ADNAN MENDERES UNIVERSITY FACULTY OF ENGINEERING … · 2020. 3. 19. · Ch8: Operational Amplifiers (Op-Amps) • The term “operational amplifier” (op amp) was coined in the

ADNAN MENDERES UNIVERSITY FACULTY OF ENGINEERING  

Department of Electrical and Electronics Engineering

EE302–Electronics-II2019–2020Spring

Assoc.Prof.Dr.OlcayÜZENGİAKTÜRK

Ch8: Operational Amplifiers (Op-Amps)

• The term “operational amplifier” (op amp) was coined in the 1940s, well before the invention of the transistor and the integrated circuit. Op-amps realized by vacuum tubes served as the core of electronic “integrators,” “differentiators,” etc., thus forming systems whose behavior followed a given differential equation. Called “analog computers,” such circuits were used to study the stability of differential equations that arose in fields such as control or power systems. Since each op amp implemented a mathematical operation (e.g., integration), the term “operational amplifier” was born.

• Op-amps find wide application in today’s discrete and integrated electronics. In the cellphones integrated op amps serve as building blocks in (active) filters. Similarly, the analog-to-digital converter(s) used in digital cameras often employ op-amps.

• The outline of this chapter is shown below:

EE302-Electronics,Assoc.Prof.Dr.OlcayÜZENGİAKTÜRK,2019-2020Fall


Ch8: Operational Amplifiers (Op-Amps)• The operational amplifier can be abstracted as a black box having two inputs and one

output. • Shown in Fig. 8.1(a), the op amp symbol distinguishes between the two inputs by the

plus and minus sign; Vin1 and Vin2 are called the “noninverting” and “inverting” inputs, respectively. We view the op-amp as a circuit that amplifies the difference between the two inputs, arriving at the equivalent circuit depicted in Fig. 8.1(b).

• The voltage gain is denoted by A0:


• It is instructive to plot Vout as a function of one input while the other remains at zero.



Ch8: Operational Amplifiers (Op-Amps)• How does the “ideal” op amp behave? Such an op amp would provide an infinite voltage

gain, an infinite input impedance, a zero output impedance, and infinite speed.

• In fact, the first-order analysis of an op-amp-based circuit typically begins with this idealization, quickly revealing the basic function of the circuit.

• We can then consider the effect of the op-amp “nonidealities” on the performance.

• The very high gain of the op-amp leads to an important observation. Since realistic circuits produce finite output swings, e.g., 2V, the difference between Vin1 and Vin2 in Fig. 8.1(a) is always small:

• In other words, the op-amp, along with the circuitry around it, brings Vin1 and Vin2 close to each other. Following the above idealization, we may say Vin1 = Vin2 if A0 =∞.


Ch8: Operational Amplifiers (Op-Amps)Non-inverting Amplifier


Ch8:Op-Amps–LinearOp-AmpCircuits

Non-inverting Amplifier



Inverting Amplifier



Inverting Amplifie``r


Integrator & Differentiator

• Our study of the inverting topology in previous sections has assumed a resistive network around the op-amp.

• In general, it is possible to employ complex impedances instead (Fig. 8.9). We can write;

• where the gain of the op-amp is assumed large. If Z1 or Z2 is a capacitor, two interesting functions result.



Integrator



Differentiator



Summing Amplifier (Voltage Adder) • The need for adding voltages arises in many applications.

• For example, in audio recording, for example, a number of microphones may convert the sounds of various musical instruments to voltages, and these voltages must then be added to create the overall musical piece. This operation is called “mixing” in the audio industry.

• For example, in “noise cancelling” headphones, the environmental noise is applied to an inverting amplifier and subsequently added to the signal so as to cancel itself.



• It is possible to implement useful nonlinear functions through the use of op-amps and nonlinear devices such as transistors. The virtual ground property plays an essential role here as well.

Ch8:Op-Amps–NonlinearOp-AmpCircuits


Precision Rectifier • The rectifier circuits suffer from a “dead zone” due to the finite voltage required to

turn on the diodes. That is, if the input signal amplitude is less than approximately 0.7 V, the diodes remain off and the output voltage remains at zero.

