东南大学电子电路基础考试样卷
2013 东南大学 电子电路实践理论考试 模电理论考试 2
考试时间：共 30 分钟
卷面总分：100 分
1、(单选题)对于下面的电路原理图，输入为 1V 的正弦波，则下面哪个输出波形是正确的
(10 分)
UO3 UO1 UO4 UO2
2、(单选题)下图为反相滞回比较器实验中的输入输出波形图，蓝线为 CH1，灵敏度档位为 1V/DIV，黄 线为 CH2，灵敏度档位为 2V/DIV，则该滞回比较器的下限阈值电平 VLT 为多少？
选择合理的输入信号的电压 将输入信号接入 CH2，输出信号接入 CH 选择示波器 X-Y 显示方式
可以用“MeБайду номын сангаасsure”功能测量
可以用“V/DIV”和“t/DIV”来测量
10、(多选题)我们可以用示波器测量双端口网络的电压传输特性，下面哪些是测量必须的正确步骤(10 分)
进行原点校准 将输入信号接入 CH1，输出信号接入 CH2 调节灵敏度旋钮使两路信号显示的高度大致相等 CHx 输入耦合方式设成 AC CHx 输入耦合方式设成 DC 选择合理的输入信号频率
(10 分)
1.2V -1.2V 0.6V -0.6V
3、(单选题)放大器输出电阻 Ro 的大小反映了放大器带负载能力的大小， 设负载为 RL ，则 RO << RL(10 分)
负载从放大器获取较大电流
负载从放大器获取较大功率
负载从放大器获取较大电压
4、(单选题)下图为方波发生电路实验中遇到的几种输出波形，请根据波形选择可能的故障原因
(10 分)
A 稳压管短路，B 正常 B 稳压管短路，A 正常 两个稳压管都短路
两个稳压管都开路
5、(单选题)请问在下图中 1 脚和 5 脚之间接的 560pF 电容的作用是
电路基础模考试题(附参考答案)
电路基础模考试题（附参考答案）一、单选题（共30题，每题1分，共30分）1、在外施激励下的一阶动态电路不产生过渡过程的条件是( )A、全响应的初始值等于稳态响应的初始值；B、稳态分量的初始值等于暂态分量的初始值；C、无法确定。
D、零输入响应的初始值等于全响应的初始值；正确答案：A2、在不改变铁心材料、尺寸及电源频率情况下，增大交流铁心线圈电压有效值时，铁心损耗（）。
A、// 无法确定。
B、//减小；C、//不变；D、//增大；正确答案：D3、Uc(0_)=Uo 的RC 串联电路接通直流电压源的充电过程中,自由分量(暂态分量)的初始值等于( )。
A、UoB、UsC、Uo-UsD、Us-Uo正确答案：C4、对称三相正弦量（包括对称三相电动势，对称三相电压、对称三相电流）的瞬时值之和恒等于( )。
A、120B、1C、-120D、0正确答案：D5、三线制Ｙ联接电路中三个线电流瞬时值之和（）。
A、不能确定B、等于零C、恒等于零D、不等于零正确答案：C6、串联电路在正弦交流电压源激励下,若，电源电压，电路的阻抗角为 ,则在下列情况中时,电路仍需经历过渡过程的情况是( )A、=0；B、=+-180C、=f ；D、无法确定。
正确答案：A7、一台三相电动机绕组星形连接，接到的三相电源上，测得线电流，则电动机每组绕组的阻抗为( ) 。
A、22B、11C、38D、66正确答案：A8、三相电路中线电压不对称,则三个线电压瞬时值的和（）。
A、等于零B、无法确定C、大于零D、小于零正确答案：A9、铁心线圈接于正弦电压源上，在不改变电压有效值、频率及线圈匝数的情况下增大铁心中的空气隙，则线圈电流的最大值将（）。
A、//减小；B、//不变；C、//增大；D、// 无法确定。
正确答案：C10、在交流铁心线圈电流的三个分量中，属无功分量的是（）。
A、//磁滞分量；B、//涡流分量；C、// 无法确定。
D、//磁化分量；正确答案：D11、三相电路中每相独立源或负载的电流叫（）。
电子电路基础期末试卷及参考答案
XX 中学XX 年XX 学期期末考试电子电路基础试卷（XX 班）（试卷满分100分 考试时间120分钟）班级 姓名 学号 得分一、填空题（每空1分，共20分）1.在时间上和数值上均作连续变化的电信号称为 模拟 信号；在时间上和数值上离散的信号叫做 数字 信号。
2.在正逻辑的约定下，“1”表示 高 电平，“0”表示 低 电平。
3.数字电路中，输入信号和输出信号之间的关系是 逻辑 关系，所以数字电路也称为 逻辑 电路。
4.在逻辑关系中，最基本的关系是 与逻辑 、 或逻辑 和 非逻辑 。
5.十进制数转换为八进制和十六进制时，应先转换成 二进 制。
6.最简与或表达式是指在表达式中 或项 最少，且 与项 也最少。
7.具有基本逻辑关系的电路称为 门电路 ，其中最基本的有 与门 、 或门 和非门。
8.两个与非门构成的基本RS 触发器的功能有 清零 、 置1 和 保持 。
9.组合逻辑电路的基本单元是 门电路 ，时序逻辑电路的基本单元是 触发器 。
二、选择题（每题2分，共30分）1.逻辑函数中的逻辑“与”和它对应的逻辑代数运算关系为（ B ）。
A 、逻辑加B 、逻辑乘C 、逻辑非 2.十进制数100对应的二进制数为（ C ）。
A 、1011110B 、1100010C 、1100100D 、11000100 3.和逻辑式AB 表示不同逻辑关系的逻辑式是（ B ）。
A 、B A + B 、B A ∙C 、B B A +∙D 、A B A + 4.数字电路中机器识别和常用的数制是（ A ）。
A 、二进制B 、八进制C 、十进制D 、十六进制 5.所谓机器码是指（ B ）。
A 、计算机内采用的十六进制码B 、符号位数码化了的二进制数码C 、带有正负号的二进制数码D 、八进制数 6.具有“有1出0、全0出1”功能的逻辑门是（ B ）。
A 、与非门B 、或非门C 、异或门D 、同或门 7.下列各型号中属于优先编译码器是（ C ）。
东南大学(934)电路考试真题(回忆版)及参考答案
4.1.1 东南大学（934）电路2008年硕士研究生入学考试试题参考答案4.2 东南大学（934）电路2009年硕士研究生入学考试试题及答案4.2.1 东南大学（934）电路2009年硕士研究生入学考试试题参考答案4.3 东南大学（934）电路2010年硕士研究生入学考试试题及答案4.3.1 东南大学（934）电路2010年硕士研究生入学考试试题参考答案4.4 东南大学（934）电路2013年硕士研究生入学考试试题（回忆版）1.网络问题，已知某一支路电流，求电压，把三角形化成星形用节点电压（好像是07 或年真题）2.黑箱问题，给了一个条件求电阻为何值事功率最大，也就是求戴维南等效，根据已知条件算出开路电压与电阻关系即可3 网络问题，已知某一电阻功率，和两个电源的变换，求支路电压，应用叠加定理算出，注意正负两个值4 功率补偿问题，以前的真题5 放大器问题正弦稳态电路6 动态电路问题已知电压电流响应求电容，和到达三分之一能量时的时间，以前的真题7 黑箱问题，求响应，10 年真题8 正弦稳态问题+理想变压器已知两电表读数相同，注意到并联谐振即可9 非线性电路问题求静态工作点时电容相当于开路10 非正弦周期问题求某电压和功率，叠加定理大题1。
陈燕书上的原题，应该也是04 年大连交通的题2。
03 年真题3.一阶电路问题，换路后右边是电感组成的一阶电路，左边由以右您所下载的资料来源于 考研资料下载中心获取更多考研资料，请访问边电感电流为受控辆的受控电压源直接接在电容两端，求某支路电流，计算时注意冲击响应4 三相电路问题，第一问求线电压，第二问是A 相故障求B，C 相电流5 非正弦周期问题，叠加定理，难点在于那个空心变压器，先戴维南化简下会简化计算6 三相电路的非对称电路，1 求线电压2 求开关断开时由来电感和三个电阻组成的回路组成的网络，应用磁链守恒和拉普拉斯都可以，和04 年真题三相电路有点像，比那题简单点。
初级电子基础试题及答案
初级电子基础试题及答案一、单项选择题（每题2分，共20分）1. 电子电路中，电流的方向是：A. 从电源正极流向负极B. 从电源负极流向正极C. 从负载流向电源D. 从电源流向负载答案：A2. 电阻的基本单位是：A. 欧姆（Ω）B. 伏特（V）C. 安培（A）D. 瓦特（W）答案：A3. 下列哪种元件不是被动元件？A. 电阻B. 电感C. 电容D. 二极管答案：D4. 在数字电路中，高电平通常用哪个电压表示？A. 0VB. 3.3VC. 5VD. 12V5. 一个完整的振荡器需要具备以下哪些条件？A. 放大器B. 反馈网络C. 电源D. 以上都是答案：D6. 晶体管放大电路中，基极电流增大时，集电极电流如何变化？A. 减小B. 增大C. 不变D. 先增大后减小答案：B7. 集成电路的英文缩写是：A. TTLB. CMOSC. ICD. CPU答案：C8. 以下哪个不是半导体材料？A. 硅B. 锗C. 铜D. 砷化镓答案：C9. 稳压二极管的主要功能是：B. 稳压C. 放大D. 开关答案：B10. 电容器的单位是：A. 法拉（F）B. 欧姆（Ω）C. 亨利（H）D. 西门子（S）答案：A二、填空题（每题2分，共20分）1. 欧姆定律的数学表达式是 V = _________。
答案：IR2. 一个电阻为10Ω，通过它的电流为0.5A时，电阻两端的电压是_________V。
答案：53. 电感的单位是 _________。
答案：亨利（H）4. 一个电容器的电容为4.7μF，其电容量是 _________法拉。
答案：4.7×10^-65. 三极管的三个主要极是发射极、基极和 _________。
答案：集电极6. 集成电路按制造工艺可分为双极型和 _________型。
7. 晶体管的放大作用是通过控制 _________极电流来实现的。
答案：基8. 一个完整的振荡器至少需要一个 _________和反馈网络。
东南大学历年电路试卷2010~2011期中A(答案)
t
15 5e 10 tV
9
t 0
六、图示电路中，已知 U=8V。求： （1） R 为何值时它消耗的功率为最大？并求出此功率。 （2） 求 R=12Ω时电流 I 和 I1 的值
I1 ＋ U － U － ＋ U － 16Ω 16Ω ＋ U － 8Ω 16Ω ＋ U － 8Ω 16Ω ＋ U － 8Ω 16Ω ＋ 8Ω 16Ω I R
i1 4A R S 10i1
i L
九、叙述并证明戴维宁定理。
共5页 第5页
共5页
第1页
i1 2 i1 2 A 20 20i 20i 10 10 I i 1A 2 3 2 7i1 20i2 7 20 3i3 4 i3 1A I 1A I i1 i2 CCVS 的功率为 P 10I i2 10 11 10W
Ix －
2Ω
U
2Ω 1.5U
共5页 第2页
五、在 t＝0 时刻，图中开关闭合，求电流 i(t)及电容的电压 u(t)。
t=0
5W
i
+ 20V -
2A
5W
+
u
- 400pF
解：开关闭合前，电容断V
