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MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Hello. Welcome. Welcome back from

spring break. Today what I want to do

is talk about op-amps. And in particular what I want to

do is talk about modularity in circuit design. How do you make circuits

that are modular? We’ll see that there’s special

problems when you think about modularity applied

to circuits. And op-amps are one solution

for helping us think about those problems. Before launching straight into

something new though, I’d like to recap where we are under the

assumption that you may not have been thinking about

this exactly currently. So last time we took our

first look at circuits. Circuits are very different from

the kinds of things we’ve thought about before. Before, in programming, in

linear systems theory, we thought about blocks that had

well-defined inputs and well-defined outputs. Circuits are different. Circuits are all connected

together. And circuit theory is thinking

about how do you organize your thoughts about complicated

interactions among things. The parts interact because they

touch each other and they share voltages. Because they touch each other,

they share currents. And you have to somehow

think about that whole complicated system. We figured out that the way you

think about that, you can think about it as three

separate enterprises. How do you think

about voltages. How do you think

about currents. And how do you think

about the elements. How do you think

about voltages? Well, the sum of the

voltages around any closed loop is zero. Done. KVL — Kirchoff’s Voltage Law. How do you think

about currents? Draw any surface. The net current that leaves

the surface is zero. KCL — Kirchoff’s Current Law. How do you think about

the elements? Well it depends on

the element. If you’re thinking about a

linear resistor, it’s Ohm’s Law, V equals IR. If you’re thinking about a

source, it could be a voltage source, then V equals v0,

some fixed number. If you’re thinking about a

current source, then current is fixed at some fixed number. So it depends. When you’re thinking about the

element laws, it depends on what element you are

thinking about. And then you just combine all

of those, and you can solve the circuit. The only problem that arises

is that those equations are highly redundant. There’s a lot more KVL equations

than you need. There’s a lot more KCL equations

than you need. And so in fact, the trick in

analyzing a circuit is to figure out the smallest number

of equations that are adequate, the number of

equations that are necessary and sufficient to

find a solution. Last time we went through three different ways to do that. We thought about what I will

think of as primitive or element voltages and currents. The idea in that technique is

you think about every element and assign to that element

its voltage and its corresponding current. Then that gives you, as in this

example if you’ve got six elements, that gives you six

unknowns, one voltage across every element, one current

through every element. So then you’ve got to dig up

six other relationships. Sorry, start again. Six elements, six voltages, six

currents, 12 unknowns, you need to come up with

12 equations. Six of them are element

equations. So six are easy. Then you have to come

up with six more. And for this particular circuit,

it turns out that there is three independent Kirchoff’s Voltage Law equations. And there are three independent

KCL, Kirchoff’s Current Law equations. 12 equations, 12 unknowns

to solve, done. That’s actually more work

than you need to do. So we looked at two other

techniques for solving the same kind of circuits. One’s called node voltages

and another is called loop currents. They are duals of each other. In the node voltage method you

figure out exactly the minimum number of voltages that I would

have to tell you in order to specify all of

the element voltages. So for example, if I told you

this voltage here, and that voltage there, and this voltage

there, and that voltage there, four of them,

that would let you calculate all of the element voltages. But now I only have four

instead of six. Four nodes instead of six

element voltages. So it’s smaller. And we talked about last time

how you could find the correct equations that go with

those node voltages. In the loop current method, you

specify the minimum number of currents that would be

sufficient to account for all of the element currents. And here in this circuit

it required three. Three is smaller than six. There were six element

currents. So again we’ve got a reduction

in the number of unknowns — 12 4, 3. So that’s kind of the idea. And just to make sure that

you’re all with it, think about this problem. How many of the following

expressions are correct? Feel free to talk to

your neighbor. At the end of about 45 seconds,

I’ll ask you to raise your hand with a number of

fingers, (1) through (5) — indicating which is the

correct answer. Dead silence. You’re allowed to talk. OK, so how many of the

relations are true? Everybody raise your hand. Indicate a number fingers that

tells me how many of the answers are correct. Come on. Blame it on your neighbor. You can say anything, right? It’s always your neighbor’s

fault. That’s very good. It’s about 95% correct,

almost all correct. This first equation,

what is this? Give that a name. KVL. Is it correct? Sure. You need to figure out which

path it corresponds to. It goes through– The only v’s are over here. So therefore it must be

something that has to do with this circuit. So this says that there’s

1, 2, 6, and 5. That’s path 1, 2, 6, and 5. So all you really need to do

is check the polarities. So if we took all of the

variables to the same side, then we’d have minus v1 plus

v2 plus v6 plus v5. If we think about going a loop

that goes through the minus v1, that would be this way. Plus v2 plus v6 plus v5. So that’s right. Easy. What’s the point of

this equation? v6 equals e1 minus e2,

what’s that say? Why’d I ask that? Yeah. AUDIENCE: It’s saying

that the– PROFESSOR: Exactly, it’s

intended to make you think through the relationship

between the primitive variables, the element

voltages, and the node voltages. If I told you the node voltages

e1 and e2, you could trivially compute v6. You could also compute

any of the other v’s. If I told you to find v– what is that? My eyes are not very good. v4, if I told you to find v4,

that would be e1 minus ground. We assign the number 0 to

ground, so that would be e1. So the idea, the reason I ask

number (2), is to make you think about how you relate

the node voltages to the element voltages. How about number (3),

what’s that? Well it’s highly related

to number (2). But it’s different from

number (2), because? AUDIENCE: Ohm’s Law? PROFESSOR: Ohm’s Law. So equation (3) is the way Ohm’s

