东南大学单级低频电压放大电路(基础)(DOC)

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 o Figure 2-3、( 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 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 , 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 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 the yparameters 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 markthe 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 the current i and theaverage 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 Z that will absorbthe 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 time and isopened 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.。

实验三单级低频放大电路1．实验目的（1）研究单管低频小信号放大电路静态工作点的意义。

（2）掌握放大电路静态工作点的调整与测量方法。

（3）掌握放大电路主要性能指标的测试方法。

2．实验涉及的理论知识和实验知识本实验体现了三极管的工作原理、放大电路的静态工作点调试方法以及放大器性能指标的基本测试方法。

3．实验仪器信号发生器、示波器、直流稳压电源、电压表4．实验电路实验电路如图3.1.1所示。

图中电位器R W是为调节晶体管静态工作点而设置的。

O图3.1.1单级共发射极放大电路5. 实验原理在电子系统中，放大电路是信号处理的基本电路。

其作用是将微弱信号增强到所需要的数值，单级低频放大电路是放大电路中最基本的结构形式，是组成各种复杂电路的单元和基础。

因此它的分析方法、电路调整技术以及参数的测量方法等具有普遍意义。

实验电路采用由NPN型硅材料三极管以及若干电阻、电容组成的共发射极放大电路，以图3.1.1所示电路为例进行研究。

（1）电路组成原则放大是最基本的模拟信号处理功能，它是通过放大电路实现的，电子技术里的“放大”有两方面的含义。

一是能将微弱的电信号增强到所需要的数值，即放大电信号，以便于测量和使用。

二是要求放大后的信号波形与放大前的波形的形状相同，即信号不能失真，否则就会丢失要传送的信息，失去了放大的意义。

因此，电路组成原则是首先要给电路中的晶体管施加合适的直流偏置，即发射结正偏、集电结反偏，使其工作在放大状态，而且还要有一个合适的工作电压和电流，即合适的静态工作点。

其次要保证信号发生器、放大电路和负载之间信号能够正常传输，即有u i时，应该有输出响应u o。

1）直流偏置原则图3.1.1所示电路采用的是电阻分压式偏置方法，通过基极偏置电阻R B1和R B2对U CC分压，获得晶体管的基极电压U BQ ，保证晶体管的发射结正偏。

U CC 是集电极电源，它通过R C 加至晶体管的集电极，保证晶体管的集电结加反向电压。

单片机笔记
Vfd荧光显示管
片内固话应用软件和系统软件
芯片制造工艺80c51低功耗
4krom
125字节RAM
两个十六位定时/计数器
64k字节总线扩展控制
可编程i/o口4*8位
可编程串行口
程序计数器16位对64kROM直接寻址PC第八位由p0输出，高八位由p2输出MCS-51高增益反相放大器
输出端XTAL1,外接晶振
输入端XTAL2，外部振荡器
32条可编程I/O口线
4个八位并行I/O端口
五个中断，两个优先级
串行通信接收RXD
串行通信发送TXD
中断申请INTO0
中断申请INTO1
外部RAM写选通WR
外部RAM读选通RD
EA=1从内部ROM开始
EA=0从外部ROM开始PROG编程脉冲输入端
PSEN访问外部程序存储器选通外部中断，0,1
T0,1，2中断
串行口中断
专用功能寄存器区sfr锁存器，定时器，串行数据缓冲器，各种控制寄存器，状态寄存器
程序状态控制字rsw
数据指针dptr，访问外部ram。

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。

单级低频电压放大电路第一篇：单级低频电压放大电路单片机笔记 Vfd荧光显示管片内固话应用软件和系统软件芯片制造工艺80c51低功耗 4krom 125字节RAM 两个十六位定时/计数器 64k字节总线扩展控制可编程i/o口4*8位可编程串行口程序计数器16位对64kROM直接寻址 PC第八位由p0输出，高八位由p2输出 MCS-51高增益反相放大器输出端XTAL1,外接晶振输入端XTAL2，外部振荡器32条可编程I/O口线4个八位并行I/O端口五个中断，两个优先级串行通信接收RXD 串行通信发送TXD 中断申请INTO0 中断申请INTO1 外部RAM写选通WR 外部RAM读选通RD EA=1从内部ROM开始 EA=0从外部ROM开始 PROG编程脉冲输入端PSEN访问外部程序存储器选通外部中断，0,1 T0,1，2中断串行口中断专用功能寄存器区sfr锁存器，定时器，串行数据缓冲器，各种控制寄存器，状态寄存器程序状态控制字rsw 数据指针dptr，访问外部ram第二篇：单级低频放大电路实验三单级低频放大电路1．实验目的（1）研究单管低频小信号放大电路静态工作点的意义。