所以
u0 u0 10V
开关闭合无限长时间后，电容断路处理，则
共5页 第3页
解：反复利用电源的等效变换，可得到除 R 之外的电路其它部分的戴维宁等效电路为：
8Ω ＋ 8V －
所以当 R=8Ω时，其消耗的功率最大，此功率为 (2)当 R=12Ω时，易求 I=8/(12+8)=0.4A 反复倒退计算，得 I1=0.0125A 七、求图中电阻能获得的最大功率。
东南大学电子电路基础考试样卷
东南大学考试卷课程名称电子电路基础考试学期08- 09- 3 得分适用专业测控技术与仪器考试形式半开卷考试时间长度120分钟1、单项选择题（共20分，每小题2分）①在PN结外加正向电压时，扩散电流_______ 漂移电流，当PN结外加反向电压时,扩散电流___ 漂移电流。
A.小于，大于B.大于，小于C.大于，大于D•小于，小于②在三极管放大电路中，下列等式不正确的是（）。
A. |E=|B I cB. | cC. I CBO 二（1 「）1 CEOD.:: :=：③下面的电路符号代表（）管。
A.耗尽型PMOSB.耗尽型NMOSC.增强型PMOSD.增强型NMOS④放大电路如图所示，已知三极管的 E -50，则该电路中三极管的工作状态为（）°A.截止B.饱和C.放大D.无法确定⑤放大电路A、B的放大倍数相同，但输入电阻、输出电阻不同，用它们对同一个具有内阻的信号源电压进行放大，在负载开路条件下测得A的输出电压小，这说明A 的（）。
A.输入电阻大B.输入电阻小C.输出电阻大D.输出电阻小⑥把差分放大电路中的发射极公共电阻改为电流源可以（）A •增大差模输入电阻B •提高共模增益C .提高差模增益D •提高共模抑制比⑦ 乙类互补对称功率放大电路（）A •能放大电压信号，但不能放大电流信号B •既能放大电压信号，也能放大电流信号C •能放大电流信号，但不能放大电压信号D •既不能放大电压信号，也不能放大电流信号⑧ 为了减小放大电路从信号源索取的电流并增强带负载能力，应引入 （ ）负反馈。
A.电压串联B.电压并联C.电流串联D.电流并联⑨已知变压器二次电压为U 2=：：j 2U 2sin’t v ，负载电阻为R L ，则桥式整流电路流⑩关于三极管高频参数，下列说法中不准确的为（ ）。
A.许二 0B. f ：. = （4：o ）f 1 C. 2、（15分）放大电路如图所示，已知电容量足够大， Vcc=18V , R Bi =75k 「R B 2=20^ 1,R E2=1.8k 「R EI =200」R C =8.2k 「R L =6.2k 「, R s =600「三极管的 3 =100 r bb ' =200门,U BEQ =0.7V 。
东南大学《电路基础》期末试卷B及答案
《电路基础》试卷B专业：姓名：共五道计算题，每题20分一、如图所示电路，a,b,c,d表示4个节点。
12R=Ω,26R=Ω,36R=Ω,求电压bU二、已知电路如图，已知1236,4,12R R R =Ω=Ω=Ω,4L R =Ω,利用戴维南定理，求该电阻上的电流i ，并画出中间过程等效电路图。
三、如图所示电路，已知()i t A =,4r =Ω， 2.4L H =，0.025C F =求电压u四、如图所示电路, 122,1R R =Ω=Ω,1,3L C X j X j =Ω=-Ω,.100()S U V =∠ 求:(1)各元件吸收的功率;(2)电源供给的功率.五、如图所示的对称三相负载，已知线电压380l U V =,负载阻抗68Z j =+,求各项负载电流和负载总功率。
试卷B 答案一、 解：由图可见，ab 间的电压6ab ad bd a b b U U U U U U =-=-=-所以 1622ab bU U I -== bc 间的电压(3)3bc bd cd b c b b U U U U U U U =-=-=--=+所以 2366bc b U U I +== 366bd b U UI ==对于节点b,根据KCL 有123I I I =+,将1I ，2I ，3I 代入上式，得 63266b b bU U U -+=+ 可解得 b U ＝3V二、解：利用戴维南定理(1)求等效电阻0R130213R R R R R R ⨯=++6124612⨯=++ 8=Ω(2)求端口开路电压OC U1212612120.54612612OC U U U ⨯⎛⎫=+=⨯+⨯+ ⎪++⎝⎭8412V =+=利用戴维南得原理图：所以 12184OC O L U i A R R ===++三、解：取电流i 的相量为：.20()I A =∠令未知电压相量为.0U .由于5/rad s ω=，根据各元件值可得 5 2.412()j L j j ω=⨯=Ω118()50.025j j C j ω==-Ω⨯ 可得各元件电压分别为：..42080()r U r I V ==⨯∠=∠ ..12202490()L U j L I j V ω==⨯∠=∠..0018201690()C U I j V j Cω==-⨯∠=∠- 由KVL,得 ....008024901690r L C U U U U =++=∠+∠+∠-0824168845()j j j V =+-=+= 将电压相量变换为正弦函数形式，得00()45)16cos(545)()u t t t V =+=+四、 解 (1)支路1的阻抗为121()Z j =+Ω,故电流..011100 4.7426.57()21S U I A Z j ∠===∠-+ 1R 吸收的有功功率和L 吸收的无功功率分别为1211214020varR L L P I R W Q I X ====支路2的阻抗为213()Z j =-Ω,故电流..022100 3.1671.57()13S U I A Z j ∠===∠-2R 吸收的有功功率和C 吸收的无功功率分别为22222210()30varR C C P I R W Q I X ===-=-(2)电路的总电流...0012 4.4726.57 3.1671.5751()I I I j A =+=∠-+∠=+ 电源供给的复功率: ..*(100)(51)5010()S S U I j j j VA ==+-=- 五、解 与线电压l U 相对应的相电压220P U V === 由于三相电路对称，因而中线两端电压'.0n n U =，根据置换定理，将'n n 短路。
精选-东南大学电子学院《电路基础》期末考试样卷及答案
Solve the following problems. (100 points)1、( 6 points)Find U ab in the circuit in Figure 1.Figure 1a b2、( 8 points)Find u o and i o in the circuit in Figure 2.u oFigure 2-3、( 6 points)In the circuit of Figure 3, readings of voltmeter ○V1，○V2and○V3 are 10V, 18V and 6V, respectively. Please determinate the reading of the voltmeter ○V.aFigure 34、( 8 points)The resonant or tuner circuit of a radio is portrayed in Figure 4, where u s1 represents a broadcast signal, given that R=10Ω，L=200μH, Us1rms=1.5mV，f1=1008kHz. If the circuit is resonant with signalus1, please determinate: (1) the value of C; (2) the quality factor Q of the circuit; (3) the current I rms; (4) the voltage U c rms.Figure 4u5、( 8 points) A balanced three-phase circuit is shown in Figure 5. Calculate the phase currents and voltages in the delta-connected load Z, if V U A ︒∠=•0220, Z =(24+j15)Ω，Z L =(1+j1)Ω.Figure 5•6、( 8 points) Find they parameters for the circuit in Figure 6.••Figure 67、( 15 points) For the circuit in figure 7, find the rms value of the current i and the average power absorbed by the circuit, given that Ω10=L ω,Ω51=R ,Ω202=R ,Ω20/1=C ω,V )302cos(50cos 10080)(︒+++=t t t u s ωω.Figure 72u s8、( 12 points) In the circuit of Figure 8, determinate the value of Z that will absorb the maximum power and calculate the value of the maximum power.Figure 8V ︒∠050ω=103•rms9、( 14 points) If the switch in Figure 9 has been closed for a long time and is opened at t = 0, find u c , i L , u k for t ≥ 0.20Figure 9u c +-10、( 15 points) The switch in Figure 10 has been in position a for a long time. At t = 0, it moves to position b. Find i 2(t) for t ≥ 0 and sketch it. (Attention: Please mark the key data in your graph.)Figure 1010ΩSolve the following problems. (100 points)1、( 6 points) Find U ab in the circuit in Figure 1.Figure 1a bSolution:U ab = 9V2、( 8 points) Find u o and i o in the circuit