山东大学《模拟电子技术》期末考试试卷及答案解析

山东大学课程名称模拟电子技术（本科）试卷B专业：电气工程与自动化、电子信息工程、通信工程试卷性质：闭卷考试时间120分钟题号一二三四五六总分分数得分阅卷人一、填空题（本题共20分，每空0.5分）PN结加正向电压时，加反向电压时，这种特性称为PN结的。场效应管是一种控制型器件，它的输入电阻，栅极电流。对共射、共集和共基三种组态放大电路，若希望电压放大倍数大，可选用组态；若希望带负载能力强，应选用组态；若希望高頻性能好，应选用组态。晶体三极管的输出特性曲线一般分为三个区，即：、、，要使三极管工作在放大区必须给发射结加、集电结加。引入电压串联负反馈可改善放大电路性能，使增益恒定性、输入电阻、输出电阻。正弦波振荡电路一般是由、和所组成，但为了保证振荡幅值稳定且波形好，常常引入环节。欲使振荡器在接通电源后能自行起振，必须满足的幅值条件。二、当集成运放引入深度负反馈时，运放工作在区；当集成运放开环或接正反馈时，运放工作在区。差分放大器的基本特点是放大、抑制.9.乙类推挽放大器的主要失真是，要消除此失真，应改用类推挽功率放大器。10.放大器级间耦合方式有三种：耦合、耦合、耦合；在集成电路中通常采用耦合。11.为了避免50Hz电网电压的干扰进入放大器，应选用滤波器；为了获得输入电压中的低频信号，应选用滤波器。12.放大电路在高频信号作用时放大倍数数值下降的原因是半导体管的电容和__电容的存在。13.常用的小功率直流电源系统是由、、和得分阅卷人四部分组成。二、分析计算题（共24分，每小题8分，）测得工作在放大电路中两个晶体管的三个电极电流如图1所示。图1图1（1）判断它们各是NPN管还是PNP管，在图中标出e，b，c极；（2）估算(b)图晶体管的值。图4图4···········································································································装订线···········································································································装订线··································································································班级姓名学号

2、二极管电路如下图(a)、(b)所示，试在图(c)中画出各电路中VO1、VO2的波形。设Vi=12sinωt(V)，且二极管具有理想特性。图2···········································································································装订线图2···········································································································装订线··································································································3、改错：改正图3所示各电路中的错误，使电路可能产生正弦波振荡。要求不能改变放大电路的基本接法（共射、共基、共集）。图3图3···········································································································装订线···········································································································装订线··································································································图3得分阅卷人三、计算题(共15分)电路如下图所示，晶体管的＝60，=100Ω。1、求解Q点、、Ri和Ro；2、设＝10mV（有效值），问＝？＝？若C3开路，则＝？＝？图4图4图4图4···········································································································装订线···········································································································装订线························································································································································································································································································································装订线··································································得分阅卷人四、计算题（共16分，每小题8分）1、电路如图5所示，试求解：（1）RW的下限值；（提示：依据起振条件）（2）振荡频率的调节范围。图5图52、设图6中运放均为理想运放，求图中电路开关S闭合和断开时的Vo值。图6图6得分阅卷人五、电路如图8所示。(本题共13分)已知VCC＝15V，T1和T2管的饱和管压降│UCES│＝1V，集成运放的最大输出电压幅值为±13V，二极管的导通电压为0.7V。图81、若输入电压幅值足够大，则电路的最大输出功率为多少？图82、为了提高输入电阻，稳定输出电压，且减小非线性失真，应引入哪种组态的交流负反馈？请从图8中直接连线画图。3、若Ui＝0.1V时，Uo＝5V，则反馈网络中电阻Rf的取值约为多少？图8图8得分阅卷人六、分析下图9所示电路，回答下列问题：（本题共12分）在下图9所示电路为光控电路的一部分，它将连续变化的光电信号转换成离散信号（即不是高电平，就是低电平），电流I随光照的强弱而变化。···········································································································装订线··································································································1、在A1和A···········································································································装订线··································································································2、试分析vO与i关系，并画出vO与vo1的传输特性曲线。图9图9答案一、选择题（本题共20分，每空0.5分）导通，截止，单向导电性电压，很大，几乎为零共射或共基，共集，共基饱和区，放大区，截止区，正向电压，反向电压提高，增大，减小放大电路，反馈网络，选频网络，稳幅，︱AF︱>1线性，非线性差模信号，共模信号交越失真，甲乙直接，阻容，变压器，直接带阻，低通极间（电容），接线（电容）电源变压器，整流电路，滤波电路，稳压电路二、分析题（本题24分，每小题8分）1、测得工作在放大电路中两个晶体管的三个电极电流如下图所示1）答：见图中标识（判断NPN管还是PNP管各1分，标出e，b，c极各2分，共6分）2、解：图（a）所示电路为同相输入的过零比较器；图（b）所示电路为同相输入的滞回比较器，两个门限电压为±UT＝±0.5UZ。两个电路的电压传输特性如下图所示：D2优先导通，D1截止、Uo钳位为-6V（理想二极管；或者-5.3V）答：4V答：RC桥式振荡电路答：集成运算放大器在信号运算方面的应用主要有：比例运算电路（同相和反相）、求和运算电路（加法和减法）、积分和微分运算电路、指数和对数运算电路等等。（2）解：(b)图晶体管的值为.....................（2分）2、（本小题8分，每个波形4分）3、解：（本小题8分，每个小题4分）（a）加集电极电阻Rc及放大电路输入端的耦合电容。（b）变压器副边与放大电路输入端之间加耦合电容，另外将副边绕组同铭端改在上端。三、计算题（共15分）解：1、Q点：..................................（4分）、Ri和Ro的分析：.........................（6分）计算题（本题15分）解：（1）由求得（2分）故由求得（2分）由求得（