Law looks when you use node equations. Ohm’s Law is a little bit

uglier when you use node equations than when you use

primitive variables. When you use primitive

variables, the same relationship would’ve simply

said that v6 is R6 times i6. Because I’m using node voltages

instead, the voltage across resistor six shows

up as a difference. How about this one? What’s that? True or false, this is KCL? AUDIENCE: False. AUDIENCE: False. AUDIENCE: –KCL. PROFESSOR: True. This is KCL, true. OK. This is KCL, false. Well, obviously we’re

numerous. And it’s not, sort of, 100%

participation, but– This is not KCL. What is that? KCL says that the sum

of current out of some closed path. This is mixed thing. i6 is one of these things. And IB and IC is one

of these things. So what is the equation for? AUDIENCE: Incorrect. PROFESSOR: Incorrect, yes. What would be the correct way

of saying equation (4)? AUDIENCE: i6 equals– i6 is– PROFESSOR: The answer to set

questions according to the theory of lectures is go

to the next slide. Here’s the correct expression. Why is this correct, and why

is that not correct? Equation (4) is intended to be

how do you relate the loop current to the element

currents? So i6 is an element current. It’s the current that

goes through R6. When we do loop currents, we

have two currents going through R6. So in the loop current view,

the total current that goes through R6 is a sum or

difference of the two loop currents that go through R6. So the element current through

R6, which is i6, is a sum or difference of loop currents. There are two loop currents

we have to worry about. Which one goes through in

the correct direction? i6 goes from left to right. So when we do the loop currents,

we need to take positive as the direction

from left to right. Well that’s the direction of

IC and is the opposite direction of IB. So if I want to use loop

currents to specify i6, it would be IC minus IB. Everybody clear on that? So equation (4) is the relation

between element currents and loop currents. So what’s equation (5)? Equation (5) is Ohm’s Law

for loop currents. Again, if I were thinking about

Ohm’s Law for R6, I would have v6 equals i6 R6. That’s the way you say it in

element voltages and currents. Over here, the current through

R6 is IC minus IB. So Ohm’s Law looks a little

bit more complicated. So the point is these three

methods represent ways of figuring out a linearly

independent set of unknowns and equations. They differ. The left hand one is probably

the easiest to think about, especially when you’re

thinking about things like Ohm’s Law. It’s the natural way to

specify the element relationships. It’s the relation between

the voltage and current through that part. That generally gives me

a large number of equations and unknowns. You can reduce the number of

unknowns by using something like node voltages

or loop currents. And that gives you fewer

equations to solve. They are completely

equivalent. They look a little different. And the reason for talking about

them is that when we think about writing a program

to solve circuits automatically, which by the way

will be the exercise in software lab this week. When we think about writing a

program, writing a program is yet a different kind

of challenge. What’s the easiest system

to automate? So the system that is easiest

for you may or may not be the easiest system to automate. So that’s the point of this

week’s software lab. We’ll do a method that’s closely

related to the node voltage method. It’s not quite node voltages. It’s a little simpler than node

voltages for a computer. It’s a little simpler

to automate. So we use a method that’s called

node voltages with component currents. OK, now I’m going to

start new stuff. That was review. What I want to think about

today is what is it about circuit design that

makes it hard. What are the issues that make

it difficult to be modular when we’re thinking

about the analysis and design of circuits. And one of the hardest things to

deal with is the idea that in a circuit, unlike in a linear

time invariant system of the type we talked about in

the previous module, in a circuit the presence of every

element affects the currents and voltages through, in principle, every other element. So if you change one thing,

you change everything. So first off, I want to just

give an example of that. Think about what would happen

if I were trying to make a circuit to control the

brightness of a light bulb. So I imagine that I’ve

got this circuit and I close the switch. Closing the switch is equivalent to adding a component. So before I close the switch,

when the switch was open, I have three elements, a voltage

source and two resistors. After I close the switch, I

have four elements, the original three plus the bulb. So the question is how would

closing the switch affect V0 and I0. Take a minute. Talk to your neighbor. Figure out how V0

and I0 change. So what’s the answer? Everybody raise your hand, show

a number of fingers equal to the answer. That’s very good. It’s 95% correct at

least, maybe 100. OK, so the answer is (2). Why is the answer (2)? How do I figure that out? AUDIENCE: So the total

