Ideal Op Amp Circuits

Ideal Op Amp Circuits The operational amplifier, or op amp as it is commonly called, is a fundamental active element of analog circuit design. It is most commonly used in amplifier and analog signal processing circuits in the frequency bandfrom0to100khz. High-frequencyopampsareusedinapplicationsthatrequireabandwidthintothe MHz range. The first op amps were vacuum-tube circuits which were developed for use in analog computers. Modern op amps are fabricated as integrated circuits that bare little resemblance to the early circuits. This chaptercoverssomeofthebasicapplicationsoftheopamp. Itistreatedasanidealcircuitelementwithout regard to its internal circuitry. Some of the limitations imposed by non-ideal characteristics are covered in the following chapter. The notation used here is as follows: Total quantities are indicated by lower-case letters with uppercasesubscripts,e.g. v I,i O,r IN. Small-signalquantitiesareindicatedbylower-caseletterswithlower-case subscripts,e.g. v i,i o,r out. Transferfunctionvariablesandphasorsareindicatedbyuppercaselettersand lower-casesubscripts,e.g. V i,i o,z in. 1.1 TheIdealOpAmp The ideal op amp is a three terminal circuit element that is modeled as a voltage-controlled voltage source. Thatis,itsoutputvoltageisagainmultipliedbyitsinputvoltage. Thecircuitsymbolfortheidealopamp is given in Fig. 1.1(a). The input voltage is the difference voltage between the two input terminals. The output voltage is measured with respect to the circuit ground node. The model equation for the output voltage is v O =A(v + v ) (1.1) where A is the voltage gain, v + is the voltage at the non-inverting input, and v is the voltage at the invertinginput. ThecontrolledsourcemodeloftheidealopampisshowninFig. 1.1(b). Figure 1.1: (a) Op-amp symbol. (b) Controlled-source model. The terminal characteristics of the ideal op amp satisfy four conditions. These are as follows: 1. Thecurrentineachinputleadiszero. 2. The output voltage is independent of the output current. i

ii IDEAL OP AMP CIRCUITS 3. The voltage gain A is independent of frequency. 4. ThevoltagegainAisverylarge,approachinginfinityinthelimit. The first condition implies that the resistance seen looking into both input terminals is infinite. The second implies that the voltage gain is independent of the output current. This is equivalent to the condition that the output resistance is zero. The third implies that the bandwidth is infinite. The fourth implies that the difference voltage between the two input terminals must approach zero if the output voltage is finite. For it to act as an amplifier, the op amp must have feedback applied from its output to its inverting input. Thatis, part ofthe outputvoltage mustbesampled by anetworkand fed backintothe inverting input. Thismakesitpossibletodesignanamplifiersothatitsgainiscontrolledbythefeedbacknetwork. To illustrate how feedback affects the op amp, consider the circuits shown in Fig. 1.2. The networks labeled N 1 and N F, respectively, are the input and feedback networks. The op amp of Fig. 1.2(a) has positive feedback whereas the op amp of Fig. 1.2(b) has negative feedback. Let aunitstep ofvoltage be appliedtotheinputofeachcircuitatt=0. Thearrowsinthefiguresindicatethedirectionsinwhichthe input voltages change, i.e. each input voltage increases. For the circuit of Fig. 1.2(a), the voltage increase atv i isfedthroughthen 1 networktocausethevoltagetoincreaseatthev + terminal. Thisisamplifiedby apositivegain(+a)andcausestheoutputvoltagetoincrease. ThisisfedbackthroughtheN F networkto furtherincreasethevoltageatthev + terminal. (Thearrowforthefeedbackvoltageisenclosedinparentheses to distinguish it from the arrow for the initial increase in voltage.) This causes the output voltage to increase further,causingv + toincreasefurther,etc. Itfollowsthatthecircuitisnotstablewithpositivefeedback. Figure 1.2: (a) Op amp with positive feedback. (b) Op amp with negative feedback. ForthecircuitofFig. 1.2(b),thevoltageincreaseattheinputisfedthroughtheN 1 networktocause thevoltagetoincreaseatthev terminal. Thisisamplifiedbyanegativegain( A)andcausestheoutput voltage to decrease. This is fed back through the N F network to cause the voltage at the v input to decrease,thustendingtocanceltheinitialincreasecausedbytheinputvoltage. Becausethev voltageis decreasedbythefeedback,itfollowsthatv O isdecreasedalso. Thusthecircuitisstable. Whennegativefeedbackisusedinanopampcircuit,thefeedbacktendstoforcethevoltageatthev inputtobe equaltothe voltage atthe v + input. Itis said thatavirtual short circuit existsbetween the two inputs. A virtual short circuit between two nodes means that the voltage difference between the nodes is zero but there is no branch for a current to flow between the nodes. There is no virtual short circuit betweenthev andv + inputstoanopampwhichhaspositivefeedback. Ifithasbothnegativeandpositive feedback, the virtual short circuit exists if the negative feedback is greater than the positive feedback. WehaveusedtheconceptofsignaltracinginthecircuitsofFig. 1.2toillustratetheeffectsoffeedback. Signal tracing is a simple concept which can be applied to any circuit to check for positive and negative feedback. Circuits which have positive feedback are unstable in general and are not used for amplifier circuits. With few exceptions, the circuits covered in this chapter have only negative feedback.

1.2. INVERTINGAMPLIFIERS iii 1.2 Inverting Amplifiers 1.2.1 The Inverting Amplifier Figure 1.3(a) shows the circuit diagram of an inverting amplifier. The input signal is applied through resistor totheinvertingopampinput. ResistorR F isthefeedbackresistorwhichconnectsfromthetheoutput to the inverting input. The circuit is called an inverting amplifier because its voltage gain is negative. This means that if the input voltage is increasing or going positive, the output voltage will be decreasing or going negative, and vice versa. The non-inverting input to the op amp is not used in the inverting amplifier circuit. Thefigureshowsthisinputgroundedsothatv + =0. Figure 1.3: (a) Inverting amplifier. (b) Controlled-source model. ForthecircuitofFig. 1.3(a),thevoltageattheinvertinginputisgivenbyv = v O /A. Forv O finite anda, itfollowsthatv 0. Eventhoughthev inputisnotgrounded, itissaidtobeavirtual groundbecausethevoltageiszero,i.e. atgroundpotential. Becausei =0,thesumofthecurrentsinto thev nodethroughresistors andr F mustbezero,i.e. i 1 +i F =0,wherei 1 =v I / andi F =v O /R F. Thuswecanwrite v I i 1 +i F =0 = + v O =0 (1.2) R F Thisrelationcanbesolvedforthevoltagegaintoobtain v O v I = R F (1.3) Theinputresistanceiscalculatedfromtherelationr in =v I /i 1. Becausev =0,itfollowsthat Theoutputresistanceisequaltotheoutputresistanceoftheopampsothat The controlled source model of the inverting amplifier is shown in Fig. 1.3(b). r in = (1.4) r out =0 (1.5) Example1 Design an invertingamplifier with an input resistanceof 2kΩ, an output resistanceof100ω, and an open-circuit voltage gain of 30(an inverting decibel gain of 29.5 db). Solution. ThecircuitdiagramfortheamplifierisgiveninFig. 1.4(a). Foraninputresistanceof2kΩ, Eq. (1.4)gives =2kΩ. Foravoltagegainof 30, itfollowsfromeq. (1.3)thatR F =60kΩ. Foran outputresistanceof100ω,theresistorr O =100Ωmustbeusedinserieswiththeoutputasshowninthe figure. Example2 CalculatethevoltagegainofthecircuitofFig. 1.4(a)ifa1kΩloadresistorisconnectedfrom theoutputtoground. ThecircuitwiththeloadresistorisshowninFig. 1.4(b).

