Ch19_OpAmpCircuit

7/27/2019 Ch19_OpAmpCircuit

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l9-1 Comparatorl9-2 SummineAmplifieril9-3 Integratorsand ifferentiator19-4 Orcillatorl9-5 ActiveFiltefi19-6 VoltageRegulators

Application $ignrnenbPuttingYourKnowledge

Ep.*Afifi EJRC$TT$uted in sucha wide varietyofapplicauon that

o (over rll ofthem in one chapteror even nexamine ome ofmenta applic.tionr to illuirate theofthe op-amp and to giveyou a foundation nop-amp circuitJ-


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Aplatn the baricoperatrioDf a compator.irclitADaryzeumminSahplifie6, avera8rngmpJifier, nd lcrting

Explainne opeDtion ofop-amp integrator and differeotiatorDncu$the operationof rveral typr ofop-amp olci rtorRecotni& and evaluatebzricop amp fitteBDercrtbehe operationof baic rerer and rhunrvottage

A faul9 FM {ereo receiver the topic ofthir applicationagignmenLAfter a preliminary heck,you decide hattheduaFpolarib/po!r'er upply hat provider112 Vto alttheop-ampr n both channeLofthe receive. r fautty.Thepowerrupply use'poritiveaDdnegativ ntegrated i.cuitvoltageregulator.After rtudying hn chapte, yo! rhoutdable o complete he applicationasignment.

study aidr for thir chapter are availabtathttpJ/\,w.prenhall.com/f loyd

"r 5ummingamplifier,r Averaging mplifier

r" Wien-bridgeo(illatorir TriangulaFwaveJciliator' Relaxation (illator

t03


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r BAS|C P-AMP tRCUtrS

ATORSOperationalamplifiers are o{len used o compare he amplitudeof one voltagewithanotherIn this application, he op-amps used n the open_looponfiguration,withthe nput voltageon one nput anda reference oltageon the other.After completing his section,you shouldbe able o. Ex?lain the basic operation of s comparator circuitr Discusszerolevel detectionr Discussnonzercleveldetection

rZero-Level etection- Oneapplicationofthe op-ampusedasacomptr.tor is to determinewhen an nputexceeds ceriain evel. Figure 19 I (a)showsa zero-leveldetectorNotice hat heinput ( ) is grcundedand he nput signalvoltage s applied o the noninverting nput(+)Because f thehigh open-loop oltagegain,a very small differcncevoltagebetween heinputs drives he amplilier irto saturation, ausinghe output voltage o go to its limit. Forexample. onsideran op-amphavingAd : 100,000.A voltagedifferenceof only 0.25mvbetween he inpuls could producean outputvoltageof (0.25mVX100,000)= 25 V iftlEop-ampwerecapable.Howevet sincemostop_3mpsaveoutputvoltage imitations of lessthana 15Y the devicewould be driven nto saturation. or many conparisonapplications'specialop-ampcomparalors re selected.

u,,

(b), t f r G U t E t t - tThe op-amp ar a zero-lryel detector

Figure 19 I (b) shows he result of a sinusoidalnpui voltageapplied o theinput of the zerc-leveldetector.When the sine wave s negative, heoutput s at its maxi.mum negativeevel. when the sinewave crosses , the anplifier is ddven to its oppositestateand he outputgoes o its ma\imun positive evel, as shown.As you can see, he zero-leveldetectorcan beused as a squaringcircuit to produceasquarewave rom a sinewave.'Non2ero-lvel Detection

The zero-leveldetectorn Figure 19-1 canbemodifled o detectvoltages ther hanzerobyconnecting fixed rcference oltage o the nverting nput (-). asshown n Figlre 19-2(a).


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col,tPARATORSr tO

moreFactical arrangements shown n paft (b) using a voltagedivider to set he reference oltageas ollows:

rmF, the output remains at the maximum negative level. W}len the input voltage ex-the referencevoltage, he outputgoes o its maximumpositive state,as shown n

u* =ujt l*u;where+vis thepositiveop-ampsupply voltage.As long asdre nput voltase(va) is less

Equation9-1

Figure 9 2(c) itha sinusoidalnputvoltage.

vt,

V*t 0

No.ze@-l*l detector.

The nput signal n Figure 19-3(a) s applied o the comparator ircuit in pan O).DIaw the output showing ts Foper relationship o the nput signal.Assume hat hemaxinurn output evels ol the op-ampare :t 12V

! FTGURE 9-2

EXAMPLE9-1


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r BA5|COP-AMPC|RCU|TS

r ! F IGURE 9- !

R , l 0 l o1 " , , , + Y = ' - - - - { + 1 5 V J I . b l \ R R , 8 . 2 t Q r . 0 t Q -As shown n Fieure 19-4, ea.h time the nput exceeds+ 1.63Y the output voltageswitches o its + 12V level, and each ime the nputgoesbelow +1.63 Y the outputswitches ack o ts 12V level.: . F IGURE 9-4

1.63

RelatedPrcblem- Determinehe eferenceoltagen Fjgure 9 3 if Rl = 22 kO andR, : 3.3kO.

Sol.dton The reference oltage s setby Rr and R .

Open ile E19-01onyour Multisim CD-ROM. Measue the output voltagewaveformand determine f it matches he waveform n Fiqure 19-4.*Answersare at the end of the chapter


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SUMMINGAMPLIFIER5 T 90

SECT|ON 9-rREYIEW 1. What i the referenceoitageor h compa.:torn FiSure9-5?2. Draw he outputwa\formor Fi8ure9-5 whena rinewa\with a 5 V peaknappliedo the npuLAnM6 are rt th end of the

. ' f r G u R E - 5

] ] .1?-2 5UMMING MPLIFIERsThe sumrninganplifier is a variationof tbe nvening op-amp onfigurarion overednChapter 8. The summing ampliffer has wo or more npuls,and ts outpurvolrage . ffipropor'lionalo the negaiiveof thealgebraic umof irs nput voltages.n rhis secrion. ouwill seehow a summingamplifierworks.andyouwill leamabout he averagingamplifer and drescalingamplifier,which arevariationsof thebasicsxmmingampUfier.Aftercompleti.ghis section,ou should eable o. Anslyze summing amplifiers, avraging amplifiers, and scaling amplifiersr Calculate umming mplilier ulput olrageor givennprts or bothunirygainandnonunity ainconditionsr Calculaie he outpurvoltage or an averaginganplifier. Calculaaehe output vollage or a scalingadderr Explain howa scalingaddercan beused n a digital-ro,analog onvefterFigure19-6 shows 1wo nputsumming rnplifier.wovohages,JNr ndyrNr,areap-plied o the nputsandproduce unents rand4, asshown.Using heconceptsf infinirejnputimpedancendviftualground, oucansee har he n\ erring nputof theop-amps

approximately Y and here s no curent from the npur.Therefbre. he oral cuffent,whichis he sun of 1ra.d 1r. s duoughRrl t : l t + 1 2

F I G U R E ' - 6rirL Two-input, nveding umming

Rl


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r BAS|COP-AMP C|RCU|T5

Equason r-,9-7

Sincevour= 1r&,

/v . " ,vo r r ( / / . )R l - : -\ ^ rUJlrhreeofrhe c. i . ro '" re qual(R

. . /Y^, . v^ ' \ ^" " _ = 1 a ' n ll9-2 vom: -(vrN,+ vN,)

+)"'& : RJ= R), then

Equation19 2 shows hat he oulputvoltage s ihe sumof the wo input voltages.A

ith inpuB- h ,

eral expressions given n Equation 19 3 for a summingamplifier with,i inputs,asin Figure19 7, where ll the esistors ft equaln value.voln: (yrNr ylN,+ vrNr .. + %N,)

%ur

' FTGURE 9-a

,i,*-'dl-jj]jDetermine he ouiput voltase n Figure 19 8.

l,iN3 +8 v

vou: (vNl+ vnu vrm): (3v rIf a fouth input of +0.5 v is added o Figure 19-8 withoutput voltage?

Open ile E 19-02on your Multisim CD-ROM.Measure he output voltageandverify that t is thesumof the npui voltages.