in Figure 2.u o Figure 2-Solution:V 4.2=o u , 0.256mA -=o i3、( 6 points) In the circuit of Figure 3, readings of voltmeter ○V1，○V2 and ○V3 are 10V, 18V and 6V, respectively. Please determinate the reading of the voltmeter ○V .aFigure 3Solution:The reading of the voltmeter ○V is 10V .4、( 8 points) The resonant or tuner circuit of a radio is portrayed in Figure 4, where u s1 represents a broadcast signal, given that R =10Ω，L =200μH , U s1rms =1.5mV ，f 1=1008kHz. If the circuit is resonant with signal u s1, please determinate: (1) the value of C ; (2) the quality factor Q of the circuit; (3) the current I rms ; (4) the voltage U c rms .Figure 4uSolution:pF 5.12)1(≈C 6.126)2(≈Q μA 150)3(=rms I mV 190)4(=Crms U5、( 8 points) A balanced three-phase circuit is shown in Figure 5. Calculate the phase currents and voltages in the delta-connected load Z, if V U A ︒∠=•0220, Z =(24+j15)Ω，Z L =(1+j1)Ω.Figure 5•Solution:A 3.69-11.74''︒∠=•B A I , V 31.28332''︒∠=•B A U6、( 8 points) Find the y parameters for the circuit in Figure 6.••Figure 6Solution:[]S y ⎥⎦⎤⎢⎣⎡=0.5 40.02 2.07、( 15 points) For the circuit in figure 7, find the rms value of thecurrent i and the average power absorbed by the circuit, given that Ω10=L ω,Ω51=R ,Ω202=R ,Ω20/1=C ω,V)302cos(50cos 10080)(︒+++=t t t u s ωω.Figure 72u sSolution:A 17.6357.2714.42.3222222120=++=++=I I I I rmsP =P 0+P 1+P 2=256+333.28+50=639.28W8、( 12 points) In the circuit of Figure 8, determinate the value of Zthat will absorb the maximum power and calculate the value of the maximum power.Figure 8V ︒∠050ω=103•rmsSolution:If Ω135jZ Z eq+==*, Z will absorb the maximum average power. The maximum average power is, W 7.2491356.1941412max=⋅=⋅=*eq OC R U P9、( 14 points) If the switch in Figure 9 has been closed for a long timeand is opened at t = 0, find u c , i L , u k for t ≥ 0.20Figure 9u c +-Solution:0 A,5.15.1)5.10(5.14040>-=-+=--t e e i t t L ， 0 V,60120)12060(12020/20/>-=-+=--t e e u t t C 0 ,V 6030902020/40>+--=-=--t e e u i u t t C L K10、( 15 points) The switch in Figure 10 has been in position a for a long time. At t = 0, it moves to position b. Find i 2(t) for t ≥ 0 and sketch it. (Attention: Please mark the key data in your graph.)Figure 1010ΩSolution: A 25.125.132t t e e i ---= 打发第三方（注：专业文档是经验性极强的领域，无法思考和涵盖全面，素材和资料部分来自网络，供参考。
电子技术基础考试试题
电子技术基础考试试题一、选择题（每题2分，共20分）1. 以下哪个不是基本的电子元件？A. 电阻B. 电容C. 电感D. 变压器2. 欧姆定律的公式是：A. V = I * RB. V = I / RC. I = V / RD. I = R / V3. 一个电路的总电阻为100Ω，当并联一个50Ω的电阻时，总电阻将：A. 大于100ΩB. 小于100ΩC. 等于100ΩD. 无法确定4. 以下哪个是半导体材料？A. 铜B. 银C. 硅D. 铁5. 一个电路的电流为2A，电压为10V，根据功率公式，该电路的功率是：A. 20WB. 5WC. 10WD. 40W6. 以下哪个不是数字逻辑门？A. ANDB. ORC. NOTD. XOR7. 一个正弦波的频率为50Hz，其周期是：A. 0.02秒B. 0.2秒C. 20秒D. 200秒8. 以下哪个是模拟信号？A. 数字电视信号B. 无线电广播信号C. 计算机数据信号D. 所有选项都是9. 一个电路的功率因数为0.8，其有效功率是视在功率的：A. 80%B. 100%C. 120%D. 200%10. 以下哪个是集成电路的分类？A. 模拟集成电路B. 数字集成电路C. 混合信号集成电路D. 所有选项都是二、填空题（每空2分，共20分）11. 电阻的单位是________。
12. 电流的国际单位是________。
13. 电压源的符号是________。
14. 一个电路的功率因数定义为________与________的比值。
15. 半导体材料的导电性介于________和________之间。
16. 逻辑门的输出状态只有两种，即高电平或________。
17. 一个周期性信号在一个周期内的平均值为________。
18. 模拟信号与数字信号的主要区别在于模拟信号是________变化的，而数字信号是离散的。
19. 集成电路按功能可分为________和________。
电子电路基础(1)--期中考试
东 南 大 学 考 试 卷课程名称电子电路基础（期中）考试学期得分适用专业06考试形式闭卷考试时间长度100分钟1、填空(20分)1.(3分) 双极型三极管是 （电流或电压）控制器件，当其工作在放大区时发射结需要加 偏置，集电结需要加 偏置。
2.(6分)电路如图1-1所示，三极管的U BE =0.7V ，β=50。
试分析①V BB =0.1V 时，三极管的工作状态__________和U CE =__________；②V BB =1V 时，三极管的工作状态__________和U CE =__________；③V BB =5V 时，三极管的工作状态__________和U CE =__________。
图1-1 图1-23.(1分)电路如图1-2所示。
电源电压u s 为正弦波，试绘出负载R L 两端电压u o 的波形。
设二极管是理想的。
4.(2分) 在有源滤波器中，运算放大器工作在 区；在施密特比较器中，运算放大器工作在 区。
共 8 页 第 1 页Ou5. (2分)测得一放大电路中三极管各极电压如图1-3所示，该管为________(NPN 还是PNP ？)，_________(硅管还是锗管？)。
图1-36.(2分) 场效应管的转移特性曲线描述了 特性，曲线上某点切线的斜率就是该状态下的_________。
7. (3分) 设某一阶有源滤波电路的电压放大倍数为.1001200A f j=+，则此滤波器为______（高通或低通）滤波器，其通带放大倍数为 ，截止频率为___________ Hz 。
8.(1分)试说明图1-4电路是什么类型的滤波器__________________。
共 8 页 第 2页①-8V-2.2V图1-4二、分析计算(80分)1. (25分) 放大电路如题图(a)所示，设三极管的β=140，R b1=2.5kΩ，R b2=10kΩ，Rc=2 kΩ，R L=1.5 kΩ，Re=750 Ω，Vcc=15V。
电子方面考试试题和答案
电子方面考试试题和答案一、单项选择题（每题2分，共20分）1. 电子电路中，三极管的主要作用是（）。
A. 放大信号B. 整流C. 滤波D. 振荡答案：A2. 在数字电路中，逻辑“与”门的输出为高电平的条件是（）。
A. 所有输入都为高电平B. 至少一个输入为高电平C. 所有输入都为低电平D. 至少一个输入为低电平答案：A3. 以下哪种元件不是被动元件？（）A. 电阻B. 电感C. 电容D. 二极管答案：D4. 集成电路中，TTL与CMOS相比，其主要缺点是（）。
A. 功耗大B. 速度慢C. 价格高D. 抗干扰能力差答案：A5. 在模拟电路中，运算放大器的理想工作状态是（）。
A. 饱和区B. 截止区C. 线性区D. 非线性区答案：C6. 以下哪种波形不是周期性波形？（）A. 正弦波B. 方波C. 三角波D. 锯齿波答案：D7. 晶体管的静态工作点设置不正确可能导致（）。
A. 截止失真B. 饱和失真C. 线性失真D. 谐波失真答案：A8. 在数字电路中，触发器的主要功能是（）。