resistance of the circuit is going to decrease because you’re

adding it in parallel. And then V0 is going to decrease

because it’s going to have a lower equivalent

resistance. And because that decreases,

you have I0 to– PROFESSOR: Yes, I think that’s

exactly correct. So the idea was that when

you add a component– Let’s think about the light

bulb being a resistor. That’s kind of pulled out of

thin air, but I sort of suggested that you might

do that here. Think about representing the

light bulb as a resistor. Then when you close the switch,

these two resistors go in parallel. When you combine two things in

parallel, the result is the same as a resistance that

has a smaller value. And then think about how that

smaller value would interact with this resistor

and that source. And you can sort of figure out

that the presence of this bulb would reduce this voltage and

increase that current. That’s kind of a high level of

reasoning given where we are. If you wanted to think through

this a little more step by step, it’s easy. You could think about figuring

out what are the voltages and currents before and after

you close the switch. Before you close the switch, you

can just ignore this, and you can calculate V0 just

from a voltage divider relationship. Right, that’s clear? So you can see that when the

switch is open, the voltage, V0, is going to be 8 Volts. And when the switch is open,

you can figure out I0 by lumping these two resistors

together to make a 3 Ohm resistor. And you see that I0 is 4 Amps. Then to figure out what happens

when you close the switch, just repeat. And the algebra is a little

more tedious. I won’t try to go through it. By the way, the answers, the

slides that I show in lecture are always posted on the web. So these slides that have the

answers, they are there. On the web there are two

handouts per lecture. There’s an electronic

version of the thing that we handed out. There’s also an electronic

version of my slides. So everything that I

showed is there. I’m not going to go through

the tedious algebra. It’s just tedious algebra. But if you go through the

tedious algebra, you find that if you represent this bulb by

resistor R, V0 becomes an expression that looks

like this. If you think about resistors,

physical resistors, if you think about light bulbs being

represented by a physical resistor, then the physical

resistor has to have a resistance between

0 and infinity. And if you think about that

expression, what could the value be if R varied between

0 and infinity? That expression is always less

than or equal to 8 Volts, showing that V0 went down when

you closed the switch. Similar tedious algebra leads to

an expression like this in R. And if you think about how

this would change as R goes from 0 to infinity, you

see that that’s always bigger than 4 Amps. So that means I0 goes up. So the point is you can think

through this in a more sophisticated way. And we hope by the end of

the course you’ll all be able to do that. Or you can think through

it in terms of just solving the circuit. Solve it in two cases when the

bulb is there and when the bulb is not. Either way, the answer is (2). The V0 went down,

and I0 went up. The point is that when I added

the element, currents that were far away from the element

still changed. And that’s a general way

circuits interact. They’re all connected. So the idea, the point of

doing this is that the addition of a new element

changed the voltages and currents through the

other element. That’s a drag if what I was

trying to do, for example, was design a brightness controller

for the flashlight bulb. Imagine that what I really

wanted to do was use a voltage divider to make 8 Volts. And what was in my head was I’d

like that 8 Volts to be across the light bulb. That’s a good idea. It just won’t work. At least It won’t work the

way this circuit worked. If I just build it like so,

there’s an interaction between the bulb and the voltage divider

circuits, so the voltage divider circuit is no

longer a voltage divider. After, in this circuit, when

I close the switch, current flows in this wire. The rule in a voltage divider

is the same current has to flow through both resistors. If the same current flows

through two resistors, then you can use the voltage divider

relationship to see how voltage partitions between

the two resistors. If current gets siphoned off

the node between the two resistors, you can’t

use the voltage divider relation anymore. That violates the premise of the

voltage divider relation. So when I close the switch, this

is no longer a voltage divider, and it no longer works

like a voltage divider. Is that clear? So what I’d really like is some

magic circuit that I can put here that isolates the

effect of the bulb on the effect of the rest

of the circuit. And that’s exactly what

an op-amp does. That’s what we’re going

to talk about next. So this magic circuit

is something that we’ll call a buffer. A buffer is a thing

that isolates the left from the right. So the buffer is going to be

something that measures the voltage on this side and

magically generates that voltage over here without

changing this side. So what we want to do now is

develop some thought tools for how you think about op-amps. Op-amps are different. And if you just look at the

picture, op-amp has to be different because there’s

too many terminals. It’s not like a resistor

that has two legs. It’s not like a V source

which has two legs. It’s not like an I source

which has two legs. It has three legs. In fact, I haven’t drawn

all the legs. There’s more than three. There’s at least five. So they’re different. And the way we think about

them are different. The key to thinking about the

way an op-amp works is to think about a new class

of elements called controlled elements. So a controlled element is an

element whose voltage current relationship depends somehow

on a voltage and current measured someplace else

in the circuit. As an example, think

about a current controlled current source. That’s depicted here. I would normally

write a current source that was a circle. That means it’s an independent

current source. That means that the

current is fixed. The little diamond thing is my

way of representing the idea that the amount of current that

comes out of this current source depends on

something else. In this case it depends on IB. And IB happens to be a current

that flows in that circuit. So the idea is that the current

in the current source depends on some other current. It’s a current controlled

current source. It’s a current source whose

current is controlled by another current. Got it? So we’ll see. Figure out, for this current

controlled current source circuit, what’s the ratio

of Vout over Vin? So what’s the answer? What’s the ratio of

Vout over Vin? 100%, wonderful. The answer is (4). Easy to get. It’s easy to get because

