Lec 8 | MIT 6.01SC Introduction to Electrical Engineering and Computer Science I, Spring 2011

Lec 8 | MIT 6.01SC Introduction to Electrical Engineering and Computer Science I, Spring 2011


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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.

3 thoughts to “Lec 8 | MIT 6.01SC Introduction to Electrical Engineering and Computer Science I, Spring 2011”

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