A. 存储一位二进制信息B. 放大信号C. 滤波D. 整流答案：A9. 以下哪种材料不适合制作半导体器件？（）A. 硅B. 锗C. 铜D. 砷化镓答案：C10. 以下哪种元件不是模拟电路中的基本元件？（）A. 电阻B. 电容C. 二极管D. 逻辑门答案：D二、多项选择题（每题3分，共15分）11. 以下哪些因素会影响三极管的放大倍数？（）A. 温度B. 电源电压C. 基极电流D. 集电极电流答案：A, C12. 在数字电路中，以下哪些逻辑门可以实现逻辑“或”功能？（）A. 与非门B. 或门C. 异或门D. 同或门答案：B, D13. 以下哪些元件属于有源元件？（）A. 电阻B. 三极管C. 电感D. 运算放大器答案：B, D14. 在模拟电路中，以下哪些因素可能导致三极管损坏？（）A. 过热B. 反向偏置C. 过压D. 反向电流答案：A, C15. 以下哪些波形属于非正弦波形？（）A. 正弦波B. 方波C. 三角波D. 锯齿波答案：B, C, D三、填空题（每题2分，共20分）16. 在数字电路中，逻辑“非”门的输出与输入的关系是________。
东南大学电工电子学试卷
共 45 页 第 1 页东 南 大 学 考 试 卷（A 卷）课程名称电工学 考试学期 11-12-2得分适用专业 机械工程及自动化 考试形式 闭卷 考试时间长度120分钟1)Birds routinely land and relax on power lines whichcarry tens of thousands of volts of electricity. Explain why these birds do not get electrocuted. (6 pts)The resistance of the air is too large for current to flow from the line to the bird to the ground.Alternately, the resistance is too large for current to divert from the wire, into the bird, and then back into the wire. It is not because of wire insulation. Birds could land on uninsulated wires just fine.6 pts for right answer. 1 pt for something vaguely correct.2) Suppose your car battery is dead, and you need to charge it using another car’s battery. You have a pair of cables 电缆which you can use to connect the terminals终端of the batteries. In order to charge your car battery, should you connect the same terminals (positive of one battery to the positive of the other, and the same with the negative) or the opposite terminals (positive of one battery to the negative of the other battery)? Why did you choose this configuration配置；结构? (6 pts)First note that real batteries have internal内部的resistance, so the universe领域will not explode爆炸if you directly connect two car batteries.Next, if the opposite terminals are connected, then P=VI and common sense tells us that both batteries will be supplying power, and the internal resistances will be consuming消耗power, so clearly no power is being delivered 递送by either battery. If the same terminals are connected,共45 页第 2 页then we can see that if one battery is of a higher voltage than the other, it will provide power to theother battery. Key to this realization实现is to know that battery voltages drop下降；终止as the battery is depleted耗尽的.6 pts for right answer. 2 pts if you try to say there is no current if two batteries are connected + to+ because voltage sources are perfectly balanced.3) A standard procedure for testing the internal resistance of a battery is the “dual双重的pulse脉冲” test. We first attach an ideal 5 mA current source between the terminals of the battery, so that current flows in the usual direction (positive to negative), and measure the voltage across the battery terminals. We then remove the 5 mA source, and attach an ideal 505 mA current source instead,and again measure the battery terminals. (6 pts)共45 页第 3 页i.Suppose that we find a 1.485V voltage with the 5 mA source, and a 1.385 with the 505 mA source, what is the internal resistance?Easy way to solve these is to subtract减去；扣掉 2nd from first, giving, or .3 pts for right answer. 2 pts if sign error or algebra 代数学mistake.ii. Assuming the battery is perfectly linear 线型的(i.e. accurately modeled 模式化；被效仿by a Thevenin 戴维宁equivalent), is it possible to find the voltage provided by the battery with no load attachedusing the data above? If so, what is it? If not, why not?共45 页第 4 页Plug塞住；用插头将与电源接通 in to one of our equations above, e.g., giving , or finally. If you did it with the current source the other way, you should still have gotten 1.486V.3 pts for right answer.Find the Thevenin or Norton equivalent circuit model.(12 pts ).共45 页第 5 页(6 pts )(6 pts )Find the Voltage v and the currents i1 and i2 for the circuit shown below. (10 pts )共45 页第 6 页SolutionLabel标注 the meshes；网状物:Since i3=2 it is not a variable.Writing KVL around the first mesh:24*(i2-i3) + 12*i1 = 0Writing KVL around the second mesh:12*i2 + 6*(i2 –i3) = 0Substituting代替，取代 for i3 and rearranging 36*i1 = 4818*i2 = 12共45 页第7 页Namelyi1 = 4/3 Ai2 = 2/3A7. Find the current i for the circuit shown below (10 pts )SolutionZeroing the 1A source gives共45 页第8 页By the current division principle, the contribution of the 2A source to i1 is2*(-1/30) / (1/30+1/5) = -2/7A.Zeroing the 2A source givesBy the current division principle, the contribution of the 1A source to i1 is1*(1/15)/(1/15+1/20) = 4/7A.Hence i1 = 4/7-2/7 = 2/7A.