iv IDEAL OP AMP CIRCUITS Figure1.4: (a)circuitforexample1. (b)circuitforexample2. (c)circuitforexample3. Solution. The voltage gain decreases when R L is added because of the voltage drop across R O. By voltage division, the gain decreases by the factor R L 1000 = R O +R L 1000+100 = 10 11 Itfollowsthattheloadedvoltagegainis(10/11) ( 30)= 27.3(aninvertingdecibelgainof28.7dB). Example 3 For the inverting amplifier circuit of Fig. 1.4(b), investigate the effect of connecting the feedback resistorr F totheloadresistorr L ratherthantotheopampoutputterminal. Themodifiedcircuitisshown in Fig. 1.4(c). Solution. Becausei 1 +i F =0,itfollowsthatv I / +v O /R F =0. Thisgivesthevoltagegainv O /v I = R F /. BecausethisisindependentofR L,itfollowsthattheoutputresistanceofthecircuitiszero. Thus thecircuitlooksliketheoriginalcircuitoffig. 1.4(b)withR O =0. WithR O 0,theopampmustput outalargervoltageinordertomaintainaloadvoltagethatisindependentofr O. Letv O bethevoltageat theopampoutputterminalinfig. 1.4(c). Byvoltagedivision,theoutputvoltageisgivenby v O v O R L R F = R L R F +R O Itfollowsthatv O islargerthanv O bythefactor1+r O /(R L R F ). Becausethisisgreaterthanunity,R O causestheopampto workharder toputoutalargeroutputvoltage. Weconcludethataresistorshould not be connected in series between the op amp output terminal and the connection for the feedback resistor. 1.2.2 The Inverting Amplifier with T Feedback Network Ifahighvoltagegainisrequiredfromaninvertingamplifier,Eq. (1.3)showsthateitherR F mustbelarge, mustbesmall,orboth. If issmall,theinputresistancegivenbyeq. (1.4)maybetoolowtomeet specifications. The inverting amplifier with a T feedback network shown in Fig. 1.5(a) can be used to obtain ahighvoltagegainwithoutasmallvaluefor orverylargevaluesforthefeedbackresistors.

1.2. INVERTINGAMPLIFIERS v Figure1.5: (a)invertingamplifierwithatfeedbacknetwork. (b)equivalentcircuitforcalculatingv O. ThesolutionforthevoltagegainissimplifiedbymakingaThéveninequivalentcircuitatthev terminal lookingtotherightthroughr 2. ThecircuitisgiveninFig. 1.5(b). Becausei 1 +i F =0,itfollowsthat v I + v OR 3 1 =0 (1.6) R 3 +R 4 R 2 +R 3 R 4 Thisequationcanbesolvedforthevoltagegaintoobtain [ v O R2 = + R ( 4 1+ R )] 2 v I R 3 (1.7) Theoutputresistanceofthecircuitiszero. Theinputresistanceis. Example 4 For the inverting amplifier with a T feedback network in Fig. 1.5(a), specify the resistor values which give an input resistance of 10kΩ and a gain of 100. The maximum resistor value in the circuit is limitedto100kω. Solution. To meetthe input resistance specification, we have =10kΩ. Let R 2 =R 4 =100kΩ. It followsfromeq. (1.7)thatR 3 isgivenby ThisequationgivesR 3 =12.5kΩ. R 3 = R 2 R 4 ( v O /v I ) (R 2 +R 4 ) 1.2.3 The Current-to-Voltage Converter The circuit diagram of a current-to-voltage converter is shown in Fig. 1.6(a). The circuit is a special case ofaninvertingamplifierwheretheinputresistorisreplacedwithashortcircuit. Becausethev terminal isavirtualground,theinputresistanceiszero. Theoutputresistanceisalsozero. Becausei 1 +i F =0and v O =i F R F,itfollowsthatthetransresistancegainisgivenby v O i 1 = R F (1.8) Figure 1.6(b) shows the current-to-voltage converter with a current source connected to its input. Because R S connectsfromavirtualgroundtoground,thecurrentthroughr S iszero. Itfollowsthati 1 =i S and v O = R F i S. ThustheoutputvoltageisindependentofR S.

vi IDEAL OP AMP CIRCUITS Figure 1.6: (a) Current-to-voltage converter. (b) Curcuit with an input current source. 1.3 Non-Inverting Amplifiers 1.3.1 The Non-Inverting Amplifier Figure1.7(a)showsthecircuitdiagramofanon-invertingamplifier. Theinputvoltagev I isappliedtothe non-invertingopampinput. AvoltagedividerconsistingofresistorsR F and connectsfromtheoutput node to the inverting input. The circuit is called a non-inverting amplifier because its voltage gain is positive. This means that if the input voltage is increasing or going positive, the output voltage will also be increasing or going positive. If the circuit diagrams of the inverting and the non-inverting amplifiers are compared, it canbeseenthatthecircuitsarethesameifv I =0. Thustheonlydifferencebetweenthetwocircuitsisthe node at which the input voltage is applied. Figure 1.7: (a) Non-inverting amplifier. (b) Controlled-source model. ForthecircuitofFig. 1.7(a),thevoltagedifferencebetweenthetwoopampinputterminalsisgivenby v + v =v O /A. Forv O finiteanda,itfollowsthatv + v. Itissaidthatavirtualshortcircuit existsbetweenthetwoinputsbecausethereisnovoltagedifferencebetweenthetwoterminals. Fori =0, theconditionthatv + =v requiresv I andv O tosatisfytheequation v + =v = v I =v O R F + (1.9) wherevoltagedivisionhasbeenusedforv. Thiscanbesolvedforthevoltagegaintoobtain The input and output resistances are given by v O v I =1+ R F (1.10) r in = (1.11)

1.3. NON-INVERTINGAMPLIFIERS vii r out =0 (1.12) The controlled source model for the non-inverting amplifier is shown in Fig. 1.7(b). Example 5 Design a non-inverting amplifier which has an input resistance of 10 kω, an open-circuit voltage gain of 20 (a decibel voltage gain of 26dB), and an output resistance of 600Ω. The feedback network is specified to draw no morethan0.1ma fromthe output of theop amp when theopen-circuit output voltage isintherange 10V v O 10V. Solution. The circuit diagram for the amplifier is shown in Fig. 1.8. To meet the input resistance specification,wehaver i =10kΩ. Forthespecifiedcurrentinthefeedbacknetwork,wemusthave0.1mA 10/(R F + ). Iftheequalityisused,weobtainR F + =100kΩ. Forthespecifiedopen-circuitvoltage gain,eq. (1.10)gives1+R F / =20orR F =19. Itfollowsthat =5kΩandR F =95kΩ. Tomeet theoutputresistancespecification,wemusthaver O =600Ω. Figure 1.8: Circuit for Example 5. Example6 Examine the effect of a connecting a resistor between the v + node and the v node in the non-inverting amplifier of the circuit for Example 5. Solution. Foranidealopamp,thevoltagedifferencebetweenthev + andv terminalsiszero. Itfollows that a resistor connected between these nodes has no current flowing through it. Therefore, the resistor has no apparent effect on the circuit. This conclusion applies also for the inverting amplifier circuit of Fig. 1.3. Withphysicalopamps,however,aresistorconnectedbetweenthev + andthev terminalscanaffectthe performance of the circuit by reducing the effective open-loop gain A. 1.3.2 The Voltage Follower The voltage follower or unity-gain buffer is a unity-gain non-inverting amplifier. The circuit diagram is shown infig. 1.9(a). Comparedtothenon-invertingamplifierofFig. 1.7(a),thefeedbackresistorR F isreplaced byashortcircuitandresistor isomitted. Becausetheoutputnodeisconnecteddirectlytotheinverting inputinsteadofthroughavoltagedivider,thecircuitissaidtohave100%feedback. Becausev + =v,it follows that v O =v I. Therefore, the circuit has unity voltage gain. The voltage follower is often used to isolate a low resistance load from a high output resistance source. That is, the voltage follower supplies the current to drive the load while drawing no current from the input circuit. Example7 Figure 1.9(b) shows a source connected to a load with a voltage follower. It is given that R S =10kΩandR L =100Ω. (a)calculatev O. (b)calculatev O ifthevoltagefollowerisremovedandthe source connected to the load. Solution. (a) With the voltage follower, there is no current through R S so that the voltage at the op ampinputisv S. Itfollowsthatv O =v S. (b)ifthevoltagefollowerisremovedandthesourceisconnected directly to the load, v O is given by v O = v S R L /(R S +R L ) = v S /101. This is a decrease in output of 20log101=40.1dB. Thisexampleillustrateshowaunitygainamplifiercanincreasethegainofacircuit.

