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# An Op Amp Tutorial

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An Op Amp Tutorial(Based on material in the book Introduction to Electroacoustics and Audio Am-pliﬁer Design, Second Edition - Revised Printing, by W. Marshall Leach, Jr.,published by Kendall/Hunt,°c 2001.)Anopamphas twoinputs andoneoutput. The circuit isdesignedsothat theoutputvoltageis proportional to the di ﬀerence between the two input voltages. In general, an op amp canbe modeled as a three-stage circuit as shown in Fig. 1. The non-inverting input is v .TheI1inverting input is v . The input stage is a di ﬀerential ampliﬁer (Q and Q ) with a currentI2 1 2mirrorload(Q −Q ). Thedi ﬀamptailsupply isthe dccurrentsourceI . Thesecond stage3 5 Qis a high-gain stage having an inverting 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 amp is stable, i.e. so that it does not oscillate. Theoutput stage is a unity-gain stage which provides the current gain to drive the load.Figure 1: Op amp model.If we assume that Q and Q are matched, that Q and Q are matched, that base1 2 3 4currents can be neglected, and that the Early e ﬀect can be neglected, we can write thefollowing equation for i :O1i =i −i = i −i = i −i (1)O1 C1 C3 C1 C4 C1 C2But i +i =I and i =I /2+i .ThusweobtainC1 C2 Q C1 Q c1i =2i −I =2i (2)O1 C1 Q c11Open-Loop Transfer ...

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An Op Amp Tutorial
(Based on material in the bookIntroduction to Electroacoustics and Audio Am-pliÞer Design, Second Edition - Revised Printing, by W. Marshall 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 voltage is proportional to the dierence between the two input voltages. In general, an op amp can be modeled as a three-stage circuit as shown in Fig. 1. The non-inverting input isvI1. The inverting input isvI2input stage is a di. The erential ampliÞer (Q1andQ2) with a current mirror load (Q3Q5). The diamp tail supply is the dc current sourceIQsecond stage. The is a high-gain stage having an inverting or negative gain. A capacitor connects the output of this stage to its input. This capacitor is called the compensating capacitor. Other names for it are lag capacitor, Miller capacitor, and pole-splitting capacitor. It sets the bandwidth of the circuit to a value so that the op amp is stable, i.e. so that it does not oscillate. The output stage is a unity-gain stage which provides the current gain to drive the load.
Figure 1: Op amp model.
If we assume thatQ1andQ2are matched, thatQ3andQ4are matched, that base currents can be neglected, and that the Early eect can be neglected, we can write the following equation foriO1:
iO1=iC1iC3=iC1iC4=iC1iC2
ButiC1+iC2=IQandiC1=IQ/2 +ic1. Thus we obtain
iO1= 2iC1IQ= 2ic1
1
(1)
(2)
Open-Loop Transfer Function
We wish to solve for the transfer function forVo/Vid, whereVidis the dierence voltage between the two op amp inputs. First, we solve for the currentIc1as a function ofVid. For the diamp, let us assume that the transistors are matched, thatIE1=IE2=IQ/2, the Early eect can be neglected, and the base currents are zero. In this case, the small-signal ac emitter equivalent circuit of the diamp is the circuit given in Fig. 2(a). In this circuit, re1andre2are the intrinsic emitter resistances given by VT2VT re1=re2=re= =(3) IEIQ Note that the dc tail supplyIQdoes not appear in this circuit because it is not an ac source. From the emitter equivalent circuit, it follows that V id Ic1=Ie1=(4) 2 (r+R) e E whereIc1=Ie1because we have assumed zero base currents.
Figure 2: Circuit for calculatingIe1. (b) Circuit for calculatingVo.
