln i d a t bi n N N V V ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101616105.1102102ln 0259.0or7305.0=bi V V (d)2/112⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛∈=d a d a bi s n N N N N e V x()()()⎢⎣⎡⨯⨯=--1914106.17305.01085.87.1122/1161616161021021102102⎥⎥⎦⎤⎪⎭⎫ ⎝⎛⨯+⨯⎪⎪⎭⎫ ⎝⎛⨯⨯⨯ or410154.0-⨯=n x cm μ154.0=m By symmetry410154.0-⨯=p x cm μ154.0=mNow snd x eN ∈=E max()()()()()14416191085.87.11101537.0102106.1---⨯⨯⨯⨯=or4max 1075.4⨯=E V/cm_______________________________________ 7.6(b) ⎪⎪⎭⎫⎝⎛-=kT E E n N Fi F i d exp()⎪⎭⎫⎝⎛⨯=0259.0365.0exp 105.110or d N 161098.1⨯=cm 3-⎪⎪⎭⎫ ⎝⎛-=kTE E n N FFi i a exp ()⎪⎭⎫ ⎝⎛⨯=0259.0330.0exp 105.110or a N 151012.5⨯=cm 3-(c)()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101615105.11098.11012.5ln 0259.0bi V695.0=V_______________________________________ 7.7200 K; 017267.0=kT ; 38.1=i n cm 3-300 K; 0259.0=kT ; 6108.1⨯=i n cm 3-400 K; 034533.0=kT ; 91028.3⨯=i n cm 3- For 200 K;()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯=2161538.1104102ln 017267.0bi V 257.1=VFor 300 K;()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=261615108.1104102ln 0259.0bi V157.1=VFor 400 K;()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2916151028.3104102ln 034533.0bi V023.1=V_______________________________________ 7.8()p n n x x W x +==25.025.0p n x x 25.075.0=3=⇒np x xa p d n N x N x =3==⇒np a dx x N N So a d N N 3=(a) ()()⎥⎥⎦⎤⎢⎢⎣⎡⨯=210105.1ln 0259.0d a bi N N V()()⎥⎥⎦⎤⎢⎢⎣⎡⨯=2102105.13ln 0259.0710.0a Nor ()⎪⎭⎫ ⎝⎛⨯=0259.0710.0exp 105.132102a Nwhich yields 1510766.7⨯=a N cm 3- 161033.2⨯=d N cm 3-2/112⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛∈=d a d a bi s n N N N N e V x ()()()⎩⎨⎧⨯⨯=--1914106.1710.01085.87.112()2/11510766.74131⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⨯⎪⎭⎫ ⎝⎛⨯ 61093.9-⨯=⇒n x cm or μ0993.0=n x m ()()()⎩⎨⎧⨯⨯=--1914106.1710.01085.87.112p x()2/11510766.74113⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⨯⎪⎭⎫ ⎝⎛⨯510979.2-⨯=cmor μ2979.0=p x m Now snd x eN ∈=E max()()()()()14416191085.87.11100993.01033.2106.1---⨯⨯⨯⨯=41058.3⨯=V/cm (b) From part (a), we can write()⎪⎭⎫ ⎝⎛⨯=0259.0180.1exp 108.13262a Nwhich yields 1510127.8⨯=a N cm 3-1610438.2⨯=d N cm 3-()()()⎩⎨⎧⨯⨯=--1914106.1180.11085.81.132n x()2/11510127.84131⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⨯⎪⎭⎫ ⎝⎛⨯510324.1-⨯=cmor μ1324.0=n x m()()()⎩⎨⎧⨯⨯=--1914106.1180.11085.81.132p x()2/11510127.84113⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⨯⎪⎭⎫ ⎝⎛⨯510973.3-⨯=cmor μ3973.0=p x m snd x eN ∈=E max()()()()()14416191085.81.13101324.010438.2106.1---⨯⨯⨯⨯= 41045.4⨯=V/cm_______________________________________ 7.9(a) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯=2101516105.11010ln 0259.0bi Vor635.0=bi V V (b)2/112⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛∈=d a d a bi s n N N N N e V x()()()⎢⎣⎡⨯⨯=--1914106.16350.01085.87.1122/115161516101011010⎥⎥⎦⎤⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛⨯ or4108644.0-⨯=n x cm μ8644.0=m Now2/112⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛∈=d a a d bi s p N N N N e V x ()()()⎢⎣⎡⨯⨯=--1914106.16350.01085.87.1122/115161615101011010⎥⎥⎦⎤⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛⨯ or41008644.0-⨯=p x cm μ08644.0=m(c)snd x eN ∈=E max()()()()()14415191085.87.11108644.010106.1---⨯⨯⨯=or4max 1034.1⨯=E V/cm_______________________________________ 7.10(a) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101617105.1104102ln 0259.0bi V80813.0=V(b)bi V increases as temperature decreases At 300=T K, we can write()2102105.1⨯=i n()()⎪⎭⎫⎝⎛-⨯⨯=0259.012.1exp 1004.1108.21919K659.4=⇒KAt 287=T K, 024778.0=kT eV()()3191923002871004.1108.2⎪⎭⎫⎝⎛⨯⨯=K n i⎪⎭⎫ ⎝⎛-⨯024778.012.1exp()()()2038103404.2105496.2659.4-⨯⨯=So 19210780.2⨯=i nThen()()()⎥⎦⎤⎢⎣⎡⨯⨯⨯=19161710780.2104102ln 024778.0bi V 82494.0=V We find()()()%100300300287⨯-bi bi bi V V V%08.2%10080813.080813.082494.0=⨯-=%2≅_______________________________________ 7.11⎪⎪⎭⎫⎝⎛=2ln i d a t bi n N N V V()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⎪⎭⎫ ⎝⎛=21516102104ln 3000259.0550.0i n T Using the procedure from Problem 7.10, we can write, for 300=T K,()2102105.1⨯=i n()()⎪⎭⎫⎝⎛-⨯⨯=0259.012.1exp 1004.1108.21919K659.4=⇒K At 300=T K,()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101516105.1102104ln 0259.0bi V68886.0=VFor 550.0=bi V V, 300>⇒T