• This drawback prohibits the use of the circuit in high-precision applications, e.g., if a small signal received by a cellphone must be rectified to determine its amplitude.

• It is possible to place a diode around an op amp to form a “precision rectifier,” i.e., a circuit that rectifies even very small signals.

• Let us begin with a unity-gain buffer tied to a resistive load [Fig. 8.22(a)].



Precision Rectifier • We note that the high gain of the op amp ensures that node X tracks Vin (for both positive

and negative cycles). Now suppose we wish to hold X at zero during negative cycles, i.e., “break” the connection between the output of the op amp and its inverting input. This can be accomplished as depicted in Fig. 8.22(b), where D1 is inserted in the feedback loop. Note that Vout is sensed at X rather than at the output of the op-amp.



Logarithmic Amplifier • Consider the circuit of Fig. 8.24, where a bipolar transistor is placed around the op-

amp. With an ideal op amp, R1 carries a current equal to Vin/R1 and so does Q1.


➢ Logarithmic amplifiers (“logamps”) prove useful in applications where the input signal level may vary by a large factor. It may be desirable in such cases to amplify weak signals and attenuate (“compress”) strong signals hence a logarithmic dependence.

➢ Note that Q1 operates in the active region because both the base and the collector remain at zero.

➢ What happens if Vin becomes negative?


Square-root Amplifier • Recognizing that the logarithmic amplifier of Fig. 8.24 in fact implements the inverse

function of the exponential characteristic, we surmise that replacing the bipolar transistor with a MOSFET leads to a “square-root” amplifier.


} Illustrated in Fig. 8.25, such a circuit requires that M1 carry a current equal to Vin/R1:

} If Vin is near zero, then Vout remains at −VTH, placing M1 at the edge of conduction. As Vin becomes more positive, Vout falls to allow M1 to carry a greater current. With its gate and drain at zero, M1 operates in saturation.


Op-Amp Nonidealities

• Our study in previous sections has dealt with a relatively idealized op-amp model (except for the finite gain) so as to establish insight. In practice, however, op amps suffer from other imperfections that may affect the performance significantly. In this section, we deal with such nonidealities.

Ch8:Op-Amps–Nonidealities


DC Offsets • The op-amp characteristics imply that Vout = 0 if Vin1 = Vin2. In reality, a zero input

difference may not give a zero output difference! Illustrated in Fig. 8.26(a), the characteristic is “offset” to the right or to the left; i.e., for Vout = 0, the input difference must be raised to a certain value, Vos , called the input “offset voltage.”

• What causes offset? The internal circuit of the op-amp experiences random asymmetries (“mismatches”) during fabrication and packaging. For example, as shown conceptually in Fig. 8.26(b), the bipolar transistors sensing the two inputs may display slightly different base-emitter voltages. The same effect occurs for MOSFETs. We model the offset by a single voltage source placed in series with one of the inputs [Fig. 8.26(c)]. Since offsets are random and hence can be positive or negative, Vos can appear at either input with arbitrary polarity.



DC Offsets • Why are DC offsets important? • Let us reexamine some of the circuit topologies studied in Section 8.2 in the presence

of op amp offsets. • Depicted in Fig. 8.27, the noninverting amplifier now sees a total input of Vin + Vos ,

thereby generating

• In other words, the circuit amplifies the offset as well as the signal, thus incurring accuracy limitations.



DC Offsets




Input Bias Current • Op-amps implemented in bipolar technology draw a base current from each input.

While relatively small (≈ 0.1–1 µA), the input bias currents may create inaccuracies in some circuits.

• As shown in Fig. 8.30, each bias current is modeled by a current source tied between the corresponding input and ground. Nominally, IB1 = IB2.



Input Bias Current • Let us study the effect of the input currents on the noninverting amplifier. As

depicted in Fig. 8.31(a), } IB1 has no effect on the circuit because it flows

through a voltage source. The current IB2, on the other hand, flows through R1 and R2, introducing an error.