共45 页第9 页Find the Thevenin and Norton equivalent circuits across terminals a and b for the following circuitWith an open circuit across terminals a and b the current through the 20 resistor must be 0.5ix and the voltage across it is then 10ix. ByKVL we also get30V = 5ix + 10ixSo ix = 2A and we calculateVTH = 10ix = 20VTo find RTH we zero the independent独立的 source to get the circuit in Figure 4.We wish to determine what resistance this circuit isequivalent 等价的to across共45 页第10 页the terminals a and b. One way to do this is to imagine connecting anindependent voltage source of voltage VS across the terminals and findingthe current drawn from this voltage source. We therefore consider thecircuit in Figure 5.KVL gives−5Ωix = VS → ix = −VS/5Ω A (1)KVL also gives20Ωi1 = VS → i1 =VS/20 Ω (2)KCL givesi = i1 + 0.5ix − ix = i1 − 0.5ix (3)共45 页第11 页Plugging (1) and (2) into (3) yieldsi =VS/20Ω+VS/10Ω=3VS/20ΩEventually,RTH =VS/i=20/3 ΩThe Thevenin equivalent circuit is depicted描述； in Figure 6.From this we get the Norton equivalent circuit in Figure 7 where we usedWe could have also determined IN by applying a short circuit across theterminals a and b and computing 计算；处理the current through the short circuit短路.This involves analyzing the circuit in Figure 8. Here wehave by KVL共45 页第12 页5ix = 30V → ix = 6AThis gives, by KCLi = ix − 0.5ix = 0.5ix = 3Awhich matches the earlier calculation计算；估计.Suppose that v1(t) = 80 cos (ωt) and v2(t) = 60 sin (ωt). Use phasors相量 to reduce the sum vs(t) =v1(t) + v2(t) to a single 单一的term of the form Vm cos (ωt +θ). Draw a phasor diagram, showing V1, V2, and Vs. State the phase relationships between each pair of these phasors. (16共45 页第13 页pts )The phasor corresponding tov1(t) = 80 cos (ωt)isV1 = 80∠0 = 80Equivalently, we get forv2(t) = 60 sin (ωt) + 60 cos (ωt −π/2)the phasorV2 = 60∠−π/2 = −j60To find the sum using phasors we add the complex numbers which yieldsV = V1 + V2 = 80 − j60 = 100e−j arctan 3/4This corresponds符合，一致 to the time signalvs(t) = 100 cos (ωt − jarctan3/4)The phasor diagram is depicted描述 in Figure 2.Note that V1 leads V2 by π/2 and leads Vs by arctan反正切 ( 4/3 ).共45 页第14 页Find an expression 表达式for v(t) of the form Vm cos (ωt +θ) when v(t) = v1(t) + v2(t) + v3(t) + v4(t) withv1(t) = 20 sin (ωt)v2(t) = 20 cos (ωt +π/6)v3(t) = 20 sin (ωt +π/3)v4(t) = −10 cos (ωt)Use phasors. (6 pts )Solution:The phasors corresponding 相应的to each component 成分；组件；[电子] 元件signal are given in Table表格 1. The complex number equivalent复数等效to each phasor is given as well.共45 页第15 页Adding all complex numbers giveswhich corresponds to the phasorwhich can be transformed变形 into the time signalFind the complex impedance阻抗 in polar form 点斜式极坐标形式极形式of the network shown below for ω = 10001/s , ω= 20001/s ,and ω = 40001/s . (6 pts )共45 页第16 页(5 pts)Both capacitors电容器 and inductors电感器 can be used to store energy for use later. Capacitors, though, are far more popular for energy storage than inductors, because they are easier to keep charged带电荷的带电的. In lab, for example, you can easily connect up a capacitor to a battery, disconnect it from the source, carry it around in your pocket to grab lunch午餐, go back to lab, and then find that it's still holding a charge even after a long time. Explain why this is relatively difficult with inductors.共45 页第17 页(5 pts) If you've ever had the batteries on a flash li ght闪光灯；手电筒 run out,you know that as soon as the ligh t starts dimming调光；变暗,your battery will very soon be dead.This is because a battery supplies its maximum vo ltage峰点电压 until right before it is depleted耗尽的.On t he axis 坐标轴on the left,I have plotted标绘的 the brightn ess亮度辉度 of a battery powered flashlight over the lif e of the battery.We can also power a flashlight using共45 页第18 页a capacitor.Assuming the brightness of a bulb 电灯泡is p roportional比例的 to the voltage across it,on the right axis,draw (qualitatively定性地) the brightness vs. Time for a capacitor powered flashlight.Draw so that bright ness is 1 at time=0,and has dropped to at least 0.1 by time=1.We just want the shape形状,not necessarily the exact values.c) (5 pts) A student is not satisfied with the amount of共45 页第19 页energy stored in his one-capacitor circuit. He wants to store more energy, so puts 10 capacitors in series串联 and charges them with a single voltage source. Is this a good idea? Why or why not?Find the Thevenin equivalent 戴维南定理of the circuit电路below at the two terminals on the far right:共45 页第20 页e) Find the Thevenin equivalent of the circuit below between terminals and [Think about the various algorithms算法 we’ve used to find Thevenin equivalents.If you get stuck被困住了,complete the rest of these tricky 棘手的复杂的problem and come back to this one and maybe some inspiration will reach you]:共45 页第21 页共45 页第22 页For the circuits below, please find expressions for the specified voltage表达式指定的电压over the indicated time 表示时间ranges in terms of the circuit parameters电路参数. Plot 绘图 the waveform 波形图on the provided axes轴线, and clearly identify. 