viii IDEAL OP AMP CIRCUITS Figure 1.9: (a) Voltage follower. (b) Circuit for Example 7. 1.3.3 Amplifier with Voltage and Current Feedback Figure 1.10(a) shows the circuit diagram of a non-inverting amplifier in which the voltage fed back to the invertinginputoftheopampisafunctionofboththeloadvoltageandtheloadcurrent. Tosolveforthe outputvoltage, itisconvenienttofirstformthe Thévenin equivalentcircuitseen by the loadresistorr L. The circuitis shown in Fig. 1.10(b). It consists of avoltage source in series with a resistor. The voltage sourcehasavalueequaltotheopen-circuitloadvoltage,i.e. theoutputvoltagewithr L. Theresistor hasavalueequaltotheratiooftheopen-circuitloadvoltagetotheshort-circuitloadcurrent,i.e. theoutput currentwithr L =0. Figure 1.10: (a) Non-inverting amplifier with voltage and current feedback. (b) Thévenin equivalent circuit seenbytheload. WithR L =,theopen-circuitloadvoltageisgivenbyv O(oc) =i 1 (R F + ). Becausethereisavirtual shortcircuitbetweenthev + andthev terminals,itfollowsthati 1 =v I /( +R 2 ). Itfollowsthatv O(oc) can be written v O(oc) =v I +R F +R 2 (1.13) WithR L =0,therecanbenocurrentthroughR F or sothatv I =v =i O(sc) R 2. Thusi O(sc) isgiven by i O(sc) = v I R 2 (1.14) Theoutputresistanceofthecircuitisgivenby r out = v O(oc) i O(sc) =R 2 +R F +R 2 (1.15) By voltage division, it follows from Fig. 1.10(b) and Eq. (1.13) that the output voltage can be written v O =v O(oc) R L +R F R L =v I (1.16) r out +R L +R 2 r out +R L

1.4. SUMMINGAMPLIFIERS ix 1.3.4 The Negative Impedance Converter Although it is not an amplifier, the negative impedance converter is an application of the non-inverting configuration. For the circuit in Fig. 1.11(a), the resistor R bridges the input and output terminals of a non-inverting amplifier. We can write r in = v I i 1 (1.17) Solutionforr in yields Thus the circuit has a negative input resistance. i 1 = v I v O (1.18) R ( v O = 1+ R ) F v I (1.19) r in = R F R (1.20) Figure 1.11: Negative impedance converters. (a) Negative input resistance. (b) Negative input capacitance. A resistor in parallel with another resistor equal to its negative is an open circuit. It follows that the output resistance of a non-ideal current source. i.e. one having a non-infinite output resistance, can be made infinite by adding a negative resistance in parallel with the current source. Negative resistors do not absorb powerfromacircuit. Instead,theysupplypower. Forexample,ifacapacitorwithaninitialvoltageonitis connected in parallel with a negative resistor, the voltage on the capacitor will increase with time. Relaxation oscillators are waveform generator circuits which use a negative resistance in parallel with a capacitor to generate ac waveforms. TheresistorisreplacedwithacapacitorinFig. 1.11(b). Inthiscase,theinputimpedanceis Z in = R F 1 jωc =jω ω 2 R F C =jωl eq (1.21) It follows that the input impedance is that of a frequency dependent inductor given by L eq = ω 2 R F C (1.22) 1.4 Summing Amplifiers 1.4.1 The Inverting Summer Theinvertingsummeristhebasicopampcircuitthatisusedtosumtwoormoresignalvoltages,tosuma dcvoltagewithasignalvoltage,etc. AninvertingsummerwithfourinputsisshowninFig. 1.12(a). Ifall

x IDEAL OP AMP CIRCUITS inputsaregroundedexceptthev Ij input,wherej=1,2,3,or4,eq. (1.3)fortheinvertingamplifiercan beusedtowritev O = (R F /R j )v Ij. Itfollowsbysuperpositionthatthetotaloutputvoltageisgivenby v O = R F v I1 R F R 2 v I2 R F R 3 v I3 R F R 4 v I4 (1.23) TheinputresistancetothejthinputisR j. Theoutputresistanceofthecircuitiszero. Figure 1.12: (a) Four input inverting summer. (b) Circuit for Example8. Example 8 Design an inverting summer which has an output voltage given by v O =3 2v I Assumethat+15Vand 15Vsupplyvoltagesareavailable. Solution. Theoutputcontainsadctermof+3V. Thiscanberealizedbyusingthe 15Vsupplyasone input. ThecircuitisshowninFig. 1.12(b). Forthespecifiedoutput,wecanwrite( 15) ( R F / )=3and R F /R 2 = 2. IfR F ischosentobe3kω,itfollowsthat =15kΩandR 2 =1.5kΩ. 1.4.2 The Non-Inverting Summer A non-inverting summer can be realized by connecting the inputs through resistors to the input terminal of a non-inverting amplifier. Unlike the inverting amplifier, however, the input resistors do not connect to a virtualground. Thusacurrentflowsineachinputresistorthatisafunctionofthevoltageatallinputs. This makes it impossible to define the input resistance for any one input unless all other inputs are grounded. The circuit diagram for a four-input non-inverting summer is shown in Fig. 1.13(a). To solve for the output voltage, itisconvenienttofirstmakenortonequivalentcircuitsatthev + terminalforeachoftheinputs. The circuit is shown in Fig. 1.13(b). Eq. (1.10)canbeusedtowritetheequationforv O asfollows: v O = v + ( 1+ R F = R 6 ) ( vi1 + v I2 + v I3 + v ) ( I4 ( R 2 R 3 R 4 R 5 ) 1+ R ) F R 2 R 3 R 4 R 6 (1.24) Theoutputresistanceofthecircuitiszero. Ifthev I2 throughv I4 inputsaregrounded,theinputresistance tothev I1 nodeisgivenby r in1 = +R 2 R 3 R 4 R 5 (1.25) The input resistance to the other inputs can be written similarly.

1.4. SUMMINGAMPLIFIERS xi Figure1.13: (a)fourinputnon-invertingsummer. (b)equivalentcircuitforcalculatingv O. Example 9 Design a two-input non-inverting summer which has an output voltage given by v O =8(v I1 +v I2 ) With either input grounded, the input resistance to the other input terminal is specified to be 10 kω. In addition, the current which flows in the grounded input lead is to be 1/10 the current that flows in the ungrounded lead. Solution. The circuit is shown in Fig. 1.14. By symmetry, it follows that R 2 =. For the input resistancespecification,wemusthave + R 3 =10kΩ. Ifv I2 isgrounded,i 2 isgivenbycurrentdivision i 2 = i 1 R 3 /(R 3 + ). Fori 2 = i 1 /10,wehaveR 3 /(R 3 + )=1/10. Itfollowsfromthesetwoequations thatr 3 =10/9.9kΩ=1.01kΩand =R 2 =9R 3 =(10/1.1)kΩ=9.09kΩ. Figure 1.14: Circuit for Example 9. Ifv I1 =v I2 =v I,itfollowsthatv O /v I =16. Thuswecanwritethedesignequation 16= v + v ( O R 3 = 1+ R ) F v I v + R 3 + /2 R 4 Itfollowsfromthisequationthat1+R F /R 4 =88. ThiscanbeachievedwithR 4 =270ΩandR F =23.5kΩ.

xii IDEAL OP AMP CIRCUITS 1.5 Differential Amplifiers 1.5.1 TheSingleOpAmpDiffAmp Adifferentialamplifier ordiff amp isanamplifierwhichhastwoinputsandoneoutput. Whenasignalis applied to one input, the diff amp operates as a non-inverting amplifier. When a signal is applied to the otherinput,itactsasaninvertingamplifier. Thecircuitdiagramofasingleopampdiffampisshownin Fig. 1.15(a). Superpositioncanbeusedtowritetheequationforv O asfollows: ( v O =v + 1+ R ) ( F R F R 2 v I2 =v I1 1+ R ) F R F v I2 (1.26) R 3 R 3 +R 2 R 3 R 3 where Eqs. (1.3) and(1.10) have been used. Figure1.15: (a)diffampcircuit. (b)equivalentcircuitforthespecialcaseofatruediffamp. Theoutputresistanceofthediffampiszero. Theinputresistancetothev I1 nodeisgivenby r in1 = +R 2 (1.27) Thecurrenti 2 whichflowsinthev I2 inputleadisafunctionofthevoltageatthev I1 input. Itisgivenby i 2 = v I2 v = 1 ( ) R 2 v I2 v I1 (1.28) R 3 R 3 +R 2 where v = v + has been used. The input resistance to the v I2 input is given by r in2 = v I2 /i 2. It can be seen that r in2 is a function of v I1. For example, v I1 = 0 gives r in2 = R 3, v I1 = v I2 gives r in2 = R 3 ( +R 2 )/( +2R 2 ),v I1 =+v I2 givesr in2 =R 3 (1+R 2 / ),etc. 1.5.2 TheTrueDiffAmp Theoutputvoltageofatruediffamp iszeroifv I1 =v I2. ItfollowsfromEq. (1.26)thattheconditionfor atruediffampis R 2 = R F (1.29) R 3 Toachievethiscondition, itiscommon tomake =R 3 andr 2 =R F. Inthiscase, theoutputvoltage can be written v O = R F R 3 (v I1 v I2 ) (1.30) The controlled-source equivalent circuit of the single op amp true diff amp is shown in Fig. 1.15(b).

