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MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So today I want to

start a new topic, circuits. No, that’s good. That’s good. Circuits are good. AUDIENCE: Yay. PROFESSOR: Thank

you, thank you. Much better. So just to provide some

perspective, I want to remind you where we are, how we got

here, and where we’re going. So at the beginning of the

course, we promised that there were several intellectual

themes that we would talk about. Probably the most important

one there is designing complex systems. That’s what we’re

really about. We would like you to

be able to make very complicated systems. How do you think about parts? How do you think about

connecting them? How do you think about things

when you want to make something that’s very

complicated? Part of that is modeling. We just finished a module

on modeling. So in order to make a complex

system, we’d like to be able to predict how it will behave

before we completely build it. Sometimes it’s impossible

to build the entire system before. Sometimes it’s impossible

to build prototypes. Sometimes you’re stuck with

going with the design at launch, and figuring

out how it works. In those cases in specific, it’s

very important to be able to model it, to have some

confidence that the thing’s going to work. We’re going to talk about

augmenting physical systems with computation. That’s the module we’re

about to begin. And we’ll conclude by talking

about how to build systems that are robust to change. So we started with the idea

of, how do you make complicated systems? And we introduced this notion

of primitives, combination, abstraction, and pattern in

terms of software engineering. We did that because that’s the

simplest possible way of getting started. It provided a very good

illustration of PCAP at the low level by thinking about

Python, by thinking about the primitive structures that Python

gives you, how those can be combined, how you can

abstract, how you can recognize patterns. But then we also built a higher

level abstraction, which was the state

machine idea. There the idea was you didn’t

have to have state machines in order to build the brain for a

robot, but it actually turns out to be easy if you

do because there’s a modularity there. You can figure out if each part

works independent of the other parts. And then you can be pretty sure

when you put it together the whole thing’s

going to work. So that was kind of our

introduction to this notion of PCAP. Then we went on to think about

signals and systems. And that was kind of our

introduction to modeling. How do you make a model

of something that predicts behavior? So we transitioned from thinking

about, how do you structure a design to how do

you think about behavior. Today we’re going to start

to think about circuits. Circuits are really going to a

more primitive physical layer. How do you think about actually

making a device? So the device that we’ll think

about is a thing to augment the capabilities — the sensor capabilities

of the robot. We’ll think about making a

light tracking system. And the idea is going to be

that you’ll build a head. The head has a neck, so you’ll

have to control the neck. The head has eyes. It will mount on top of the

robot, and that’ll let you drive the robot around

looking for light. And you’ll do that by

designing a circuit. So that’s kind of the game plan

for the next three weeks. So today, what I want to do

is introduce the notion of circuits, introduce

the theory for how we think about circuits. And then in the two labs this

week, the idea will be to become familiar with the

practice of circuits. How do you actually

build something? How do you actually make

something work? So today’s the theory. So the theory, the idea in

circuits is to think about a physical system as the

interconnection of parts and rules that connect them. In fact, the rules fall

into two categories. We’re going to think about the

currents that go through parts and the voltage that develops

across parts. And we’ll see that there’s a

way of thinking about the behavior of the entire circuit

that integrates all of those three pieces. How does the part work? How do the currents that go

through the part work? And how do the voltages

that are produced across the part — how do they work? So I’ll just start with two

very simple examples. The first is the most trivial

example you can think about. You can think about a flashlight

as a circuit. You close the switch,

current flows. Very simple idea. We will make a model of that

that looks like this. We’ll think about the battery

being a source. In this case, a voltage

source. We’ll think about the light

bulb being a resistor. We’ll have two parts. We’ll have to know the

current-voltage relationships for both of the parts. And we’ll have to know the

ramifications for those currents and voltages when

you put them together. Very, very simple. The other simple example that

I want to illustrate is this idea of a leaky tank. Here the idea that I want to get

across is that the circuit idea is quite general. When we talk about circuits,

we almost always talk about electronic circuits. But the theory is by no means

limited to electronics. So for example, if we think

about a leaky tank, we think about a pipe spewing water

into a reservoir. Maybe that’s the Cambridge

Reservoir. Maybe that’s the water coming

out of the Woburn Reservoir. Maybe that’s the demand put on

by the Cambridge people trying to take showers in

the morning — I don’t know. So we think about flow

into a tank, a reservoir, and flow out. And we can make a model for that

in terms of a circuit. In the circuit there

are through variables and across variables. In an electronic circuit, the

through variable is current. Here the through variable

is the flow rate. So it’s the flow of water in

and the flow of water out, represented here

by these things that look like currents. And the across variable for an

electronic device is voltage. The across variable for this

kind of a fluidic device, the across variable is pressure. So we think as this thing gets

