Effects of Op-Amp Finite Gain and Bandwidth

Transcription

1 Effects of Op-Amp Finite Gain and Bandwidth Open-Loop Transfer Function Inouranalysisofop-ampcircuitsthisfar,wehaveconsideredtheop-ampstohaveaninfinitegain and an infinite bandwidth. This is not true for physical op-amps. In this section, we examine the effects of a non-infinite gain and non-infinite bandwidth on the inverting and the non-inverting amplifier circuits. Fig. 1 shows the circuit symbol of an op-amp having an open-loop voltage-gain transfer function As). The output voltage is given by As)V + V ) 1) wherecomplexvariablenotationisused. WeassumeherethatAs)canbemodeledbyasingle-pole low-pass transfer function of the form As) 1+s/ω 0 2) where isthedcgainconstantandω 0 isthepolefrequency. Mostgeneralpurposeop-ampshave a voltage-gain transfer function of this form for frequencies such that Ajω) 1. Figure 1: Op-amp symbol. Gain-Bandwidth Product Figure 2 shows the Bode magnitude plot for Ajω). The radian gain-bandwidth product is defined asthefrequencyω x forwhich Ajω) 1. Itisgivenby ω x ω 0 A ω 0 3) where we assume that >> 1. This equation illustrates why ω x is called a gain-bandwidth product. Itisgivenbytheproductofthedcgainconstant andtheradianbandwidthω 0. Itis commonlyspecifiedinhzwiththesymbolf x,wheref x ω x /2π. Manygeneralpurposeop-amps haveagain-bandwidthproductf x 1MHzandadcgainconstant Itfollowsfrom Eq. 3) that the corresponding pole frequency in the voltage-gain transfer function for the general purposeop-ampisf / ) 5Hz. Non-Inverting Amplifier Figure 3 shows the circuit diagram of a non-inverting amplifier. For this circuit, we can write by inspection As)V i V ) 4) V +R F 5) 1

2 Figure2: Bodeplotof Ajω). Simultaneous solution for the voltage-gain transfer function yields V i As) 1+As) / +R F ) 1+R F / 1+1+R F / )/As) For s jω and 1+R F / )/Ajω) << 1, this reduces to /V i 1+R F / ). This is the gainwhichwouldbepredictediftheop-ampisassumedtobeideal. 6) Figure 3: Non-inverting amplifier. WhenEq. 2)isusedforAs),itisstraightforwardtoshowthatEq. 6)canbewritten V i f 1+s/ω 0f 7) wheref isthegainconstantwithfeedbackandω 0f istheradianpolefrequencywithfeedback. Thesearegivenby f 1+ / +R F ) 1+R F / 8) 1+1+R F / )/ ω 0f ω 0 1+ A ) 0 9) +R F It follows from these two equations that the radian gain-bandwidth product of the non-inverting amplifier with feedback is given by f ω 0f ω 0 ω x. This is the same as for the op-amp withoutfeedback. Fig. 4showstheBodemagnitudeplotsforboth /V i andajω). Thefigure showsthatthebreakfrequencyontheplotfor /V i liesonthenegative-slopeasymptoteofthe plotforajω). 2

3 Figure4: Bodeplotfor /V i. Example 1 At very low frequencies, an op-amp has the frequency independent open-loop gain As) Theop-ampistobeusedinanon-invertingamplifier. Thetheoreticalgain is calculated assuming that the op-amp is ideal. What is the highest theoretical gain that gives an errorbetweenthetheoreticalgainandtheactualgainthatislessthan1%? Solution. Thetheoreticalgainisgivenby1+R F / ). Theactualgainisalwayslessthanthe theoreticalgain. Foranerrorlessthan1%,wecanuseEq. 6)towrite < R F / )/ ) Thiscanbesolvedfortheupperboundonthetheoreticalgaintoobtain 1+ R ) F 1 < Example 2 An op-amp has a gain-bandwidth product of 1MHz. The op-amp is to be used in a non-inverting amplifier circuit. Calculate the highest gain that the amplifier can have if the halfpoweror 3dBbandwidthistobe20kHzormore. Solution. Theminimumbandwidthoccursatthehighestgain. Forabandwidthof20kHz,we canwritef Solutionforf yieldsf 50. Example 3 Two non-inverting op-amp amplifiers are operated in cascade. Each amplifier has a gain of 10. If each op-amp has a gain-bandwidth product of 1 MHz, calculate the half-power or 3 db bandwidth of the cascade amplifier. Solution. Eachamplifierbyitselfhasapolefrequencyof10 6 /10100kHz,correspondingto a radian frequency ω 0f 2π 100,000. The cascade combination has the voltage-gain transfer function given by ) V i 1+s/ω 0f The half-power frequency is obtained by setting s jω and solving for the frequency for which /V i /2. Ifweletxω/ω 0f,theresultingequationis ) x

