第五章集成运算放大电路

1第五章

集成运算放大电路

2集成电路：60年代发展起来的一种新型器件，把众多晶体管、电阻、电容及连线制作在一块半导体芯片（如：硅片）上，做成具有特定功能的独立电子线路。外型一般用金属圆壳或双列直插结构。

集成电路具有性能好，可靠性高，体积小，耗电少，成本低等优点。3JackKilby的第一块集成电路

45采用65纳米工艺制造，内部集成2.91亿个晶体管6集成运放：是一种模拟集成电路，早期实现各种数学运算，主要用于模拟计算机；现在广泛应用于各种电子系统中。75.1集成运算放大器的特点1.级间采用直接耦合方式(集成工艺不能制作大电容和电感)；2.尽可能采用有源器件代替无源器件(避免使用大电容、大电阻)；3.利用对称结构改善电路性能

(采用对称结构的差动放大器，抑制工作点漂移，解决零漂现象)。8

图5.1.1集成运算放大器组成框图差动放大器负载为有源负载的共射放大器射随器或互补射随器95.2电流源电路1、为各级电路提供稳定的直流偏置电流2、可作为有源负载代替集电极电阻RC。电流源的作用：10

单管电流源电路IC0IBUCER1R2ICR3－UEEICRo

(a)晶体管的恒流特性(b)电流源电路(c)等效电流源表示法11图5.2.1镜像电流源一、镜像电流源12图5.2.2多路镜像电流源13IrRrIC1IC2UCCIC3RrIrIC2IC3V3V2V1UCC图5.3.2多集电极晶体管镜像电流源(a)三集电极横向PNP管电路(b)等价电路集成电路中多路镜像电流源的实现14二、比例电流源

图5.2.4比例电流源15若β>>1，则IE1≈Ir,IE2≈IC216三、微电流电流源RrIrIC2V2V1UCCR2图5.2.5微电流电流源当β1>>1时，

IE1≈Ir,IE2≈IC2已知Ir=1mA,要求IC2=10μA时17

四、威尔逊电流源图5.2.6威尔逊电流源18

若三管特性相同，则β1=β2=β3=β,19当环境改变（如温度升高）

IC3↑→IE3↑→IC2↑→IC1↑→(Ir固定)IB3↓→IC3↓20六、有源负载放大器UCCV3V2uoV1uiRr图5.3.1有源负载放大器(a)共射电路(b)等效电路21图5.3.1有源负载放大器(b)等效电路(b)交流小信号等效电路225.4差动放大电路5.4.1零点漂移现象图5.4.1零点漂移现象（a）测试电路（b）输出电压的漂移23零点漂移现象1.静态时，由于温度变化，电源波动等因素的影响，会使工作点电压(即集电极电位)偏离设定值而缓慢地上下飘动。2.

对直接耦合放大电路，这种飘动会逐级放大，会使后级放大器进入截止区和饱和区，这样整个电路将无法正常工作。3.差动放大器电路能有效地克服零点漂移。24抑制零点漂移现象的措施：（1）电路中引入直流负反馈，稳定静态工作点，减小零漂。（2）利用热敏元件对放大管进行温度补偿。（3）采用“差动放大电路”。255.4.2差动放大器的工作原理及性能分析、电路形成原理图5.4.2差动放大电路的形成26图5.4.2差动放大电路的形成27UCCRCUC2RLRCUC1Ui1Ui2RE－UEEV1V2－＋Uo

图5.4.2基本差动放大器当Ui1=Ui2=0时则流过RE的电流I为故有二、差动放大电路的静态分析28UCCRCUC2RLRCUC1Ui1Ui2RE－UEEV1V2－＋Uo

图5.4.2基本差动放大器静态时，差动放大器两输出端之间的直流电压为零。（a）

双端输出29

图5.4.2基本差动放大器（b）

单端输出IBQ1=IBQ2ICQ1=ICQ2UCEQ2≈UCC+0.7-ICQ1RC

UCEQ1≈-UCQ1+0.730二、差动放大电路的动态分析UCCRCUC2RLRCUC1Ui1Ui2RE－UEEV1V2－＋Uo共模信号:Ui1和Ui2所加的信号大小相等,极性相同