ahead of that thing, as there’s more stuff coming in

than going out, the height goes up and the pressure

builds. Same idea. So the point is that we’ll

develop the theory for circuits in terms

of electronics. But you should keep in the back

of your mind that it’s a lot more general idea than

just electronics. In fact, there are two

completely distinct reasons why we even bother

with circuits. One is that they’re very

important to physical systems. If you’re designing a power

network, of course you have to think about the way the power

network, the power grid works as a circuit. That’s obvious. In electronics, of course, if

you’re going to build a cell phone, you have to know how

the parts interconnect electronically. That’s obvious. But probably the biggest use for

circuits these days is not those applications, although

those are very important. Circuits are also used

as models of things. So many models for complex

behaviors are in fact circuit models. So in terms of electronics, the

idea is that we want to get on top of electronics. We want to understand how

circuits work, so we can understand things like that. If you look at how complex

processors have got over my professional life, we start with

my professional life down at about 1,000 transistors

per processor. And today, we’re up at

about a billion. That’s enormous. Even in the stone ages when we

were designing things that had a thousand parts, we still had

trouble thinking about those thousands parts all at once. We still need PCAP. We still needed ways of

combining the activities of many things into a conceptual

unit that was bigger. Here, it’s just impossible

if you don’t have that. So that’s one of the reasons

we study circuits. And the other reason is here. So here I’m showing a

model for the way a nerve cell works. This model is taken

from 6.021. The idea, this comes from the

study of the Hodgkin-Huxley model for neural conduction,

arguably the most successful mathematical theory

in biophysics. Which explains the completely

non-trivial relationship between how the parts from

biology works and the behavior in terms of propagated action

potentials works. So the idea is that we

understand how this biological system works because

we think about it in terms of a circuit. That’s the only successful way

we have to think about that. So what I want to do then is

spend today figuring out circuits At the very most

primitive level. The level that I’m going to

talk about in terms of circuits is roughly analogous

to the level that we talked about with Python when we were

thinking about how Python provides utilities for

primitives, combinations, abstractions, and patterns. So I’m going to start at the

very lowest level and think about, what are the basic

primitives, the smallest units we’ll ever think about

in terms of circuits? And what are the rules by

which we combine them? So I’ll start with the very

simplest ideas, the very simplest elements. We will oversimplify things

and think about the very simplest kind of electronic

elements as resistors that obey Ohm’s Law, V equals iR. Voltage sources, things that

maintain a constant voltage regardless of what you do. And current sources, things

that maintain a constant current regardless

of what you do. These things are, as I said,

analogous to the primitive things that we looked

at in Python. They’re also analogous to the

primitive things that we looked at in system functions. Can somebody think of, when

we were doing difference equations, what were the

primitives that we started with — when we started to study

difference equations? What’s the most primitive

elements that we thought about? AUDIENCE: Delay. PROFESSOR: Delay, yeah. So we thought about

things like– so we had delay. Anything else? AUDIENCE: Gain. PROFESSOR: Gain. Anything else? Add. So we had exactly three

primitives. And we got pretty far with

those three primitives. We learned the rules for

interconnection. We didn’t really make a

big deal out of it. We didn’t formalize it, but the

rules for interconnection were something like

every node has to have exactly one generator. You can’t connect the

output of this to this, that’s illegal. Every node has to

have one source. And every node can source

lots of inputs. That was kind of the rules

of the interconnect. The interconnects here will be a

little bit more complicated. So those are the elements

that we’ll think about. And the first step’s going to be

to think about, how do they interconnect? The simplest possible

interconnections are trivial. In the case of the battery, you

hook up the voltage source to the resistor. The voltage source makes

the voltage across this resistor 1 Volt. If we say the resistor is 1

Ohm, then there’s 1 Amp current period. Done. Easy. Similarly, if we were to hook