V + 8 V ) = - 1 0 Va 10kfl rcsistor.what s the

1 0 K )

l0 lo


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5UMMING AMPIIFIER5 T 9

[isumming mplifierwith GainGreaterThanUnityWhenRris larger han he input resistors. he amplifier hasa gainofR//& whereR is thevalueof each nput resistor The generalexFession or theoutput s

v-r= -F(y,, , +yrN,+ ., + yN,,)Asyoucansee, heoutput s thesumot'all rhe nputvoltages ulripliedby a constant e-terminedby the ratio R//R

Detemine the output voltage or the sunming amplifier n Figule 19 9.

Equation 9-4

l ' iNr o2v

R/= 10kA andR : Rr= R, : 1.0 O.Therefore,R, r o r.-ov . - t v . . . _ v - . \ ' - - - . n . \ ' _ 0 . 5 \ I _ t 0 { 0 . 7 v l 7 vR ' ' ' " 1 . 0 t o -

Determineheoutput oltagen Figure 9 9 if the wo nput esisrors re2.2kO andthe eedbackesistors 18kO.

Open ile El9-03 on yourMultisim CD-ROM. Measure heoutputvoltageandverify that t matches hecaiculatedvalue.

EXAMPLE

r.0ko

::AveragingmplifierAn averagingamplifter,which s a variation f r sunrming mplifier. dn prcducehe Imalhemaricalveragei |he npulvolrage,. hi. is doneb] ,erring he alioR,/Requat o -the reciprocalof thenumberof inpurs.Youknow that rhe average f severalnumbers s ob-tained by first adding he numbersand hendividing by ihequanrityof numbers ou have.Examinationof Equation 9-4 and a liule thoughtwill convinceyou that a summingam-plifierwil do hesane.Example 9-4will illustate.


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I BA5|COP-AMPC|RCU|T5

t - t oI'lNr lv

Solufion Since he nput esistors reeqwl, R - 100kO. Theoutput oltagesR.v-r: t(v^, + vN, + vrN3 ylN1)2 5 t Q . , tt 0 o k l r 2 \ - J \ ' 1 V ) - ; ( 2 v l 0 . 5 V

A simple calculationshows hat heaverage f the four input values s the sane asYourbutofopposite ign.- l V + 2 V 3 V + : l V 2 V4 - -

Related Prcbkm Specify he changes equired n the averagingamplifier n Figure I 9 I0 in order ohandleive npurs.

Open ile E19-04onyour Multisim CD ROM. Measure he output voltageandverify that t is the average f the nput voltages.

Show hat he amplifiern Figure19 l0produces noutputwhosemagnitudes theaverage f the nput voltages.

5calingAdderA different weight can be assignedo each npui of a summingamplifier. oming aI adder by simply adjusting he valuesof the input resistors.As you haveseen, hevoltage anbeexpresseds

Equatione-5 """,: -(f ^, *.&u,* * * f,u,-)For example, f an nput voltage s to havea weight of I, thenR : Rr Or, fa weight of

The weighi of a parricular nput is set by the ratio of Rr o the resistanceor thatis requhed,R = 2Rl The snaller fte valueof the nput resistanceR, thegreaterhe

r00to


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. :rUMMING AMPLIFIERS

EXAMPI.EI9-5

> F I G U R E ? - I !

For thescaliDgadder n Figurclind theoutput voltage. l9-11, determine heweight of each nput voltageand

YN1= 3v

r4N, +2v

YN=+8v

10 ol0 koR.Weightof input1:& = 1

, 0.10,213

^ R/ i0 kOwerenr npur: E = -*"'"",^.,"",,,,.&lglq=- " ' ' - ' - ' - R , 47koTheoutpuroltage.

/ R , & & \y.- = -\4y^, + nryrN, 4yN3,/= -11(3v) + 0.1(2v)+ 0.213(8v)l- ( 3 V+ 0 . 2 V 1 . 7 V ) : 4 . 9 VRelated ,oblem Detemineheweight f eachnputvoltagen Figure19- I if Rr= 22kn, R, : 82K),Rr= 56kO, andRr: 10kf). Also indyour.

Open ile Ela-05 on)our MuhisimCD-ROM.Measureheourpurollage trdcomDareo the calculated alue.


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r BA5|C P-AMP |RCUTT5

rq\DD!F-q-EENTIATORS

-

An op-amp ntegratorsimulatesmathematicalntegration,which is basically asummingprocesshat deteminesthe aotalareaunder he curve of a function. An op-amp differentiatorsimulatesmathematical ifferentiation,which is aprocessofdetermining he instantaneousate of changeof a function. The integratoNanddifferntiatorsshown n thjs sectionare dealized o showbasicpdnciples.Practicalintegratorsoften havean additional esistoror other chcuity in parallelwith $efeedback apacitor o preventsaturatior.Practicaldifferentiatorsmay nclude a seiesrcsistor o reducehish ftequencynoise.After completing his section,you shouldbe able or Explaitr the operation of op-amp integrato$ and differentiators. Recognizean ntegrator. Detemine the rate of changeof ihe integratoroutputvoltage. Recognizea differentiatorr Determine hedifferentiatoroutput voltage

TheOp-Amp ntegratorAn ideal integrator is shown r Figure I9 I 2. Notice that the fedbackelement s a ca-pacitor that forms an RC circuit with the input resistor Although a large-value esistor snomally used n parallelwith the capacitor o linit thegainat low tuequencies,t doesnotat'fect he basic operationand s nor shown or purposes fthis analysis.: r F T G U R E9- t2An i dealop-amp integratoi

How a Capacitor Charya To understand ow the ntegmtorworks, t is impo ant o re-view how a capacitor harges.Recall hat he charge0 on a capacitor s proportional o thechargingcuffent and he time.Q : I C

Also, in termsof the voltage. he chargeon a capacitor sQ : C V C

From these wo relationships.he capacitorvoltagecan be expressed sv, \ ; .F

This expression as he orm of an equation or a straight ine beginningat zero with acon-start slope of 1y'C Rememberrom algebra hat thegeneral ormula for a straight ine ,s) : ru + h Tnthis ase,l,= Vc,n - IclC, x : t, ar|,d = o.Recall hat the capacitorvoltage n a simple RC circuit is not linear but is exponential.This is becausehe chargingcurent continuouslydecreases s the capacitorcbargesandcauseshe rate of changeof thevoltage1ocontinuouslydecrease. he key dling aboutus-


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INTEGRATOR5ND

ingan op-ampwith an RC circuit to form an nte$ator is rhar he capacitor'schargingcur-rent s madeconstant, husproducinga straightline (iinear) voltage atherthan an expo-nentialvoltage.Now lefs seewhy this is true.In Figure 19- 13, he nverting nput of the op-amp s at virtual ground 0V), so rhe volt-ageacrossRr equals 4,. Therefore, he input curent isv,"

t s t G U R E t - 1 3 h

DIFFERENTIATORS 91

Cudentr in an intetratoi

If yi, is a consiantvoltage, hen i, is also a constantbecausehe nverting nput alwaysremainsat 0 V keeping a constantvoltageaffoss Ri Because f the very high inpur im-ledanceof the op-amp, here s negligiblecunenr from rhe nverting npur. A1l of the npurcuffent s through hecapacitor.as ndicatd n Figure ]9 13,so

'Ihe Capacitot VoLtaEe Since4. is constant, o s 1.' The constant c charges hecapacitor linearly andproduces Linear oltageacrossC Theposirivesideof rhecapacitor s heldat0 V by thevirtual groundof theop amp.The voltageon rhe negarive ide of rhe capacitor decreasesinearly fiom zero as hecapacitorcharges, s shown n Figurc 19 14.This\olt ee is called a negativ nmp.TheOutput Voltage V,/ is ihe sameas he voltageon the negarive ide of the capacirorWhenaconstant nput voltage n the form of a srepor pulse(apulsehasa constantamplitudewhen high) is applied, he ouDut mmpdecreasesegativelyunril rheop-ampsaruratesat ts maximum egativeevel.This s ndicatedn Figure19-15.