I rigged this question to be easy. I rigged it to be easy because

you can sort of figure out everything that’s going

on on the left. And then you can figure

out everything that’s going on on the right. And so there’s no sort

of complicated coupling between the two. So what’s going on on

the left, well, you can solve for IB. IB is just Vi over 1,000 Ohms. Then you can take that value

of IB and use that to solve for what’s going on over here. Over here the out is this

current, 100 IB times this resistance, 5 Ohms. But we just found out that

IB is VI over 1,000 Ohms. Substitute it in,

you get half Vi. So the answer is number (4). The ratio of Vout to

Vin is a half. So the point is these are a

different kind of element, controlled sources,

dependent sources. But they’re not too

hard to work with. They sort of look different

structurally. So think about what’s

going on here. The controlled current source,

the current controlled current source, I have to think of that

as a box, because the current source has to know

the value of IB. So somehow this box, this

thing that’s doing this current controlled current

source, it knows about IB, and it knows about the

current source. So they’re somehow linked. So that’s why I put

a box around it. And then think about the

equations that characterize that component. So now we’ve got a component

that’s got four wires coming out of it. The components we had before

had two wires coming out of them, resistors, current

sources, voltage sources. This kind of a component has

four wires coming out of them. We call this kind of a component

a two-port because we think of the left port

and the right port. That’s compared to resistor,

which we would call a one-port. We think about this

being a two-port. And there’s now two equations. Now I’ve got, for that two-port,

I’ve got two voltages and two currents. There’s the voltage across

the left part. And there’s the voltage

across the right part. And there’s current through

the left part. And there’s the current through

the right part. So with the elements we’ve

thought about before, I had one voltage across it and

one current through it. Now I’ve got two voltages across

it and two currents. It’s kind of twice as big. Not surprisingly, it takes

twice as many equations. Ohm’s Law was a single

equation for one resistor, a one-port. V equals V0 was one

equation for one component, a voltage source. Here, I’ve got one component

that has four wires, four unknowns. And I get two equations. So for this particular dependent

source, I know that the voltage across this pair

of terminals is 0, because it’s essentially a short circuit

between the two. Short circuit just means

connected with a wire. And I know that the current I2

is related to the current over there this way. The idea then is that this

current controlled current source can be represented by a

two-port, two voltages, two currents related by

two equations. It’s kind of structurally twice

as difficult to think about as a one-port. Functionally, it’s different

from having two one-ports, because the two one-ports

are coupled. And that’s the important part

is the coupling between the two that you can’t model with a

simple resistor and a simple constant source. So when we think about an

op-amp, a good first model for an op-amp is to think about

it as a voltage controlled voltage source. And so that’s depicted here. I want to think about the

op-amp, which I’ll symbolically write this way. This means something that has

two inputs, a plus input and then a minus input, and

a single output, Vout. I can think about

it as a voltage controlled voltage source. This I mean to be the element

representation. This is how I’ll draw it

when I make a schematic diagram of a circuit. This is the functional form. This is the way I’ll

think about it when I’m analyzing it. I’ll think about the op-amp as

being a voltage controlled voltage source. This voltage source adopts a

voltage that is some number K times the difference voltage

Vplus minus Vminus. And the trick in op-amps is that

K is typically a very big number, typically bigger

than 10 to the 5th. We’ll see in a minute

why that’s a frightfully useful component. Let’s just walk through an

example to see how you would solve a circuit that has this

kind of a voltage controlled voltage source in it. So think about this circuit

where I’m applying a voltage to the plus lead of an op-amp,

and I’m wrapping the output back through the minus lead,

through two resistors. And what I want to do is

analyze that circuit by thinking about the op-amp

as a voltage controlled voltage source. So I can see just by the way

it’s wired up that the voltage at the plus lead– So I’m going to be thinking

node voltages. Node voltages tend to be easy. I’m thinking node voltages. So I’m going to define

all my voltages referenced to a ground. So I’ll call this node ground. That’s what the funny

symbol means. Then this node voltage, the

voltage at the plus terminal, is just the same as

the input voltage. This voltage at the minus

terminal looks like a voltage divider relationship. If you look at my model, it’s

very clear that I1 is 0, because there’s no connection

here between the Vplus and the Vminus. So I1 is 0. That means the total current

that flows into the plus lead of the op-amp is 0. The total current that flows

into the minus lead of the op-amp is 0. So because there’s no current

flowing in the plus and minus leads, I can calculate the

voltage at this minus terminal as a voltage divider. And it’s just R1 over the sum

of R1 and R2 times V0. Then according to the voltage

control voltage source model, Vout should be K times

the difference between Vplus and Vminus. So I just substitute in for

Vplus — this guy, Vi, and for Vminus — this guy. Do some algebra. I get this expression