Ib1=Ib2，Ic1=Ic2

Uc1=Uc2输出电压Uo=031UCCRCUC2RLRCUC1Ui1Ui2RE－UEEV1V2－＋Uo差模信号:Ui1和Ui2所加的信号大小相等,极性相反Δ

Uc1=－ΔUc2

Uo=Δ

Uc1－Δ

Uc2=2

Uc1

321、共模抑制特性UCCRCUC2RLRCUC1Ui2RE－UEEV1V2－＋Uo＋Uic1－－＋Uic2Ui1UiC2RE－UEE2RE－UEE33双端输出时，负载RL上的电流为零，相当于RL开路每个管子的射极相当于各接有2RE的电阻RCUoc2RCUoc1UicV1V2－＋Uoc＋－2RE图5.4.4基本差动放大器的共模等效通路R2E341.共模电压放大倍数Auc双端输出时的共模电压放大倍数RCUoc2RCUoc1UicV1V2－＋Uoc＋－2RER2E35差动电路能够克服零点漂移现象的根本原因： 共模信号一般指由于外界影响（β，T，UCC），引起工作点的漂移，折算到输入端就是一种共模信号，双端输出时，只要对称性好，则UOC=0，可以完全抑制外界的干扰。单端输出时，由于RE的调节作用使输出大为减少。RCUoc2RCUoc1UicV1V2－＋Uoc＋－2RER2E36

2.共模输入电阻

3.共模输出电阻

双端输出时为

单端输出时为RCUoc2RCUoc1UicV1V2－＋Uoc＋－2RER2E37共模电压放大倍数通常满足(1+β)2RE>>rbe，所以又可简化为图5.4.5单端输出38通常满足(1+β)2RE>>rbe，所以又可简化为图5.4.5单端输出结论：由于射极电阻2RE的存在，使单端输出时的共模电压放大总倍数大为减小。即差动放大器对共模信号不是放大而是抑制，且RE↑→抑制作用越强。39UCCRCUC2RLRCUC1Ui1Ui2RE－UEEV1V2－＋Uo2、差模放大特性＋Uid1Uid2－－Uid=Uid1-Uid2＋RE上只有静态电压，而不产生差模信号电压。双端输出时，负载RL的中点电位为0。40图5.4.6基本差动放大器的差模等效通路－＋Uod1＋－Uod2＋－RL2RL2V2V1＋＋Uid1Uid2－－Uid=Uid1-Uid2＋－RCRCUod41

1.差模电压放大倍数在双端输出时－＋Uod1＋－Uod2＋－RL2RL2V2V1＋＋Uid1Uid2－－＋－RCRC42所以式中：结论：双端输出时的差模电压放大倍数等于单边共射放大器的电压放大倍数。－＋Uod1＋－Uod2＋－RL2RL2V2V1＋＋Uid1Uid2－－＋－RCRC43UCCRCUC2RLRCUC1Ui1Ui2RE－UEEV1V2－＋Uo单端输出时:信号只从一端输出。44或RCUC2RLRCUC1Ui1Ui2REV1V2－＋Uod1Uid=Uid1-Uid2式中：45结论：单端输出时的差模电压放大倍数为单边共射电路电压放大倍数的一半，且两输出端信号的相位相反。RCUC2RLRCUC1Ui1Ui2REV1V2－＋Uod－＋Uod1＋－Uod2＋－RL2RL2V2V1＋＋Uid1Uid2－－＋－RCRC462.差模输入电阻－＋Uod1＋－Uod2＋－RL2RL2V2V1＋－RCRCUodUid1Uid2Iid47－＋Uod1＋－Uod2＋－RL2RL2V2V1＋－RCRCUodUid1Uid2Iid

3.差模输出电阻双端输出时为单端输出时为48三、共模抑制比KCMR差模：需要放大的有用信号，尽可能的放大。共模：无用的干扰信号，需要抑制。为了衡量差动放大电路对差模信号的放大和对共模信号的抑制能力，通常用共模抑制比来衡量。49KCMR实质上是反映实际差动电路的对称性。理想情况下：在双端输出理想对称的情况下， Auc=0，KCMR→∞。为了定量分析，通常用单端输出的KCMR。50四、对任意输入信号的放大特性UCCRCUC2RLRCUC1Ui1Ui2RE－UEEV1V2－＋Uo＋＋~~5152UCCRCUC2RLRCUC1RE－UEEV1V2－＋Uo＋~~＋~~5354在实际应用中，当只有一路信号源接到差动放大器的两个输入端时：如两端都不接地，这种接法称为双端输入；如信号源一端接地，这种接法称为单端输入。55差放特性的几点结论：1、差动放大电路的性能只与输出端的接法有关，与输入端的接法无关；RCUC2RLRCUC1Ui1Ui2REV1V2－＋Uod－＋Uod1＋－Uod2＋－RL2RL2V2V1＋＋Uid1Uid2－－＋－RCRC562、双端输出的差模电压放大倍数等于半边差模等效电路的电压放大倍数，即与单管共射放大电路相同。单端输出差模电压放大倍数仅是半边差模等效电路电压放大倍数的一半；RCUC2RLRCUC1Ui1Ui2REV1V2－＋Uod－＋Uod1＋－Uod2＋－RL2RL2V2V1＋＋Uid1Uid2－－＋－RCRC573、双端输出的输出电阻为2RC，单端输出的输出电阻仅是双端输出的一半；4、无论是双端输入还是单端输入，差模输入电阻均等于半边差模等效电路输入电阻的两倍。共模输入电阻远大于差模输入电阻。RCUC2RLRCUC1Ui1Ui2REV1V2－＋Uod－＋Uod1＋－Uod2＋－RL2RL2V2V1＋＋Uid1Uid2－－＋－RCRC585.4.3具有电流源的差动放大电路