4 Thisequationreducesto1+x 2 2. Solutionforxyieldsx Itfollowsthatthehalf-power frequencyis kHz64.4kHz. Inverting Amplifier Figure 5a) shows the circuit diagram of an inverting amplifier. Fig. 5b) shows an equivalent circuitwhichcanbeusedtosolveforv. Byinspection,wecanwrite As)V 10) Vi V + V ) o R F ) 11) R F These equations can be solved for the voltage-gain transfer function to obtain 1/)As) R F ) V i 1+1/R F )As) R F ) R F / 1+1+R F / )/As) Forsjωand 1+R F / )/Ajω) <<1,thevoltage-gaintransferfunctionreducesto /V i R F /. Thisisthegainwhichwouldbepredictediftheop-ampisassumedtobeideal. 12) Figure5: a)invertingamplifier. b)equivalentcircuitforcalculating. When Eq. 2) is used for As), it is straightforward to show that the voltage-gain transfer function reduces to f 13) V i 1+s/ω 0f wheref isthegainconstantwithfeedbackandω 0f istheradianpolefrequencywithfeedback. Thesearegivenby f 1/) R F ) 1+1/R F ) R F ) R F / ω 0f ω 0 1+ R F R F 14) 1+1+R F / )/ ) 15) )ω R F Notethatf isdefinedhereasapositivequantitysothatthenegativesignfortheinvertinggain isretainedinthetransferfunctionfor /V i. Let the radian gain-bandwidth product of the inverting amplifier with feedback be denoted by ω x. It follows from Eqs. 14) and 15) that this is given by ω x f ω 0f ω x R F /R F + ). This is less than the gain-bandwidth product of the op-amp without feedback by the factor R F /R F + ). Fig. 6showstheBodemagnitudeplotsfor /V i andforajω). Thefrequency labeledω 0f is the break frequency for thenon-invertingamplifier with thesame gain magnitude as the inverting amplifier. The non-inverting amplifier with the same gain has a bandwidth that is greaterbythefactor1+ /R F ). Thebandwidthoftheinvertingandthenon-invertingamplifiers isapproximatelythesameif /R F <<1. Thisisequivalenttotheconditionthatf >>1. 4

5 Figure6: Bodeplotfor /V i. Example 4 An op-amp has a gain-bandwidth product of 1MHz. Compare the bandwidths of an invertingandanon-invertingamplifierwhichusetheop-ampforf 1,2,5,and10. Solution. Thenon-invertingamplifierhasabandwidthoff x /f. Theinvertingamplifierhas a bandwidth of f x /f ) R F / +R F ). If we approximate f of the inverting amplifier by f R F /, its bandwidth reduces to f x /1+f ). The calculated bandwidths of the two amplifiers are summarized in the following table. f Non-Inverting Inverting 1 1 MHz 500 khz khz 333 khz khz 167 khz khz 91 khz Forthecaseoftheidealop-amp,theV inputtotheinvertingamplifierisavirtualgroundso that the input impedance Z in is resistive and equal to. For the op-amp with finite gain and bandwidth,thev terminalisnotavirtualgroundsothattheinputimpedancediffersfrom. WeusethecircuitinFig. 5a)tosolvefortheinputimpedanceasfollows: Z in V i I 1 V i V i V )/ 1 V / ) /V i ) To put this into the desired form, we let V / 1/As) and use Eq. 12) for /V i. The equationforz in reducesto [ Z in + R F 1+As) RF R ) ] 1 1 F s 17) R F ω 0 whereeq. 2)hasbeenused. ItfollowsfromthisequationthatZ in consistsoftheresistor in serieswithanimpedancethatconsistsoftheresistorr F inparallelwiththeseriescombinationof aresistorr 2 andaninductorlgivenby 16) R 2 R F 18) L R F ω 0 19) 5

6 TheequivalentcircuitforZ in isshowninfig. 7a). If, itfollowsthatr 2 0and L 2 0sothatZ in. TheimpedancetransferfunctionforZ in isoftheformofahigh-pass shelving transfer function given by Z in s)r DC 1+s/ω z 1+s/ω p 20) wherer DC isthedcresistance,ω p isthepolefrequency,andω z isthezerofrequency. Theseare given by R DC + R F 1+ 21) ω p R 2+R F L ω z R 2+R F L TheBodemagnitudeplotforZ in isshowninfig. 7b). ω 0 1+ ) 22) ω 0 1+ A ) 0 +R F 23) Figure7: a)equivalentinputimpedance. b)bodeplotfor Z in. Example 5 At very low frequencies, an op-amp has the frequency independent open-loop gain As) The op-amp is to be used in an inverting amplifier with a gain of WhatistherequiredratioR F /? ForthevalueofR F /,howmuchlargeristheinputresistance than? Solution. ByEq. 12),wehave ThiscanbesolvedforR F / toobtain R F / R F / )/ ) R F /1000) ByEq. 17),theinputresistancecanbewritten R in 1+ R ) F/ 1005 )

7 Example 6 Anop-amphasadcgain andagainbandwidthproductf x 1MHz. The op-amp is used in the inverting amplifierof Fig. 5a). The circuit element values are 1kΩ andr F 100kΩ. Calculatethedcgainoftheamplifier,theuppercutofffrequency,andthevalue of the elements in the equivalent circuit for the input impedance. In addition, calculate the zero andthepolefrequenciesinhzfortheimpedancebodeplotoffig. 7b). Solution. Thedcvoltagegainis f. Eq. 14)canbeusedtocalculatef toobtain f k 100k)/1k k 100k)/100k ByEqs. 3)and15),theuppercutofffrequencyf 0f isgivenby f of ω 0f 2π k 100k ) 9.91kHz 100k The element values in the equivalent circuit of Fig. 7a) for the input impedance are as follows: 1kΩ,R F 100kΩ,R 2 0.5Ω,and L15.9mH whereeqs. 18)and19)havebeenusedforR 2 andl. ThepoleandzerofrequenciesintheBode impedance plot are f p f 0 1+ ) f x 1+ )1.00MHz ) f z f kHz +R F 7