确定 the key parameters参数 in your graph图表. (3A) Consider the circuit of Fig. 4. The switch is open for t <0, closed for 0≤t < t1, and open for t ≥t1, where t1 = 3RC. Find and plot the voltage v C (t).共45 页第23 页(3B) Consider the circuit of Figure 5, in which > -1.共45 页第24 页The switch is open for t <0, and closed for t 0. vC(0-) = V0. Find and plot the voltage v1(t).共45 页第25 页The circuit shown in Figure 4 has been at rest with theswitch open for a long time. At t = O the switch is closed.共45 页第26 页Sketch 素描；略图；梗概the current ic集成电路 through the capacitor for t > O on the axes below. Using the circuit parameters R1, R2, C, and Vo,indicate on your sketch (i) the initial value初始值 ic(Ot), (ii) the final value ic(oo), and (iii) the time constant时间常数.共45 页第27 页Problem 1F: 5 pointsThe capacitor in the circuit in Figure 6 has an initial初始最初的voltage V O on its terminals at t =0−when a step of voltage Vu(t) is applied at t = 0. Find an expression for the voltage across the capacitor v C for t> 0.共45 页第28 页Consider the current divider circuit分流电路 shown below:共45 页第29 页1a. How many branches分支 are there in the circuit?1b. How many nodes节点 are there in the circuit?2. Derive an expression派生一个表达式 for the output current 输出电流(Iout, shown above) as a function of theinput current (Iin) and the two resistors (R1 and R2). Use any technique you’d like.3. How should the resistors be chosen such that twice as much current flows throughR1 compared to R2?4. Using the equation 方程等式you obtained in #2, show what happens to Iout输出电流 if R2 is replacedwith a short circuit (i.e. R2 = 0). Repeat if R2 is replaced with an open circuit.(think about what R2 should be replaced with here) Now suppose the following values are used: Iin = 3 A, R1= 5 , R2 = 20 , and we wish to find the Thevenin共45 页第30 页equivalent circuit across the two indicated terminals below:5. Find Voc and Isc across the two terminals indicated on the circuit.6. Draw the Thevenin equivalent circuit for the two terminals indicated above.7. Suppose a load is now attached across the two terminals of the Theveninequivalent circuit:a. Write an equation for I in terms of V.b. Sketch the IV characterstics伏安特性. Use I as the“y-axis Y轴” and V as the“x-axis”. Label标注all relevant points相关点.共45 页第31 页8. Suppose the load has an IV characteristic given as follows:a. What are the values of I and V if the load is attached to the Thevenin equivalent circuit ] 等效电路?b. Looking at the IV characteristic, what is the load? (describe the load as much as you can)For problems 9 through 12, consider the following circuit:9. Using KVL and mesh analysis, write down the equations needed to solve the circuit. Convert转变the resistor voltages into currents.10. The dependent source requires one additional equat共45 页第32 页ion附加方程to be generated产生. Write down that equation by looking at the 3 resistor.11. Solve the equations simultaneously同时地to determine the two mesh currents网孔电流.12. To check your answer, confirm确认that power is conserved 守恒in the circuit. Find the power dissipated浪费的in each resistor. Find the power generated生成的(or dissipated in each source). Be clear in your answers whether power is being generated or dissipated in each circuit element. For problems 1–7, consider the following circuit:1. Find the transfer转移function, i.e. Vout输出电压()/Vin(), in terms of R, L, and .2. What is the magnitude大小of the transfer function传递函数, i.e. | H() |, in dB?3a. What is | H() | equal to (in dB) for very low共45 页第33 页frequencies频率?3b. What happens to | H() | at high frequencies? What kind of a filter滤波器do we have here?4. Derive the equation for the break frequency推导出方程为截止频率. The break frequency is definedas the frequency where | H() | = –3.0103 dB.5. Sketch the magnitude Bode波特图plot for this filter滤波器. Label标注all critical points andslopes. Show how you obtained the value for any non-zero slopes零斜坡.6. Suppose R = 1 , L = 2 H, and Vin = 3 cos(5t + 30o). Find Vout(t).7. Draw a phasor diagram, indicating the voltage and current phasors of all thecircuit elements电路参量.For problems 8–12, consider the following circuit:共45 页第34 页8a. Find the current that appears on the output side ofthe transformer变压器.8b. What is the voltage that appears across the capacitor. (phasor form is acceptable,assume an angular frequency角频率of 1 rad/s)9. Find the voltage that appears on each side of the transformer.10. Find the voltage that appears across the current source.11. Replace the transformer and input circuitry with an equivalent current sourceand inductor. In other words, draw a circuit which will have a current source,an inductor, and a 3 F capacitor all in series. Label the values for the current source and the inductor.Check your answer to #11 by comparing the amount of power共45 页第35 页generated by the currentsource in each circuit.12. Find the amount of power generated by the current source in the new circuit.Compare this with the amount of power generated by the current source in theoriginal circuit原电路 (they should be the same)For problems 4–7, consider the following circuit:4. Find the transfer function, i.e. Vout ()/Vin (), in terms of R, L, and (or f).5. What is the magnitude of the transfer function, i.e. | H() |?