在差动放大电路中，特别是在单端输出电路中，我们希望发射极电阻RE的阻值越大越好，这样可以有效地抑制工作点漂移，提高共模抑制比。59UCCRCUC2RLRCUC1Ui1Ui2RE－UEEV1V2－＋Uo增大RE→流过RE的电流I减小→每管的静态集电极电流IEQ减小保持电源电压－UEE不变60UCCRCUC2RLRCUC1Ui1Ui2RE－UEEV1V2－＋Uo增大RE→造成电源－UEE过大保持电流I不变61RCRCUi1V1V2－＋UoUi2－UEEIUCCRCUC2RCUC1Ui1V1V2－＋Uoc－UEEV3UB3R1R2R3Ui2UCC图5.4.9具有电流源的差动放大器电路(a)用单管电流源代替RE的差动电路(b)电路的简化表示恒流源62静态工作点的估算：RCUC2RCUC1Ui1V1V2－＋Uoc－UEEV3UB3R1R2R3Ui2UCC63图5.4.10场效应管差动放大电路64

上述章节讨论的电压增益往往是小信号情况下的分析，在大信号情况下，差动电路的工作情况又如何？5.4.4差动放大器的大信号分析65图5.4.11简化的差动放大器RCRCuidV1V2－＋uo－UEEIUCC－＋iC2iC1一、传输特性

定义：差动放大器输出电流或输出电压与差模输入电压之间的函数关系。66图5.4.11简化的差动放大器RCRCuidV1V2－＋uo－UEEIUCC－＋iC2iC167——电流传输特性68——电压传输特性69iC1,iC2IiC1iC2iC1iC2－6UT/－4UT－2UT02UT4UT6UTuidQI2(a)电流传输特性曲线70(b)电压传输特性曲线图5.4.12差动放大器的传输特性曲线uoIRC－6UT/－4UT－2UT02UT4UT6UTuid－IRC71两管集电极电流之和恒等于I传输特性具有非线性特性当uid=0时，差动电路处于静态，这时iC1=iC2=ICQ=I/21.在静态工作点附近，当|uid|≤UT，即室温下，uid在26mV以内时，传输特性近似为一段直线。2.当|uid|≥4UT,即uid超过100mV时，传输特性明显弯曲，而后趋于水平。72(RB)(RB)V1V2RR－UEEI(a)串接R(RB)的线性区扩展电路

图5.4.13扩展差动电路的线性区范围73iC1,iC2I①②I/2R↑(RB↑)0uid

图5.4.13扩展差动电路的线性区范围(b)线性区扩展后的电流传输特性曲线74二、差动放大电路正常工作的前提条件差放电路输入电压的幅值是有限制的差模输入电压所受的限制75共模输入电压所受的限制当共模输入电压为正，且超过差分对管的集电极电压时，则差分对管进入饱和区；当共模输入电压为负，且负电压低于电流源晶体管的基极电位，电流管进入饱和区。即共模输入电压应满足UB3<uic<UC1RCUC2RCUC1Ui1V1V2－＋Uoc－UEEV3UB3R1R2R3Ui2UCC762.电流源电流I小于差放管的集电极临界饱和电流ICS(临界)两差放管的静态工作点应该设置在交流负载线（由于差放电路是直接耦合，交、直流负载线重合）中点偏低的位置，即ICQ≈IEQ<ICS(临界)/277三、差动放大电路作模拟乘法器uid<<2UT≈50mV78795.4.5差动放大器的失调及温漂一、差动放大器的失调当输入信号为零时，由于两晶体管参数和电阻值不可能做到完全对称，因而使得输出不为零。这种现象，称为差动放大器的失调。