1.5. DIFFERENTIALAMPLIFIERS xiii Example10 For the diff amp circuit of Fig. 1.15(a), it is given that =R 3 =10kΩ and R 2 =R F = 20kΩ. Solvefortheoutputvoltage,theinputresistancetothev I1 terminal,andtheinputresistancetothe v I2 terminalforthethreecases: v I1 =0,v I1 = v I2,andv I1 =+v I2. Solution. Because R F /R 3 = R 2 /, the output voltage is given by Eq. (1.30). It follows that v O = 2(v I1 v I2 ). Theinputresistancetothev I1 nodeis30kω. Asdescribedabove,theinputresistancetothe v I2 terminalisafunctionofv I1. Forv I1 =0,itis10kΩ. Forv I1 = v I2,itis6kΩ. Forv I1 =+v I2,itis 40kΩ. 1.5.3 Differential and Common-Mode Voltage Gains Figure1.16showsasingleopampdiffampcircuitwiththreesourcesatitsinput. Thetwoinputvoltages aregivenby v I1 =v CM + v D (1.31) 2 v I2 =v CM v D 2 (1.32) Thevoltagev CM iscalledthecommon-modeinputvoltage becauseitappearsequallyatbothinputs. The voltage v D is called the differential input voltage because one-half of its value appears at each input with opposite polarities. Figure 1.16: Diff amp with differential and common-mode input sources. It is often convenient to analyze diff amp circuits by expressing the input voltages as common-mode and differentialcomponents. Thevoltagesv CM andv D canbeexpressedintermsofv I1 andv I2 asfollows: v D =v I1 v I2 (1.33) v CM = v I1+v I2 2 (1.34) These two equations can be used to resolve any two arbitrary input voltages into differential and commonmodecomponents. Forexample,v I2 =0givesv D =v I1 andv CM =v I1 /2,v I2 = v I1 givesv D =2v I1 and v CM =0,v I2 =v I1 givesv D =0andv CM =v I1,etc. ByEq. (1.26),theoutputvoltageofthediffampinFig. 1.16canbewritten v O = ( v CM + v ) D R 2 2 +R 2 ( R 2 = v CM 1 R F +R 2 R 2 R 3 ( 1+ R F R 3 ) + v D 2 ) ( R F R 3 ) RF R 3 ] v CM v D 2 [ 1+ R 2(R 3 +R F ) R F ( +R 2 ) ] (1.35)

xiv IDEAL OP AMP CIRCUITS This equation can be used to define the differential and common-mode voltage gains, respectively, as follows: A d = v O = R [ F 1+ R ] 2(R 3 +R F ) v D 2R 3 R F ( +R 2 ) (1.36) A cm = v ( O R 2 = 1 R ) F v CM +R 2 R 2 R 3 (1.37) IfR F /R 2 =R 3 /,theseequationsgivea d =R F /R 3 anda cm =0. 1.5.4 The Common-Mode Rejection Ratio Foratruediffamp,thecommon-modevoltagegainiszero. Inpractice,itisdifficulttoachieveacommonmode gain that is exactly zero because of resistor tolerances. A figure of merit for the true diff amp is the ratio of its differential voltage gain to its common-mode voltage gain. This is called the common-mode rejectionratio orcmrr. Ideally,itisinfinite. TheCMRRofthecircuitinFig. 1.16isgivenby CMRR= A d A cm = [ R F 2R 3 1+ R2(R3+RF) R F(+R 2) Thisisoftenexpressedindecibelsbytherelation20log(CMRR). ] [ ] (1.38) R 2 +R 2 1 RFR1 R 2R 3 Example11 ForthediffampinFig. 1.17,solveforv O,thecurrenti,theresistanceseenbythegenerator, v I1,v I2,andthecommon-modeinputvoltage. Solution. ByEq. (1.30),v O isgivenby Figure 1.17: Circuit for Example 11. v O = R F R 3 v D Because there is a virtual short circuit between the inverting and the non-inverting op amp inputs, it follows thatiisgivenby i= v D 2R 3 Thusthegeneratorseestheresistance2R 3. Tosolveforv I1 andv I2,wecanwrite v I1 =i(r 3 +R F )= v D (R 3 +R F )= v ( ) D RF +1 2R 3 2 R 3 v I2 =v I1 v D = v D 2 ( ) RF 1 R 3

1.5. DIFFERENTIALAMPLIFIERS xv Thecommon-modecomponentofv I1 andv I2 isgivenby v CM = v I1+v I2 2 = R F v D R 3 2 = v 0 2 Thus the op amp forces a common-mode voltage at the two diff amp inputs that is equal to one-half the output voltage. 1.5.5 The Switch Hitter The single op amp diff amp circuit of Fig. 1.18(a)is known as a switch hitter. The signal applied to the non-inverting op amp input is taken from the wiper of a potentiometer. To solve for the output voltage asafunctionofthepositionofthewiper,wedenotethepotentiometerresistancefromwipertogroundby xr p, where0 x 1. Byvoltage division, itfollowsthatthevoltageatthenon-invertingopampinput is v + = xv I. The circuit is redrawn in Fig. 1.18(b) with separate sources driving the inverting and the non-inverting inputs. By superposition of the two sources, the output voltage can be written v O =2v + v I =(2x 1)v I (1.39) whereeqs. (1.3)and(1.10)havebeenused. Itfollowsthatthevoltagegainofthecircuitis2x 1. This hasthevalues 1forx=0,0forx=0.5,and+1forx=1. Thusthecircuitgaincanbevariedfrom 1 through0to+1asthepositionofthepotentiometerwiperisvaried. Figure 1.18: (a) Switch hitter. (b) Equivalent circuit. 1.5.6 TheTwoOpAmpDiffAmp AtwoopampdiffampisshowninFig. 1.19. Bysuperposition,theoutputvoltageofthiscircuitisgiven by v O = R F2 R 3 v 01 R F2 R 2 v I2 = R F1R F2 R 3 v I1 R F2 R 2 v I2 (1.40) whereeq. (1.3)hasbeenused. Thecircuitoperatesasatruediffampifeitheroftwoconditionsissatisfied. Theseare =R F1 andr 3 =R 2 or =R 2 andr F1 =R 3. Undereithercondition, theexpressionfor the output voltage reduces to v O = R F2 R 2 (v I1 v I2 ) (1.41) Theinputresistancetothev I1 inputis. Theinputresistancetothev I2 inputisr 2. Theoutputresistance is zero. Thetwoopampdiffamphastwoadvantagesoverthesingleopampdiffamp. First,theinputresistance toeitherinputisnotafunctionofthevoltageattheotherinput. Thusthecommon-modevoltageatthe inputscannotbe afunctionoftheoutputvoltageasitiswith thesingleopamp diffamp. Second, when thecircuitisusedasatruediffamp,thedifferentialvoltagegaincanbevariedbyvaryingasingleresistor withoutsimultaneouslychangingthecommon-modevoltagegain. ThisresistorisR F2. Thesingleopamp diffampdoesnothavethisfeature.

xvi IDEAL OP AMP CIRCUITS Figure1.19: Twoopampdiffamp. Example12 Design a two op amp diff amp which has a differential voltage gain of 20, a common-mode voltagegainof0,andaninputresistancetoeachinputof10kω. Solution. ForthecircuitofFig. 1.19,theinputresistancespecificationscanbemetwith =R 2 =10kΩ. For the differential gain specification, it follows from Eq. (1.41) that R F2 /R 2 = 20. Thus we must have R F2 =200kΩ. Foracommon-modegainofzero,wemusthaveeither =R F1 andr 3 =R 2 or =R 2 andr F1 =R 3. Becausewehavealreadyspecifiedthat =R 2,wemusthaveR F1 =R 3. Thevaluefor theseresistorsisarbitrary. WespecifyR F1 =R 3 =200kΩ. 1.5.7 The Instrumentation Amplifier ThediffampcircuitofFig. 1.20isknownasaninstrumentationamplifier. Insomeapplications,itiscalled anactivetransformer. Tosolveforv O,weusesuperpositionoftheinputsv I1 andv I2. Withv I2 =0, the v terminalofopamp2isatvirtualgroundandopamp1operatesasanon-invertingamplifier. ByEq. (1.10), its output voltage is given by ( v O1 = 1+ R F1 ) v I1 (1.42) Becausethereisavirtualshortcircuitbetweenthev + andv inputsofopamp1,thevoltageatthelower node of is v I1. It follows that op amp 2 operates as an inverting amplifier. By Eq. (1.3), its output voltage is given by v O2 = R F1 v I1 (1.43) Opamp3operatesasatruediffamp. ByEq. (1.30),itsoutputvoltageisgivenby Similarly,forv I1 =0,v O isgivenby By superposition, the total output voltage is v O = R F2 (v O1 v O2 )= R ( F2 R 2 R 2 v O = R F2 R 2 v O = R F2 R 2 ( ( 1+2 R F1 1+2 R F1 1+2 R F1 ) v I1 (1.44) ) v I2 (1.45) ) (v I1 v I2 ) (1.46) Thisisthevoltageoutputofatruediffamp. Theinputresistancetoeachinputoftheamplifierisinfinite. The output resistance is zero.