after some algebra. And then I say, yeah but I

know that K’s really big. So if K’s really big, KR1

is big compared to R1. So I can ignore that. In fact, K is so big that for

any reasonable choice of R1 and R2, KR1 is even

bigger than R2. So that means this reduces to

this kind of a fraction. So the response of this circuit,

this op-amp circuit– So what did I do? I just took the op-amp circuit,

and I modelled it as a voltage controlled voltage

source, plugged through the equations, and found out that

the ratio of the output voltage to the input voltage

is R1 over R1 plus R2 divided by R1. So the idea then is that

this simple circuit works like an amplifier. It’s an amplifier in the sense

that I can make the output voltage bigger than

the input voltage. That’s a very useful thing. In fact you’ll find useful ways

to use that when you do the design lab this week. So this as an amplifier. So here’s a question. Make sure you follow

what I just did. How could I choose the

components R1 and R2 so that I make Vout equal to Vi? AUDIENCE: Why is this 0? PROFESSOR: There’s no

wire connecting. AUDIENCE: Oh. PROFESSOR: So there’s nowhere

for current to go. OK so how would I choose the

components R1 and R2 in order to make the output voltage equal

to the input voltage? Wonderful. So the idea is that all of these

manipulations have the same effect. All you need to do

is look at the expression that we developed. The expression was R1

plus R2 over R1. If you substitute, R1 goes to

infinity, R2 equals 0, or the two at the same time, you get

one in all of those cases. So that’s a way that you can

turn this amplifier circuit, that in general would make the

output bigger than the input, into something that makes the

output equal to the input. Now I want to turn to

a simplification. I just dragged you

through the math. Thinking about the op-amp as a

voltage controlled voltage source, there’s actually

a shortcut. The shortcut is something we

call the ideal op-amp. So what I want to do in this

slide is drag you through the math one more time, but

then we’re done. The idea is that if you have

an op-amp, a voltage controlled voltage source, if

you represent an op-amp as a voltage controlled voltage

source, and if K is very big, the effect will be to make the

difference between the positive and negative

terminals of the op-amp quite small. You can see that here by way

of a simple example. So let me think about this case

with, again, the voltage controlled voltage

source model. So here I’ve got Vout. According to the voltage

controlled voltage source model, Vout should be K

times the difference between the two inputs. The positive input

is clearly Vi. The negative input is Vo. One equation, two unknowns,

solve for the ratio. The ratio of Vout to Vi

is K over (1 plus K). That’s the answer. Take that answer and back

substitute to figure out how big was the difference between

Vplus and Vminus. That’s just algebra. And what you see is that if this

is the answer, then Vplus minus Vminus can be written as

a fraction of Vi or a similar fraction of Vo. K is big. K is essentially the

same as K plus 1. K is in the denominator

of both. What that says is the difference

between the plus and the minus leads, the voltage

between the plus and the minus leads, is very small

if K is very large. OK? So we call that the ideal

op-amp relationship. The utility of that is that it

makes solving the op-amp circuits much, much

easier than what we’ve just been doing. I’ve been solving the circuits

by thinking about the op-amp as a voltage controlled

voltage source. That’s fine. But if I additionally know

that K is very big, I can shortcut it. I can say, look, the effect of

the op-amp is going to be to make the positive and negative

inputs the same. If it didn’t do that, think

of what it would mean. If K is a big number, and if the

output voltage is K times the difference, if the

difference is anything other than epsilon, Vout has

to be infinity. I mean if K is very

big, right? If K is very big, the only way

the output could be some reasonable number like a Volt

would be if the difference between Vplus and Vminus

is very small. OK, well let’s work backwards. Let’s start with the assumption

that Vplus minus Vminus is very small. And that lets us solve the

circuits very quickly. So for example, the same circuit

that took, previously, a few lines to get the answer

to, if I just take as a rule that Vplus has to be Vminus,

it’s a one step. Vplus equals Vminus, OK, Vi

equals Vo, period, done. It’s a very simple way to think

about the answer to an op-amp circuit. So if the op-amp can be

represented by a voltage controlled voltage source and if

K is very large, then Vplus is roughly Vminus. Shortcut, ideal op-amp

assumption. So, use that or ignore

it depending on what your mood is. And figure out the voltage

relationship for this slightly more complicated circuit. So what’s the answer? Yes? No. How many are done? How many are not done? OK, take a minute. This is supposed to be easy. Think ideal op-amp. OK, what’s the output? Everybody raise your hand,

what’s the output? More hands, more hands,

more hands. OK, tiny number of hands, but

those who showed hands are about 100% correct. I don’t know how

to grade that. Small participation, 100%

among those who did participate. So how do I think about this? What’s step (1)? All right, according to the

theory of lectures, what is step (1)? AUDIENCE: Look at the

previous slide. PROFESSOR: Look at the

previous slide, yes. What was the previous slide? OK The previous slide

had to do with? AUDIENCE: The ideal op-amps. PROFESSOR: Ideal op-amps. What’s ideal op-amps say? Vplus equals Vminus. What happens here if Vplus