Figure 2(b) shows the equivalent circuit which we use to calculateVoassume that. We Reqis the eective load resistance on the current2Ic1this case, the current which. In ßows through the compensating capacitorCcis given by V V V o1id o2 Io1= 2Ic1+ =(5) Reqre+REKReq where we have used Eq. (4) and the relationVo1=Vo2/Kvoltage. The Vo2is given by · ¸ Io1Vo2VidVo21 Vo2=Vo1++ = (6) Ccs K re+REKReqCcs If we assume that the output stage has a gain that is approximately unity, thenVo'Vo2. LetG(s) =Vo/Vid. It follows from Eq. (6) thatG(s)is given by VoVo2KReq1 G(s) ='=×(7) VidVidre+RE1 + (1 +K)ReqCcs This is of the form A G(s) =(8) 1 +s/ω1 whereAandω1are given by KReq1 A=ω1= 2πf1=(9) re+RE(1 +K)ReqCc
2
Figure 3: Asymptotic Bode magnitude plots. (a) Without feedback. (b) With feedback.
Gain Bandwidth Product
The asymptotic Bode magnitude plot for|G(jω)|the pole3(a). Above is shown in Fig. frequencyω1, the plot has a slope of1dec/dec or20frequency at whichdB/dec. The |G(jω)|= 1is called the unity-gain frequency or the gain-bandwidth product. It is labeled ωxin theÞgure and is given by
K1 1 ωx= 2πfx=Aω1=' 1 +K re+RECc(re+RE)Cc
(10)
where the approximation holds forKÀ1. It follows that an alternate expression forG(s) is A G(s) =(11) 1 +sA/ωx For maximum bandwidth,fxshould be as large as possible. However, iffxis too large, the op amp can oscillate. A value of1MHz is typical for general purpose op amps.
Example 1An op amp is to be designed forfx= 4MHz andIQ= 50µA. IfRE= 0, calculate the required value forCc.
Solution.Cc= 1/(2πfxre) =IQ/(4πfxVT) = 38.4pF, where we assume thatVT= 0.0259 V.
Slew Rate
The op amp slew rate is the maximum value of the time derivative of its output voltage. In general, the positive and negative slew rates can be di1simple model of Fig. erent. The predicts that the two are equal so that we can write
dvO SR≤ ≤+SR dt
whereSRis the slew rate. To solve for it, we use Eqs. (5) and (6) to write µ ¶ Io1Vo2Vo21 Vo2=Vo1+ = + 2Ic1+ C s K KR C s c eq c
3
(12)
(13)
This can be rearranged to obtain · µ ¶¸ 1 1 2Ic1 sVo2=1 + 1 + (14) K R C C eq c c If we assume thatKis large and letVo2'Vo, this equation reduces to 2Ic1 sVo'(15) Cc Thesoperator in a phasor equation becomes thed/dtoperator in a time-domain equation. Thus we can write dv2i o c1 =(16) dt Cc It follows that the slew rate is determined by the maximum value ofic1total collector. The current inQ1is the sum of the dc value plus the small-signal ac value. Thus we can write iC1=IQ/2 +ic1current has the limits. This 0iC1IQ. It follows that the small-signal ac component has the limitsIQ/2ic1IQ/2we can write. Thus IQdvo+IQ ≤ ≤(17) Ccdt Cc It follows that the slew rate is given by IQ SR=(18) Cc Example 2Calculate the slew rate of the op amp of Example 1. Solution.SR=IQ/Cc= 1.30V/µs.
Relations between Slew Rate and Gain-Bandwidth Product
IfCc(10) and (18), we obtain the relationis eliminated between Eqs. µ ¶ IQRE SR= 2πfxIQ(re+RE) = 4πfxVT1 +(19) 2VT This equation clearly shows that the slew rate isÞxed by the gain-bandwidth product if RE= 0. IfRE>0, the slew rate and gain bandwidth product can be speciÞed independently. Example 3Emitter resistors with the valueRE= 3kare added to the input diamp in the op amp of Example 1. Iffxis to be held constant, calculate the new value of the slew rate and the new value ofCc. Solution.SR= 2πfxIQ(re+RE) = 5.07V/µs.Cc=IQ/SR= 9.86pF. The slew rate is greater by a factor of3.9andCcis smaller by the same factor. The above example illustrates how the slew rate of an op amp can be increased without changing its gain-bandwidth product. WhenREis added,ωxdecreases. To makeωxequal to its original value,CcIt can be seenmust be decreased, and this increases the slew rate. from Eq. (19) that the slew rate can also be increased by increasingIQthis causes. However, ωxto increase. To makeωxequal to its original value,REmust also be increased. Therefore, the general rule for increasing the slew rate is to either decreaseCc, increaseIQ, or both. All of these makeωxbringincrease. To ωxback down to its original value,REmust be increased. The change inREdoes not aect the slew rate.