A F I G U R E 9 - ' 4 , : . T G U R E9 - i 5A rinearampwltage n poducedaco$ C by h conrtant haint A co.rtlnt input @ltage

producer a


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r aAstcOP-AMP |RCU|T5

Equation 9-6

Rate of Chan4eofthe Odfpat The rateatwhich the capacitorcharges, nd dErefore heslope of the output ramp, s set by the ratio /./C, asyou haveseen.Since1c = v,,/Ra therateof change r slope f the ntegmtor's utput oltages A%,//ArLV."t _ Vi"Ar RrC

(a) Determine he rate of changeof the outpui voltage n responseo the first inputpulse n apulsewaveform,as shown or the ntegrator n Figlre 19-16(a).Theoutputvoltage s iniiially zero.(b) Describe he output after the first pulse.Draw the outputwaveform.

;:"_f--l__J

t 0v

Solution (^) The rate of changeof the outputvoltageduring he rime that the nputpulse sHIGH sLV",, R,C (10kOX0.01pF)Y,, 5v - -50 kv/s = -50 mv/ps

(b) The rate ofchange was ound to be 50 mv&s in paft (a).when the nput is at+5 V the output s a negative-goinganp. when the nput is at 0Y the output s aconstantevel. n 10O,us,hevoltage ecreases.4y'., = (-50 mv&9(100!s) = -5 V

Therefore, he negative-going amp reaches 5 V at the end ofthe pulse.Theout-putvoliage hen remainsconstantat -5 V for the time that the nput is zero.when the nextpulseoccurs,dle ouFut againbecomes negative-going ampwhich reaches 10v at the end of the secondpulse.The output will remain at- 10 V thereafterbecauset has eachedts maximumnegative imit. The waye-fo ns areshownn Figue 19-16(b).

t -16

,,;i -l-_--J- l-;


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INTEGRATOR5 ND DIFFERENTIAT ORS 9I

Related Prcblen Modiry the ntegator in Figure 19-16 to make he oulput change rom 0 to -5 V in50 "s with thesamenDut.

Open ile E19-06onyour Multisim CD-ROM. Observehe ourputvoltagewavefbrmand compare o the wavefom in Figure 19 l6(b).

The Op-Amp DifferentiatorAn dealdifferntiator s shownn Figure19 17.Noticehow heplacementfthe capac- *itor and resistordiffers from theirplacementn the ntegrator.The capacitor s now the n-put element.A differentiatorproducesan output hat is propo ional to the rate of changeof the nput voltage.Although a small-value esistor s normally used n serieswith the ca-pacitor o limit thegainat high frequencies,i doesnoi affeci he basic operationand s noishown or DurDosesf this analvsis.

Toseehowthe differentiator works,let's apply apositive-going amp voliage o the n-putas ndicated n Figure t9 18. n this case.lc = 4^and he vohageacross he capacitoris equal o yl^ at all times(vc : vr,) because f virtual groundon the nverting nput.FIom the basic orrnula,yc = (/c/Or, the capacitorcu[ent is

! F I G U R E ' - I 7 . i ,F t G U a E 9 - l EAn ide.l oP-amPdifferenbator. A differentiatorwith : ranp input.

Since he cunent ai the nverting nput is negligible,1F Ic Bothcunentsareconstantbe-cause he slope of the capacitorvoltage(ycl, is constant.Theoutputvoltage s also constantand equal o the voltageacrossRr because ne sideof the feedback esistor s always0V (vifual ground).u:(+)'v,",: rRRr=hRlv".': (v'\Rc

Theoutput s negativewhen te input is apoliiive-going rampandpositivewhen he nputis a negative-goingamp,as llustraGdn Figure19 19.During hispositive lope ftbeinput, he capacitor s charging rom the nput sourceand he constant unent thrcugh hefeedback esistor s in the direciion shown.Dudng the negativeslopeof the nput, the cur-rent s in the oppositedirection becausehe capacitor s dischalging.

Equation I 9-7


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r BAS|COP-AMPC|RCU|T5

of ! differentiatorMth aot poritiveand ngatiw rampr

aw) on th nput.

)a lGURE t -20

;a,'\,-lNotice n Equation 9-7 that he ermyc, is the slopeof the nput. f the slope ncreases,

%d increases.f the slopedecreases,%,, decreases. o, he output voltage s proportionalto the dope (rateof change)of tbe input. The constantof Foponionality is the time con'stant, P.

Determine he output voltageof the op-ampditrerentiator n Figure I 9 20 for thetrianqular-wavenDut shown.

Solutlb' Startingaft : 0, the nput voltage s apositive-goingamp ranging tom -5 V to +5 V(a + l0 V change) n 5 rls.Then it changes o a negative-going amp ranging from+5 V to -5 V (a -10V change)n 5 rs.The time constant sR/a: (2.2kO)(0.001lrD 2.2/^

Determine he slopeor rateof change ycl, of thepositive-going amp and calculatethe output voltageas ollows:v . 1 0 v ^ - . ,: - = t y / u . s

v.,,: (!t)n,c : (2vtt!s)2.2s: -4.4vLil(ewise, ie slopeof the negative-going anp is 2 V/ts. Calculate he outputvoltage. 1+,c : - (-2 v / p.s)2.2t = 4.4v


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oSG|LLATORS r t1

Finally,Figure19-21showsheoutpui oltagewavefom elativeo the nput.F I G U n E 9 - 2 1

v , 0

Related Prcblem What would the output voltagebe fthe feedback esistor n Figure 19 20 s changedto 3.3kO?

1 What n the feedbackelement n an op-amp intglator?2. For a con'tant inputvoltage to an integrator, lvhy ii the voltaS a.rclr the3. Whdt i the feedbackelement n an op-amp differenbi.tor?4, How ir the ouqut ofr diffrnuator elated o the input?

l 9 -4 0SCtLLATOR5Feedback scillatorswere ntroduced n Chapter17 and heprinciplesof operationwere discussed. Iso, several ypes of oscillator circuits were coveredusing discretetransistors.Inhissection, everalypes f leedback scillatorsmplernentedith opampsare ntrcduced.The Har{ey and Colpitis oscillatorsdiscussedn Chapter17canalso be mplementedwith op-alnps.The relaxarionoscillator s another ype which isintroducedn thissection.Aftercompleringhissection,ou should e able o! Discuss he operation of several ypes of op-amp oscillators. Identifi a Wien-bridgeoscillator and analyze ts operationt ldentify a trianguiar-wave scillator and analyze ts operationr Identifi a relaxa.ionoscillatorand analyze ls operation

The Wien-BridgeOscillatorOne ypeof sinusoidalscillators theWedb.ldge osillator. fundamentalan of rheWien-bddgescilator sa ead-lag ircuit ike haishownn Figurc 9 22(a).Rl andCl together brm the agporlionof the circui! R, and C, form the eadportion-Theoperationoflhiscircult s asfollows. t lower requencies,he ead ircuitdominatesue othehigh eaciance f C:. As the requencyncreases.c decreases,husallowing heoutput oltage

sEctroNNEytEW

ffi


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r 8A5tCOp-AMP C|RCU|T5

nd tsrponre ll, :i

IEquation 9-8

Equatjon 9-9

to increase.At somespecified requency, he esponse f the ag circuit takesover.and hedecreasing alue of Xcr causeste ouFut voltage o decrease.The response uNe for the eadlag circuir shown n Figxre 19 22(b) ndicates har houtput voltagepeaksat a frequencyt At rhispoint,rhe au,enuationy,a, y,,) of rhe circuilis J4f Rr = R?andXrI = X., as staredby dre ollowing equarion:

Ihe fomula for the resonant requencys, _ 1" zonc

To summarize,he ead-lag ircuit hasa resonanttequency, at which the phaseshift tbrough thecircuit is 0' and he attenuations 14. elowt the leadcircuir dominaresand heoutput eadshe nput.Abovet the agcircuirdominatesnd heourputlags heinput.TheBatk Ci,cuit The eadlag circuit is used n thepositive eedback oopof an op-amp,asshownn Figure19 23(a).A voltage ivider s usedn thenegativeeedbackoop.TheWien-bridgeoscillatorcircuit canbe viewedasa noninvertingamplifierconfigurationwiththe nput signal edback uom he output hrough he eadlag circuit. Recall hat he closed-loopgainof theanplifier is deteminedby thevolragedivider.