K At 380=T K, 032807.0=kT eV Also()()()3191923003801004.1108.2659.4⎪⎭⎫⎝⎛⨯⨯=in⎪⎭⎫ ⎝⎛-⨯032807.012.1exp2410112.4⨯=Then()()()⎥⎦⎤⎢⎣⎡⨯⨯⨯=24151610112.4102104ln 032807.0bi V 5506.0=V 550.0≅V_______________________________________ 7.12(b) For 1610=d N cm 3-,⎪⎪⎭⎫⎝⎛=-i d Fi F n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=1016105.110ln 0259.0or3473.0=-Fi F E E eVFor 1510=d N cm 3- ()⎪⎪⎭⎫⎝⎛⨯=-1015105.110ln 0259.0FiF E Eor2877.0=-Fi F E E eV Then28768.034732.0-=bi V or0596.0=bi V V_______________________________________ 7.13(a) ⎪⎪⎭⎫⎝⎛=2ln i da t bi n N N V V ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯=2101612105.11010ln 0259.0or456.0=bi V V (b)()()()⎢⎣⎡⨯⨯=--1914106.1456.01085.87.112n x2/116121612101011010⎥⎥⎦⎤⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛⨯ or71043.2-⨯=n x cm (c)()()()⎢⎣⎡⨯⨯=--1914106.1456.01085.87.112p x2/116121************⎥⎥⎦⎤⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛⨯ or31043.2-⨯=p x cm(d)snd x eN ∈=E max()()()()()14716191085.87.111043.210106.1---⨯⨯⨯=or2max 1075.3⨯=E V/cm_______________________________________ 7.14Assume silicon, so2/12⎪⎪⎭⎫⎝⎛∈=d s D N e kT L()()()()()2/12191914106.1106.10259.01085.87.11⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=---d Nor2/1510676.1⎪⎪⎭⎫⎝⎛⨯=d D N L(a)14108⨯=d N cm 3-, μ1447.0=D L m (b)16102.2⨯=d N cm 3-,μ02760.0=D L m(c)17108⨯=d N cm 3-,μ004577.0=D L m Now(a)7427.0=bi V V (b)8286.0=bi V V (c)9216.0=bi V V Also()()()⎢⎢⎣⎡⨯⨯=--1914106.11085.87.112bi n V x2/117171081108⎥⎥⎦⎤⎪⎪⎭⎫ ⎝⎛+⨯⎪⎪⎭⎫ ⎝⎛⨯⨯dd N NThen(a)μ096.1=n x m (b)μ2178.0=n x m (c)μ02730.0=n x m Now(a)1320.0=n D x L(b)1267.0=nDx L (c)1677.0=nD x L_______________________________________ 7.152/1max2⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫⎝⎛+∈=E d ada sbi NN N N eVWe find()()()71419100904.31085.87.11106.122---⨯=⨯⨯=∈s e (a)(i) For1710=a N ,1410=d N ;6350.0=bi V V(ii) 1510=; 6946.0=V(iii) 1610=; 7543.0=V(iv) 1710=; 8139.0=V(i) For 1710=a N , 1410=d N ; max E 410443.0⨯=V/cm(ii) 1510=; 41046.1⨯=V/cm(iii) 1610=; 41060.4⨯=V/cm(iv) 1710=; 4102.11⨯=V/cm (b) (i) For1410=a N ,1410=d N ;4561.0=bi V V(ii) 1510=; 5157.0=V(iii) 1610=; 5754.0=V(iv) 1710=; 6350.0=V(i) For 1410=a N ,1410=d N ;4max 10265.0⨯=E V/cm (ii) 1510=;410381.0⨯=V/cm(iii) 1610=; 410420.0⨯=V/cm(iv) 1710=; 410443.0⨯=V/cm(c) max E increases as the doping increases,and the electric field extends further into the low-doped side of the pn junction._______________________________________7.16(a) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯=2101516105.110105ln 0259.0bi V 6767.0=V(b) ()2/12⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛++∈=d a d a R bis N N N N e V V W(i) For 0=R V ,()()()⎩⎨⎧⨯⨯=--1914106.16767.01085.87.112W()()2/1151615161010510105⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⨯+⨯⨯510452.9-⨯=cm or μ9452.0=W m (ii) For 5=R V V,()()()⎩⎨⎧⨯+⨯=--1914106.156767.01085.87.112W()()2/1151615161010510105⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⨯+⨯⨯410738.2-⨯=cm or μ738.2=W m(c) ()WV V R bi +=E 2max(i)For 0=R V ,()44max 1043.1109452.06767.02⨯=⨯=E -V/cm (ii)For 5=R V V,()44max 1015.410738.256767.02⨯=⨯+=E -V/cm _______________________________________ 7.17(a) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101617105.1104102ln 0259.0bi V8081.0=V (b)()2/112⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛+∈=d a d a R bis n N N N N e V V x ()()()⎩⎨⎧⨯+⨯=--1914106.15.28081.01085.87.1122/1161716171041021104102⎪⎭⎪⎬⎫⎪⎭⎫ ⎝⎛⨯+⨯⎪⎪⎭⎫ ⎝⎛⨯⨯⨯4102987.0-⨯=cm or μ2987.0=n x m()2/112⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛+∈=d a a d R bis p N N N N e V V x()()()⎩⎨⎧⨯+⨯=--1914106.15.28081.01085.87.1122/1161717161041021102104⎪⎭⎪⎬⎫⎪⎭⎫ ⎝⎛⨯+⨯⎪⎪⎭⎫ ⎝⎛⨯⨯⨯61097.5-⨯=cmor μ0597.0=p x m()2/12⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛++∈=d a d a R bis N N N N e V V W()()()⎩⎨⎧⨯+⨯=--1914106.15.28081.01085.87.112()()2/116171617104102104102⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⨯⨯⨯+⨯⨯4103584.0-⨯=cm or μ3584.0=W mAlso μ3584.0=+=p n x x W m(c) ()()4max 103584.05.28081.022-⨯+=+=E W V V R bi 51085.1⨯=V/cm(d) ()()2/12⎭⎬⎫⎩⎨⎧++∈=d a R bida s N N V V N N e A C()()()()()⎩⎨⎧+⨯⨯⨯=---5.28081.021085.87.11106.110214194()()2/116171617104102104102⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⨯+⨯⨯⨯⨯121078.5-⨯= F or 78.5=C pF_______________________________________ 