输入失调电压——UIO

输入失调电流——IIO。补偿方法：人为的在输入端加补偿电压或电流，80图5.4.16差放的输入失调电压和输入失调电流令

1=，

2=+

81图5.4.17差动放大电路的调零电路射极调零集电极调零82

射极调零电路的差动放大器的差模增益和输入电阻为：83

二、失调的温度漂移

调零电路可在某一特定温度下，使输出为0。但失调会随着温度的改变而发生变化，所以仍存在零点的温度漂移现象。 调零电路可以克服失调，但不能克服温漂。84UIO的温漂与该温度下的UIO的大小成正比。85IIO的温漂主要取决于β的温度系数和IIO本身。86IC

=IC1+IC2

=

1IB

+

2(1+

1)IB

=[

1+

2(1+

1)]IB

用两只同类型的双极型晶体管按图形式连接，即一组电极并联，一组电极串联，两只晶体管的电流符合电流流通方向，便组成一个三端等效复合器件。通常把这种双管复合器件称为达林顿复合管或达林顿对。

=IC/IB

=

1+

2(1+

1)

1

2ICIBIE

1

2

IC1IC2IB25.5复合管及其放大电路8788电流放大倍数Ai=

1

2

输入电阻若忽略rbb’

得89Ro=RC

905.6集成运算放大器的输出级电路RL<RE

正向输出电压达到最大值接近UCC

负向输出电压接近因此负向跟随范围比正向跟随范围小由于915.6集成运算放大器的输出级电路UCCUoV1V2RL－UCC－＋＋Ui－

图4―21互补对称型射极92交越失真产生的原因及波形硅管导通电压，约为0.5V,在0.5～-0.5V之间，两管的输出电流近似为零。93RCVD1VD2V1V2V3UCCUi＋－RL－UEERCV1V2V3UCCUi＋－AI1I2R1R2BRL－UEEV4图5.6.3克服交越失真的互补电路(a)二极管偏置方式(b)模拟电压源偏置方式945.8集成运放的外部特性及其理想化5.8.1集成运放的模型(a)旧符号(b)新符号

从外部看，可以认为集成运放是一个双端输入、单端输出、具有高差模增益、高输入电阻、低输出电阻、能较好地抑制温漂的差动放大电路。95图5.8.2电压传输特性96

在非线性区，输出被限幅，输出电压不是UOH就是UOL。在线性区，曲线的斜率为电压放大倍数，即uo=Aud(u――u+)

由于Aud非常大，可达几十万倍，因此集成运放电压传输特性的线性区间非常窄。975.8.2集成运放的主要性能指标一、静态参数输入失调电压（inputoffsetvoltage）

输入失调电流（inputoffsetcurrent）

输入偏置电流（inputbiascurrent）

输入失调电压温漂

输入失调电流温漂

98一、静态参数最大差模输入电压（maximumdifferentialmodeinputvoltage）

最大共模输入电压（maximumcommonmodeinputvoltage）

99二、动态参数开环差模电压放大倍数（openloopvoltagegain）

差模输入电阻（inputresistance）

共模抑制比（commonmoderejectionratio）

100二、动态参数－3dB带宽（-3dBbandwidth）

单位增益带宽（unitgainbandwidth）

转换速率（压摆率）（slewrate）

1015.8.3理想集成运放一、理想化条件（1）开环差模电压放大倍数（2）差模输入电阻（3）差模输出电阻

（5）频带宽度（4）共模抑制比102二、线性状态下理想运放的特性（1）虚短特性（2）虚断特性

103作业5.25.4(1)(2)(3)5.65.8104R1R2ICR3－UEERBR3rbeβIbrce＋－UoIbIo图4―2单管电流源电路105图4―8威尔逊电流源106(1)(2)(3)RrR3rbeβIb3rce＋－UoIorbeβIb2Ib2rbeβIb1Ib1Ib3107(4)(5)(6)108109以平面工艺为基础的半导体集成电路的制造工艺1.由外延、氧化、光刻、扩散和薄膜淀积（或叫蒸铝）五种基本技术组成平面工艺。2.采用PN结隔离技术和介质隔离技术。3.NPN型晶体三极管是最基本的器件；PNP型晶体三极管有纵向和横向两种结构，由于发射区不是高掺杂的，因而，它们的β值极低，约在2～20之间。横向PNP型管的β值更低，典型值为3～5。1104.电阻通常有扩散电阻和金属膜电阻两类。扩散电阻就是杂质半导体的体电阻，由标准N+扩散流程形成的N+区电阻的阻值一般为20～100Ω，由标准P扩散流程形成的P区电阻的阻值一般为100～20KΩ，阻值的误差较大，约为±20%。利用标准的薄膜沉积流程在二氧化硅表面层上淀积一层金属膜作为电阻。1115.一般电容都是PN结（特别是发射结）在反向偏置时的结电容，也可用MOS电容，这是以SiO2为介质的电容器。一般电容量不宜超过100pF。112对单级放大器，当在晶体三极管的发射结上加上恒压源及正弦信号电压时，由于伏安特性的非线性，将使集电极电流ic中除了直流和基波分量之外，还包含各次谐波分量，若要求二次谐波振幅为基波振幅的2.5%，则允许信号电压的振幅为：Usm≤2.6mVT=300K时，kT/q=0.026V113114115115Chapter5OperationalAmplifier116116OperationalAmplifier-Chapter5