R 1+ R '

v.,,, IV," 3

, 1 R,/(& + R) R,Thecircuit s redrawnn Figure19 23(b) o show har heop,amps connectedcrossthebridgecircuit.One eg ofthe bridge s the eadlag circuit,and heother s the voh-agedivider.

Potitive FeedbackConditont fot O'ci ation As you know, for thecircuit ro produceasustained jnusoidaloutpui (oscillate), he phaseshifr around he positive feedback oopmust be0' and dregainaround he oop musrbe uniiy (1).The0' phaseshift condition smet when he frcquency s becausehe phaseshifr tkough the lead lagcircuit is 0o andthere s no inversion rom the noninverting nput (+) of rhe op-anp to the ourput. This isshownn Figure19 24(a).The unity gaincondition n the feedbackoop is mer when

This offsets he N attenuation f the eadlag circuit, rhusmaking rhe oial gainaround hepositiveedbackoopequalto1,asdepictedn Figure19 24(b).Toachieve closedloopgainof3,

& = 2 R z


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osc[rAToRS r 9

/ | lFrGuREt-2: tO) Wien bridge ciruir conbins a

voltagedivider and a ead-lagcircuitTheWien-bidte orcillator .hemati. rhown n hrc eguivalent orm!.

< FtGUtE 9-24Conditrbnr or o$illet on in thewien-bnld$ clrcuit

. R r+ R ,

(a) Thepn6e shift dound the oop is 0', Lb)Thevolraee ainmund the oop s I

2R" + R, 3R.

Stad-Up Conditiont Initially, the closedloop gainof the amplifier itself must be morethan three(Ad> 3) until the output signal builds up to a desired evel.Thegainof the am-plifler must hen decreaseo 3 so hat he total gainaround he oop is I and he output sig-nalstaysat the desired evel, thus sustainingoscillation.This is illustrated n Figure 19-25.A JFETstabilizedWien-bridgeoscillator s shown n Figur 19 26. Thegainof the op-amp s controlled by drc components hown n the blue box, whichinclude heJFET.TheJFBTh drain-sourceesistance epends n thegatevollage.With no output signal, hegateis at zero volts, causing he drain-source esistanceo be at the minimum. With this condi-tion, the oop gain is greater han l. Oscilations begin and rapidly build to a large outputsignal. Negativeexcu(sionsof the output signalforward-bias r, causingcapacitorC3 ocharge o a negativevoltage.This voltage ncreaseshe &ain-source esistance f the JFETand educes hegain (andhence he outpuo.This is classicnegative edback t work. Withtheproperselectionof components,hegaincan be stabilizedat the required evel.

v"'3(.J=1


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r SA5|COP-AMPC|RCU|TS

' , , I n r 'a l l ) . ope , r Ar i r c r rb ' | !aJse .ou ,pu 'bu ldLp . (b) Loop gainoi I causs a sustai.edcon$ant oulput.i . F IGUtE l9 -2 5Oicillatortart-upcondiuonr.

, . FTGURE t -26selfitartingwin-b dgeorcillatoruing!JFETin he negativeeedback

F T G U R E9 - 2 7

Determine he frequencyof oscillation or the JFET stabilizedWien bridgeoscillatorin Figure19-27.

10ko

I0tQ t0 o T


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o5ctLtAToRs I 9

Jorufior For the eadlag circuit, Rr : R, : R :resonanthequency sJ,

l0 kQ ard Cr = C: = C = 0.0011FTheI

2r(10ko)(0.001 F) = 15.9 IIzRelabd Problem If Cr and C?are changed o 0.01 rF,what s the iequency of oscillation n Figure19-211

::ATriangular-WavescillatorOnepracticalmplemeniation fa triangular-ware oscillator utilizesan op ampcomparat:r -!' .hown n FiCureq-28. Ite operarion. a. follows. o begin. .sumeiar theourpulolr -ageof the comparators at ts maximumnegativeevel.Thisoutput s connectedo the nvert-ing input of the intgratof hroughR , Foducing a positive,goingamp on the oueui of theintegrator.When the ramp voltage caches he upper riggerpoint (JTP), the comparatorswitcheso ts ma{imumpositiveevel.Thispositiveevelcauseshe ntegator rarnp o change!o a neganve-goingircction.TherdmpcontinuesD this directionunril dle ower riggerpoint(tjIP) of the cornparators reached. t thispoint, he compamtor utputswitches ack o thenaximum negativeevel and he cycle repeats. his action s illustraied n Figure19,29.

' . :FTGUREt-28A trian8ulatsw&eorcill.tor urint

vrreSince he comparator roduces squarewaveoutput, he circuit in Figure 19-28 can beusedasboth a triangllar-wave oscillatorand a square-wave scillator.Devicesof this typearc corunonly known as t ction generatorsbecnnseheyproducemore than one outputfunction. The output anplitude of thesquarewave s set by the output swing of the com-parator,and he resistorsR, and R3set he amplitudeof the tdangular output by establish-ing the UT? and LTP voltagesaccording:to he following formulas:

/ R .v!.rP +r1_,1; l

. . atGuRE 9-2 'Wawfomr for th circuit n FituEl9-28.

Equafion 9-10

Equation 9-11.-: ".-("")


21/44

2 . BASICOP-AMPCIRCUIT5

where he comparatorouipur evels, +y,ea and -I"_ are equal.The frequencyof borlwaveformsependsn heRLCtime onstanr s.rJell s heanptirude-settingesisrors,,and R3.By varyingRr. the frequencyof oscilaiion can be adjustedwidour changing heouFurarnpl'tude.r /R^ 'Equa ton 9 -12 . , " l ^ i : )-" . . ' " , /

Detemine he tequency f thecircuit n Figure19-30changedo make he frequency20kHz? Towhatvaluemust Rr beFtGUtE t r -30

" | ( R , \ |' lRrC \Rrl \ - L4(roko)oo; (3r!l1okJ =s's krr'To make/ - 20kHz,

^ t (a ' \ | t \ / l3kn \R o1cln./ o,rnu",,rr.orur-,/ l i ,nJ nttuoRelatedPrcblem What s the amplirudeof the rriangularwave n Figure 19-30 if the conparatoroutputi s : t l 0 V ?

10Kl

A Square-Wave Relaxation OJcillatorFl The basic square-wavescillator shown n Figue 19-31 is a type of relaxation oscillatorbecausets operation s basedon ihe chargingand dischargingof a capacitor.Norice thar

r : F I G U i E l 9 - ! lA ryuarc-wave ela\ationorctllator.


22/44

theop amp's nvedng input is the capacitor oltageand hc noninverting nput is aportionoftbe oulpul edback hrough esistors zandRr. When hecircxit s first tumedon, hecapacitors uncharged, nd hus the nveriing nput is at 0V This mates he outputaposr-livemaxinum,and hecapacitor e$ns o chargeoward ,,,, hroughRr.When hecapac-itor voltage eachesa valueequal1() he feedbackvoltage on the noninveting inpul, theop ampswitcheso themaxinulnnegalive iate. t thispoint, he capacitor eginso dis-charge rom +Vr toward vr Whenthe capaciior oltage eachesvj the op ampswirchesack o themaximum ositive tatc. hisaclion ortinues o epeat, sshownnFigure 9 32.and square-waveutput oltages obiained.

ACTIVEFIITERS I 92

F T G U R E9 - 3 2

u,,,

Wavefom, fo. the rqlare-wave

Thercarewo eedba(koop! n a Wien-bridgercillatorWhat ' thpurpdeofeach?What i a tunction enator, ndbatically hatdoer tdo?Uponwhatpdnciple oes .elaxauon rcillator perate?

oN l 9 -4REVIEW 1 .2.

t.