7.18(a) ⎪⎪⎭⎫⎝⎛=2ln i d a t bi n N N V V ⎪⎪⎭⎫ ⎝⎛=2280ln idt n NV We find⎪⎪⎭⎫ ⎝⎛=tbi i dV V n N exp 8022()⎪⎭⎫⎝⎛⨯=0259.0740.0exp 105.12103210762.5⨯=1510684.2⨯=⇒d N cm 3- 1710147.2⨯=a N cm 3- (b)()2/112⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛+∈=d a d a R bi s n N N N N e V V x()()()⎩⎨⎧⨯+⨯=--1914106.110740.01085.87.1122/1151710684.210147.21180⎭⎬⎫⎪⎭⎫ ⎝⎛⨯+⨯⎪⎭⎫ ⎝⎛⨯410262.2-⨯=cm or μ262.2=n x m()2/112⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛+∈=d a a d R bis p N N N N e V V x()()()⎩⎨⎧⨯+⨯=--1914106.110740.01085.87.1122/1151710684.210147.21801⎭⎬⎫⎪⎭⎫ ⎝⎛⨯+⨯⎪⎭⎫ ⎝⎛⨯61083.2-⨯=cmor μ0283.0=p x m(c) ()WV V R bi +=E 2max()()4100283.0262.210740.02-⨯++= 41038.9⨯=V/cm (d) ()()2/12⎭⎬⎫⎩⎨⎧++∈='d a R bi da s N N V V N N e C()()()()⎩⎨⎧+⨯⨯=--10740.021085.87.11106.11419()()2/11517151710684.210147.210684.210147.2⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⨯+⨯⨯⨯⨯91052.4-⨯='C F/cm 2_______________________________________ 7.19(a) ()()a bi a bi N V N V -3()⎥⎥⎦⎤⎢⎢⎣⎡-⎥⎥⎦⎤⎢⎢⎣⎡=22ln 3ln i a d t i a d t n N N V n N N V()⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡+=22ln ln 3ln i a d t i a d t n N N V n N N V()()()3ln 0259.03ln ==t V 02845.0=V (b) ()2/12⎭⎬⎫⎩⎨⎧+∈≅'R bi a s V V N e CSo ()()732.13332/1==⎭⎬⎫⎩⎨⎧=''a a a a N N N C N C(c) For a larger doping, the space charge width narrows which results in a larger capacitance._______________________________________ 7.20(a) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101715105.1104104ln 0259.0bi Vor766.0=bi V V Now()2/1max 2⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛+∈+=E d a da s R bi N N N N V V eor ()()()()()⎢⎢⎣⎡⨯+⨯=⨯--1419251085.87.11106.12103R bi V V()()⎥⎦⎤⨯+⨯⨯⨯⨯17151715104104104104 or()R bi V V +⨯=⨯91010224.1109 so that()53.73=+R bi V V V which yields 8.72=R V V(b) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101716105.1104104ln 0259.0bi Vor826.0=bi V V We have()()()()()⎢⎢⎣⎡⨯+⨯=⨯--1419251085.87.11106.12103R bi V V ()()⎥⎦⎤⨯+⨯⨯⨯⨯17161716104104104104 so that()008.8=+R bi V V Vwhich yields 18.7=R V V(c) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101717105.1104104ln 0259.0bi Vor886.0=bi V V We have()()()()()⎢⎢⎣⎡⨯+⨯=⨯--1419251085.87.11106.12103R bi V V ()()⎥⎦⎤⨯+⨯⨯⨯⨯17171717104104104104 so that()456.1=+R bi V V V which yields 570.0=R V V_______________________________________ 7.21 (a)()()()()2/12/122⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛++∈⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛++∈=dB a dBa R biB s dA a dA a R biA s N N NN e V V N N N N e V V B W A Wor()()()()()()2/1⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⋅++⋅++=dA dB dB a dA a R biB R biA N N N N N N V V V V B W A W We find()()()()7543.0105.11010ln 0259.02101518=⎥⎥⎦⎤⎢⎢⎣⎡⨯=biA V V()()()()8139.0105.11010ln 0259.02101618=⎥⎥⎦⎤⎢⎢⎣⎡⨯=biB V VWe find()()⎢⎢⎣⎡⎪⎪⎭⎫⎝⎛++⎪⎭⎫ ⎝⎛=16181518101010108139.57543.5B W A W2/115161010⎥⎥⎦⎤⎪⎪⎭⎫ ⎝⎛⨯or()()13.3=B W A W(b)()()()()()()()()RbiB RbiA R biB R biA V V V V A W B W B W V V A W V V B A ++⋅=++=E E 22 ⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛=8139.57543.513.31or()()316.0=E E B A (c)()()()()()()2/12/122⎥⎦⎤⎢⎣⎡++∈⎥⎦⎤⎢⎣⎡++∈=''dB a R biB dB a s dA a R biA dA a s j j N N V V N N N N V V N N B C A C2/1⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛++⎪⎪⎭⎫ ⎝⎛++⎪⎪⎭⎫ ⎝⎛=dA adB a R biA RbiB dB dA N N N N V V V V N N2/1151816181615101010107543.58139.51010⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛++⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=or()()319.0=''B C A C j j_______________________________________ 7.22(a) We have()()()()()()2/12/122100⎥⎦⎤⎢⎣⎡++∈⎥⎦⎤⎢⎣⎡+∈=''d a R bi da s d a bi d a s j j N N V V N N N N V N N C Cor()()2/113.3100⎪⎪⎭⎫ ⎝⎛+==''bi Rbi j j V V V C CFor 10=R V V, we find()1013.32+=bi bi V V or137.1=bi V V (b)()n p p x x W x +==2.02.0Thenad n p N Nx x ==25.0Now⎪⎪⎭⎫ ⎝⎛=2ln i da t bi n N N V V so()()⎥⎥⎦⎤⎢⎢⎣⎡⨯=262108.125.0ln 0259.0137.1a NWe can then write()⎥⎦⎤⎢⎣⎡⨯=0259.02137.1exp 25.0108.16a N which yields161023.1⨯=a N cm 3- and151007.3⨯=d N cm 3-_______________________________________ 7.23()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=261516108.1105102ln 0259.0bi V162.1=VRbi V V C +∝'1So()()1221R bi R bi R R V V V V V C V C ++=''5.0162.1162.150.12++=R V()662.1162.150.122R V +=which yields 58.22=R V V_______________________________________ 