1.5. DIFFERENTIALAMPLIFIERS xvii Figure 1.20: Instrumentation amplifier. The instrumentation amplifier can be thought of as the cascade connection of two amplifiers. The first stageconsistsofopamps1and2. LetitsvoltagegainbedenotedbyA 1. Thesecondstageconsistsofop amp3. LetitsvoltagegainbedenotedbyA 2. Thetwogainsaregivenby A 1 = v O1 v O2 v I1 v I2 =1+2 R F1 (1.47) A 2 = v O v O1 v O2 = R F2 R 2 (1.48) ItcanbeseenthatA 1 representstheratioofadifferentialoutputvoltagetoadifferentialinputvoltage. The instrumentation amplifier is used in applications where a true diff amp is required with a very highcommon-moderejectionratio. Apotentiometerconnectedasavariableresistorinserieswith can be used to adjust the voltage gain without simultaneously changing the common-mode rejection ratio. A potentiometerconnectedasavariableresistorinserieswithr 3 canbeusedtooptimizethecmrr. Todo this experimentally, the two inputs are connected together and a common-mode signal voltage applied. The potentiometerinserieswithr 3 isadjustedforminimumoutputvoltage. Example 13 Design an instrumentation amplifier which has a differential voltage gain of 100 (a decibel gainof40db)andacommon-modevoltagegainofzero. Solution. Thegainof100mustbedividedbetweenthetwostagesofthecircuit. Itisconvenienttogive theinputstage,consistingofopamps1and2,againof10andthesecondstage,consistingofopamp3,a gainof10. UsingEqs. (1.47)and(1.48),wecanwritethetwodesignequations 1+2 R F1 =10 and R F2 R 2 =10 Withtwoequationsandfourunknowns,itisnecessarytoassignvaluestotwooftheresistors. LetR F1 = R F2 =10kΩ. ItfollowsthatR 2 =1kΩand =(10/4.5)kΩ=2.22kΩ. 1.5.8 The Differential Output Amplifier Figure 1.21 shows the circuit diagram of a differential output amplifier. This circuit has two output voltages whichhaveoppositepolarities. Thatis,ifv O1 ispositive,v O2 willbenegative,andviceversa. Becausethe lowernodeofresistor isatvirtualground,eq. (1.10)canbeusedtowriteforv O1 ( v O1 = 1+ R F1 ) v I (1.49)

xviii IDEAL OP AMP CIRCUITS Becausethereisavirtualshortbetweentheinvertingandnon-invertinginputstoopamp1,theuppernode of seestheinputvoltagev I. ThusEq. (1.3)canbeusetowriteforv O2 v O2 = R F2 v I (1.50) Figure 1.21: Differential output amplifier. Inmostapplicationsofthedifferentialoutputamplifier,theconditionv O2 = v O1 isdesired. Whenthis is satisfied, the amplifier is said to be a balanced differential output amplifier. This requires the condition 1+R F1 / =R F2 / whichreducesto R F1 =R F2 (1.51) Inthiscase,theoutputvoltagescanbewritten The differential output voltage is given by v O1 = v O2 = R F2 v I (1.52) v O1 v O2 = 2R F2 v I (1.53) Example 14 Design a balanced differential output amplifier with an open-circuit voltage gain of 4, an input resistanceof10kω,andabalancedoutputresistanceof600ω. Theamplifieristodrivea600Ωload. Ifthe maximumpeakoutputvoltagefromeachopampis±12v,calculatethemaximumpeakloadvoltageandthe output level in dbm for a sine wave input signal. (The dbm is the decibel output power referenced to the powerp ref =1mW.) Solution. ThecircuitisshowninFig. 1.22. Fortheinputresistancespecification,wehaveR i =10kΩ. Foranopen-circuitvoltagegainof4,itfollowsfromEq. (1.53)that2R F2 / =4. Thiscanbesatisfiedby choosingr F2 =20kΩand =10kΩ. Eq. (1.51)givesR F1 =10kΩ. Toachievea600Ωbalancedoutput resistance, we must have R O1 =R O2 and R O1 +R O2 =600. It follows thatr O1 =R O2 =300Ω. If the voltageoutputofopamp1peaksat+12v,thevoltageoutputfromopamp2peaksat 12Vandthepeak loadvoltageisv P =24 600/(600+600)=12V,wherevoltagedivisionhasbeenused. Theoutputlevel indbmisgivenby ( ) ( v 2 OutputLevel =10log P /2R L 12 2 ) /1200 =10log =20.8dBm 0.001 P ref

1.6. OP AMP DIFFERENTIATORS xix 1.6 Op Amp Differentiators 1.6.1 The Ideal Differentiator Figure 1.22: Circuit for Example 14. A differentiator is a circuit which has an output voltage that is proportional to the time derivative of its inputvoltage. Fig. 1.23givesthecircuitdiagramofanopampdifferentiator. Thecircuitissimilartothe invertingamplifierinfig. 1.3withtheexceptionthatresistor isreplacedbyacapacitor. Itfollowsthat Eq. (1.3) can be used to solve for the voltage gain transfer function of the differentiator by replacing with the complex impedance of the capacitor. The voltage gain transfer function is given by V o V i = R F (1/C 1 s) = R FC 1 s (1.54) Figure 1.23: Ideal differentiator. Because a multiplication by s in the complex frequency domain is equivalent to a differentiation in the timedomain,itfollowsfromtheaboveequationthatthetimedomainoutputvoltageisgivenby v O (t)= R F C 1 dv I (t) dt (1.55) ThusthecircuithasthetransferfunctionofaninvertingdifferentiatorwiththegainconstantR F C 1. Because the gain constant has the units of seconds, it is called the differentiator time constant. The output resistance ofthecircuitiszero. TheinputimpedancetransferfunctionisthatofthecapacitorC 1 tovirtualground given by Z in = 1 (1.56) C 1 s With s = jω, it follows that Z in 0 as ω becomes large. This is a disadvantage because a low input impedance can cause large currents to flow in the input circuit.

xx IDEAL OP AMP CIRCUITS 1.6.2 The Modified Differentiator Withs=jω,itfollowsfromEq. (1.54)thatthemagnitudeofthevoltagegainofthedifferentiatorisωR F C 1. For large ω, the gain can get very high. This is a disadvantage in circuits where out-of-band high-frequency noisecanbeaproblem. Tolimitthehigh-frequencygain,aresistorcanbeusedinserieswithC 1 asshown in Fig. 1.24(a). This also has the advantage that the high-frequency input impedance does not approach zero. Athigh frequencieswherec 1 isashort, the gain magnitude islimited tothe value R F / and the inputimpedanceapproaches. Thevoltagegaintransferfunctionofthecircuitwith isgivenby V o R F = V i +1/C 1 s = FC 1 s 1+ C 1 s (1.57) Thisisoftheformofthetransferfunctionofadifferentiatormultipliedbythetransferfunctionofalow-pass filterwhichhasapoletimeconstant C 1. Figure1.24: (a)modifieddifferentiator. (b)bodeplotfor V o /V i. The Bode magnitude plot for the transfer function of Eq. (1.57) is given in Fig. 1.24(b). For ω << 1/ C 1,theasymptoticplotexhibitsaslopeof+1dec/decwhichistheproperslopeforadifferentiator. In thisband,thevoltagegainisgivenbyv o /V i = jωrf C 1. Atω=1,themagnitudeofthegainisR F C 1. Forω>>1/ C 1,theasymptoticslopeis0andthemagnitudeofthegainshelvesatthevalueR F /. It followsthatthecircuitwith actsasadifferentiatoronlyforfrequenciessuch thatω<<1/ C 1. The inputimpedancetransferfunctionofthecircuitwith isgivenby Z in = + 1 C 1 s = 1+C 1 s C 1 s (1.58) Thisisoftheformofaconstantmultipliedbythereciprocalofahigh-passtransferfunction. Fors=jω, itfollowsthat Z in asωbecomeslarge. Example15 Design a modified differentiator which has a time constant of 10ms and a pole frequency of 1kHz. Fora1Vpeaksine-waveinputsignalat100Hz,calculatethepeaksinewaveoutputvoltageandthe relative phase of the output voltage. Solution. The circuit is shown in Fig. 1.24(a). For the gain constant specification, we must have R F C 1 =0.01. Ifwe letc 1 =0.1µF, it followsthat R F =100kΩ. For the pole frequency of1000hz, we must have C 1 = 1/2π1000. This gives = 10,000/2π = 1.59kΩ. From Eq. (1.57), the voltage gain magnitudeatf=100hzisgivenby V o V i = R FC 1 (j2π100) 1+j2π100 C 1 = 0.01 2π100 =6.25 1+0.1 2