I9 -5 ACTIVEFILTERSFiliers are usually categonzed y the manner n which fie outputvoltageMries withthe iequency of the input voltage.The categories f active ilteff that we $'i11 xamine;n thisseciion re ow pass, igh-pass,ndband-pass.he ermdclit meanshatagainelements used:n thiscase, nop-ampAftercompletinghis section,ou should e able or Recognizeand evaluate basic op-amp ffltenr Evaluate inglepoleand wo-poleow-passilterst Evaluate ingle-polend wo polehigh-passltersi Determine he resonantrequencyof a cedain ype of band-passi1ter

Low-Pass Active FiltersFigure19-33shows basic ctive fltor and tsresponseu e. Notice hat he nputcir-cuit s a single ow passRCcircuit,andunityga is provided y theop ampwi& a rcgaiive eedbackoop.Simplystaied.his s a voltageollowerwith annC filler belweenheinputsignal nd henoninvertingnput.

6


23/44

. BA5|C P-AMP |RCU|TS

i F I G U R E t - 3 3single-pole,ctrtveoy-pas ilter and erponreure.

The voltageat the noninverring nput, V+. is as olows:-. / x. .\,,+ = | ;...:: \\/el4)''"

Sincehegainoffte op-amps l. theoutpui oltages equal o y+./ vEquatione-13 v". - ( !-\v r t n _ x i l , , ,

A filter with oneRCcircuir thatproduces 20 dB/decade oll-off beginningatt is saidtobeasi gle-poleorfnt-o erJiker.The term"- 20 dB/decade"means hat the volragegaindecreasesy ten times(-20 dB) when he tequency ncreases y ten times(dcad).Low-Patt Two-Pote tlterr Therc are several ypesof active lilrers and they can hayevarying numben of poles,bur we will use a rwo-pote ilter ro illusrrare.Figure 19 34(a)showsa two-poie (second-order)ow-pass ilrer. Since eachRC circuit in a filter is con-sidered o be one-pole, he two,pole filrer uses wo RC circuits to producea roll off rateof 40 dB/deade, s ndicared n Figure 19 34(b). The active ilrer in Figure 19 34 hasunitygainbelowL becauseheop-amps connecredsa voltage-follower

! FTGURE 9- !4Tvo-pole, .ctte lov-pa$ filter rnd itr ideal rerponre curve.

One of the RC circuits in Figwe 19-3zl(a) s forned by Rr and Cr, and rheother by X,andCr. The critical hequencyof this lilter canbe calculatedusing rhe ollowing fomula:

Ar(dB)

Equatlon9-t4 2"\/ R"RC,c,


24/44

ACTIVEFILTERST 925

Figure19-35(a) showsan exampleof a two-pole ow-pass ilter with valueschosenoa responsewith a critical frequencyof 1 kHz. Note that Cr - 2C, andRl : R, be-these elationships csult in a gain of 0.70? (-3 dB) ati For cdtical ftequenciesthanI kIIz, ihe capacitancealuescan be scalednverselywith the hequency Forex-ample, sshown n Figurc 19-35(b) and(c), to geta 2 kHz liltet halve he valuesofCr andc,: for a 500Hz filter. double he values.

] | FtGURE t -45(c) t=500l l2

2C2=0.015Frc , =20.0075,uF

t5ko 15r


25/44

| BAS|COP-AMPC|RCUTTJ

Solve for Cr, and then detemine CjI8.'Rl:I 0.'70'7: - = -v22nRf, znxt ': 2C2 2\0.001'7pF)

= 0.0017,Fo.'70'7

2nQ2ka)Q kJtz)= 0.0034FRelatpd Problem Determinei in Figure 19-36 for R, = R, : 27 kO, C, : 0.001pR and C: = 500pF.

Open ile El9-10 onyourMultisim CD ROM.Verily that he critical requencyis 3 kHz.

czcl

High-Pass Acb've Filtertln Figure 19 37(a), a high-passactive ilter with a 20 dB/decadeoll-off is shown No-rice that the nput circuit is a singlehigh-passRC circuit and hatunity gain s providedbythe op-ampwith negative eedback.The response uve is shown n Figue 19-37(t).

cain (dB)

3t,,,vre---l

(a )

20

i F T G U R E t - 3 75ingle-pole, active high-pa$ fi ltd rnd rcrponre c!re.

Ideally, a high-pass ilter passes ll frequencies bovet without Limit. as ndicated nFiglre l9-38(a): In practice,of course,such s not the case.All op-amps trierently haveintemal RC circuits hat init theamplifier's response t high frequencies.Such s the casewith the aciive highipass ilter There s an upper requency irnit to its response,which, ineffect, makes his type of filter a band-passllter with a very wide bandwidthndrer than arue high-pass ilter. as ndicated n Figure 19-38(1r). n many applications, he intemalhigh-ftequency utoff is so muchgeater than he ilter's critical Fequency hat he ntemalhigh-frequency utoff can beneglected.


26/44

,, :r n ),,- ' \ ^ ' F + x ' , 1 ' ^If tle internalcritical flequenciesof theop-amparc assumedo be muchgrcater han he

Thevoltageat the noninvering inpui is as ollows:v+:l - ;-_ j ] ] lv ,"\ vR' + xil

sewitha critical requency f 1kHz. Note hatR, : 2RrandCr = C, becausehese

Y--l

theop-amp s connectedas a voltage-followerwith unirygain,rhe outputyoltage s

; ofthe ilter thegainwil roll offat 20 dB/decadesshownn Figue l9-38(b).is s a single-pole ilter becauset hasone RC circuii.Two-PoLeilte6 Figure 19-39 showsa two-poleacrivehigh-passiLter.Noticeit is identical o the correspondingow-pass ype,except or ihe positionsof the resis-andcapacitors. his filter hasa roll-offrate of 40 dB/decade elowt and he criti-Aequencys the sameas or the ow-pass llrer given n Equation 19-14Figure 9 40 shows two-polehigh-passilter with values hoseno produce re-

High-pa$ fflte. lPonre.

Equat'lon9-15

a F I G U R Et - 4 0

ACTIVE FILTERS T 92

l F I G U R E9 . 3 8

*l

Trc-pole, hith-pa$ ilter (f" = I kHz)-


27/44

r BAS|COP-AMPC|RCUTT5

relationshipsesult n a gainof0.707 ( 3 dB) ali. For frequencies ther han kHz,theresistancealues anbe scalednversely, swasdonewlth thecapacitorsn the ow-

For he ilter of Figure1941, calculaiehe csistancealuesequiredoprcducecritical tequency f5.5 kHz.F I G U R E 9 - 4 I

v ' ' - l4.0422l1F

Jotution The capacitorvalueshavebeenpreselectedo be 0.0022 lF each.Since heseditrerfrom the I kHz reference ilter, you cannotuse he scalingmethod1{)getthe resistorvalues. seEquation9 14andsquare othsides.I' z"\/R,R,, ,

f : = |An'RB{C,Since R, = 2Rt and C, : C? C

4r'(2R?)c'

^, 1'' 8"' C:/,'

f t -Solve or Rr, and hen deternine Rr.

o.'70'7 = 9.3 O2z(0.0022 F)(5.5kHz)R2 2R' = 2(9.3 O) = 18.6 O

RelatedPrcblen Determine, n Figure1941 for Rr = 9.3kQ andR, : 18.6 O if Cr andC, aIchanged o 4700pF

Open ile 819-11onyourMuliisimCD-ROM.Verify hat he critical requencys5.5kHz.

\/t2ncl 2ncf"

EXAMPLE


28/44

ACTIVEFILTERj T '29

l,Band-ParsilterUling a High-Pass/Low-ParrombinaUonOneway to implementa bandpass ilter is ro usea cascaded nangement f a high pass i Iter olowed by a ow-pass ilter, asshown n Figwe 19-42(a).Eachof the ilters shown satwo-poleconligurationso hat he roll-offrates ofthe response uNe are 40 dB/decade,sindicated n the composite esponseurveof part(b).The critical frequencyof each ilter ischosen o that he response urvesoverlap,as ndicated.The critical frequercyof the highpassilter is lowerthan hat of the ow-pass ilte/The owerfrcquency, r , of ahepassbands setby the critical frequencyof the high-passfilter The upper requency,tz, of fte passbands thecritical ftequencyof the ow-pass il-ter ldeally. the center requency,r, of thepassbands thegeonetric average ftr andIr.The ollowing formulasexpresshe kee tuequenciesf the bandpassilter in Figure t 9-42.

z"',/n,n ,c,t

z"t/ n.acj\,'EJ"

vr,-l

Equation 9-16

Equation t-17Equation19-18

: F t G U n Et - 4 2Eand-pag ilter fomed by conbintng a t*o-pole, hith-pa$ filter with a two-pole loe-pas fflter (ltdos notmatter n which dd the ilter.rc c,caded.)

c4


29/44

r BA5|COP-AMPCTRCU|T5

(a) Determine he bandwidthand center requency or the filter in Figure 19,43.(b) Drawtbe rcsponse urve.