} Using superposition and setting Vin to zero, we arrive at the circuit in Fig. 8.31(b),


Ch8:Op-Amps–NonidealitiesInput Bias Current ➢ Fig. 8.31(b), which can be transformed to that in Fig. 8.31(c) if IB2 and R2 are replaced with their Thevenin equivalent.

➢ Interestingly, the circuit now resembles the inverting amplifier, thereby yielding;

(if the op amp gain is infinite.)


Input Bias Current • The error due to the input bias current appears similar to the DC offset effects

illustrated in Fig. 8.27, corrupting the output. However, unlike DC offsets, this phenomenon is not random; for a given bias current in the bipolar transistors used in the op-amp, the base currents drawn from the inverting and noninverting inputs remain approximately equal. We may therefore seek a method of canceling this error.

• For example, we can insert a corrective voltage in series with the noninverting input so as to drive Vout to zero (Fig. 8.32). Since Vcorr “sees” a noninverting amplifier, we have




Input Bias Current



Input Bias Current • Vcorr depends on IB2 and hence the current gain of transistors. Since β varies with process

and temperature, Vcorr cannot remain at a fixed value and must “track” β. Vcorr can be also obtained by passing a base current through a resistor equal to R1||R2, leading to the topology shown in Fig. 8.33.

} Here, if IB1 = IB2, then Vout = 0 for Vin = 0. (take the finite gain of the op amp into account and prove that Vout is still near zero.)

} Observe that the input bias currents have an identical effect on the inverting amplifier. Thus, the correction technique shown in Fig. 8.33 applies to this circuit as well.

} In reality, asymmetries in the op-amp’s internal circuitry introduce a slight (random) mismatch between IB1 and IB2. Problem 8.53 (in the book) studies the effect of this mismatch on the output in Fig. 8.33.


Finite Input and Output Impedances • Actual op-amps do not provide an infinite input impedance or a zero output impedance—

the latter often creating limitations in the design. We analyze the effect of this nonideality on one circuit here.

• Consider the inverting amplifier shown in Fig. 8.42(a), assuming the op amp suffers from an output resistance, Rout .

• How should the circuit be analyzed? We return to the model in Fig. 8.1 and place Rout in series with the output voltage source [Fig. 8.42(b)].

• We must solve the circuit in the presence of Rout.



Ch8:Op-Amps–NonidealitiesFinite Input and Output Impedances • Recognizing that the current flowing through Rout is equal to

(−A0vX − vout )/Rout , we write a KVL from vin to vout through R2 and R1:



Speed Limitations • Finite Bandwidth: Our study of op-amps has thus far assumed no speed limitations.

In reality, the internal capacitances of the op-amp degrade the performance at high frequencies.

• Another critical issue in the use of op amps is stability; if placed in the topologies seen above, some op-amps may oscillate. Arising from the internal circuitry of the op-amp, this phenomenon often requires internal or external stabilization, also called “frequency compensation.”

• Slew Rate: In addition to bandwidth and stability problems, another interesting effect is observed in op amps that relates to their response to large signals. The slewing is a nonlinear phenomenon


❑ Thedifferential-pairordifferential-amplifierconfigurationisthemostwidelyusedbuilding block in analog integrated-circuit design. For instance, the input stage ofevery op-amp is a differential amplifier. Also, the BJT differential amplifier is thebasisofavery-high-speedlogiccircuitfamily,calledemitter-coupledlogic(ECL).

❑ Whydifferential?❑ Basically,therearetworeasonsforusingdifferential inpreferencetosingle-ended

amplifiers.

– First, differential circuits aremuch less sensitive to noise and interference than single-endedcircuits.

– The second reason for preferring differential amplifiers is that the differentialconfiguration enables us to bias the amplifier and to couple amplifier stages togetherwithouttheneedforbypassandcouplingcapacitorssuchasthoseutilizedinthedesignofdiscrete-circuitamplifiers.

Ch1:DifferentialandMultistageAmplifiers


❑ Figure shows the basic MOS differential-pairconfiguration.

❑ Itconsistsoftwomatchedtransistors,Q1andQ2,whosesources are joined together and biased by a constant-currentsourceI.