up the resistor to a current source, we would get something

equally easy. Except now the current source

would guarantee that the current through the resistor

is an amp. Therefore, the voltage across

the resistor, by Ohm’s law, would be a Volt. So we would end up with the same

solution for a completely different reason. Here the voltage

is constrained. Here the current’s

constrained. Just to make sure everybody’s

with me, figure out, what’s the current i that goes

through this resistor? Slightly more complicated

system. Take 20 seconds, talk to your

neighbor, figure out a number between (1) and (5). OK, so what’s the answer? Everybody raise your hand with

a number (1) through (5). Come on, everybody vote. Come on. You can blame it on your

neighbor, that’s the rules. You talk to your neighbor,

then you can blame dumb answers on your neighbor. OK, about 80% correct I’d say. So how do you think

about this? What’s going to be

the current? How would you calculate

the current? What do I do first? Shout. If you shout, and especially

if my head’s turned away I don’t know who you are. AUDIENCE: Kirchhoff’s law. PROFESSOR: Kirchoff’s

law, wonderful. Which one? There’s two of them. AUDIENCE: [UNINTELLIGIBLE]. PROFESSOR: [UNINTELLIGIBLE]. What loop do you want to use? AUDIENCE: Left. PROFESSOR: Left side. So if we use the left side loop,

we would conclude that there’s a volt across

the resistor. So the current would be? AUDIENCE: 1 Amp. PROFESSOR: An amp. Where’s the current come from? AUDIENCE: The voltage. PROFESSOR: The voltage source

just like before. So not quite. So the voltage source

establishes this voltage would be 1. That makes this current be 1. That would be consistent with

the current coming out of here, except we have to also

think about that 1 Amp source. So the question is, what’s

does the 1 Amp source do? Nothing? It’s just there sort of

for decoration or for [UNINTELLIGIBLE] so that we

can make an interesting question to ask in lecture? Maybe. So where’s the current? Where’s the 1 Amp that

goes through the resistor come from? AUDIENCE: [UNINTELLIGIBLE]

on the right. PROFESSOR: It comes

from the right. It comes from the current

over here. So the idea is that if this

current, ignore the voltage over here for the moment. If this current flowed through

the resistor, then you’d have 1 Amp going through there, and

you’d have 1 Volt generated by that current which just happens

to be exactly the right voltage to match

the voltage from the voltage source. So if you think about this, the

voltage guarantees that this is 1 Volt, but so

does the current. In order to simultaneously

satisfy everything, all you need to do is have all of this

current go around and come down through that resistor. That will generate the volt, so

there’s no propensity for more current to flow out of the

source because the source is 1 Volt and it’s facing a

circuit that’s already 1 Volt. So the idea was to try to give

you something that’s relatively simple that you can

think through on your own, but not trivial. So the answer was 1 Amp. But the 1 Amp was not for

the trivial reason. The 1 Amp is because the current

from the right flows through the resistor and

makes the voltage be 1. So the right answer is 1. But for the reason that

you might not have originally thought. But more importantly, I wanted

to use that as a motivation for thinking about, how do we

think about bigger circuits? So when the simple circuit,

like two parts, it’s no problem figuring out what the

answer’s going to be. But when the circuit has even

three parts, it may require more thinking. And you may want to have a

more structured way of thinking about the solution. Yes? AUDIENCE: What would have

occurred if the current provider on the right

side was 2 Amps? PROFESSOR: Great question. Had this been 2 Amps, you can’t

violate this voltage. So that would have

been 1 Volt. So that would have been 1 Amp

through the resistor. So then you’re left with the

problem with this guy’s pushing 2 and that guy’s

only eating 1. But the rules for the voltage

source say eat or source however much current is

necessary in order to make the voltage equal to 1. So the excess amp goes through

the voltage source. So the voltage source is, in

fact, being supplied power rather than supplying

power itself. Had this been 2 Amps, some of

the power from this source would have gone into

the resistor. And some of the power from

this source would have actually gone into the

voltage source. So if the voltage source were,

for example, a model for a rechargeable battery,

that rechargeable battery would be charging. Does that make sense? So if there had been a mismatch

in the conditions, you still have to satisfy all

the relationships from all of the sources. AUDIENCE: What if the

voltage was larger? PROFESSOR: The same thing

would have happened. Except now the flowing

current would be in the opposite direction. Let’s say that if the voltage

here had been 2 Volts, then the voltage would have required

that there is 2 Amps flowing here. 1 Amp would come from here,

but another Amp would come from here. This voltage source will supply

whatever current is necessary to make its

voltage law real. Ok. In fact, what we’ll do now is

turn toward a discussion of more complicated systems that

will let you go back and in retrospect, analyze all those

cases that we just did. And you’ll be able to see

trivially how that has to be the right answer. So what I want to do now is

generate a formal structure for how you would

solve circuits. Yes? AUDIENCE: So did we know of

anything that could generate a current without generating a

voltage, like in real life? PROFESSOR: Can anything generate

a current without generating a voltage? That’s a tricky question. If you think about something as