（2）掌握放大电路静态工作点的调整与测量方法。

（3）掌握放大电路主要性能指标的测试方法。

2．实验涉及的理论知识和实验知识本实验体现了三极管的工作原理、放大电路的静态工作点调试方法以及放大器性能指标的基本测试方法。

3．实验仪器信号发生器、示波器、直流稳压电源、电压表4．实验电路实验电路如图3.1.1所示。

图中电位器RW是为调节晶体管静态工作点而设置的。

RB1 RW 100kΩ Ucc(+12V)RC 2kΩ Rs + 3kΩ usRB2 18kΩ R1 6.2kΩ 9013 + C2 10μF + + RL 2kΩ uO CE 47μF RE 1kΩ 图3.1.4 输入电阻测量原理图图3.1.5 输出电阻测量原理图在被测放大电路前加一个电阻R，输入正弦信号，用示波器分别测量R两端对地的电压us和ui。

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 o Figure 2-3、( 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 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 , 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 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 the yparameters 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: Pleasemark 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 the current i andthe 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 Z that willabsorb 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 time and isopened 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: Pleasemark the key data in your graph.)Figure 1010ΩSolution: A 25.125.132t t e e i ---=打发第三方。

实验7 单管低频电压放大电路实验目的1.认识实验所用的电子元器件（双极型晶体管、电阻、电位器和电容器等）。

2.学习看图接线，能熟练使用常用电子测量仪器。

3.学习单管电压放大电路的基本测试方法。

4.了解双极型晶体管电压放大电路中引入负反馈后对其工作性能的影响。

实验原理1. 单管低频电压放大电路介绍Oe CC图3.7a.1单管低频电压放大电路阻容耦合分压偏置共发射极电压放大电路如图3.7a.1所示。

该电路中的双极型晶体管T 是电路中的放大器件，它能把输入回路（基极—发射极）中微小的电流信号在输出回路中（集电极—发射极）放大为一定大小的电流信号。

输出回路中得到的较大输出电流是源自直流电源，双极型晶体管在电路中实际上起着电流控制作用。

电源提供放大电路能量，还为双极型晶体管的集电极提供反向偏置，使其处于放大工作状态；并通过基极电阻R CC U B1 和R B2的分压，提供合适的基极电压，调节电位器R P 的阻值可以改变基极电流，从而改变集电极电流。

集电极电阻R C 可以将集电极电流的变化变换为集电极电压的变化，在输出回路中得到放大的电压信号。

发射极电阻R E 对集电极电流的直流分量有负反馈的作用，稳定了静态工作电流。

发射极电容C E 对集电极电流的交流分量提供了交流通路，起了分流交流作用。

C 1、C 2能够分隔直流电位，通过交流分量电流，起到隔直流通交流的作用；它们分别把交流信号电流输入基极以及把放大后的交流信号电压送到负载端，而不影响晶体管的直流工作状态。

2. 静态工作点Q 的估算当外加输入信号为零时，在直流电源的作用下，基极和集电极回路的直流电流和电压分别用I BQ 、U BEQ 、I CQ 、U CEQ 表示，并在其输入和输出特性上各自对应一个点，称为静态工作点。

此时电路的直流通路如图3.7a.2所示。

假设，则有 1BQ BE 0.7I I U >>，＝V B2BQCC B1B2R U R R ≈+ ， BQ BE EQ E U U I R −=， EQ BQ 1I I β=+ 由于CQ EQ I I ≈， CEQ CC CQ C E ()U U I R R ≈−+R 1.5k 1I TCC图3.7a.2电压放大电路的直流通路3. 静态工作点的选择放大器静态工作点Q 的位置对放大器放大信号有很大影响，从图3.7a.3的输出特性图上能直观地看到。