7.24(a) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101615105.1104102ln 0259.0bi V6889.0=V()()2/12⎭⎬⎫⎩⎨⎧++∈='=d a R bida s N N V V N N e A C A C()()()()()⎩⎨⎧+⨯⨯⨯=---R V 6889.021085.87.11106.110514194()()()2/116151615104102104102⎭⎬⎫⨯+⨯⨯⨯⨯RV C +⨯=-6889.0102806.612(i) For 0=R V , 567.7=C pF (ii) For 5=R V V, 633.2=C pF(b) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=261615108.1104102ln 0259.0bi V157.1=V()()2/12⎭⎬⎫⎩⎨⎧++∈='=d a R bi da s N N V V N N e A C A C()()()()()⎩⎨⎧+⨯⨯⨯=---R V 157.121085.81.13106.110514194()()()2/116151615104102104102⎭⎬⎫⨯+⨯⨯⨯⨯RV C +⨯=-157.1106457.612(i) For 0=R V ,178.6=C pF (ii) For 5=R V V, 678.2=C pF_______________________________________ 7.25()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101517105.1105102ln 0259.0bi V7543.0=V(a) ()()2/12⎭⎬⎫⎩⎨⎧++∈='=d a R bida s N N V V N N e A C A C()()()()()⎩⎨⎧+⨯⨯⨯=---107543.021085.87.11106.110814194()()()2/115171517105102105102⎭⎬⎫⨯+⨯⨯⨯⨯1210904.4-⨯=C F()22121f C L LCf ππ=⇒=()()[]26121025.1210904.41⨯⨯=-πL310306.3-⨯=H 306.3=mH(b)(i) For 1=R V V, 14.12=C pF()()[]2/11231014.1210306.321--⨯⨯=πf 51094.7⨯=Hz 794.0=MHz (ii) For 5=R V V, 704.6=C pF()()[]2/112310704.610306.321--⨯⨯=πf 610069.1⨯=Hz 069.1=MHz _______________________________________ 7.26()2/1max 2⎥⎦⎤⎢⎣⎡∈+≅E s d R bi N V V eLet 75.0≅bi V V(a) ()25105.2⨯()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯+⨯=--14191085.87.111075.0106.12d N 161088.1⨯=⇒d N cm 3- (b) ()2510()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯+⨯=--14191085.87.111075.0106.12dN 151001.3⨯=⇒d N cm 3-_______________________________________ 7.27()()()p n p x x W x +==20.020.0()()n p x x 2.08.0= p n x x 4= ()p d n d p a x N x N x N 4==d a N N 4=⇒ (a) ⎥⎥⎦⎤⎢⎢⎣⎡=2ln i da t bi n NN V V ()()⎥⎥⎦⎤⎢⎢⎣⎡⨯=262108.14ln 0259.0dN()()2/12⎭⎬⎫⎩⎨⎧++∈='=d a R bida s N N V V N N e A C A C()()()()()⎩⎨⎧+⨯⨯=---221085.81.13106.11014194bi V()()2/1254⎪⎭⎪⎬⎫⨯d dN N210724.2106.02012+⨯=⨯--bi dV NBy trial and error,1510504.1⨯=d N cm 3-, 1510016.6⨯=a N cm 3-, 10.1=bi V V (b) From part (a),510724.2106.02012+⨯=⨯--bi dV NBy trial and error,1510976.2⨯=d N cm 3-, 161019.1⨯=a N cm 3-,135.1=bi V V_______________________________________ 7.28(a) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯=2101415105.110105ln 0259.0bi V or5574.0=bi V V (b)2/112⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛∈=d aadbis p N N N N e Vx()()()⎢⎣⎡⨯⨯=--1914106.15574.01085.87.1122/11514151410510110510⎥⎥⎦⎤⎪⎭⎫ ⎝⎛⨯+⎪⎪⎭⎫ ⎝⎛⨯⨯or61032.5-⨯=p x cmAlso2/112⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛∈=d a d a bi s n N N N N e V x()()()⎢⎣⎡⨯⨯=--1914106.15574.01085.87.1122/11514141510510110105⎥⎥⎦⎤⎪⎭⎫ ⎝⎛⨯+⎪⎪⎭⎫ ⎝⎛⨯⨯ or41066.2-⨯=n x cm(c) For μ30=n x m, we have()()()⎢⎢⎣⎡⨯+⨯=⨯---19144106.11085.87.1121030R bi V V2/11514141510510110105⎥⎥⎦⎤⎪⎭⎫ ⎝⎛⨯+⎪⎪⎭⎫ ⎝⎛⨯⨯which becomes()R bi V V +⨯=⨯--7610269.1109 We find4.70=R V V_______________________________________ 7.29An p n + junction with 1410=a N cm 3-, (a) A one-sided junction and assume bi R V V >>. Then 2/12⎥⎦⎤⎢⎣⎡∈≅a R s p eN V xor()()()()()1419142410106.11085.87.1121050---⨯⨯=⨯R Vwhich yields 193=R V V (b)⎪⎪⎭⎫⎝⎛=⇒=d a p n a d n p N N x x N N x x so()⎪⎪⎭⎫ ⎝⎛⨯=-1614410101050n x 41050.0-⨯=cm μ50.0=m (c)()4max 105.5015.19322-⨯=≅E W V R or4max 1065.7⨯=E V/cm_______________________________________ 7.30(a) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101517105.1102102ln 0259.0bi V7305.0=V(b) ()2/12⎥⎦⎤⎢⎣⎡+∈⋅≅'=R bi d s V V N e A C A C=()()()⎩⎨⎧+⨯--R bi V V 2106.110195()()()}2/115141021085.87.11⨯⨯⨯-Rbi V V C +⨯=-1310287.1(i) For 1=R V V, 1410783.9-⨯=C F (ii) For 3=R V V, 1410663.6-⨯=C F (iii) For 5=R V V, 1410376.5-⨯=C F_______________________________________ 7.31(a) ()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯=2616108.1108ln 0259.0d bi N V 20.1=()()⎪⎭⎫⎝⎛⨯=⨯0259.020.1exp 108.11082616d N151036.5⨯=⇒d N cm 3-(b) ()()2/12⎭⎬⎫⎩⎨⎧++∈='=d a R bi da s N N V V N N e A C A C()()⎩⎨⎧+⨯=⨯--0.120.12106.11010.11912A()()()()()2/115161516141036.51081036.51081085.81.13⎭⎬⎫⨯+⨯⨯⨯⨯⨯-51056.7-⨯=⇒A cm 2 (c) ()()()⎩⎨⎧+⨯⨯=⨯---R bi V V 2106.11056.71080.019512()()()()()2/115161516141036.51081036.51081085.81.13⎭⎬⎫⨯+⨯⨯⨯⨯⨯-Rbi V V +⨯=⨯--88101585.2100582.1R R bi V V V +==+⇒20.1161.496.2=R V V_______________________________________ 7.32 Plot_______________________________________ 7.33(a) ⎪⎪⎭⎫⎝⎛=2ln i dO aO t bi n N N V V (c) p-region()saO