6 01 Fall Semester 2007 Lecture 9 Notes 1 MASSACHVSETTS INSTITVTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6 01 Introduction to EECS I Fall Semester 2007 Lecture 9 Notes Op Amps So far we have considered circuits with resistors and voltage sources Now we are going introduce a new component called an operational amplifier or op amp for short We are studying op amps because they are a very important circuit element as well as because they will allow us to explore a sequence of models of how they work These models vary in complexity and fidelity The simplest is the easiest to use for basic circuit designs but does not capture some important behavioral properties The more complex models give us a more complete picture but are often unnecessarily complicated There is no right model of an op amp it all depends on the question that you are trying to answer Basic model Figure 1 a shows a diagram of our simplest op amp model The basic behavioral model is that it adjusts vout in order to try to maintain the constraint that v v and that no current flows in to n or n nin nout n n nout n a Basic op amp model RF RI b A non inverting amplifier circuit Figure 1 Basic op amp models The best way to understand why we might want such a device is to see how it behaves in some small circuit configurations 6 01 Fall Semester 2007 Lecture 9 Notes 2 n nout n n nout n1 nnin RI n2 RF RI RF RI a An inverting amplifier b A voltage summer Figure 2 Basic op amp models Non inverting amplifier Not surprisingly a primary use of an op amp is as an amplifier Here is an amplifier configuration shown in figure 1 b Let s see if we can figure out the relationship between vin and vout The circuit constraints tell us that v iI RI 1 v vout iF RF 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 that vout vin RF RI RI This is cool We ve arranged for the output voltage to be greater than the input voltage and we can arrange just about any relationship we want by choosing values of RF and RI We can think intuitively about how it works by examining some cases First if RF 0 then we ll have vout vin so there s not a particularly interesting change in the voltages This is still a useful device called a voltage follower which we ll study a bit later Now let s think about a more interesting case but simplify matters by setting RF RI We can look at the part of the circuit running from Vout through RF and RI to ground This looks a lot like a voltage divider with v coming out of the middle of it Because v needs to be the same as vin and it is vout being divided in half then vout clearly 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 thinking of the path through the resistors as the terminal of the resulting circuit Let s figure out the relationship between vin and vout for this one The circuit constraints tell us that vin v iI RI v vout iF RF 6 01 Fall Semester 2007 Lecture 9 Notes 3 iF i I 0 v v v 0 Solving we discover that vout vin RF RI If RF RI then this circuit simply inverts the incoming voltage So for example if vin is 10V with respect to ground then vout will be 10V Again we can see the path from nin through the resistors to nout as a voltage divider Knowing that v has to be 0 we can see that vout has to be equal to vin If we want to scale the voltage as well as invert it we can do that by selecting appropriate values of RF and RI Voltage summer A voltage summer1 circuit as shown in figure 2 b can be thought of as having three terminals with the voltage at nout constrained to be a scaled inverted sum of the voltages at n1 and n2 You should be able to write down the equations for this circuit which is very similar to the inverting amplifier and derive the relationship vout RF v1 v2 RI Voltage follower Figure 3 a shows a basic voltage follower circuit What will it do We can see from basic wiring constraints that v Vc vout v Adding in the op amp constraint that v v then we can conclude that vout Vc So we ve managed to make a circuit with the same voltage at nout as at the positive terminal of the voltage source What good is that We ll see in the next section Voltage controlled voltage source model Let 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 seems like a good strategy we can use one with two equal resistances to make 7 5V and then connect it to the motor as shown in figure 3 b But what will the voltage vmotor end up being It all depends on the resistance of the motor If the motor is offering little resistance say 100 then the voltage vmotor will be very close to 0 2 So this is not an effective solution to the problem of supplying 7 5V to the motor In figure 4 a we have used a voltage follower to connect the voltage divider to the motor Based on our previous analysis of the follower we expect the voltage at nout to be 7 5V at least before we connect it up to the motor But our simple model of the op amp doesn t let us understand how 1 2 As in thing that sums not as in endless summer Go 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 15V 10K nmotor n Rmotor Motor nout VC 10K n a A voltage follower b A motor connected to a voltage divider Figure 3 15V 10K n 10K nout n Rmotor i nout nMotor K v v n a Motor connected to a voltage follower ngnd b Op amp as a voltage controlled voltage source Figure 4 6 01 Fall Semester 2007 Lecture 9 Notes 5 15V 10K n nout n K v v VC i n i nout K v v 10K Rmotor Motor n a Voltage follower b Motor controller with buffer Figure 5 connecting the motor to the output …
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