6.01, Fall Semester, 2007—Lecture 9 Notes 1MASSACHVSETTS INSTITVTE OF TECHNOLOGYDepartment of Electrical Engineering and Computer Science6.01—Introduction to EECS IFall Semester, 2007Lecture 9 NotesOp-AmpsSo far, we have considered circuits with resistors and voltage sources. Now we are going introducea new component, called an operational amplifier or op-amp, for short. We are studying op-ampsbecause they are a very important circuit element, as well as because they will allow us to explore asequence of models of how they work. These models vary in complexity and fidelity. The simplestis the easiest to use for basic circuit designs, but do es not capture some important behavioralproperties. The more complex models give us a more complete picture, but are often unnecessarilycomplicated. There is no right model of an op-amp: it all depends on the question that you aretrying to answer.Basic modelFigure 1(a) shows a diagram of our simplest op-amp model. The basic behavioral model is that itadjusts voutin order to try to maintain the constraint that v+≈ v−and that no current flows into n+or n−.-+n-n+nout(a) Basic op-amp model-+n-ninnoutRIRF(b) A non-inverting amplifier circuitFigure 1: Basic op-amp modelsThe best way to understand why we might want such a device is to see how it behaves in somesmall circuit configurations.6.01, Fall Semester, 2007—Lecture 9 Notes 2-+n-ninnoutRIRFn+(a) An inverting amplifier-+n-n1noutRIRFn+RIn2(b) A voltage summerFigure 2: Basic op-amp modelsNon-inverting amplifier Not surprisingly, a primary use of an op-amp is as an amplifier. Hereis an amplifier configuration, s hown in figure 1(b). Let’s see if we can figure out the relationshipbetween vinand vout. The circuit constraints tell us thatv−= iIRI(1)v−− vout= iFRF(2)iI+ iF= 0 (3)vin= v−(4)The KCL equation 3 has no term for the current into the op-amp, because we assume it is zero.Equation 4 is the op-amp contraint. So, we find thatvout= vinRF+ RIRI.This is cool. We’ve arranged for the output voltage to be greater than the input voltage, and wecan arrange just about any relationship we want, by choosing values of RFand RI.We can think intuitively about how it works by examining some cases. First, if RF= 0, then we’llhave vout= vin, so there’s not a particularly interesting change in the voltages. This is still a usefuldevice, called a voltage follower, which we’ll study a bit later.Now let’s think about a more interesting case, but s implify matters by setting RF= RI. We canlook at the part of the circuit running from Voutthrough RFand RIto ground. This looks a lot likea voltage divider, with v−coming out of the middle of it. Because v−needs to be the same as vin,and it is voutbeing divided in half, then voutclearly has to be 2vin.Inverting amplifier Figure 2(a) shows a very similar configuration, called an inverting amplifier.The difference is that the + terminal of the op-amp is connected to ground, and the we’re thinkingof the path through the resistors as the terminal of the resulting circuit. Let’s figure out therelationship between vinand voutfor this one. The circuit constraints tell us thatvin− v−= iIRIv−− vout= iFRF6.01, Fall Semester, 2007—Lecture 9 Notes 3iF− iI= 0v+= v−v+= 0Solving, we discover thatvout= −vinRFRI.If RF= RI, then this c ircuit simply inverts the incoming voltage. So, for example, if vinis +10Vwith respect to ground, then voutwill be −10V. Again, we can see the path from ninthrough theresistors, to nout, as a voltage divider. Knowing that v−has to be 0, we can see that vouthas tobe equal to −vin. If we want to scale the voltage, as well as invert it, we can do that by selectingappropriate values of RFand RI.Voltage summer A voltage summer1circuit, as shown in figure 2(b), can be thought of as havingthree terminals, with the voltage at noutconstrained to be a scaled, inverted, sum of the voltagesat n1and n2. You should be able to write down the equations for this circuit, which is very similarto the inverting amplifier, and derive the relationship:vout= −RFRI(v1+ v2) .Voltage follower Figure 3(a) s hows a basic voltage follower circuit. What will it do? We cansee from basic wiring constraints that:v+= Vcvout= v−Adding in the op-amp constraint that v+= v−, then we can conclude that vout= Vc. So, we’vemanaged to make a circuit with the same voltage at noutas at the positive terminal of the voltagesource. What good is that? We’ll see in the next section.Voltage-controlled voltage-source modelLet’s start by thinking about using a variable voltage to control a motor. If we have a 15V supply,but only want to put 7.5V across the motor terminals, what should we do? A voltage divider seemslike a good strategy: we can use one with two equal resistances, to make 7.5V, and then connect itto the motor as shown in figure 3(b). But what will the voltage vmotorend up being? It all dependson the resistance of the motor. If the motor is offering little resistance, say 100Ω, then the voltagevmotorwill be very close to 0.2So, this is not an effective solution to the problem of supplying 7.5Vto the motor.In figure 4(a), we have used a voltage follower to connect the voltage divider to the motor. Basedon our previous analysis of the follower, we expect the voltage at noutto be 7.5V, at least beforewe connect it up to the motor. But our simple model of the op-amp doesn’t let us understand how1As in thing that sums, not as in endless summer.2Go back and review the discussion of adding a load to a voltage divider, if this doesn’t seem clear.6.01, Fall Semester, 2007—Lecture 9 Notes 4-+n-n+nout+-VC(a) A voltage followernmotor10K+15VRmotorMotor10K(b) A motor connected to a voltage div iderFigure 3:-+n-n+noutRmotorMotor10K+15V10K(a) Motor connected to a voltage followern-n+noutK(v+− v−)ngndi(b) Op-amp as a voltage-controlled voltage sourceFigure 4:6.01, Fall Semester, 2007—Lecture 9 Notes 5n-n+ nout+-VCK(v+− v−)i(a) Voltage followern-n+ noutRmotorMotor10K+15V10KK(v+− v−)i(b) Motor controller with bufferFigure 5:connecting the motor to the output of the follower will affect the behavior of the voltage divider orwhat exactly will happen to the motor.So, now, we need a somehwat more sophisticated model of the op-amp, which is s hown schematicallyin figure 4(b). The constraint model relates the voltages at all four terminals:vout− vgnd= K(v+− v−) ,where K is a very large gain, on the order of 10,000, and
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