s eN x dx d ∈-=∈=E ρor1C xeN saO +∈-=E We have0=E at spaO p x eN C x x ∈-=⇒-=1Then for 0<<-x x p()p saOx x eN +∈-=E n-region, O x x <<0()sdOs eN x dx d ∈=∈=E 21ρ or212C xeN sdO +∈=En-region, n O x x x <<()sdOs eN x dx d ∈=∈=E ρ2 or32C xeN sdO +∈=EWe have 02=E at n x x =sn dO xeN C ∈-=⇒3so that for n O x x x <<, we have ()x x eN n sdO-∈-=E 2 We also have 12E =E at O x x = Then()O n sdO s O dO x x eNC x eN -∈-=+∈22which gives⎪⎪⎭⎫⎝⎛-∈-=22O n s dO x x eN CThen for O x x <<0 we have⎪⎪⎭⎫⎝⎛-∈-∈=E 221O n s dO s dO x x eN x eN _______________________________________ 7.34(a)()()()dx x d x dx x d s E -=∈-=ρφ22For μ12-<<-x m, ()d eN x +=ρ So1C x eN eN dx d s d s d+∈=E ⇒∈=EAt μ2-=x m O x -≡, 0=ESosOd x eN C ∈=1Then()O sdx x eN +∈=EAt 0=x , ()()10-=E =E x , so ()()410210-⨯+-∈=E sdeN ()()()()()41415191011085.87.11105106.1---⨯⨯⨯⨯=or()410726.70⨯=E V/cm(c) Magnitude of potential difference is⎰E =dx φ()dx x x eN O sd⎰+∈=222C x x x eN O sd+⎪⎪⎭⎫ ⎝⎛⋅+∈=Let 0=φat O x x -=, thens O d O O s d x eN C C x x eN ∈=⇒+⎪⎪⎭⎫ ⎝⎛-∈=22022222 Then we can write()22O sdx x eN +∈=φAt μ1-=x m()()()()()[]24141519110211085.87.112105106.1---⨯+-⨯⨯⨯=φor863.31=φV Potential difference across the intrinsicregion()()()44210210726.70-⨯⨯=⋅E =d φ or45.152=φVBy symmetry, the potential difference acrossthe p-region space-charge region is also 3.863 V. The total reverse-bias voltage is then()2.2345.15863.32=+=R V V_______________________________________ 7.35(a) Bcrits B eN V 22E ∈=or()()()()()40106.121041085.87.1121925142--⨯⨯⨯=E ∈=B crit s B eV NThen 1610294.1⨯==a B N N cm 3- (b) ()()()()()20106.121041085.87.11192514--⨯⨯⨯=B NOr 161059.2⨯==a B N N cm 3-_______________________________________ 7.36()()()()()80106.121041085.87.1121925142--⨯⨯⨯=E ∈=B crit s a eV N151047.6⨯=cm 3-_______________________________________ 7.37(a) For 1610=d N cm 3-, from Figure 7.15, 75≅B V V (b) For 1510=d N cm 3-,450≅B V V_______________________________________ 7.38(a) From Equation (7.36),()2/1max 2⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛+∈+=E d a d a s R bi N N N N V V eSet crit E =E max and B R V V =()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101616105.1102102ln 0259.0bi V7305.0=VThen()()()()⎪⎩⎪⎨⎧⨯+⨯=⨯--141951085.87.11106.12104R bi V V()()2/116161616102102102102⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⨯+⨯⨯⨯⨯ 77.51=+⇒B bi V V V So 04.51=B V V (b)()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101515105.1105105ln 0259.0bi V6587.0=V Then()()()()⎪⎩⎪⎨⎧⨯+⨯=⨯--141951085.87.11106.12104R bi V V()()2/115151515105105105105⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⨯+⨯⨯⨯⨯ 1.207=+⇒R bi V V So 206≅R V V_______________________________________ 7.39For a silicon n p + junction with15105⨯=d N cm 3- and 100≅B V V, then, neglecting bi V we have 2/12⎥⎦⎤⎢⎣⎡∈≅d B s n eN V x()()()()()2/1151914105106.11001085.87.112⎥⎦⎤⎢⎣⎡⨯⨯⨯=-- or()41009.5min -⨯=n x cm μ09.5=m _______________________________________7.40We find()()()()933.0105.11010ln 0259.02101818=⎥⎥⎦⎤⎢⎢⎣⎡⨯=bi V VNowsnd x eN ∈=E maxso()()()()14181961085.87.1110106.110--⨯⨯=n xwhich yields61047.6-⨯=n x cm Now()2/112⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛+∈=d a d a R bis n N N N N e V V x Then()()()⎢⎣⎡⨯⨯=⨯---191426106.11085.87.1121047.6()⎥⎥⎦⎤⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛+⨯18181818101011010R bi V V which yields468.6=+R bi V V V or54.5=R V V_______________________________________ 7.41Assume silicon: For an p n + junction ()2/12⎥⎦⎤⎢⎣⎡+∈=a R bi s p eN V V xAssume R bi V V << (a) For μ75=p x m()()()()()1519142410106.11085.87.1121075---⨯⨯=⨯RVwhich yields31035.4⨯=R V V (b) For μ150=p x m()()()()()1519142410106.11085.87.11210150---⨯⨯=⨯RVwhich yields41074.1⨯=R V VNote: From Figure 7.15, the breakdown voltage is approximately 300 V. So, in eachcase, breakdown is reached first._______________________________________ 7.42Impurity gradien2241810102102=⨯⨯=-a cm 4- From Figure 7.15, 15≅B V V_______________________________________ 7.43(a) For the linearly graded junction ()eax x =ρ Then()ss eaxx dx d ∈=∈=E ρ Now122C x ea dx eax s s +⋅∈=∈=E ⎰At O x x += and O x x -=, 0=E So⎪⎪⎭⎫⎝⎛∈-=⇒+⎪⎪⎭⎫ ⎝⎛∈=2202112O s O s x ea C C x ea Then()222Osx x ea -∈=E (b)()22332C x x x ea dx x O s +⎥⎦⎤⎢⎣⎡⋅-∈-=E -=⎰φSet 0=φ at O x x -=, thens O O O s eax C C x x ea ∈=⇒+⎥⎥⎦⎤⎢⎢⎣⎡+-∈-=332032233。