❑ A constant-current source is usually implemented by aMOSFETcircuitofthetypeshownbelow.

Ch1:TheMOSDifferentialPair

❑ Assume that the current source is ideal and that it has infinite outputresistance.

❑ Althougheachdrain is shownconnected to thepositive supply througharesistance RD, in most cases active (current-source) loads are employed.However,theessenceofthedifferential-pairoperationwillbeexplainedbyutilizingsimpleresistiveloads.

❑ Whatever typeof load isused, it isessential that theMOSFETsnotenterthetrioderegionofoperation.


❑ To seehow thedifferential pairworks, consider first thecasewhenthetwogateterminalsarejoinedtogetherandconnected to a voltage VCM , called the common-modevoltage.

Ch1:OperationwithaCommon-ModeInputVoltage



❑ Now, letusvary thevalueof thecommon-modevoltageVCM.We see that, as long asQ1 andQ2remaininthesaturationregion,thecurrentIwilldivide equally between Q1 and Q2 and thevoltages at the drainswill not change. Thus thedifferential pair does not respond to (i.e., itrejects)common-modeinputsignals.



Exercise Solve the same exercise to find the input common-mode range for the case in which two drain resistances are increased by a factor of 2.Ans. -0.28V to 1.0 V


Solutionbcde


Ch1:OperationwithaDifferentialInputVoltage❑Next we apply a difference or differential input voltage by grounding the gate of Q2 and

applying a signal vid to the gate of Q1, as shown in figure below. ❑ If vid > 0 vGS1 > vGS2 iD1 >iD2 and the difference output voltage (vD2 – vD1) > 0

❑If vid < 0 vGS1 < vGS2 iD1 < iD2 and the difference output voltage (vD2 – vD1) < 0

❑From the above, we see that the differential pair responds to difference-mode or differential input signals by providing a corresponding differential output signal between the two drains.

❑ What is the value of vid that causes the entire bias current I to flow in one of the two transistors?

❑In the positive direction, this happens when vGS1 reaches the value that corresponds to iD1 = I, and vGS2 is reduced to a value equal to the threshold voltage Vt , at which point vS = –Vt. The value of vGS1 can be found from


Ch1:OperationwithaDifferentialInputVoltage


Ch1:OperationwithaDifferentialInputVoltage

❑ Exercise: For the MOS differential pair specified in previous example, find

(a) the value of vid that causes Q1 to conduct the entire current I, and the corresponding values of vD1 and vD2;

(b) the value of vid that causes Q2 to conduct the entire current I, and the corresponding values of vD1 and vD2;

(c) the corresponding range of the differential output voltage (vD2 – vD1).

Ans. (a) +0.45 V, 0.5 V, 1.5 V; (b) –0.45 V, 1.5 V, 0.5 V; (c) +1 V to –1 V

❑ To use the differential pair as a linear amplifier, we keep the differential input signal vid small.

❑ As a result, the current in one of the transistors (Q1 when vid is positive) will increase by an increment ΔI proportional to vid , to ( I/2 + ΔI ).

❑ Simultaneously, the current in the other transistor will decrease by the same amount to become (I/2 − ΔI ). A voltage signal -ΔIRD develops at one of the drains and an opposite-polarity signal, ΔIRD, develops at the other drain. Thus the output voltage taken between the two drains will be 2ΔIRD, which is proportional to the differential input signal vid.


❑ We now derive expressions for the drain currents iD1 and iD2 in terms of the input differential signal vid ≡ vG1 – vG2.

❑ The derivation assumes that the differential pair is perfectly matched and neglects channel-length modulation (λ = 0).

❑ Thus these expressions do not depend on the details of the circuit to which the drains are connected, and these connections are not shown.

❑ Assume both MOSFET are in saturation region at all times.

Ch1: Large-Signal Operation


Ch1:Large-SignalOperation

❑ These two equations can be used to obtain the normalized plots, and versus shown below.

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ADNAN MENDERES UNIVERSITY FACULTY OF ENGINEERING … · 2020. 3. 19. · Ch8: Operational Amplifiers (Op-Amps) • The term “operational amplifier” (op amp) was coined in the