is equal to Vminus? Well what’s Vplus? AUDIENCE: 0. PROFESSOR: So what’s Vminus? AUDIENCE: 0. PROFESSOR: 0. So if Vminus is 0,

what do I do now? That bad, that hard huh? Well, I got a [UNINTELLIGIBLE]

here in which three different currents can flow. How big is the current that

can come into this node? What’s the sum of all the

currents that can come into that node? Well one could come

in this way. One could come in that way. One could come in that way. How much current flows in the

minus lead of the op-amp? AUDIENCE: 0. AUDIENCE: 0. PROFESSOR: None. No current goes into

the minus. No current enters the op-amp

through the minus lead. So it’s the sum of

three currents. How big is this current? How big is the current that

flows in that lead? V1 — V1 over 1, right? Ohm’s Law. So this node is 0. That node is V1. The voltage across this resistor

is V1 minus 0. The voltage across is V1. The current is V1

over R. R is 1. So the current that flows

in this leg is V1. How much current flows

in this leg? How much current flows

in this leg? AUDIENCE: V0. PROFESSOR: V0. The idea is that the total

current that flows into this node is V1 plus V2 plus V0. Solve for V0. V0 is minus V1 minus V2. Right? Got it? The idea was that this is really

easy to solve if you use the ideal op-amp

approximation. You can see that this is 0. Therefore, this is 0. So you have a single

KCL equation. And the result is that this

circuit looks like an inverting summer. It computes the sum of V1 and V2

and then takes the negative of that and presents

that at the output. What I’m trying to motivate is

that there’s a whole different level of reasoning that you

can do when you have this element, which is an op-amp. Here what we’ve done is we’ve

made something that performs a numerical operation. It presents, at the output,

the negative sum of the two inputs. OK another problem. Determine R so that V0

is twice V1 minus V2. So how should I choose R? Not very many hands. This is the perfect nano

quiz practice, right? This looks like a perfect nano

quiz question, right? Smile. So what’s the answer? OK we’re down to about

half correct. This is harder. Basically you do exactly the

same thing, it’s just that it’s a little algebraically

messier. So not surprisingly, the first

idea is to think about the ideal op-amp. We’ll think about what was the

voltage here, what was the voltage there, and then

we’ll equate them. What’s the voltage

at the plus lead? Well that’s easy. That’s just a voltage

divider here. Since no current flows in

here, I can compute the voltage relative to this ground

as R over (1 plus R). That’s showed here. Times V1. This one’s a little bit trickier

because I’ve got two sources, V2 and V0,

each pumping current into this place. The way I thought about it was

start with V2 and then add to it this component, which can

be thought of as a voltage divider here. So how big is the voltage

at Vminus? Well it’s V2 plus a voltage

divider here, which is given by this. That’s a little tricky. If you’re not comfortable with

that step, you could also solve for that voltage using

the node method. The node method will give

you the same answer. The answer is that the voltage

at the Vminus port is two thirds of V2 and one

third of V0. That sort of looks right because

the resistors are in a ratio of 2:1. And then using the ideal op-amp

assumption, we equate the two, do some more algebra,

and we get some relationship and figure out that R is 2. The point is that this is a relatively complicated circuit. You could have done it if I had

asked you to do it with the voltage controlled

voltage source model. But the algebra is already

hard here with the ideal op-amp assumption, and with the

voltage controlled voltage source it’s even harder. So the idea is that the ideal

op-amp assumption, the idea that Vplus is equal to Vminus,

makes the work of calculating these responses significantly

easier. Everybody’s with the

bottom line? We started with the idea that

when you add an element to a circuit, in general, adding an

element changes voltages and currents throughout

the circuit. We wanted a way to make the

design more modular, a way of adding a component without

changing everything else. We thought about this

op-amp thing. The model for the op-amp

was voltage controlled voltage source. We inferred this ideal

op-amp model. The ideal op-amp model

was great for calculating the response. It has this one problem. It seems to say these circuits

are identical. If I literally believe the ideal

op-amp assumption that all the op-amp does is magically

make the plus and minus terminals the same, I

would conclude that this circuit where the input comes

in the plus and the output wraps around to the minus

generates precisely the same input-output relationship as

this one, where the input comes in the minus and the

output wraps around the plus. The ideal op-amp assumption

simply says for both of those, Vi equals Vo. So I would assume from the ideal

op-amp model I get the result that these two circuits

work exactly the same way. Somehow that sounds wrong. I’ve got this part, and

I can wire it up the right way and it works. Or I can flip the two wires,

and it still works. Conservation of badness

doesn’t let that happen, right? If you do something bad,

it should break. The ideal op-amp assumption

seems to lead to a bogus result. It seems to say these

two are the same. So what I’d like to do is think

about that for a moment. We want to be comfortable with

the assumption that we make. The ideal op-amp assumption

makes analysis really easy, but we’d like to understand

exactly what we’re assuming. This just doesn’t sound right. OK, so let’s back up. Let’s not do the ideal