5.1 WhatisanOpAmp?5.2 IdealOpAmp5.3 ConfigurationofOpAmp5.4 CascadedOpAmp5.5 Application–Digital-toAnalogConverter117117

5.1WhatisanOpAmp(1)Itisanelectronicunitthatbehaveslikeavoltage-controlledvoltagesource.Itisanactivecircuitelementdesignedtoperformmathematicaloperationsofaddition,subtraction,multiplication,division,differentiationandintegration.118118

5.1WhatisanOpAmp(2)Atypicalopamp:(a)pinconfiguration,(b)circuitsymbol119

5.1WhatisanOpAmp(3)120120

5.1WhatisanOpAmp(4)TheequivalentcircuitOfthenon-idealopampOpAmpoutput:voasafunctionofVdvd=v2–v1;vo=Avd=A(v2–v1)121121ParameterTypicalrangeIdealvaluesOpen-loopgain,A105to108

∞Inputresistance,Ri105to1013

∞

Outputresistance,Ro10to100

0Supplyvoltage,VCC5to24V

5.1WhatisanOpAmp(5)Typicalrangesforopampparameters122122

5.1WhatisanOpAmp(6)FortheopampcircuitofFig.5.44,theopamphasanopen-loopgainof100,000,aninputresistanceof10k

,andanoutputresistanceof100

.Findthevoltagegainvo/viusingthenonidealmodeloftheopamp.123123

5.2IdealOpAmp(1)

Anidealopamphasthefollowingcharacteristics:Infiniteopen-loopgain,A≈∞Infiniteinputresistance,Ri≈∞Zerooutputresistance,Ro≈0124124

5.2IdealOpAmp(2)

Example1:Determinethevalueofio.*Refertoin-classillustration,textbookAns:0.65mA125125

5.3ConfigurationofOpamp(1)Invertingamplifierreversesthepolarityoftheinputsignalwhileamplifyingit126126

5.3ConfigurationofOpamp(2)

Example2Refertotheopampbelow.Ifvi=0.5V,calculate:(a)theoutputvoltage,voand(b)thecurrentinthe10kresistor.Ans:(a)-1.25V;(b)50μA*Refertoin-classillustration,textbook127127

5.3ConfigurationofOpamp(3)Non-invertingamplifierisdesignedtoproducepositivevoltagegain128128

5.3ConfigurationofOpamp(4)

Example3Fortheopampshownbelow,calculatetheoutputvoltagevo.Ans:-1V*Refertoin-classillustration,textbook129129

5.3ConfigurationofOpamp(4)Giventheopampcircuitshownbelow,expressvointermsofv1andv2.130130

5.3ConfigurationofOpamp(5)SummingAmplifierisanopampcircuitthatcombinesseveralinputsandproducesanoutputthatistheweightedsumoftheinputs.131131

5.3ConfigurationofOpamp(6)

Example4Calculatevoandiointheopampcircuitshownbelow.Ans:-3.8V,-1.425mA*Refertoin-classillustration,textbook132132

5.3ConfigurationofOpamp(7)Differenceamplifierisadevicethatamplifiesthedifferencebetweentwoinputsbutrejectsanysignalscommontothetwoinputs.133133

5.3ConfigurationofOpamp(1)Invertingamplifierreversesthepolarityoftheinputsignalwhileamplifyingit134134

5.3ConfigurationofOpamp(6)

Example5DetermineR1,R2,R3andR4sothatvo=-5v1+3v2forthecircuitshownbelow.Ans:R1=10kΩR2=50kΩR3=20kΩR4=20kΩ*Refertoin-classillustration,textbook1351355.4CascadedOpAmp(1)Itisahead-to-tailarrangementoftwoormoreopampcircuitssuchthattheoutputtooneistheinputofthenext.136136