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Figure 4: (a) Ampli Þlter.
er with feedback. (b) AmpliÞer with feedback preceded by a low-pass
Closed-Loop Transfer Function
Figure 4(a) shows the op amp with a two resistor voltage divider connected as a feedback network. The output voltage can be written
V=G(s) (VV) =G(s) (VbV) o i f i o
wherebis the gain of the voltage divider given by
R1 b= R1+RF
Note that0b1. Eq. (20) can be solved forVo/Vito obtain
VoG(s)Af = = Vi1 +bG(s+) 1 s/ω1f
(20)
(21)
(22)
where Eq. (11) is used forG(s)dc gain. The Afand the pole frequencyω1fare given by
A1 Af=' 1 +bAfb 1 +bAωx ω1f= 2πf1f=ωx='bωx A Af
(23)
(24)
where thefin the subscript implies “with feedback” and the approximations assume that bAÀ1. It can be seen from these equations that
Afω1f=Aω1=ωx
(25)
Fig. 3(b) shows the Bode plot for|Vo/Vi|for two values ofAf. Asbis increased,Af decreases and the bandwidthω1fincreases so that the product of the two remain constant. This illustrates whyωxis called the gain-bandwidth product.
Example 4An op amp has the gain bandwidth productfx= 8the upperMHz. Calculate 3dB frequencyfuif the op amp is operated at a voltage gain of21.
Solution. The upper3dB frequency is equal to the pole frequency of the closed-loop transfer function. Thusfu=f1f=fx/Af= 381kHz.
5
Figure 5: (a) No slewing step response. input voltage.
Transient Response
(b) Step response with slewing.
(c) Dierential
Let the input voltage to the op amp in Fig. 4(a) be a step of amplitudeV1. We can write vI(t) =V1u(t), whereu(t)The Laplace transform ofis the unit step function. vI(t)is Vi(s) =V1/s. The Laplace transform of the output voltage is given by
V1Af Vo(s) = s1 +s/ω1f
(26)
The time domain output voltage is obtained by taking the inverse Laplace transform to obtain vO(t) =AfV1[1exp (ω1ft)]u(t)(27) A plot ofvO(t)5(a).is shown in Fig. The maximum time derivative ofvO(t)occurs att= 0and is given by
dvOd Af =AfV1{[1exp (ω1ft)]u(t)}|=V1=ωxV1 ¯ t=0 dt dtω1f max
(28)
If the derivative exceeds the slew rate of the op amp , the output voltage will be distorted as shown in Fig. 5(b), where the non-slewing response is shown by the dashed line. The maximum value ofV1before the op amp slews is given by
SR SR V1max= = =IQ(re+RE) ω2πf x x
Example 5Calculate the maximum value ofV1for the op amps of Examples 2 and 3.
(29)
Solution. For Example 1,V1max=SR/ωx= 51.7mV. For Example 2,V1max= 202mV. This is greater by about a factor of3.9the same as the ratio of the two slew rates., i.e.