op-amp assumption. Let’s Instead say that we use

the voltage controlled voltage source model. So what I’ve done here

is substitute into the left circuit. So this circuit is shown here

and the other over here. All I’ve done is connected the

input either to the plus or to the minus port and wrapped

the other one around. But now I’m analyzing it using

the voltage controlled voltage source model. And I’ve done the

tedious algebra. And I don’t think there’s

any mistakes in the tedious algebra. In one case I get K over

(1 plus K), which for large K is about 1. In the other case I get minus K

over (1 minus K), which for K large is about 1. The ideal op-amp assumption says

that it doesn’t matter. The voltage controlled

voltage source model says it doesn’t matter. And Freeman thinks this doesn’t

make any sense. So what’s going on here? What’s going on is that we’ve

actually made an enormous leap in thinking about the op-amp

even as a voltage controlled voltage source. The two models that we talked

about, the voltage controlled voltage source and the ideal

op-amp model, are great for calculating answers. They are not so good at

providing a mechanism for how the op-amp is actually

working. Think about just the ideal

op-amp assumption. It’s great to think, OK, the

op-amp is going to do whatever magic is necessary to make

these two leads the same. Well how does it do that? The thing about the ideal op-amp

assumption is that it doesn’t tell you the mechanism

by which that happens. What does the op-amp actually

do in order to get Vplus equal to Vminus? What’s actually going on

inside the op-amp? The ideal op-amp assumption is

very good for analysis and not very good with mechanism. What’s the mechanism? What’s the op-amp

actually doing? What it’s really doing is

moving charge around. 8.02 — charge. So if you’re going to have a

change in voltage, there has to be motion of charge. And that’s what’s missing from

the voltage controlled voltage model and from the ideal

op-amp model. How do you think about

moving charge around? Well it’s the same as

moving water around. Somehow, I think it’s all those

years of playing with water when I was a little kid,

I find my intuition about water is better than my

intuition about charge. So let me start by thinking

about the intuition for water. The way a water tank works is

if the flow in is different from the flow out, the

height changes. That’s continuity. So if the water is conserved,

if there’s not significant evaporation over the duration

of this experiment, or if molecules of water do not

spontaneously disappear or appear, under either of those

two assumptions, then the change in the height is

proportional to the difference between the rate at which it’s

coming in and the rate at which it’s coming out. If the rate at which it’s going

out is equal to the rate at which it’s going in, there’s

no change in height. If it’s coming in faster,

it’s going up. If it’s coming in slower,

it’s going down. All easy, right? Charge works the same way. The thing that accumulates

charge in an electronic circuit is a capacitor. And there’s a direct analogy

between thinking about the way the flux of water generates

height and the way flux of charge generates volts. So we can think about the flux

of charge, that’s current. If there’s a net flux, if

there’s a bigger current into a capacitor then there is

out, the voltage on that capacitor goes up. If there’s a smaller current in

than goes out, the voltage on the capacitor goes down

just like the height in a tank of water. We can make a much more

realistic model for the way an op-amp works by explicitly

making a representation for how the charge flows. And that’s shown here. I’m going to take the voltage

controlled voltage source model but explicitly make a

representation for the output charging up. What’s the op-amp do? The op-amp senses the voltage

at the input and the output and does something to change

the voltage at the output. The thing it does is if the

positive voltage is greater than the negative voltage,

it pumps charge into the output node. If the opposite occurs,

it sucks charge out. Now it’s important, this is not

an accurate depiction of what’s inside an op-amp. This is the model

for an op-amp. I don’t want to lead any of

you to think that this is literally what’s in an op-amp. This is not literally

what’s in an op-amp. Here is much more literally

what’s in an op-amp. This is the schematic

diagram of a 709 which is a Widlar circuit. It’s a complicated transistor

circuit. It’s ingenious. This is how you build a

circuit that has the remarkable property of the

ideal op-amp circuit. It’s not perfectly obvious

from here. And even here this doesn’t give

Widlar enough credit. Here’s what he really did. He designed masks to shield

semiconductor materials so that you could turn them into

transistors so that it would turn into an op-amp. We’re not going to

worry about this. This is two levels of

abstraction more complex than we’re going to worry about. We’re just going to say, I don’t

know what’s in there, but this is the way

it behaves. This is intended to be a

behavioral model for how an op-amp works. And in fact, that’s

an important idea that we use circuits– I use circuits on a daily basis

not because I design semiconductor devices

but because I work on biological issues. I actually study hearing. And we make circuit models for

the way biological parts work. And that helps us to understand

the way the biological part works. For example, here is a model

taken from 6.021, where we try to understand how a nerve

propagates an action potential by making a circuit model. That’s the same thing

we’re doing here. We’re making a circuit model

for how the op-amp works. it’s not intended

to be literally what’s inside the op-amp. It’s intended to be a model that

let’s us think about the behavior of the op-amp. This lets us understand now why

it’s different when you flip the wires. Let’s start with the original

configuration where I put the voltage in the plus lead, and I

wrap the output back around to the minus lead. What’s going to happen? What’s the op-amp actually do? Imagine that things