第5章常用半导体元件习题5。

1晶体二极管一、填空题:1．半导体材料的导电能力介于和之间,二极管是将封装起来,并分别引出和两个极。

2．二极管按半导体材料可分为和 ,按内部结构可分为＿和，按用途分类有、、四种。

3．二极管有、、、四种状态，PN结具有性，即。

4．用万用表（R×1K档）测量二极管正向电阻时，指针偏转角度，测量反向电阻时，指针偏转角度。

5．使用二极管时，主要考虑的参数为和二极管的反向击穿是指。

6．二极管按PN结的结构特点可分为是型和型。

7．硅二极管的正向压降约为 V,锗二极管的正向压降约为 V;硅二极管的死区电压约为V,锗二极管的死区电压约为 V。

8．当加到二极管上反向电压增大到一定数值时，反向电流会突然增大，此现象称为现象。

9．利用万用表测量二极管PN结的电阻值，可以大致判别二极管的、和PN结的材料。

二、选择题：1。

硅管和锗管正常工作时，两端的电压几乎恒定，分别分为( ).A。

0。

2-0.3V 0。

6—0.7V B. 0.2—0.7V 0.3-0。

6VC.0.6-0。

7V 0.2-0.3V D。

0。

1—0.2V 0.6-0.7V的大小为( ).2。

判断右面两图中，UABA. 0。

6V 0.3V B。

0。

3V 0.6VC. 0.3V 0。

3V D。

0。

6V 0.6V3.用万用表检测小功率二极管的好坏时，应将万用表欧姆档拨到( ）Ω档.A。

1×10 B. 1×1000 C. 1×102或1×103 D。

1×1054. 如果二极管的正反向电阻都很大，说明（ ) .A. 内部短路 B。

内部断路 C。

正常 D. 无法确定5。

当硅二极管加0。

3V正向电压时，该二极管相当于（ ) 。

A. 很小电阻 B。

很大电阻 C。

短路 D。

开路6．二极管的正极电位是—20V，负极电位是-10V，则该二极管处于（）。

A．反偏 B．正偏 C．不变D。

断路7．当环境温度升高时，二极管的反向电流将（）A．增大 B．减小 C．不变D。

Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V xt x m ,,2222ψ⋅+∂ψ∂- ()tt x j ∂ψ∂=, Assume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m exp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j exp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu exp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u kSetting ()()x u x u 1= for region I, the equation becomes:()()()()021221212=--+x u k dx x du jk dxx u d α where222mE=α Q.E.D.In Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu exp This equation can be written as:()()()2222x x u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV OSetting ()()x u x u 2= for region II, this equation becomes()()dx x du jk dxx u d 22222+ ()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O α where again222mE=α Q.E.D._______________________________________3.3We have()()()()021221212=--+x u k dx x du jk dxx u d α Assume the solution is of the form: ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp The first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexp and the second derivative becomes()()[]()[]x k j A k j dxx u d --=ααexp 2212 ()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differential equation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k ()[]0exp =+-⨯x k j B α We find that00= Q.E.D. For the differential equation in ()x u 2 and the proposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2()[]x k j D +-+βexp for 0<<-x b .The first boundary condition is ()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dx dudx du which yields()()()C k B k A k --+--βαα()0=++D k β The third boundary condition is ()()b u a u -=21 which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp()()[]b k j D -+-+βexp and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp()[]0exp =+-b k j D β The fourth boundary condition isbx a x dx dudx du -===21 which yields()()[]a k j A k j --ααexp()()[]a k j B k j +-+-ααexp ()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexp and can be written as ()()[]a k j A k --ααexp()()[]a k j B k +-+-ααexp()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________ 3.5(b) (i) First point: πα=aSecond point: By trial and error, πα729.1=a (ii) First point: πα2=aSecond point: By trial and error, πα617.2=a_______________________________________3.6(b) (i) First point: πα=aSecond point: By trial and error, πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a_______________________________________ 3.7ka a aaP cos cos sin =+'αααLet y ka =, x a =α Theny x x xP cos cos sin =+'Consider dy dof this function.()[]{}y x x x P dy d sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydxx sin sin -=- Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-'For πn ka y ==, ...,2,1,0=n 0sin =⇒y So that, in general,()()dk d ka d a d dy dxαα===0 And 22 mE=α Sodk dEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221 α This implies thatdk dE dk d ==0α for an k π= _______________________________________ 3.8(a) πα=a 1π=⋅a E m o 212()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J 34E E E -=∆1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________3.9(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o ()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πoE19105172.2-⨯=J o E E E -=∆11919105172.2104114.3--⨯-⨯= 2010942.8-⨯=Jor 559.0106.110942.81920=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka . From Problem 3.5, πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV_______________________________________3.10(a) πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 12E E E -=∆1919104114.310830.7--⨯-⨯= 19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV (b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J 34E E E -=∆1818103646.1109242.1--⨯-⨯= 1910597.5-⨯=Jor 50.3106.110597.51919=⨯⨯=∆--E eV_____________________________________3.11(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα727.0=a oπ727.022=⋅a E m o o()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=Jo E E E -=∆11919108030.1104114.3--⨯-⨯= 19106084.1-⨯=Jor 005.1106.1106084.11919=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6,πα515.12=aπ515.1222=⋅a E m o()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J23E E E -=∆191810830.7103646.1--⨯-⨯= 1910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV_______________________________________3.12For 100=T K, ()()⇒+⨯-=-1006361001073.4170.124gE164.1=g E eV200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV 600=T K, 032.1=g E eV_______________________________________3.13The effective mass is given by1222*1-⎪⎪⎭⎫⎝⎛⋅=dk E d mWe have()()B curve dkE d A curve dk E d 2222> so that ()()B curve m A curve m **<_______________________________________ 3.14The effective mass for a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p We have that()()B curve dkEd A curve dk E d 2222> so that ()()B curve m A curve m p p **<_______________________________________ 3.15Points A,B: ⇒<0dk dEvelocity in -x directionPoints C,D: ⇒>0dk dEvelocity in +x directionPoints A,D: ⇒<022dk Ednegative effective massPoints B,C: ⇒>022dkEd positive effective mass _______________________________________3.16For A: 2k C E i =At 101008.0+⨯=k m 1-, 05.0=E eV Or ()()2119108106.105.0--⨯=⨯=E J So ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4o m m 488.0=* For B: 2k C E i =At 101008.0+⨯=k m 1-, 5.0=E eV Or ()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C 3711025.1-⨯=⇒CNow ()()37234121025.1210054.12--*⨯⨯==C m 321044.4-⨯=kg or o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_______________________________________ 3.