1.7. THEINTEGRATOR xxi Fora1Vpeakinputsinewaveat100Hz,itfollowsthatthepeakoutputvoltageis6.25V. Itfollowsfrom Eq. (1.57)thatthephaseoftheoutputsignalwithrespecttotheinputsignalisgivenby ϕ=+90 o tan 1 (2π100 C 1 )=84.3 Aperfectdifferentiatorwould haveaphase of +90. Thusthereisaphaseerrorof 5.7. Notethat the negative sign in Eq. (1.57) does not affect the phase. This is because a negative sign indicates an inversionwhereasaphaseshiftisassociated withashiftintime. Ifasinewaveisobservedonthescreen ofanoscilloscope,aninversionwouldflipthesinewaveaboutthetimeaxis. Aphaseshiftwouldshiftthe position of the zero crossings along the time axis. 1.7 The Integrator 1.7.1 The Ideal Inverting Integrator Anintegrator isacircuitwhichhasanoutputvoltagethatisproportionaltothetimeintegralofitsinput voltage. The circuit for the integrator can be obtained by interchanging the resistor and the capacitor in the differentiator of Fig. 1.23. The circuit is shown in Fig. 1.25. The voltage gain transfer function is obtained fromeq. (1.3)byreplacingR F withthecompleximpedanceofthecapacitorc F toobtain V o = (1/C Fs) = 1 V i C F s (1.59) Figure 1.25: Inverting integrator. Because a division by s in the complex frequency domain is equivalent to an integration in the time domain, it follows from this equation that the time domain output voltage is given by v O (t)= 1 t v I (τ)dτ (1.60) C F Thusthecircuithasthetransferfunctionofaninvertingintegratorwiththegainconstant1/ C F. Because C F hastheunitsofseconds,itiscalledtheintegratortimeconstant. Theinputresistancetothecircuit is. Theoutputresistanceiszero. 1.7.2 The Modified Inverting Integrator Atzerofrequency,C F isanopencircuitandtheopampintheintegratorcircuitlosesfeedback. Fornon-ideal opamps,thiscancauseundesirabledcoffsetvoltagesattheoutput. Toprovidefeedbackatdc,aresistor canbeusedinparallelwithc F asshowninfig. 1.26(a). AtlowfrequencieswhereC F isanopencircuit, themagnitudeofthevoltagegainislimitedtothevaluer F /. Thetransferfunctionforthevoltagegain oftheintegratorwithr F isgivenby V o = R F (1/C F s) = 1 V i C F s R FC F s 1+R F C F s (1.61)

xxii IDEAL OP AMP CIRCUITS whereeq. (1.3)hasbeenused. Thisisoftheformofthetransferfunctionofanidealintegratormultiplied bythetransferfunctionofahigh-passfilterwhichhasthepoletimeconstantr F C F. TheBodemagnitude plotfor the transfer function is given in Fig. 1.26(b). For ω <<1/R F C F, the plotexhibits a slope of0. Forω>>1/R F C F,theslopeis 1dec/decwhichistheproperslopeforanintegrator. Itfollowsthatthe circuitwithr F actsasanintegratoronlyforfrequenciessuchthatω>>1/r F C F. Figure1.26: (a)modifiedinvertingintegrator. (b)bodemagnitudeplotfor V o /V i. Example16 Designamodifiedintegratorwhichhasatimeconstantof0.1sandapolefrequencyof1Hz. Fora1V peaksine-waveinputsignalat10hz,calculatethepeaksine-waveoutputvoltageandtherelative phase of the output voltage. Solution. ThecircuitisshowninFig. 1.26(a). Forthetimeconstantspecification,wehave C F =0.1. If wetakec F =0.1µF,itfollowsthat =1MΩ. Forthepolefrequencyof1Hz,wemusthaveR F C F =1/2π. This gives R F = 1/ ( 2π 0.1 10 6) = 1.59MΩ. From Eq. (1.61), the gain magnitude at f = 10Hz is given by V o V i = 1 j2π10r FC F j2π10 C F 1+j2π10R F C F = 1.59 =0.158 1+10 2 Fora1Vpeakinputsinewaveat100Hz,itfollowsthatthepeakoutputvoltageis0.158V. ItfollowsfromEq. (1.61)thatthephaseoftheoutputsignalwithrespecttotheinputsignalisgivenby ϕ= tan 1 (2π10R F C F )= 84.3 Aperfectintegratorwouldhaveaphaseof 90. Thusthereisaphaseerrorof+5.7. Asisdiscussedin Example15,thenegativesigninEq. (1.61)indicatesthattheoutputsignalisinvertedwithrespecttothe input signal and does not represent a phase shift. 1.7.3 The Non-Inverting Integrator The circuit diagram of a non-inverting integrator is shown in Fig. 1.27(a). The voltage output from the op ampisfedbacktobothitsinvertinginputandtoitsnon-invertinginput. Thusthecircuithasbothpositive and negative feedback. To solve for the voltage gain transfer function, it is convenient to make two Norton equivalentcircuitsatthev + node,onelookingtowardtheinputthroughtheleftrandtheotherlooking towardtheoutputthroughtherightr. ThecircuitobtainedisshowninFig. 1.27(b),wherethetwoparallel

1.7. THEINTEGRATOR xxiii resistorsarecombinedintoasingleresistorofvaluer/2. BecausethereisavirtualshortbetweentheV + andthev inputs,wecanwrite ( )( V o 2 = Vi R +V o R R 2 1 ) 1 =(V i +V o ) (1.62) Cs 1+RCs/2 This equation can be solved for the voltage gain transfer function of the circuit to obtain V o = 2 V i RCs (1.63) This is the transfer function of a non-inverting integrator with the gain constant 2/RC. The time constant of the integrator is RC/2. Figure 1.27: (a)non-invertingintegrator. (b) Equivalent circuitfor calculating V o. (c)equivalentcircuit forz in. TheinputcurrentinthecircuitofFig. 1.27(a)canbesolvedforasfollows: I i = V i V + = V i V = V ( i 1 1 ) R R R RCs (1.64) wherev =V o /2hasbeenused. Thisequationcanbesolvedfortheinputimpedancetransferfunctionto obtain Z in = V i = R ( R 2 Cs ) I i R+( R 2 (1.65) Cs) TheequivalentcircuitwhichhasthisimpedanceisaresistorRinparallelwithanegativeinductor R 2 C. The equivalent circuit is given in Fig. 1.27(c). Because the inductor is a short circuit at zero frequency, it followsthattheinputimpedancetothecircuitiszeroforadcsource. Example17 Thenon-invertingintegratorofFig. 1.27(a)hasthecircuitelementvaluesR=1kΩandC= 1µF. Forasinewaveinputsignal,calculatethevoltagegainofthecircuitatthefrequencyf=100Hz. In addition, calculate numerical values for the circuit elements in the equivalent circuit for the input impedance. Solution. Thevoltagegainatf=100HziscalculatedfromEq. (1.63)asfollows: V o 2 = V i 10 3 j2π100= j3.17 10 6

