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An Op Amp Tutorial(Ba s ed on ma terial in the book Introduction t o Electroacousti cs and Audio Am-plifier Design, Second Edition - Revised Printing, by W. M arshall Leach, Jr.,published by Kendall/Hunt,c° 2001.)An op amp has two inputs and one output. The circuit is designed so that the output voltageis proportional to the difference bet w een the tw o input voltages. In general, an op amp canbe modeled as a three-stage circuit as sho w n in Fig. 1. The non-inve rting input is vI1.Thein v e rting input is vI2. The input stage is a differential ampl ifier (Q1and Q2) with a currentmirror load (Q3−Q5). The diff amp tail supply is the dc current source IQ. The second stageis a high-gain stage having an in v erting or negative gain. A capacitor connects the outputof this stage to its input. This capacitor is called the compensating capacitor. Other namesfor it are lag capacitor, Miller capacitor, and pole-splitting capacitor. It sets the bandwidthof the circuit to a value so that the op am p is stable, i.e. so t hat it does not oscillate. Theoutput stage is a unity -gain stage which provides the curren t gain to drive the load.Figure 1: Op amp model.If we assume that Q1and Q2are matched, that Q3and Q4are matched, that basecurrents can be neglected, and that the Early effect can be neglected, w e can write thefollowing equation for iO1:iO1= iC1− iC3= iC1− iC4= iC1−iC2(1)But iC1+ iC2= IQand iC1= IQ/2+ic1.ThusweobtainiO1=2iC1− IQ=2ic1(2)1Open-Loop Transfer FunctionWe wish to solve for the transfer function for Vo/Vid,whereVidis the difference voltagebetw e en the t wo op amp inputs. First, w e solve for the current Ic1as a function of Vid.Forthe diff amp, let us assume that the transistors are matched, that IE1= IE2= IQ/2,theEarly effect can be neglected, and the base currents are zero. In this cas e, the sm all-sig nalac emitter equivalent circuit of the diff amp is the circuit given in Fig. 2(a). In this circuit,re1and re2are the intri nsic emitter resistances given b yre1= re2= re=VTIE=2VTIQ(3)Note that the dc tail supply IQdoes not appear in this circuit because it is not an ac source.From the em itte r equivalen t circuit, it follo w s thatIc1= Ie1=Vid2(re+ RE)(4)where Ic1= Ie1becausewehaveassumedzerobasecurrents.Figure 2: Circuit for calculating Ie1. (b) Circuit for calculating Vo.Figure 2(b) shows the equiv alent circuit which w e use to calculate Vo. We assume thatReqis the effective load resistance on the current 2Ic1. In this case, the current which flowsthrough the compensating capacitor Ccis giv en b yIo1=2Ic1+Vo1Req=Vidre+ RE−Vo2KReq(5)wherewehaveusedEq.(4)andtherelationVo1= −Vo2/K. The voltage Vo2is giv en b yVo2= Vo1+Io1Ccs=−Vo2K+·Vidre+ RE−Vo2KReq¸1Ccs(6)If w e assume that the output stage has a gain that is approx imately unity, then Vo' Vo2.Let G (s)=Vo/Vid. It follows from Eq. (6) that G (s) is given b yG (s)=VoVid'Vo2Vid=KReqre+ RE×11+(1+K) ReqCcs(7)This is of the formG (s)=A1+s/ω1(8)where A and ω1are giv en byA =KReqre+ REω1=2πf1=1(1 + K) ReqCc(9)2Figure 3: Asymptotic Bode magnitude plots. (a) Without feedback. (b) With feedback.Ga in Band width Pr oduc tThe asymptotic Bode magnitude plot for |G (jω)| isshowninFig. 3(a). Abovethepolefrequency ω1, the plot has a slope of −1 dec/dec or −20 dB/dec. The frequ ency at which|G (jω)| =1is called the unity-gain frequency or the gain-bandwidth product. It is labeledωxin the figure and is given byωx=2πfx= Aω1=K1+K1re+ RECc'1(re+ RE) Cc(10)where the approximation holds for K À 1. It follows that an alternate expression for G (s)isG (s)=A1+sA/ωx(11)For maxim um bandwidth, fxshould be as large as possible. However, if fxis too large,the op amp can oscillate. A value of 1 MHz is t ypical for general purpose op amps.Example 1 Anopampistobedesignedforfx=4MHz and IQ=50µA. If RE=0,calculate the required value for Cc.Solution. Cc=1/ (2πfxre)=IQ/ (4πfxVT)=38.4 pF, where w e assume that VT=0.0259V.Slew RateThe op amp slew rate is the maximum value of the time derivativ e of its output v oltage. Ingeneral, the positive and negative slew rates can be different. The simple mo del of Fig. 1predicts that the two are e qua l so that we can write−SR ≤dvOdt≤ +SR (12)where SR is the slew rate. To solve for it, w e use Eqs. (5) an d (6) to wri teVo2= Vo1+Io1Ccs=−Vo2K+µ2Ic1+−Vo2KReq¶1Ccs(13)3This can be rearr anged to obtainsVo 2·1+1Kµ1+1ReqCc¶¸=2Ic1Cc(14)If w e assume that K is large and let Vo2' Vo, this equation reduces tosVo'2Ic1Cc(15)The s operator in a phasor equation becomes the d/dt operator in a time-domain equation.Thus w e can writedvodt=2ic1Cc(16)It follow s that the slew rate is determined by the maximum value of ic1. The total collectorcurren t in Q1is the sum of the dc value plus the small-signal ac value. Thus w e can writeiC1= IQ/2+ic1. This current has the limits 0 ≤ iC1≤ IQ. It follows that the small-signalac component has the limits −IQ/2 ≤ ic1≤ IQ/2.Thuswecanwrite−IQCc≤dvodt≤+IQCc(17)It follow s that the slew rate is giv en bySR =IQCc(18)Example 2 C alculate the slew rate of the op am p of Example 1.Solution. SR = IQ/Cc=1.30 V/µs.Relations between Slew Rate and Gain-Bandwidth ProductIf Ccis eliminated between Eqs. (10) and (18), we obtain the relationSR =2πfxIQ(re+ RE)=4πfxVTµ1+IQRE2VT¶(19)This equation clearly shows that the slew ra te is fixed by the gain-bandwidth product ifRE=0.IfRE> 0, the slew rate and gain bandwidth product can be specified independen tly.Example 3 Emitter resistors with the value RE=3kΩ are added to the input diff amp inthe op amp of Example 1. If fxis to be held constant, calculate the new value of the slewrate and the new value of Cc.Solution. SR =2πfxIQ(re+ RE)=5.07 V/µs. Cc= IQ/SR =9.86 pF. The slew rate isgreater by a facto r of 3.9 and Ccis smaller by the same factor.The above example illustrates how the slew rate of an op amp can be increased withoutc hanging its gain-bandwidth product. When REis added, ωxdecreases. To mak e ωxequalto its original value, Ccmust be decreased, and this increases the slew rate. It can be seenfrom Eq. (19) that the slew ra te can also be increased by increasing IQ. How ever, this causesωxto increase. To make ωxequal to its original value, REmust also be increased. Therefore,the general rule for incre asing the slew rate is to either decrease Cc,increaseIQ,orboth.All of these make ωxincreas e. To bring ωxback down to its original value, REmust beincreased. The c


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GT ECE 4435 - An Op Amp Tutorial

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