共45 页第36 页6. What is | H() | equal to for very low frequencies? For very high frequencies?What happens at resonance谐振? (i.e. = o =LC1) What kind of a filter滤波器do we have here?7. Sketch the magnitude Bode plot for this filter. Label all critical points andslopes斜率. Use R = 2 , L =41H, and C =81F. Show how you obtained the valuefor any non-zero slopes.In order to measure the voltage across a resistor in a circuit, a voltmeter电压表is used.Although we assume an ideal voltmeter in most cases, a real voltmeter has a finite有限的resistance (R3) which can have an effect on your circuit. Consider the followingmeasurement on a voltage divider分配共45 页第37 页Vin, R1, and R2 are just the standard elements of a voltage divider.1a. How many nodes are there in the circuit?1b. How many meshes are there in the circuit?2. Write a KCL equation at node A. Convert the currents into voltages.3. Solve the equation for Vout, so that you have an equation for Vout based onVin, R1, R2, and R3.4. Show that as R3 approaches infinity无限大(i.e. R3 ), you get the standardequation for a voltage divider.For problems 5 through 11, consider the following circuit:共45 页第38 页Find the Norton equivalent circuit, as follows:5. Short circuit the two terminals and redraw the circuit without the 2 resistor.Explain why the 2 resistor can be dropped from the circuit.放弃了电路6. Use node analysis to find Isc. (You will need to come up with one additionalequation to eliminate消除ix and get a numerical数字的answer) 7. Now consider the circuit when the two terminals are open circuited. Using meshanalysis, find ix under these conditions.8. Using your results from #7 (or using node analysis, which will take more time),find Voc, the voltage across the two terminals in an open circuit condition.共45 页第39 页9. Confirm that power is conserved in the circuit under open-circuit conditions.Find the power dissipated in each resistor. Find the power generated (ordissipated in each source). Be clear in your answers whether power is beinggenerated or dissipated in each circuit element.10a. Draw the Norton equivalent circuit. Label values for all circuit elements.10b. Draw the Thevenin equivalent circuit. (Note: you already have everything youneed)11a. Assume that a circuit element (that draws some current I and has somevoltage drop V) is attached to the terminals of the Norton equivalent circuit.Using KCL, write an equation for I in terms of V.11b. Graph the IV relationship (i.e. the load line). Use I as the “y-axis” and V as the共45 页第40 页“x-axis”. Label all relevant points.For problems 1–5, consider the following circuit:Use Vin = 100 v, R1 = 1000 , R2 = 250 , R3 = 1000 , and C = 1 F. At t = 0, theswitch is moved down, disconnecting R1 and connecting R2 to the circuit. We want to find the voltage v(t), and current i(t), across the capacitor as a function of time. 1a. Redraw the circuit, showing how the circuit looks just before the switch ismoved. Assume a long time has already elapsed before the switch is moved.1b. What is the continuity连续性condition for this circuit共45 页第41 页at t = 0?2. Using the circuit generated in #1a, find v(0–) and i(0–)3a. Find v(). Explain how you get this result.3b. Write a KCL equation for the circuit after the switch as been moved at t = 0.Convert the currents into voltages. (you should end up with a first-orderdifferential equation for v(t))4a. What form should the solution for v(t) take?4b. Substitute替代this solution into the differential equation generated in #3b. Use theresults of the substitution, along with the results for #1b and #2 to find all theconstants defined for v(t). (Clearly show your work) 5. Sketch curves of i(t) and v(t) as a function of time (t ). Indicate allcritical and asymptotic values.For problems 6–10, consider the following second-order共45 页第42 页circuit:The switch is opened at t = 0. The goal is to find the current through the inductor, i(t).6a. Assume that the switch has been closed for a long period of time. Find i(0–).6b. What should i(0+) be? Explain.7a. Find i(). Explain how you get this result. Draw the corresponding相应的circuitat this steady state condition.7b. Since this is a second-order circuit, another initial 初始/boundary condition is needed.Find that other value. Show your work.8. Derive导出the second-order differential equation for this circuit. Start by writing aKVL equation for the circuit after the switch has been opened. Then convert共45 页第43 页voltages in each passive element into currents.9a. What form does the solution for i(t) take?9b. Substitute your solution into the differential equation. Solve the correspondingcharacteristic equation.