generating a current, then the voltage is not necessarily

determined by that part. So that’s kind of illustrated

here. If this guy is generating a

current, this guy is not actually the element that is

controlling its own voltage. In general, if you want to speak

simultaneously about the current and voltage across the

device, you have to know what it was connected to. Each part– we’ll get to this in a moment

in case some of you are worried about launching ahead. We will cover this. This is very good motivation for

figuring out what’s going to happen in the next

three slides. So each part gets to tell you

one relationship between voltage and current. Generally speaking, that’s not

enough to solve for voltage and current. Voltage and current is

like two unknowns. Each element relationship

is one equation. So the current source gets to

say current equals x, current equals 1 Amp. It doesn’t get to tell you

what the voltage is. So being a little more physical

to try to address your question more physically,

there are processes that can be extremely well modeled

as current generators. In fact, many electronic

semiconductor parts, like transistors, work more like

a current source than like anything else. So there are devices that behave

as though they were current sources, but they don’t

simultaneously get to tell you what is their current

and what is their voltage. They only get to tell you

what is their current. So let’s think about now, if

you had a more complicated system, how could you

systematically go about finding the solution? As was mentioned earlier,

there’s something called Kirchhoff’s law. And in fact, there’s

two of them. Kirchhoff’s voltage law and

Kirchhoff’s current law. Kirchhoff’s voltage law, in its

most elementary form, says that if you trace the path

around any closed path in a circuit, regardless of

what the path is– every closed path– the sum of the voltages going

around that closed path is 0. So for example in this

circuit, the red path illustrates one closed path

through the circuit. It goes up through the voltage

source, down through this resistor, and then down

through that resistor. Kirchhoff’s voltage law says the

sum of the voltages around that loop is 0. That’s written mathematically

here, minus v1 for here, plus v2 for here, plus v4

for here is 0. OK, where do the signs

come from? The signs came from the

reference directions that we assigned arbitrarily

to the elements. Before I ever do a circuits

question, I always assign a reference direction. Every voltage has a

positive terminal and a negative terminal. And I must be consistent in

order to apply these rules. These rules only work if I

declare a reference direction and stick with it. If midway through a problem

I flip it, I’ll get the wrong answer. So the minus sign has to do with

the fact that as I trace this path, I enter the minus

part of this guy, but the plus part of that guy and that guy. So the sign of v1 is negated

relative to the others. A different way to think about

that is here, we can think that v1 is the sum

of v2 and v4. That’s sometimes more intuitive

because if you started here, going through

this path you would end up with a voltage that is v1 higher

than where you started. Whereas starting here, you would

end up with a voltage here that’s v4 higher than

where you started. And then by the time you got to

here, it would be v2 plus v4 higher than where

you started. You start one place and on one

route, you end up v1 higher. And in the other route, you

get v4 plus v2 higher. So it must be the case that v1

is the same as v2 plus v4. Those are absolutely

equivalent ways of thinking about it. So those laws are equivalent. If you think about it

a path, you think about some of the paths– no. The path coinciding with the

negative direction of some of the elements and the positive

direction of others. OK, how many other

paths are there? Take 20 seconds, talk

to your neighbor. Figure out all of the

possible paths for which KVL has to apply. OK, so everybody raise your

hand and show a number of fingers equal to the number

of KVL equations less two. Oh, very good. Virtually 100% correct. Why do you all say (5)? Which is to say 7. Why do you all say 7? So there’s 3 obvious ones. I was expecting a

couple of 3’s. This was supposed to be– OK, yeah, I do plot

against you. I was expecting some 3’s. So there’s 3 obvious paths

that are analogous to the first one we looked at. If I call the first path A, then

there’s B and C which are the excursions around here. And you can write the equations