I, r . rGuRE | 9-4315kOY , o 1 15kO0.01tF

SoLution (a, Thecritical frcquencyofthe high-pass ilter isi

The critical hequencyofthe

r-, L= --:==_L=r.ss*n,"" z-:/44c"ct z'{rs rr4lr: roylo.or,F)(0.00a7F)BW= f,z - f. ' = 1.s5 Hz 450Hz : l.l kHz.t - \En" : r,{+so ,1rss n4 : ezz ,(b) The esponseuNe s shownn Figure19 44.

= 450Hz2"{25 kox5okf)XoJrpFXo0t ,F)Iow pass ilter is

Related Problem Describehowyou cin increase hebandwidrhof the filter in Figure 19-43 wirhourchangingrr.


30/44

VOLTAGE EGUIATOR5 I 93

sEcTtoN t9-5REVIEU/ 1 ,3 .

In termrofcircuit componenB.what doe! the termpole .efer o?What9pe of rrponrecharacterizerhe ringl-pol ow-pan filter?Howdoe'a hieh-pa$ ilter differ n itr implmentation iom a corerponding ow-l{ the reinrance alue(ofa hieh-pa$ ilter dredoubled,whai happen! o the

I9 -6 VOLTAGE EGULATORSTwo undamentalypesoflinearvollage eguiatorsrc inlroducedn thissection. neis theseriescgulator nd hcothers theshunt eglrlaior.After completinghissection,oushould e able or Descdbe th operation of basic series and shunt voltage regulators. Explainhowabasicop-amp eriesegulator orksr Discussshol1circuit and overloadprotectionr Explain how a basic op amp shunt egulatorworks

Baric Series RegulatorA simple cprcsentation f a ineaf series gulator is shown n Figufe 19 45(a), tnld the ba Effisiccomponentsre shownn theblolk diagramn pan b).Notice hat hecontrol lementsin sedcs ith he oadbetweennputandoutput. heoutput ample ircuit enses changentheoutputvoltage.The enor detector ompareshe samplevoltagewith a rcfercnce oltageandcauseshe contrcL lement o compensaten oder to maintaina constant utputvoltage.

F I G U R E 9 - 4 5Blockdiagi!d of a three-terdinal, rder voltage egulatol

RegulatingActionAbas;copampsees egulatorircuii s shownn Figu 19+6.Itsoperation s illustrated r Figurc I9 47 and s as ollows.The csistirvoltagedivider ormedbyR, andRr scnscs ny changen theoutputvoltage.When he outprit ries o decrease,sshown


31/44

op-amp lerier regulator

(a)When vrNor lS. dercds. voE aitcmpls o decre.se.Thefeedbackvoltaee.vru, xho altenpts b datlse, and$ a rc$nt, theop mp\ ourputvoxaec vB attenpl! to inoeaseithus conpensanngfor theauempteddccreasen vour by increasing hc 0r enittervollage.Cha.ecs n vour aF exagger.tedbr illustation

(c) When VrNor n. incredts, vour ateDpts to incFase.Tbefedb.ck voltage.vFB,ako alteDpts o jncEase,and as arerult,vB applied o lhe b.se of lhe conhl transistoi ltcmps ro dccrease,thus conpensaiine or the anenpted ncreAe in vour byd*reasi.g thc gr eniner voltage,

(b)WhenVr \ (orRr) sabilizes t s new oNervalne,lhe oltagesE1ufr io their orignralv.lues, thus *ping vom constlnt as aesDlt ol thc negalive eedback.

{d) When vN (or R.) {xbilizes a1 ts new higher aluc, tne voltagesEturn o their riginal alues,hus eepingvoul const.nl s aFsultol ihenceativeedbac(.

: FIGURE t -47jilurvation o{ rerier tulatoraction hat keP3 orconrtantwhen yN o, Rr change,


32/44

VOLTAGE EGULATORS 93

in part(a),because f a deqeasen vN or because f an ncrcd-sen ./L decreasen R.), apro-portionalvoltagedecreases applied o the op-amp's nveting input by thevoltagedividerSince he zenerdiode,Dr, holds he oiherop-amp nput at a nea y constanteferencc oltage(vRF),a smalldifierencevoltage enorvoltage) s developed crcss heop itmp's upurs. hisdifrerence oltage s amplilied,and he op-amp'soutputvoltage y, incrcr.ses. his increaseis applied o the baseof Cr, causing he emittervoltage.vour, to incrcase ndl the voltage othe nvening nput againequals le retbrencezener) oltage.This actionolTsetsheattempteddecrea-senoutputvollage, huskeepingt nearlyconstant, sshown npad(b).Thepower ransistor01, s usuallyusedwilh a heatsink becauset musthandleall of the oadcurnt.The oppositection occurswhen he output ries o increase, s llustratedn Figure1947(c) and d).Percentegulation rs discussedn Chapter 6.Theop-ampr Figure 9 .16 s actuxlly onnectedsa noninvertinsmplifiern whichthe reference oltage,yREF,s the nput xt the noninverling eflninal. and he RzlRlvoltagedivider oms thenegativeeedbackircuit.Theclosedloop oltage ain sR.A , , : : + l Equa t i on9 -19

Therefbre, he regulatedoutput voltage s/R. \Yo|r : l; + I lvEF Equauon19-20\^r /

From this analysisyoucan see hat the outputvoltage s determined y thezenervoltageand heresistors , .nd Rr. t is relatively ndependeni f the nput voltage.and herefore.egulation s achievedas ongas$e inprt voltageand oad currentarewithin specified imits).

Detcrminc he outputvoliage br the regulator n Figure 1948FTGURE ' -48

Jolatto, vsr, = 5.1Y the zenervoltage.Therefore,/R \ / lokov' ' ln ' ' /u" ' \ 'uo"

RebkAtuoblen The.followinghangesremade n thecircniiofFiglre 19-48:A 3.3V zener eplacethe5.1V zener,ir : 1.8kO,,R? 22 kO, and& = 18kf,).What s theoutputvoltage'1

Shot-Citcuit or Ove oad PrcEction lf an excessiye mourt of loadcrnenl is drawn,the series-passansisior can bequickly damaged r destroyed.Most regulatorsuse sometypeof protectioniom excess unent n the olm of a cufrentlimiting echanism.

+ r)s.rv {z)s.r= ro.zv


33/44

r BA5|COP-AMPCIRCUIT5

9 - 4 t

Figufe 19-49 shows one method of current iimiiing to prcvent overloads called co,?rrdrt-cutent limiting. The cuJrcnt.limiting circuit consisis of tlansistor O and resisiance Ra

The oad cunent hroughRacreates voltage rom baseo emitterof Or.When4-reachesa predeterminedmaximum value, he voltage drop acrcssRa s sufiicient to forward-biasthebase-efitterjunctiont' 02, huscausingt to conduct. nough r base unent s di-veded nto the collectorof 0, so that 1L s lim;ted to its maximum value L(,.-r Since hebase-io-emitter oltageof A cannotericeed bout0.7 V for a silicon transistor thevoltageacross a s held o thisvalue, nd he oadcunnt s im;ted o, 0.7rq^*.t= &

BaricShuntRegulatorAs you haveseen, he controlelement n the series cgulator s the series-passransistorAsimple representation f a shunt ype of linear regulator s shown n Figure 19-50(a), andthe basic components re shownn the block diagram n pa( (b).