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C 3921025.6-⨯=⇒C()()39234221025.6210054.12--*⨯⨯-=-=C m31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C 382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________ 3.18(a) (i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν1410429.3⨯=Hz(ii) 141010429.3103⨯⨯===νλc E hc 51075.8-⨯=cm 875=nm(b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm_______________________________________ 3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0=k , and is negativearound 2π±=k . _______________________________________ 3.20()[]O O k k E E E --=αcos 1 Then()()()[]O k k E dkdE ---=ααsin 1()[]O k k E -+=ααsin 1 and()[]O k k E dk E d -=ααcos 2122Then221222*11 αE dk Ed m o k k =⋅== or212*αE m =_______________________________________ 3.21(a) ()[]3/123/24lt dn m m m =*()()[]3/123/264.1082.04oom m =o dn m m 56.0=*(b)o o l t cnm m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cn m m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lh hh dp m m m +=*()()[]3/22/32/3082.045.0o om m +=[]o m ⋅+=3/202348.030187.0o dp m m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=*()()()()om ⋅++=2/12/12/32/3082.045.0082.045.0 o cp m m 34.0=*_______________________________________ 3.23For the 3-dimensional infinite potential well, ()0=x V when a x <<0, a y <<0, and a z <<0. In this region, the wave equation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψ Use separation of variables technique, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=⋅+XYZ mEDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ mEz Z Z y Y Y x X XLet01222222=+∂∂⇒-=∂∂⋅X k x X k x X X xx The solution is of the form: ()x k B x k A x X x x cos sin +=Since ()0,,=z y x ψ at 0=x , then ()00=X so that 0=B .Also, ()0,,=z y x ψ at a x =, so that ()0=a X . Then πx x n a k = where ...,3,2,1=x n Similarly, we have2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅From the boundary conditions, we find πy y n a k = and πz z n a k = where...,3,2,1=y n and ...,3,2,1=z n From the wave equation, we can write022222=+---mE k k k z y xThe energy can be written as()222222⎪⎭⎫ ⎝⎛++==a n n n m E E z y x n n n z y x π _______________________________________ 3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by()332a dk k dk k g T ⋅=ππ where222 mEk =We can then writemEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121 Substituting these expressions into the density of states function, we have()dE E mmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233 ππ Noting thatπ2h=this density of states function can be simplified and written as()()dE E m h a dE E g T ⋅⋅=2/33324π Dividing by 3a will yield the density of states so that()()E h m E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k a n E m n ==*π Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n *⋅=21dE Em dk n⋅⋅⋅=*2211 Then()dE Em a dE E g n T ⋅⋅⋅=*2212 π Divide by the "volume" a , so ()Em E g n *⋅=21πSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o n m m 08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E h m 22/332/33224+*-⋅⋅=π()()2/332/323224kT hm n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV()()19106.10259.0-⨯= 2110144.4-⨯=J Then ()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3- or 19100.6⨯=c g cm 3-(ii) At 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯= 21105253.5-⨯=J Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3- or 191024.9⨯=c g cm 3-(b) GaAs, o nm m 067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π ()()2/3542102288.1kT ⨯=(i) At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m 3- or 171027.9⨯=c g cm 3-(ii) At 400=T K, 21105253.5-⨯=kT J ()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3-181043.1⨯=c g cm 3-_______________________________________ 3.27(a) Silicon, o p m m 56.0=* ()()E E h mE g p-=*υυπ32/324()dE E E h mg E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E hm 32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT hmp-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.4310969.2-⨯⨯=υg2510116.4⨯=m3-or 191012.4⨯=υg cm 3- (ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m3-or 191034.6⨯=υg cm 3- (b) GaAs, o p m m 48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3- or 191027.3⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υg cm 3-_______________________________________ 3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1 For c E E =; 0=c g1.0+=c E E eV; 4610509.1⨯=c g m 3-J 1-2.0+=c E E eV; 4610134.2⨯=m 3-J 1-3.0+=c E E eV; 4610614.2⨯=m 3-J 1- 4.0+=c E E eV; 4610018.3⨯=m 3-J 1- (b) ()E E h m g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4 For υE E =; 0=υg1.0-=υE E eV; 4510634.5⨯=υg m 3-J 1-2.0-=υE E eV; 4510968.7⨯=m 3-J 1-3.0-=υE E eV; 4510758.9⨯=m 3-J 1-4.0-=υE E eV; 4610127.1⨯=m 3-J 1-_______________________________________ 3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pncmm