xxiv IDEAL OP AMP CIRCUITS FromEq. (1.65),itfollowsthattheinputimpedancecircuitconsistsofa1000Ωresistortogroundinparallel withanegativeinductortogroundhavingthevalue 1000 2 10 6 = 1H. 1.8 Low-Pass Amplifiers 1.8.1 The Inverting Low-Pass Amplifier Thissectioncoversseveralofthemanyopampcircuitswhichhaveavoltagegaintransferfunctionthatisof the form of single-pole low-pass and low-pass shelving transfer functions. Fig. 1.28(a) shows the circuit of an invertinglow-passamplifier. ThevoltagegainisobtainedfromEq. (1.3)byreplacingR F withr F (1/C F s). Itisgivenby V o = R F (1/C F s) = R F V i 1 1+R F C F s (1.66) Thisisoftheformofagainconstant R F / multipliedbyalow-passtransferfunctionhavingapoletime constant R F C F. The Bode magnitude plot for the transfer function is given in Fig. 1.28(b). The input resistanceofthecircuitis. Theoutputresistanceiszero. Figure1.28: (a)invertinglow-passamplifier. (b)bodemagnitudeplotfor V o /V i. Example 18 Design an inverting low-pass amplifier circuit which has an input resistance of 10 kω, a lowfrequencyvoltagegainof 10,andapolefrequencyof10kHz. Solution. ThecircuitdiagramoftheamplifierisshowninFig. 1.28(a). Foraninputresistanceof10kΩ, wehave =10kΩ. ThevoltagegaintransferfunctionisgivenbyEq. (1.66). Foralow-frequencygainof 10,wehaveR F =10 =100kΩ. Forapolefrequencyof10kHz,wehaveC F =1/2π10 4 R F =159pF. Asecond invertinglow-passamplifiercircuitisshown infig. 1.29(a). ThecurrentsI 1, I 2, andi F are given by V i I 1 = (1.67) +(1/Cs) R 2 1/Cs I 2 =I 1 R 2 +1/Cs =I 1 1 1+R 2 Cs (1.68) I F = V o R F (1.69) whereitisassumedthatthev opampinputisatvirtualgroundandcurrentdivisionhasbeenusedfor I 2. ThevoltagegainofthecircuitcanbeobtainedfromtherelationI 2 +I F =0toobtain V o = R F 1 V i +R 2 1+ R 2 Cs (1.70)

1.8. LOW-PASSAMPLIFIERS xxv Thisisoftheformofagainconstant R F /( +R 2 )multipliedbyalow-passtransferfunctionhavinga poletimeconstant( R 2 )C. TheBodemagnitudeplotforthetransferfunctionisgiveninFig. 1.29(b). Figure1.29: (a)invertinglow-passamplifier. (b)bodemagnitudeplotfor V o /V i. (c)bodemagnitudeplot for Z in. Theoutputresistanceofthecircuitiszero. Theinputimpedanceisgivenby Z in = + 1 Cs R 2=( +R 2 ) 1+( R 2 )Cs 1+R 2 Cs (1.71) Thistransferfunctionisintheformofalow-passshelvingfunctionhavingapoletimeconstantR 2 C and azerotimeconstant( R 2 )C. TheBodemagnitudeplotoftheimpedanceisgiveninFig. 1.29(c). The low-frequencyimpedanceis +R 2. Asfrequencyisincreased,theimpedancedecreasesandshelvesatthe value. Example 19 Specify the circuit element values for the circuit of Fig. 1.29(a) for an inverting voltage gain ofunityandapoletimeconstantof75µs. Whatisthepolefrequencyinthevoltage-gaintransferfunction? Solution. Let C = 0.01µF and R 2 =. It follows from Eq. (1.31) that ( R 2 )C = ( /2)C = 75 10 6. Solution for and R 2 yields = R 2 = 15kΩ. For an inverting voltage gain of unity, we must have R F = +R 2 = 30kΩ. The pole frequency in the transfer function has the frequency f =1/ ( 2π75 10 6) =2.12kHz. 1.8.2 The Non-Inverting Low-Pass Amplifier Figure 1.30(a) shows a non-inverting low-pass amplifier consisting of a non-inverting amplifier with a RC low-passfilteratitsinput. Thevoltagegaintransferfunctionofthecircuitisgivenby V o = V + V o = 1/Cs ( 1+ R ) ( F = 1+ R ) F 1 (1.72) V i V i V + R+1/Cs 1+RCs where voltage division and Eq. (1.10) have been used. This is ofthe form of a gain constant 1+R F / multiplied by the transfer function of a low-pass filter having a pole time constant RC. The Bode magnitude

xxvi IDEAL OP AMP CIRCUITS plotforthetransferfunctionisgiveninfig. 1.30(b). Theoutputresistanceofthecircuitiszero. Theinput impedance is given by Z in =R+ 1 Cs =R 1+RCs (1.73) RCs ThisisoftheformofaresistorRmultipliedbythereciprocalofahigh-passtransferfunction. Figure1.30: (a)non-invertinglow-passamplifier. (b)bodemagnitudeplotfor V o /V i. Example20 The non-inverting amplifier of Fig. 1.30(a) is to be designed for a voltage gain of 12. The inputlow-passfilteristohaveacutofffrequencyof100khz. Specifytheelementvaluesforthecircuit. Solution. Tomeetthecutofffrequencyspecification, itfollowsfromeq. (1.72)thatRC =1/ ( 2π10 5). EitheravalueforRoravalueforC mustbespecifiedbeforetheothercanbecalculated. LetC=510pF. ItfollowsthatR=3.12kΩ. Foragain of12, wemusthave1+r F / =12. Ifwechoose =1kΩ, it followsthatr F =11kΩ. 1.8.3 The Non-Inverting Low-Pass Shelving Amplifier The circuit diagram of a non-inverting low-pass shelving amplifier is shown in Fig. 1.31(a). The voltage gainisobtainedfromeq. (1.8)byreplacingR F withr F (1/C F s). Itisgivenby V o =1+ R ( F (1/C F s) = 1+ R ) F 1+RF C F s (1.74) V i 1+R F C F s Thisisoftheformofagainconstant1+R F / multipliedbyalow-passshelvingtransferfunctionhavinga poletimeconstantr F C F andazerotimeconstant(r F )C F. TheBodemagnitudeplotforthevoltage gain is shown in Fig. 1.31(b). The low-frequency gain is 1+R F /. As frequency is increased, the gain decreases and shelves at unity. Example21 ThecircuitofFig. 1.31(a)istobedesignedforalow-frequencygainof2(a6dBboost). The zero frequency in the transfer function is to be 100 Hz. Specify the circuit element values and calculate the frequencyatwhichthevoltagegainis3db. Solution. For a low-frequency gain of 2, it follows from Eq. (1.74) that 1+R F / = 2, which gives R F =. Forthezerointhetransferfunctiontobeat100Hz,itfollowsthatR F C F =1/(2π100). Ifwe choosec F =0.1µF,itfollowsthat =R F =31.8kΩ. Withs=j2πf,thevoltagegaintransferfunction can be written V o =2 1+jf/100 V i 1+jf/50

1.9. HIGH-PASSAMPLIFIERS xxvii Figure1.31: (a)non-invertinglow-passshelvingamplifier. (b)bodemagnitudeplotfor V o /V i. Atthe3dBboostfrequency,wehave V o /V i 2 =1/2. Thisconditiongives 1+(f/100) 2 1+(f/50) 2 = 1 2 Thiscanbesolvedforf toobtainf =100/ 2=70.7Hz. 1.9 High-Pass Amplifiers 1.9.1 The Inverting High-Pass Amplifier This section covers several of the many op amp circuits which have a voltage gain that is of the form of high-pass and high-pass shelving transfer functions. Fig. 1.32(a) shows an inverting high-pass amplifier circuit. ThevoltagegainisobtainedfromEq. (1.3)byreplacing with +1/(C 1 s). Itisgivenby V o R F = V i +1/(C 1 s) = R F C 1 s 1+ C 1 s (1.75) This is of the form of a gain constant R F / multiplied by a high-pass transfer function having a pole time constant C 1. The Bode magnitude plot for the voltage gain is given in Fig. 1.32(b). The output resistance of the circuit is zero. The input impedance transfer function is given by Z in = + 1 C 1 s = 1+C 1 s C 1 s (1.76) Thisisoftheformofaresistance multipliedbythereciprocalofahigh-passtransferfunction. Example22 Design an inverting high-pass amplifier circuit which has a gain of 10 and a pole time constant of 500µs. The input impedance to the circuit is to be 10kΩ or higher. Calculate the lower halfpower cutoff frequency of the amplifier. Solution. The circuit is shown in Fig. 1.32(a). The voltage-gain transfer function is given by Eq. (1.75). For the gain specification, we must have R F / = 10. For the pole time constant specification, we must have C 1 = 500 10 6. Because there are three unknowns and only two equations, one of the circuit elements must be specified before the others can be calculated. Eq. (1.76) shows that the lowest value of theinputimpedanceis. Thuswemusthave 10kΩ. If 10kΩ, itfollowsthatc 1 0.05µF. LetuschooseC 1 =0.033µF. Itfollowsthat =15.2kΩandR 2 =152kΩ. Thelowerhalf-powercutoff frequencyisf=1/ ( 2π 500 10 6) =318Hz.