5.4CascadedOpAmp(2)

Example6Findvoandiointhecircuitshownbelow.Ans:350mV,25μA*Refertoin-classillustration,textbook137137

5.4CascadedOpAmp(3)

Example7Ifv1=1Vandv2=2V,findvointheopampcircuitshownbelow.Ans:8.667V*Refertoin-classillustration,textbook138138

5.4CascadedOpAmp(3)

Example7Ifv1=1Vandv2=2V,findvointheopampcircuitshownbelow.Ans:8.667V*Refertoin-classillustration,textbook139139

5.4CascadedOpAmp(3)ExampleDeterminethevoltagetransferratiovo/vs

intheopampcircuitbelow,whereR=10k

.–++–RRRRRVoVs++--1401405.5Application(1)Digital-toAnalogConverter(DAC):itisadevicewhichtransformsdigitalsignalsintoanalogform.Four-bitDCA:(a)blockdiagram(b)binaryweightedladdertypewhereV1–MSB,V4–LSBV1toV4areeither0or1V141141

5.5Application(2)

Example8Forthecircuitshownbelow,calculatevoifv1=0V,v2=1Vandv3=1V.Ans:-0.75V*Refertoin-classillustration,textbookHW8Ch5:9,13,19,27,33,41,47,52,58,91142143IntroductionThebasicoperationalamplifierorOpAmpisaveryimportantcircuittostudybecauseitissowidelyused.OpAmpswereinitiallymadefromdiscretecomponentslikevacuumtubesandresistorswithtransistorseventuallyreplacingvacuumtubesInthemid1960’sthefirstintegratedcircuitopampswereproduced(mA709)IthadquiteafewtransistorsandresistorscomparedtothediscreteversionsitreplacedItscharacteristicswerepoorbytoday’sstandardsItwasexpensive,butItwasallonasinglesiliconchipanditusheredinanewerainanalogelectroniccircuitdesignItwasmorereliablethandiscreteopampsEngineersstartedusingtheminlargequantitiesandthepricedroppeddramaticallyfromovertendollarseachto30centseachandthereliabilitycontinuedtoincreaseTheopampisaveryversatilecircuitandcanbeusedinnumerousothercircuits.Theintegratedcircuitopampcharacteristicscomeprettyclosetothatofan“ideal”amplifier.Thenear“ideal”behaviormakesitmucheasiertodesignwithopampsBytheendofthisgroupoflecturesessionsthereadershouldbeabletodesignsomefairlycomplexcircuitsusingopampsWewillstartbytreatingaopampasa“buildingblock”anddescribeitsterminalcharacteristicsandsaveadetaileddiscussionaboutwhatisinsideanopampforalatercourse(EGRE307)144TheOpAmpTerminalsTheterminalsofacircuitcanbedividedupintodifferentgroupsbasedontheirfunctions,forexamplethepowersupplyterminals,thesignalterminals,etc.Theopampcircuithasthreesignalterminals,twoinputs(#1,#2)andoneoutput(#3)asshownonthefigurebelowTheamplifierrequiresdcpowertooperate.Mostopampsrequiretwodcpowersuppliestooperate.Mostofthetimethetwosuppliesareequalinmagnitudebutoppositeinpolarity,forexampleapositivevoltagesupply+V(relativetoground)andanegativevoltagesupply-V(relativetoground)connectedtoterminals#4and#5asshown.outputinputs-+12345+V-Voutputinputs-+123145OpAmpTerminals,continuedAnopampcanuseasinglepowersupply(+10Vandgroundforexample)butthesignal“reference”ofthecircuitforsymmetricalbehaviorwillthenbeatonehalftheuppersupplyvoltageor+5Vinthiscase.Thereferencegroundingpointinop-ampcircuitsisjustthecommonterminalofthetwopowersupplies.Thatis,noterminaloftheopamppackageisphysicallyconnectedtogroundInmanycasesthepowersupplyconnectionswillnotbeexplicitlyshownontheschematic.Theywillbeomittedtoreducethe“clutter”buttheconnectionsarestillimpliedwhenyoudrawthetriangularopampsymbol.Anopampmayhaveotherterminalsforfrequencycompensationandforremoving(ornulling)outoffsetvoltages.Outputsignalgroundat+5Vrelativetorealgroundinputs-+12345+10V0VOutputSignalgroundisatsamelevelasrealgroundinputs-+12345+5V-5V+10Vor146ExerciseWhatistheminimumnumberofterminalsrequiredforasingleopamp?Five-TwoInputs,Twopowerconnections(either+/-,symmetricor+andGround,non-symmetric)andonesignaloutputreferencedtoground(symmetric)or1/2thepositivesupplyinanon-symmetricconfigurationWhatistheminimumnumberofterminalsrequiredforanintegratedcircuitpackagewhichcontainsfouropamps(assumeallfoursharepowersupplies)?Fourteen,fourpairsofinputs,twopowerconnectionsandfouroutputs147TheIdealOpAmpTheopampisdesignedtosensethedifferencebetweenthevoltagesignalsappliedtoitstwoinputs(v2-v1),andmultiplythisbyanumberA,andcausetheresultingvoltageA(v2-v1)toappearattheoutputterminalrelativetoground.Notethatv1andv2arealsoimpliedtoberelativetogroundasshowninthefigurebelow.Theidealopampisnotsupposedtodrawanycurrentfromthesignalsource(itwilldrawcurrentfromthepowersupplyterminalwhicharenotshown).TheinputimpedanceissupposedtobeinfiniteTheoutputissupposedtoactasanidealvoltagesourceindependentofthecurrentdrawnbyanyloadplacedontheopamp.TheoutputimpedanceissupposedtobezeroA(v2-v1)v1v2I2=0I1=0Commonterminalofthepowersupplies213Theoutputhasthesamesign(isinphasewith)theinputv2andisoutofphase(oppositesign)asv1.Forthisreasoninputterminal#1iscalledtheinvertinginput(distinguishedbya“-”sign)andinputterminal#2iscalledthenon-invertinginput(distinguishedbya“+”sign).Theopamponlyrespondstothedifferencesignalv2-v1andignoresanysignalcommontobothinputs.148CommonSignalvs.DifferentialSignalThepartofthesignalthatisthesameforbothinputsiscalledthecommonmodecomponentandisnotamplifiedIfbothinputsseethesamesignaltheDIFFERENCE(v2-v1)iszeroandthereisnothingtoamplifyandtheoutputiszero.Thismeansthattheamplifierhasrejected(notamplified)whatthetwosignalshadincommonAnidealopamphasinfinitecommon-moderejectionForthetimebeingwewillconsidertheopamptobeadifferentialinputsingle-ended-output(outputfromterminal3toground)Vp-pVp-p12oneachsideoutofphaseVp-pv2v1A(v2-v1)00oneachsideinphaseVp-p12149TheIdealOpAmpcontinuedThegain,Aiscalledthedifferentialgain(basedonthedifferencebetweenterminals2and1)Thegain,Aisalsosometimescalledthe“openloop”gainasopposedtolateronwhenweaddafeedbackpathfromoutputtoinputandlookatthe“closedloop”gainOpampsaredirect-coupled(DC)ordirectcurrent(DC)amplifiers.DCamplifiershavemanyapplicationsbutcanalsoposesomepracticalproblemsIDEAL