just the same. They each involve

three voltages. And they each go through, some

starting at the negative side and some starting at

the positive side. So those are in some sense,

the obvious ones. But there are others too. So one way to think about it,

what I’d like you to do is enumerate all the paths

through the circuit. I should have said all the paths

through the circuit that go through each element

one or fewer times. I don’t want you to go through

the same element twice. So here’s another path that

would go through elements at most one time. So up through here, over through

here, which didn’t go through any elements. Down through that element,

across that, down through here, et cetera. And you get an equation

for that. Here’s another. Here’s another. Here’s another. And if you try to think about a

general rule, a general rule is something like, how many of

those panels can you make and piece together where the loop

goes through the perimeter? You’re not allowed to go through

an inner place because if you went through an inner

node, you’d have to go through it twice. If you wanted the path to go

through an inner element, you’d have to go through

that element twice. So in fact, the answer is 7. There are 7 different paths

according to Kirchhoff’s voltage law, all of which the

sum of the voltages around those paths has to be 0. The problem is, of course, that

those equations are not all linearly independent. So if you just had a general

purpose equation solver– and by the way, we’ll write

one of those in week 8 for solving circuits. If you just passed those 7

equations into a general purpose equation solver, it

would tell you there’s something awry with your

equations because they’re not linearly independent. So you can, however, think about

linearly independent in particularly simple cases. This network is a particular

kind of network that we call a planar network. A planar network is one that I

can draw on a sheet of paper without crossing wires. So I can draw this network

without crossing wires. I’ll call it planar. And it turns out that

Kirchhoff’s voltage laws for the innermost loops are always

independent of each other. That’s kind of obvious because

as you go to a — so each loop contains at least

one element that some other loop didn’t have. So that’s kind of the reasoning

for why it works. So if you think about this

particular loop, which we included in the 7, you can think

about that as being the sum of the loops this way, the

A loop and the B loop. Because if you write KVL for the

A loop and KVL for the B loop and add them, you end up

deriving KVL for the more complicated path. And if you think about what’s

going on, it’s not anything terribly magical. This path is the same as the A

path added to that path, where I went through this element down

when I did the A path and up when I did the B path. So those parts canceled out. That was the rule that I was

talking about how I don’t really want to go through the

same element twice when I’m applying KVL. So the idea then is that there’s

a systematic way, an easy way to figure out

all the KVL loops. You just think about

all the possible paths through the circuit. You do have to worry about

linearly independent. In the case of planar networks,

that’s pretty straightforward. Planar networks, you can always

figure out the linearly independent KVL equations by

looking at the smallest possible loops. The loops with small area. OK, so that’s half of it. That’s KVL. The other Kirchhoff’s law is

KCL, Kirchhoff’s Current Law. There we are thinking of

the flow of current. So the flow of current is

analogous to the flow of incompressible fluid. Water, for example. If you trace the amount of water

that flows through a pipe that goes into a Y, then

the sum of the flows out has to equal the flow in. If that weren’t true, the water

would be building up. So we think about pipes as

transporting the flow of water without allowing it to

build up anywhere. That’s precisely how we

think about wires in electrical circuits. The wires allow the transport

of electrons but don’t allow the buildup of electrons. OK, do electrons build up? Sure. But in our idealized world, we

say they don’t build up in the wires, they build

up in a part. And we’ll have a special

part that allows the electrons to build up. So we’re not excluding the

possibility that they build up, we’re just saying that in

this formalism, we don’t allow the electrons to build

up in the wires. So for the purpose of the wires,

current in is equal to the current out. The net current in is 0. So we will think then, about

the circuit having nodes. The nodes are the places where

more than one element meets, two or more elements meet. And we will apply KCL

at each node. So for example, in this simple

circuit where I would have three parts connected in what

we would call parallel, they share a node at the

top and they share a node at the bottom. So even though it looks like

there’s multiple interconnects up here, we say that’s

one node. And we would say that the sum of

the currents into the node is equal to the sum of

the currents out. So if I lab all of the possible

currents that come out of that node, I would

have i1, i2, i3. i1 goes through the first

one, the second one and the third one. And so I would conclude from

Kirchhoff’s current law that the sum of i1, i2,

and i3 is 0. OK. Easy, right? As I said, we’re going to make

an abstraction where the electrons don’t build

up in the wires. They don’t even build

up in the parts. They do get stored

in the parts. That’s a little confusing,

we’ll come back to that. If they don’t build up in the

parts, then the current that goes in this leg has to

come out that leg. If that’s true, then i1 is i4,

i2 is i5, i3 is i6, and we end up with another equation down

here, which turns out to be precisely the same as

the one at the top. Everybody’s happy with that? So we’re thinking about this

just the way we would think about water flow. If there’s water flow into a

part, it better be coming out. If there’s water flow in a pipe,

the water that goes into the pipe better come out

of the pipe someplace. So here is an arbitrary network made out of four parts. How many linearly independent