: f r G U t E | 9 - 5 0Block diatGfl ofa the-terhinal ,hunt rgulatoi

In the bas;cshunt regulator, the contlol elements a transistor Or) in parallel with theload.asshownn Figure19 sl. A seriesesistorRr) s n series ift the oad.Theopera-tion of the circuit is similar to ihat of the series egulator.except hai regulation s achievedby controllinghecurrenthrough reparallelransistor !.When be outputvoltage ries o decrcase ue o a changen input voltageor loadcurent,as shown n Figue 19 52(a), he aftempted ecreases sensed y Rr and Raand applied o

egulator ith conrtant-

Equation19-21

E


34/44

VOI.TAGEREGULATORs T

a F | G U R E9 - 5 t

935

vN Baric p-amp hut regulator

{ F T G U R E9 - 5 2segue.ce frerponrer henYourtrie to dec@re ar a Erult of a

1 ,t , s)0,

l',t i .

;l:i' |::

,!liitjrl sanple


35/44

T BAsICOP-AMP IRCUITs

' F I G U R E 9 - 5 3

the op amp'snoninvening nput.The resulting difference n voltage educes he op-amp'soutput(yB),diving Or less, hus reducing ts colector curent (shuntcurrent)and ncreas-ing its intemalcollector-to-emitter esistance,... Since '.?actsasa voltagedivider with Rr,this actionoffsets heatiempted ecreasen your andmaintainst at analrnostconstantevlThe oppositeaction occurs when the output tries to increase,as shown in FiSure19 52(b).With IL and volr constant,a change n the input voltageproducesa changenshuntcunent(1s)as ollows:

^ . AVu lWirh a constantyL\ andVo!r, a change n load currentcauses n opposjtechangen shunt

A/s : N'This brnula says hat f 1Lncreases,sdedeases,ndviceversa.The shunt egulator s lessefficient than he series ypebut offers nherentshof-circuitprotection. f the output s shorled your = 0), dre oad cunent s limited by the series e-sistor,Rl, to a maximum value(Is : 0).

y,"

In Figure 19-53,whatpower atingmustRl havef the maximum nputvoltage s 12 5 V?

sofuabn The worst-case owerdissipation n Rl occurswhen he output s shofi-circuitedandVour 0:when vrN= 12.5Y thevoltage ropped cross r syIN vour = I2.5V

Thepower issipationn Rl isP 4 = ' J - ' i " J t - ' ' *Therefore,a resistorof at teast 10w shouldbe used.

Retated rcblem InFigure 9 53,Rr ischangedto33Q.wharmusibethepowerrat ingofRr i f ihemarimum inputvoltage s 24 V?


36/44

T BA5ICOP-AMPCIRCUIT5

2;M'19],2

F t G U t E 9 - 5 5

In d op amp compealor when he nlut voltagoexceeds specjfied eference oltage, heThe outputvoltageof a summlnganplifier is proportional10 he sum of the nlut voltages.AD avragingmplitier is a sunning amplifierwilh a closed oop gainequal o the reciprccalofIn a scaLing dder,a diftierentweight cu be assignedo eachnpu! rhusmatirg the npulcondbute more or conlribule eli to the output.The integral ol a step s a ramp.The derivaliveof a rdp is a step.In a Wien-biidge osclllatoa he closed oop gainnust be equal o 3 in order o haveunlty Saindound lhe positnr feedbackoop.In fitter tmiDolog), a sjngle nC cicuit is called a ',ld.Eachpole n a filter causeshe output o roll olT(de.rease)at a rareof -20 dB/decade.Tso polef,lteB rclI otr at a m.ximum Ete of 40 dB/decad.h a series ohage regulator, he control elemeri is a tlosistor jn seneswith the oad.In a shuntvollage egulator,hecontrolelements a tdsistof (orener diode) n par"llel wii,'ra load.The eminals on a *nee-teminal voltage egulatorde lnput voltage,ourputvoltage,andground.

t ie lbo ld te ln l in the .haPte ra rede fined in thend-of -bookg |o$aryActive tilter A frequencyselecnve ircuit consislingof activedvicessuchas banslsron or op amlscombinedwilh rcacnve RC)circuits.Averaging amplifier An amplifier wilh several nputs thlt producesan ourpul \'olrage hat is tlemalhematicalareraSe f the nput voltages.


37/44

FORMUIA5 I 93

Comparator A circuit thatcompms two nput volta8esdd produces noulpul n eitherof two staindicating heBrarer or les than relationshipof the nluts.DiITerentiator A cjrcuit that produces n output thal app.oachcshe mathenaticalderivanveof lhinlut, whlcb is rhe ate of chdge.Integmtor A circuit thatp.oduceran output hat approacheshc nathemrical integralof rhe npuRelaxationoscillator A type of oicillatoa generauynonsinusoidal,whoseonerarion s basedon thchargingand discha.Sing f a capacirorScaling dderA specialtypef summing nplifierwirhwelghrednpurs.Sris egulator A type of voltage egDlalorwith the conrrolelemenr n seriesbetween he npurddoutput.Shunt regulator A type of voltage egulatofwith the controletementbetween heoutput andgrcunSummingampliffer A! anpliner wilh severalnpuls thatproduces r ourpur oltage prcporionaltheatgebraicun of the nputvolrages.Tliangular rvaveosciUator A type of oscillator that p.oducesa tiangular waveoutput voltage.Wien-bridge oscillalor A sinusoldaloscillator hat enploys a leadlag circuir h the fee.lbacl loop

@19-lr9-2l9-319-.1

l9-5l9-6

t9-7l9-a19-9l9-10

l9-lt

t9-12

l9-1319-14

l9-15

r , " " "=a,*^q{+r , ) Compamtor eference ohage

= -f,tu,* + u,* * * u^,1 Adderith dn

-(vNr + vrrt- (vrN1+ %N:+ yrN3+. . .+ y,N,) n input dder

u"- = -(.!u,. *"&u,*- .f n") Adderith ainLV,tu _ _ VL,AI R,C

/ v-\v",,,_ !y',, 3. _ I" ' - 2nRc"*="-(t)Y . , " = v ^ l : l_ 1 / r , \" ,rxrc\x'l/ x - \\ vR ' - x i /

IztYRB,C1C'

- ' \ v f : + \ 2 / '

Rate of change n irtegrator

Difterertialor outputwith mmp ryDtLeadJag din alterualion)ltLe.d lag resonantrcquencyUpler triggerpoinl, fiangular'wave oscillatorLo\rer r.iggerpoint. trianSuldrvave oscillatorFrequencyf osciliation, iangula.waye scillator

Outpul of onepolefiller (low-pas only)Crilical fteque.cy of two pote iller (low-pass)

Oulputofone-pole ilter (highpassonlt)


38/44

r BA5|COP-AMPClRCUlr5

t9-16t9-17

19-r8l9-19t9-2tlt9-21

z.'/na,caI'. , - z"^,/ A"c,c,

1,=xj iR.A " r = + 1t R . \v o l r = \ j + r J v n s

" 0.7

Lower criticnl frequencyof a bandpass ilterUpper cntical frequencyof a bmd_pas filter

center tequency fa bandpassilterClosedoopvoltaSerin, voltageegulator

Series egulatoroutlut voltageFor coNtet-cuftent limiting

an,."o-.,.,r'"na r.r..r'"p,""1. ThePuryoseof a comParators to(a) amPIry an npur voltage

(b) delect he occurenceof a changingnput voltage(c) prodDce chege in output when an nput \oltage equalsa reierencevoltage(d) maintalna constantouFut when he dc nput voltagechanges

2 To usea compararoror zeroletel detection. he nverting nput is comected o(c) a ositve refercncevolrase (d) a nesatile reference oltdge

3. In a 5Vleve] deleclorcircuit.(a) fte noninvening nlut is connected 0+5 v(b) the nvenjng nput is connected 0+5 v(c) the nput signal s linited to a 5 V peakvalue(d) !h nputsiSnalmustbe idingona +5 dc elel

4. In a certain ourinput summinganplifier, al1 he nput resislon de 2.2 kO and he feedbackesistor is 2.2kO. r all the nput \ollages are2Y the outputvoltage s(a) 2v (b) 10V (c) 2.2Y (d) l lv

5. Thegainof theanplifier n Question is(a) I (b) 2.2 (. ) 4 (d) mknosn

6. To converta snmmingamplif,er o an avengineamplifier(a) all input resistos must bea ditrerent alue(b) the ratio R/R must equal hereciprocalof tle numberof inpurs(c) themtio4n mustequal he nunber of inplts(d) answeK a) and(b)(a) the npul resisto6 de all the sme valueG) rhe nput resislors re all diflerert laluos(c) the nput rcsistoNhavevalues hal dependon theassiSned eightof each npul(d) the radoR/ must berhe sme fbf eachnlut

8. The eedbackalh n anop-anp nteSratoronsisrsf

(b) the dc s!'pplyvouage

(c) a resislor and a capacitorD series (d) a rcsistorand a caPacirorn pafallel9. Thefeedbackpath n anop dp diilerentiator consjslsof(c) a resistoranda capacitorn series {d) aesistor ed acapacitor n FraUel


39/44

PROBIEMs 94

10. Theop-up compdator tcun uses

l l .