相应的特征方程10. Finish determining i(t) by using the results in #6b and #7b to find any constantsyou have defined in your solution for i(t).For problems 1–5, consider the following circuit:Use Vs = 5 v, R1 = 5 , R2 = 10 , and L = 1 H. After a long time has elapsed, theswitch is opened at t = 0. We want to find the voltage v(t), and current i(t), across共45 页第44 页R2 as a function of time.1a. Redraw the circuit, showing how the circuit looks just before the switch isopened.1b. Using this result, find v(0-).2a. What is the continuity/boundary condition for this circuit at t = 0?2b. Find the numerical value of the boundary condition.3. Write a KVL equation for the circuit after the switch as been opened at t = 0.(note: do not use the circuit you drew in #1a). Convert the voltages into currents.4. Solve the differential equation generated in #3 for i(t). Clearly show how youobtained all of the numerical values in i(t).5. Sketch curves of i(t) and v(t) as a function of time (t ). Indicate allcritical and asymptotic渐近的values.共45 页第45 页。
电路基础题库及答案详解
电路基础题库及答案详解一、选择题1. 电路中最基本的元件是：A. 电阻B. 电容C. 电感D. 电源答案：A2. 欧姆定律描述的是：A. 电流与电压的关系B. 电流与电阻的关系C. 电压与电阻的关系D. 电流、电压和电阻的关系答案：D3. 串联电路中，总电阻等于：A. 各电阻之和B. 各电阻的倒数之和C. 各电阻之积D. 各电阻的平方和答案：A4. 并联电路中，总电阻的倒数等于：A. 各电阻的倒数之和B. 各电阻之和C. 各电阻之积的倒数D. 各电阻的平方和的倒数答案：A5. 一个电路中，如果电流为2A，电阻为5Ω，则电压为：A. 10VB. 5VC. 2VD. 0.4V答案：A二、填空题6. 电流的单位是________。
答案：安培7. 电压的单位是________。
答案：伏特8. 电阻的单位是________。
答案：欧姆9. 欧姆定律的公式为________。
答案：V = IR10. 串联电路中，总电阻R等于各电阻R1, R2, ..., Rn之和，即R = R1 + R2 + ... + Rn。
三、简答题11. 什么是基尔霍夫电流定律（KCL）？答案：基尔霍夫电流定律指出，在一个电路节点上，流入节点的电流之和等于流出节点的电流之和。
12. 什么是基尔霍夫电压定律（KVL）？答案：基尔霍夫电压定律指出，在一个闭合电路中，沿着闭合路径的电压降之和等于电压升之和。
四、计算题13. 一个由三个电阻串联组成的电路，R1 = 10Ω，R2 = 20Ω，R3 =30Ω，求总电阻。
答案：总电阻R = R1 + R2 + R3 = 10Ω + 20Ω + 30Ω = 60Ω。
14. 如果上述电路两端的电压为120V，求通过电路的总电流。
答案：根据欧姆定律，总电流I = V / R = 120V / 60Ω = 2A。
五、应用题15. 一个并联电路由两个电阻组成，R1 = 4Ω，R2 = 6Ω，求总电阻。
答案：总电阻 R = 1 / (1/R1 + 1/R2) = 1 / (1/4Ω + 1/6Ω)= 2.4Ω。
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东 南 大 学 考 试 卷
课程名称 电子电路基础 考试学期 08-09-3
得分
适用专业 测控技术与仪器
考试形式
半开卷
考试时间长度 120分钟
1、单项选择题（共20分，每小题2分）
①在PN 结外加正向电压时，扩散电流 漂移电流，当PN 结外加反向电压时，扩散电流 漂移电流。
A. 小于，大于
B. 大于，小于
C. 大于，大于
D. 小于，小于
②在三极管放大电路中，下列等式不正确的是（ ）。
A.C B E I I I +=
B. B C I I β≈
C. CEO CBO I I )1(β+=
D. βααβ=+ ③下面的电路符号代表（ ）管。
A.耗尽型PMOS
B.耗尽型NMOS
C.增强型PMOS
D.增强型NMOS
④放大电路如图所示，已知三极管的05=β，则该电路中三极管的工作状态为（ ）。
A. 截止
B. 饱和
C. 放大
D. 无法确定
⑤放大电路A 、B 的放大倍数相同，但输入电阻、输出电阻不同，用它们对同一个具有内阻的信号源电压进行放大，在负载开路条件下测得A 的输出电压小，这说明A 的（ ）。
A. 输入电阻大
B. 输入电阻小
C. 输出电阻大
D.输出电阻小
⑥把差分放大电路中的发射极公共电阻改为电流源可以（ ）
A ．增大差模输入电阻
B ．提高共模增益
C ．提高差模增益
D ．提高共模抑制比 ⑦乙类互补对称功率放大电路（ ） A ．能放大电压信号，但不能放大电流信号 B ．既能放大电压信号，也能放大电流信号 C ．能放大电流信号，但不能放大电压信号
D ．既不能放大电压信号，也不能放大电流信号
⑧为了减小放大电路从信号源索取的电流并增强带负载能力，应引入（ ）负反馈。
A. 电压串联
B. 电压并联
C. 电流串联
D. 电流并联 ⑨已知变压器二次电压为t U u ωsin 222=V ，负载电阻为R L ，则桥式整流电路流过每只二极管的平均电流为（ ）。
A. L R U 29
.0 B. L R U 2
C. L R U 2
45.0 D. L
R U 2
2 ⑩关于三极管高频参数，下列说法中不准确的为（ ）。
A. ββf f T 0=
B. βαβf f )1(0+=
C. βf f T >>
D. βαf f <
2、（15分）放大电路如图所示，已知电容量足够大，Vcc=18V ，R B1=75k Ω，R B2=20k Ω，R E2=1.8k Ω，R E1=200Ω，R C =8.2k Ω，R L =6.2k Ω，R S =600Ω，三极管的β=100，Ω=200'bb r ，U BEQ =0.7V 。
试：（1）计算静态工作点（I BQ 、I CQ 、U CEQ ）；（2）画出放大电路的小信号等效电路；（3）计算电压放大倍数A u 、输入电阻R i .和输出电阻R o 。
（4）若)mV ( sin 15t u s ω=，求u o 的表达式。
3、（8分）判断图中电路引入了哪些反馈(包括局部反馈)；指出反馈元件，说明是正反馈还是负反馈？是直流反馈还是交流反馈？若为交流反馈请说明反馈类型。
u o
cc
+
-
4、（9分）图示电路中，运放为理想器件，其最大输出电压Uom=±14V ，稳压管的稳压值U Z =±6V ，t=0时刻电容C 两端电压u C =0V ，试求开关S 闭合时电压u O 1、u O 2、u O 3的表达式。
5、（8分）某滤波器传输函数为7
10
10.5f
j U U A i
o u ＋=
=
∙
∙
∙
，试分析：（1）求转折点频率；
（2）求通带增益分贝值；（3）画出其幅频特性渐近波特图；（4）此为 通滤波器。
U 1=1V
6、（8分）晶体振荡电路如图所示，试画出交流通路，说明它属于哪种类型的晶体振荡电路，并指出晶体在电路中的作用，求振荡信号的频率。
7、（9分）单相桥式整流电容滤波稳压电路如图所示，若u 2=24sin ωt V ，稳压管的稳压值U Z =6V ，试： 1．求U o 的值。
2. 若电网电压波动（u 1上升），说明稳定输出电压的物理过程。
3．若电容C 断开，画出u I 、u o 的波形，并标出幅值。
U I
U O
8、（9分）放大电路如下图所示。
已知场效应管的跨导为g m ，三极管的电流放大倍数为β，电容值足够大，试：
（1）画出放大电路的小信号等效电路（设电容对于交流信号可视为短路）。
（2）写出放大电路电压放大倍数、输入电阻和输出电阻的表达式。
-o
u CC
V +++
-
9、（8分）电路如图所示，试画出输出电压u O和电容C两端电压u C的波形，求出它们的最大值和最小值以及振荡频率。
-
+
10、（6分）电路如下图所示，试分析：
(1)为了使u O引回到V2基极的反馈成为负反馈，请在图中标出运放
A的同相端和反相端；
(2)在接成负反馈的情况下，欲使电路的A u f=60，则R f=?
(3)在上述情况下，若运放的A u o变化10%，问A u f的值变化多少？
O。