g g υ_______________________________________3.30 Plot_______________________________________ 3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66=(ii) ()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________ 3.32()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F exp 11(a) kT E E F =-, ()()⇒+=1exp 11E f()269.0=E f (b) kT E E F 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f(c) kT E E F 10=-, ()()⇒+=10exp 11E f ()51054.4-⨯=E f_______________________________________ 3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F exp 1111or()⎪⎪⎭⎫ ⎝⎛-+=-kT E E E f F exp 111(a) kT E E F =-, ()269.01=-E f (b) kT E E F 5=-, ()31069.61-⨯=-E f(c) kT E E F 10=-, ()51054.41-⨯=-E f_______________________________________3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp F f 61066.5-⨯=kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +; ()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+; ()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =; ⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kTE -υ; ()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯= kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=_______________________________________3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1 ()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So ()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp ()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c +-=-+υOr midgap c F E E E E =+=2υ_______________________________________ 3.3622222ma n E n π =For 6=n , Filled state()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eV For 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eV Therefore 8.1240.9<<F E eV_______________________________________ 3.37(a) For a 3-D infinite potential well()222222⎪⎭⎫ ⎝⎛++=a n n n mE z y x π For 5 electrons, the 5th electron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x π()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--E eV For the next quantum state, which is empty, the quantum state is 2,2,1===z y x n n n . This quantum state is at the same energy, so 35.2=F E eV(b) For 13 electrons, the 13th electronoccupies the quantum state 3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE 1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eVThe 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is at the same energy, so 746.5=F E eV_______________________________________ 3.38The probability of a state at E E E F ∆+=1 being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111or()⎪⎭⎫ ⎝⎛∆+=-kT E E f exp 11122so ()()22111E f E f -= Q.E.D. _______________________________________3.39(a) At energy 1E , we want01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F FThis expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F F or()⎪⎪⎭⎫⎝⎛-=kT E E F 1exp 01.01Then()100ln 1kT E E F += orkT E E F 6.41+= (b)At kT E E F 6.4+=, ()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F which yields()01.000990.01≅=E f_______________________________________ 3.40 (a)()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kT which yields 740=T K_______________________________________ 3.41 (a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b) At 1000=T K, 08633.0=kT eV Then()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫ ⎝⎛-+=E for 99.7% (d)At F E E =, ()21=E f for all temperatures_______________________________________ 3.42(a) For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =, 82.030.012.12=-=-E E F eV Then()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b) For 4.02=-E E F eV,72.01=-F E E eVAt 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E f_______________________________________ 3.43(a) At 1E E =()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f F or()61032.9-⨯=E fAt 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.012.1expor()191066.11-⨯=-E f (b) For 4.02=-E E F ,02.11=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________ 3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛⨯kT E E kT F exp 1or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kT E E kT E E kT dE E df F F (a) At 0=T K, For()00exp =⇒=∞-⇒<dE dfE E F()0exp =⇒+∞=∞+⇒>dEdfE E FAt -∞=⇒=dEdfE E F(b) At 300=T K, 0259.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()65.91110259.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1-(c) At 500=T K, 04317.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________ 3.45(a) At midgap E E =,()⎪⎪⎭⎫⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kT E kTE E E f g F2exp 11exp 11Si: 12.1=g E eV, ()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for()101007.4-⨯=E fGe: 66.0=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E for()61093.2-⨯=E f GaAs: 42.1=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0242.1exp 11E for()121024.1-⨯=E f(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._______________________________________3.46(a) ()⎥⎦⎤⎢⎣⎡--=kT E E f F F exp ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 108or ()810ln 60.0+=kT()032572.010ln 60.08==kT eV ()⎪⎭⎫⎝⎛=3000259.0032572.0Tso 377=T K(b) ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 106()610ln 60.0+=kT()043429.010ln 60.06==kT ()⎪⎭⎫⎝⎛=3000259.0043429.0Tor 503=T K_______________________________________ 3.47(a) At 200=T K,()017267.03002000259.0=⎪⎭⎫⎝⎛=kT eV⎪⎪⎭⎫ ⎝⎛-+==kT E E f F F exp 1105.019105.01exp =-=⎪⎪⎭⎫ ⎝⎛-kT E E F()()()19ln 017267.019ln ==-kT E E F 05084.0=eV By symmetry, for 95.0=F f , 05084.0-=-F E E eVThen ()1017.005084.02==∆E eV (b) 400=T K, 034533.0=kT eV For 05.0=F f , from part (a),()()()19ln 034533.019ln ==-kT E E F 10168.0=eVThen ()2034.010168.02==∆E eV _______________________________________。