12.

@AnJwe4toodd-nUmbeedPob |eml .Gat t leendo f thebook '

(c) resenerariveeedback (d) no feedbackUnily gain andzero phase hift around he feedback oo! are conditiors that desoibe(a) an active ilter (b) a comldator(c) e oscillalor (d) d irtegrator or ditreenllalorThe npul tequency fa single'pole,ow-passctive i1terncreaseslom 1.5 Hz o 150 Hzff the critical freqDency L5 kHz, rhegaindefe$es by(a) 3dB (b) 20dB (c) a0dB (d) 60dB

sEcT loN 9 -BASICROBtE\,rs

l. Detemine theoutput evel (maximumlositivc or maximumnegative) or eachcompeator nFlgDre9-56.-' *{\-'- +f--''-+-.2-" -.,--4,..'-"'I(a) (b) (c)

2. A certajno! mp hasan open oop gainof 80,000.The ndinum saturated utput evelsof thipaticuld deviceaE 1 12V when he dc supplyvoltagese I 15V If a differentialvoltageof0. 5 mV ms is alplied betwee! tle inputs.what s thepeak-to-peakalueof lhe outpul?

3. Draw the output voltagewavefom for eachcircuit in Figure 19-57sirh respect o the nput.Show oltageevels.

> FTGUIE 19-56

__L

F FTGURE 9-5 t

sEcTtoN 9-2F f l G U t E t t - 5 8

(a)

Summing Amplifirj4. Derernineheoutpnt ollageor each ircuit n Figure

+ 1 5

19 58.

t0ko

10ko

(b)


40/44

r BA5|COP-AMPCIRCUTT5

t - 6 0

9 - 6 r

7. Fjnd theoulput volbge when rhe nput voltagesshown n Figure I 9-60 de applied o thescalingadderwhat is the curent through1i /

v r=+2v

10. A tliangularwaveform s applied o the nPut of the dealditre@ntialor n FiSue 19 62asshown.Determinewhat the outlut shouldbe, and draw ts waveforn in relaiion 10 he npur'

Determine hefollowing in Figure 19 59:(a) v^r md vp (b) curent diough & (c) vdrFin.l the value of Rr. ecessaryo goduc anoulput that s 5 times he sum of the nputs nFigm 19 59.

8. DelerDinehevalues f the npDt esistor cqunedn a six_point caling dder o hal helowest eiShtednputs I and ach uccessjvenpul as weigbl{r? thepreviousneUseRr: 100kc).SECTION | 9-3 rntegrator and Diffrentiaton

9, Deteminehe ateof clangeof theoutput oltagen responseo thestepnPut o the dealintegrabr Figue19-61.

22kA

l0ko

33kO

9 1 k O


41/44

r BAS|COP-AMPC|RCU|T5

9-66

rc^ l l

t 500

56 O

pF l50Op!

c32200pI

l 5 k Q l 5 t oc1I000 F

4.7kO

20. De,enine he bandwrLlthnd.enrer iequen.)of edch ilrer n FiSurea 60

ISCTION t 9-6 Voltage Rgulators

21. Detemine the outputvoltage or the seies rcgulator n Figure i9 57.22. If nr in Figure19-67 s doubled.what haplers to the output voltage?23. If the zenrvoltage s 2.7 V insteadof 2 V in Figure 19 57, whal is the ouDDtvoltage?24. A series oltagee8ulator irhconstant-cunentinidngis shownn Figure19-68.Determinethe value of nl if the oad curent is to be imiled io a maxinud valueof 250 nA. whal powerranng must Rabave?25. ff Xi (delemined in Pioblem 24) is halved,whar s the na{inum load curent?

r F T G U R Et - 6 8


42/44

ANSWERS r 9

26. In the shDnt gulator of Fjgure 19 69, when he oad curent increases, oesCr condudmo27. Assune that 1L enaiDsconstdl andyN increases y Mn Figure 19-69. What s the chag28. With a conslanl nlut voltageof 18Y the load resisrancen Figure 19,69 is varied rcn1.0 kfi to 1.2kO. Negl@ringdy chdge in oulput voltage,how much does h shut curert

thronghOr changel

in the colletor c!rcnt of 0r ?

' FTGURE t-69

Pt9-29

.iiMU_LXtSllruispuFlGiHoonNG'PRo.BtEll|5CD-ROMib cncuiB ihown n FEu 19-70.29. Olen file P 9-29 anddetemine if th@ is a faDlt. so, dentifr it.30. Open ile P19-30ard delermire f thee is a fault. n so, dentify it.3r. Open ile P19-31dd determine f there s a fault. lJ so, dentify it.32. Open ile P19 32 md detemine if there s a fault. IJ so, dentify it.

. . .FtGUtEP19-30


43/44

r 8A5tCOP-AMPCtRCUTT5

-lrlr0.5

1000Ez

Pt9-31 Ptg 32: . F IGURC 9-70 (Cont . )

sEcTroN 9-t

sEcTtoN | 9-2

SECTION 9 -3

SECTIONEVIEWSr. (10k,cyl0 kO) 5 - 1.36v2. Se igue19 71.

r -F T G U R Et - t t

Sum.DingAmplifi6rul. Thesumnary oints lhe erminal f theop-ampherethe nput esisrorsre comonly2. Ry'R 1/5= A.23 . 5 k OIntogratoB and Dt#rentiatorl. The feedbackelement n an ntegratof s a capacilor2. The capacitorvolrage s lined becausehe capacitorcurent is constant.3. Tlle feedbackelenent in a difiereltialor is a resistor4. The outpu!of a diferentiator is proporrional o rhe ate of chrnge of rhe npDt.

Orcillator'r. The negative eedback oop sels he closedloop gain:theposltive edbackoop s1she2. A tuction generators anoscillator nslrumert thatprodu.es tore rhe one ype of oulput3. The basisof a relaxationoscillator s the chdgine and dischargingof a calacitor

sEcTroN 9-4


44/44

AN5WERS

SECTION | 9-5 Active FiltE1. A Pole s a single SCcir'nir2, A single-poleow-pass requency esponses flat ftom dc to the criltcal frcquency3. Th n and Cpositiols de intrchdged.4. Tte critical frequency s halved

SECTION 19-6 Voltags Rgulttort1. I! a shunt eguhtd, the conirol elements tu Pdallel with ihe oad rather hd in srieswiththe oad.2. A shunt egulatorhas nherentconnt imitin8, but it is lessefftciert than a Sriescgutatorb@ause al1of the oad orent must bebwased though the @ntrol element

I Appllctuon A$ignnGntl. The tuse rating shouldbe I A2. Thesede optional capacito6 hatprevert oscillations3, The 7912 s a negative-voltagegulator'

l9- l 1.96V19-2 -10.5V19-3 -5.73V1+-4 Add a 100kO input resisbr andchaDgeRr,to20 kO.l9-5 0.451.12;0.18rou = -303V19-6 Chdge C lo 5000 F.l9-7 Sme waveformburwith dplitude of 6.6V19-a 1.59kIIz19--9 6.06v peakto-leakr9-r0 8.34lHz19-11 2.57 YJlz19-12 Increaset: by reducing he resistora Voi capacilorvalues.19-13 7.33V19-14 17.5W

.. .9"4

r. (c)8. (b)

2, (^\e. (a)

3. (b)r0. (d)

4. (d)11. (c) 12. (c)

6. O)

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Ch19_OpAmpCircuit

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