xxviii IDEAL OP AMP CIRCUITS Figure1.32: (a)invertinghigh-passamplifier. (b)bodemagnitudeplotfor V o /V i. 1.9.2 The Non-Inverting High-Pass Amplifier Figure 1.33(a) shows a non-inverting high-pass amplifier circuit. The voltage gain transfer function is given by V o = V + V ( o R = 1+ R ) ( F = 1+ R ) F RCs (1.77) V i V i V + R+1/(Cs) 1+RCs wherevoltagedivisionandeq. (1.10)havebeenused. Thisisoftheformofagainconstant(1+R F / ) multiplied by a single-pole high-pass transfer function having a pole time constant RC. The Bode magnitude plot for the voltage gain is given in Fig. 1.33(b). The output resistance of the circuit is zero. The input impedance transfer function is given by Z in =R+ 1 Cs =R 1+RCs RCs ThisisoftheformofaresistanceRmultipliedbythereciprocalofahigh-passtransferfunction. (1.78) Figure1.33: (a)non-invertinghigh-passamplifier. (b)bodemagnitudeplotfor V o /V i. Example 23 Design a non-inverting high-pass amplifier which has a gain of 15 and a lower cutoff frequency of20hz. Theinputresistancetotheamplifieristobe10kΩinitspassband. Solution. ThecircuitisshowninFig. 1.33(a). Intheamplifierpassband,C isashortcircuit. Tomeet the input resistance specification, we must have R = 10 kω. The voltage-gain transfer function is given by Eq. (1.77). Foralowerhalf-powercutofffrequencyof 20Hz, wemusthaverc =1/(2π20). Solution for C yields C = 0.796 µf. For the gain specification, we must have 1+R F / = 15 or = R F /14. If R F =56kΩ,itfollowsthat =4kΩ.

1.10. THE OP AMP AS A COMPARATOR xxix 1.9.3 The Non-Inverting High-Pass Shelving Amplifier The circuit diagram of a non-inverting high-pass shelving amplifier is shown in Fig. 1.34(a). The voltage gainisgivenbyeq. (1.10)with replacedby +1/(C 1 s). Itfollowsthatthegaincanbewritten V o R F =1+ V i +(1/C 1 s) = 1+(R F+ )C 1 s 1+ C 1 s (1.79) Thisisoftheformofahigh-passshelvingtransferfunction havingapoletime constant C 1 andazero timeconstant(r F + )C 1. TheBodemagnitudeplotforthevoltagegainisshowninFig. 1.34(b). Itcan be seen fromthe figure that the gain at low frequencies is unity. At high frequencies, the gain shelves at 1+R F /. Figure1.34: (a)non-invertinghigh-passshelvingamplifier. (b)bodemagnitudeplotfor V o /V i. Example24 Design a high-pass shelving amplifier which has unity gain at low frequencies, a pole in its transferfunctionwithatimeconstantof75µs,andazerowithatimeconstantof7.5µs. Whatarethepole and zero frequencies and what is the gain at high frequencies? Solution. The circuit is shown in Fig. 1.34(a). The voltage-gain transfer function is given by Eq. (1.79). Forthepoletimeconstantspecification, we musthave C 1 =7.5µs. Forthe zerotimeconstant specification, we musthave ( +R F )C 1 =75µs. Because there are three circuit elements and only two equations,wemustspecifyoneelementinordertocalculatetheothertwo. LetC 1 =0.001µF. Itfollowsthat =7.5kΩandR 2 =75kΩ =67.5kΩ. Thezerofrequencyisf z =1/ ( 2π 75 10 6) =2.12kHz. Thepolefrequencyisf p =1/ ( 2π 7.5 10 6) =21.2kHz. Thegainathighfrequenciesis1+R F / = 1+67.5/7.5=10. 1.10 TheOpAmpasaComparator 1.10.1 The Inverting Comparator A comparator is an active circuit element which has two input terminals and one output terminal. The output voltage exhibits two stable states. The output state depends on the relative value of one input voltagecomparedtotheotherinputvoltage. Theopampisoftenusedasacomparator. Fig. 1.35(a)shows the circuit diagram of an op amp used as an inverting comparator. The voltage applied to the non-inverting inputisthedcreferencevoltagev REF. Theoutputvoltageisgivenby v O =A(V REF v I ) (1.80)

xxx IDEAL OP AMP CIRCUITS whereaisthevoltagegainoftheopamp. Foranidealopamp,weassumethatA. Thisimpliesthat v O forv I <V REF andv O forv I >V REF. However,aphysicalopampcannothaveaninfinite outputvoltage. LetusdenotethemaximumvalueofthemagnitudeoftheoutputvoltagebyV SAT. Wecall V SAT thesaturationvoltage oftheopamp. Figure1.35: (a)invertingcomparator. (b)plotofv O versusv I. For an ideal op amp that exhibits saturation of its output voltage, the output voltage of the inverting comparator circuit in Fig. 1.35(a) can be written v O =V SAT sgn(v REF v I ) (1.81) wheresgn(x)isthesignum orsignfunction definedbysgn(x)=+1forx>0andsgn(x)= 1forx<0. Theplotofv O versusv I forthecircuitisgiveninfig. 1.35(b). 1.10.2 The Non-Inverting Comparator Fig. 1.36(a) shows the circuit diagram of an op amp used as a non-inverting comparator. The output voltage ofthecircuitisgivenby v O =V SAT sgn(v I V REF ) (1.82) Thegraphofv O versusv I isgiveninfig. 1.36(b). Figure1.36: (a)non-invertingcomparator. (b)plotofv O versusv I. 1.10.3 The Comparator with Positive Feedback or Schmitt Trigger Positive feedback is often used with comparator circuits. The feedback is applied from the output to the non-invertinginput ofthe op amp. Thisis in contrast tothecircuits covered in the precedingsections of this chapter where feedback is applied to the inverting input. (The non-inverting integrator is an exception. This circuit uses feedback to both op amp inputs.) Fig. 1.37(a) gives the circuit diagram of an inverting op amp comparator with positive feedback. The circuit is also called a Schmitt trigger. It is named after OttoH.Schmittwhowasagraduatestudentwhenheinventeditin1934. ThecapacitorC F inthefigureis assumedtobeanopencircuitinthefollowing. Thiscapacitorisoftenusedtoimprovetheswitchingspeed

1.10. THE OP AMP AS A COMPARATOR xxxi ofacomparatorbyincreasingtheamountofpositivefeedbackathighfrequencies. Ithasnoeffectonthe input voltage at which the op amp switches states. Figure1.37: (a)invertingcomparatorwithpositivefeedback. (b)plotofv O versusv I. TheoutputvoltagefromthecircuitofFig. 1.37(a)canbewritten v O =V SAT sgn(v + v I ) (1.83) Becausev O hasthetwostablestatesv O =+V SAT andv O = V SAT,itfollowsthatv + canhavetwostable states given by V A =V REF R F R F + V SAT R F + (1.84) V B =V REF R F R F + +V SAT R F + (1.85) where superposition and voltage division have been used for each equation. For v I < V A, it follows that v O =+V SAT. Forv I >V B,itfollowsthatv O = V SAT. ForV A <v I <V B,v O canhavetwostablestates, i.e. v O =±V SAT. Thegraphofv O versusv I isgiveninfig. 1.37(b). The value of v O for V A < v I < V B depends on whether v I increases from a value less than V A or v I decreasesfromavaluegreaterthanv B. Thatis,thecircuithasmemory. Ifv I <V A initiallyandv I begins toincrease,v O remainsatthe+v SAT stateuntilv I becomesgreaterthanv B. Atthispointv O switchesto the V SAT state. If v I >V B initially andv I begins to decrease, v O remains atthe V SAT state until v I becomeslessthanv A. Thenv O switchestothe+v SAT state. Thepathforv O onthegraphinfig. 1.37(b) is indicated with arrows. The loop in the graph is commonly called a hysteresis loop. Example25 The Schmidt trigger circuit of Fig. 1.37(a) has the element values R F = 1MΩ and = 33kΩ. IfV REF =3VandtheopampsaturationvoltageisV SAT =12V,calculatethetwothresholdvoltages V A andv B. Solution. By Eqs. (1.84) and(1.85), we have 1 V A =3 1+0.033 1 V B =3 1+0.033 12 0.033 1+0.033 =2.52V +12 0.033 1+0.033 =3.29V