Input Stage Overload
For the step input signal to the op amp with feedback in Fig. voltage is given by
4(a), the dierential input
V1 vI D(t) =vI(t)bvO(t[1 +) = bA0exp (bωxt)]u(t) 1 +bA
6
(30)
It follows thatvID(0) =V1andvID() =V1/(1 +bA). A plot ofvI D(t)is shown in Fig. 5(c). The peak voltage occurs att= 0. If the op amp is not to slew, the diamp input stage must not overload with this voltage. If base currents are neglected, the emitter and collector currents inQ1andQ2can be writtenµ ¶ µ ¶ vBE1vB E2 iE1=iC1=ISexpiE2=iC2=ISexp(31) VTVT whereISis the BJT saturation current. The dierential input voltage can be written
vI D= (vB E1vB E2) + (iE1iE2)RE
With the relationiE2=IQiE1, these equations can be solved to obtain µ ¶ iC1 vI D=VT(2ln + iC1IQ)RE IQiC1
(32)
(33)
The same equation holds foriC2exceptvI Dis replaced withvI D. BothiC1andiC2must satisfy the inequality0iIQ. At either limit of this inequality, one transistor in the diamp is cut ous consider the di. Let -amp active range to be the range for whichiC1andiC2satisfy0.05IQi0.95IQis the. This 5%to95%range for the currents. When the diamp is operated in this range, it follows from Eq. (33) thatvI D must satisfy vI D(max)vI D≤ −vI D(max)(34) wherevI D(max)is given by vI D(max)=VTln 19 + 0.9IQRE(35) IfvI Dlies in this range, neither transistor in the diamp can cut oand the op amp cannot exhibit slewing.
Example 6CalculatevI D(max)forIQ= 50µA forRE= 0and forRE= 3k.
Solution. ForRE= 0, we havevI D(max)=VT= 76ln 19 .3mV. ForRE= 3000,vI D(max)= 6 76.3mV+0.9×50×10×3000 = 211mV. These values are greater than the values of V1maxcalculated in Example 5 because the analysis here is based on the large signal behavior of the BJT.
Full Power Bandwidth
Figure 6 shows the output voltage of an op amp with a sine wave input for two cases, one where the op amp is not slewing and the other where the op amp is driven into full slewing. The full slewing waveform is a triangle wave. The slew-limited peak voltage is given by the slope multiplied by one-fourth the period, i.e.
T SR VPslew=SR×= 4 4f
(36)
whereT= 1/f. When the op amp is driven into full slewing, an increase in the amplitude of the input signal causes no change in the amplitude of the output signal. If the frequency is doubled, the amplitude of the output signal is halved.
7
Figure 6: Non-slewing and full slewing output voltage waveforms.
Let the input voltage to the op amp be a sine wave. If the op amp does not slew and is not driven into peak clipping, the output voltage can be writtenvO(t) =VPsinωt. The time derivative is given bydvO/dt=ωVPcosωtmaximum value of. The |dvO/dt|occurs at ωt=nπ, wherenis an integer, and is given by|dvO/dt|=ωVP. For a physical op amp max , this cannot exceed the slew rate, i.e.ωVP< SRfollows that the maximum frequency. It that the op amp can put out the sine wave without slewing is given by
SR fmax= 2πVP
Conversely, the peak output voltage without slewing is given by
SR V= P(max) 2πf
(37)
(38)
LetVclipIf an op amp is drivenbe the op amp clipping voltage at midband frequencies. at this level and the frequency is increased, the op amp will eventually slew and the maxi-mum output voltage will decrease as the frequency is increased. The full power bandwidth frequencyfFPBWis deÞned as the highest frequency at which the op amp can put out a sine wave with a peak voltage equal toVclipis given by. It
SR fFPBW= 2πV clip
(39)
Figure 7 shows the peak output voltage versus frequency for a sine wave input signal. At low frequencies, the peak voltage is limited to the op amp clipping voltageVclip. As frequency is increased, the peak voltage becomes inversely proportional to frequency when the op amp is driven into full slewing and is given bySR/4f. TheÞgure also shows the peak voltage below which the op amp does not slew. It is given bySR/2πf.
Example 7The op amps of Examples 2 and 3 have clipping voltages of±13V. Calculate the full power bandwidth frequency if the op amps are not to slew at maximum output.
6 Solution. For the op amp of Example 2,fFPBW= 1.3×10/(2π13) = 15.9thekHz. For 6 op amp of Example 3,fFPBW= 5.07×10/(2π13) = 62.1kHz.
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