had been stable. This was 0. The output was 0. Everybody was happy. And now all of a sudden the

input voltages steps up. What happens? The input voltage steps up. The voltage source suddenly

generates a huge positive voltage. And that starts to put current

into the capacitor that represents the voltage at the

output of the op-amp. So as a result of this voltage

being higher than where it was, current is being dumped

by the op-amp into this capacitor, and the output

voltage starts to increase. That’s shown it red. As the capacitor voltage gets

bigger, as the output voltage increases, the difference

between Vplus and Vminus gets smaller, and the rate at which

charge is flowing into the capacitor slows. And in fact, as the voltage at

the output gets very close to the voltage at the input, the

rate of current into the capacitor goes to 0. And the output voltage

stabilizes at the input voltage. The same thing happens in

reverse if I were to change input voltage and make

it go negative. If the input voltage went

negative, then K times Vplus minus Vminus would be a big

negative number and it would suck current out. The voltage controlled voltage

source would suck current out of the capacitor and make

the voltage fall. The voltage would fall. The absolute difference between

the plus and minus port would get smaller. The current flowing

would become less. And it would again stabilize

when the output is equal to the input. Contrast that to what would

happen if I put the input into the minus port. If I put the input into the

minus port, and the input goes through a step, then the input

going positive makes the voltage controlled voltage

source generate a big negative voltage. It sucks current out. So the input went positive. And the output goes negative. It goes the wrong way. That’s bad. So here, if I put the input

into the minus lead, a positive transition of the

input leads to the output going the wrong way. And as it goes the wrong way,

the drive to make it go the wrong way gets bigger. It’s a runaway system. It’s positive feedback. So the idea is that by

supporting the input into the negative lead instead of into

the positive lead, leads to a positive feedback situation in

which a small change at the input makes the output go the

wrong way, convinces the op-amp that things are getting

worse, so it makes even more current, which makes it go

even more the wrong way. And the same thing happens

with the flip situation. The idea then is that

by thinking about the flow of current– What’s the op-amp

actually doing? The op-amp is actually sensing

the difference between the voltage at the positive port and

the minus port, and it’s sourcing current that

changes the output voltage to go up or down. You need to wire the op-amp so

that ultimately the output equals the input. Otherwise the ideal op-amp

condition that Vplus equals Vminus will ever be attained. So the left one is what we would

refer to as a stable feedback situation. You can think about that as

analogous to thinking about a ball in a valley. So we have a valley,

and we have a ball. The ball’s going to roll

down here eventually. It’s stable. Bop it a little bit to the

right, it’ll roll back into the valley. Bop it a little to the

left, it’ll roll back into the valley. That’s as opposed to, when we’ve

wired up the wrong way, it’s like we have an unstable

equilibrium. It’s as though we’re trying

to put the ball there. Bop it a little to right, it

runs away to the right. Bop it a little to the left,

it runs away to the left. That’s unstable. There is a metastability

plane. If you actually balanced it

exactly, exactly, exactly at the right place, it

would stay there. That’s what the solution

is telling you. That was the minus K

over (1 minus K). The fact that the equations

had a solution is correct, it’s just an unstable

solution. The tiniest little disturbance

will make it go awry. The idea then is that

the ideal op-amp assumption is fine. It’s a good thing to use. It doesn’t tell you about

the mechanism. If you think a little bit more

about the way the physics works, you have to wire the

circuit up so that it has negative feedback in order to

get a stable equilibrium. So if you understand that, and

you just make sure that you’ve hooked it up so that it has

negative feedback, then you can use the ideal op-amp

assumption. Everything’s just fine. And in the interest of time,

I’m going to just skip over this because we’ve already

talked about all of this. It’s just another example. The last thing I want to mention

is just that in order to work this magic,

the op-amp has to get power from somewhere. And so that means that it’s not

a three-terminal device. It’s not the kind of device I’ve

been drawing so far that has got power pins too. The way the op-amp works is it

takes power from those power pins to be able to force the

output to a voltage that’ll make the input the Vplus

equal to Vminus. That’s the mechanism

by which it works. And that’ll have important

consequences when we build the robot head, because what it

means is that we’re not going to be able to generate arbitrary

voltages at the output of an op-amp. We’re going to be limited to

generating voltages that are between the power rails. So if we were to supply plus

and minus 10 Volts to the power pins, we would not be

expecting to be able to generate 20 Volts

at the output. So that’s an important

implication when we do the design lab. In summary, what we’ve done is

we’ve showed how circuits can be a pain to make modular,

because in principle, adding one component changes voltages

and currents everywhere. But there’s a way using op-amps

to have this buffering idea that lets us logically

separate parts of circuits so the one can control the other. Have a good day.

Thanks for sharing this with us

after this lecture every doubt i have is cleared,thanks mit

Funny how I pay for tuition at my university but actually learn from MIT's online courseware.