opampsamplifyfromzerofrequencytoinfinitefrequency,theyaresaidtohaveinfinitebandwidthA(v2-v1)v1v2I2=0I1=0Commonterminalofthepowersupplies213Thegainofanidealopampshouldapproachinfinity,butthenhowcouldwemakeuseofanidealopampsinceanydifferencetimesinfinitywouldbeinfinity?Weusuallydon’tuseopampsinanopenloopconfigurationFeedbacklowersthegaintopracticallevelsThevalueforAofa“real”opampmightbeontheorderofamillion150Exercise2.2Consideranopampthatisidealexceptthatitsopen-loopgainA=103.Theopampisusedinafeedbackcircuitandthevoltagesappearingattwoofitsthreesignalterminalsaremeasured.Ineachofthefollowingcases,usethemeasuredvaluestofindtheexpectedvalueofthevoltageatthethirdterminal.A)v2=0V,v3=2Vv1=-{(v3/A)-v2}=-{(2/1000)-0}=-0.002VB)v2=5V,v3=-10Vv1=-{(v3/A)-v2}=-{(-10/1000)-5}=5.01VC)v1=1.002V,v2=0.998Vv3=A(v2-v1)=1000(0.998-1.002)=4.0VD)v1=-3.6V,v3=-3.6Vv2={(v3/A)+v1}={(-3.6/1000)+3.6}=3.6036V151Exercise2.3Theinternalcircuitofaparticularopampcanbemodeledbythecircuitshowntotheright.Expressv3asafunctionofv1andv2.ForthecasewhenGm=10mA/V,R=10kWandm=100,findthevalueoftheopen-loopgainA.3v1v2mvdvdRGmv1Gmv221152TheAnalysisofCircuitsContainingIdealOpAmpsTheInvertingConfigurationResistorR2isconnectedfromtheoutputterminal,3,backtotheinvertingornegativeterminalinputterminal,1.WesaythatR2appliesnegativefeedback,ifR2wereconnectedbetweentheoutput,3,andtheotherinput,2,wewouldsaythatitwaspositivefeedback.Wehavegroundedinputnumber2andappliedtheinputsignalbetweenterminaloneandground.Thisisknownasasinglesidedinputsinceterminal2doesnotcontributeanyinformationtothesignal(sinceitisgrounded).Theoutputistakenatterminal3relativetogroundandiscalledasinglesidedoutput.Notethattheimpedanceatterminal3(output)ofanidealopampisinfinite,thusthevoltagevo

willnotdependonthevalueofthecurrentthatmightbesuppliedtoaloadplacedacrossvo.outputinput-+1234R2vIR1+vo-153Closed-LoopGain(G)AnalysisGisdefinedasWecandrawtheequivalentcircuitfortheopamp.Assumetheopampispoweredupandworkingandtheoutputisfinite.Withafiniteoutputandinfinitegainthedifferencebetweentheinputsisnegligiblysmall.Sinceterminal2isheldatgroundthenterminalonealsohastobeatground(forcedtherebythecircuitNOTconnectedtoground).Usingourdefinitionwewrite-+123R2vIR1+vo--+123R2vIR1+vo-0i1i2+_A(v2-v1)v2-v1Wesaythatthetwoinputterminalsaretrackingeachotherinpotential154VirtualGroundThecircuitforcesthetwoinputterminalstobeatvirtuallythesamepotential,theyare“virtually”shortedtogether.Aphysicalconnectionforchargemovementdoesnotexistbutsincetheytrackeachotherinpotentialislikethereisaconnection(avirtualshortcircuit)Ifterminal2isgroundedthenwecanrefertoterminal1asa“virtual”ground.Itisforcedtobeatzerovoltseventhoughitisnotdirectlyconnectedtogrounditactslikeitis.155BacktotheanalysisSincewenowknowv1wecanuseOhm’slawandfindthecurrentflowingthroughtheresistorR1.Sincethiscurrentcannotflowintotheinputterminalofanidealopamp(infiniteinputresistance)itmustflowthroughtheresistorR2tothelowimpedanceterminal,3(output).WecannowapplyOhm’slawtoR2anddeterminevo.ThenegativesignfortheclosedloopgainindicatesthattheamplifierprovidessignalinversionIfweapplya50mVpeak-to-peaksine-waveinputsignaltheoutputwillbeanamplifiedsine-wavewitha180degreephaseshift(inversion).156TheEffectofFiniteOpen-LoopGainHowdoourresultschangeifwelettheopenloopgain,A,befinite?IfwelettheoutputbedenotedbyvO,thenthevoltagedifferencebetweentheinputsisvO/A.Sincethepositiveinputisgroundedthenegativeinputmustbeat-vO/A.WecannowfindthecurrentthroughresistorR1TheinfiniteinputresistanceoftheopampforcesallofthecurrenttoflowthroughR2.WecanwriteanotherexpressionwhichcontainsvO.Collectingtermsandsolvingfortheclosed-loopgainG,wegetNote:AsAapproachesinfinity,Gapproachestheidealvaluethatwepreviouslyderived157InputandOutputResistanceAsAapproachesinfinitythevoltageattheinvertingterminalapproacheszero(ourvirtualground)InordertominimizetheeffectofAonGweshouldmakeIfweassumeanidealopampwithaninfiniteinputresistancetheoverallinputresistanceoftheclosedloopcircuit(containingtheidealopamp)willbeThevalueoftheopenloopgain,A,hasverylittleeffectontheinputresistanceForahighinputresistanceweneedR1high,butifwewantthegaintoalsobehighwemightrequireR2tobeverylarge(maybeimpracticablylarge).Canyouthinkofawaytoincreasetheinputresistance?Theoutputoftheinvertingconfigurationistakenattheterminalofanidealvoltagesource(whosevalueis(v2-v1)A)whichcansupplyinfinitecurrent,Rout=V/Iiszero.+vI_+_Puttingthepreviousresultstogetherwehavethefollowingmodeloftheinvertingclosedloopamplifiercircuit(usinganidealopamp)Ri=R1Ro=0158Example2.1ConsidertheinvertingamplifierconfigurationshownbelowwithR1=1kWandR2=100kW. A)Findtheclosed-loopgainforthe