KCL equations are there? So how many linearly independent

KCL equations are in that network? Everyone raise your hand, some

number of KCL equations. OK, I’m seeing a

bigger variety. I see (1)’s, (2)’s, and (3)’s. I don’t see any (4)’s. That’s probably good. So how do you think about the

number of linearly independent KCL equations? So the first thing to do

is to label things. So you have to have reference

directions before you can sort of think about things. So we have four elements. We would be expecting to see

four element currents. The same current that goes

into an element has to come out of it. So there’s element current

1, 2, 3, and 4. There are three nodes,

so we might be expecting three KCL equations. Here’s one node from which you

would conclude that the sum of i1 and i2 better be 0. Here’s a node from which you

would conclude that the current in i2 better

be i3 plus i4. And here is a node from which

you would conclude that i1 plus i3 plus i4 is 0. So I can write one KCL equation

for every node, that’s not surprising. But if you look at those

equations, you’ll see that they’re not linearly

independent. In fact, if you solve

this one for i2– it’s already solved for i2. Stick that answer up here, you

get i3 plus i4 added to i1 is 0, which is just the same

as that equation. So of those three equations,

only two of them are linearly independent. The answer to that

problem was (2). And there’s a pattern. So think about the pattern in

terms of figuring out the number of linearly independent

KCL equations that are in a slightly more complicated

network. So what’s the answer

here, how many KCL equations are in this network? Wow. Well, I’m not getting any of the

answers I would have said. What does that mean? Ah, I’m forgetting to add 2. That’s my problem. OK, now I’m getting some of

the answers that I would expect to get. OK, got it. I confused myself. OK, the vast majority say (1). How do you get that? Which is 3. So again, you think of how in

this circuit there are four nodes, A, B, C, D. So we can

think about writing a KCL equation for each one. If we go to A, A has three

currents coming out of it — 1, 2, 3. So the sum of those has

to be 0, et cetera. And if you think about those

equations, they’re not linearly independent either. If you work through the math,

you see that there’s exactly one of those equations that

you can eliminate. So you’re left with

three linearly independent KCL equations. And so there’s a pattern

emerging here. Somebody see the pattern? 1 minus. Can somebody prove

the pattern? So there’s a pattern here. The pattern is take the number

of nodes and the number of independent KCL equations

as one less. So the challenge is,

can you prove it? And by the theory

of lectures– AUDIENCE: Yes. PROFESSOR: Yes. And by a corollary of the theory

of lectures, the way you would prove it is? AUDIENCE: On the next slide. PROFESSOR: On the next slide. Exactly. So how do I prove it? Yeah? AUDIENCE: Whenever you take

minus 1, you just add all the [UNINTELLIGIBLE] together

[UNINTELLIGIBLE]. PROFESSOR: Yeah. So there’s something special

about the last one. Why should there be something

special about the last one. AUDIENCE: Because the

circuit’s closed. PROFESSOR: Because the

circuit’s closed. That’s right. So the idea is to sort

of generalize the way we think about KCL. So we start with a circuit. We think about having

four nodes here. It’s certainly the case that

KCL holds for each node. So here’s KCL for that node. But now if you think about KCL

for this node, and then add them, that looks like

a KCL equation. But it applies to

a super node. Imagine the node defined by the

black box, and think about the net currents into or

out of the black node. This current i2, which leaves

the red node, enters the green node, but doesn’t go through

the surface of the black node at all. That’s exactly the current

that’s subtracted out when we added the red equation to

the green equation. Does that make sense? So KCL says, oh, if all the

currents at a node have to sum to 0, and if elements have the

same current coming out and going in, then if you draw a

box around an element, what goes into the element

is the same as what comes out of the element. It doesn’t change the net

current through the surface. So the generalization of the KCL

equation, KCL says the sum of the currents into

a node is 0. The generalization says take any

closed path in a circuit, the sum of the currents going

across that closed path is 0. So if we apply that rule again,

think about node 3. If we add the result of node 3

to the black node, which was the sum of 1 and 2, we get

the new green curve. We get the new green equation. And what that says is the sum

of the currents going across the green super node– OK, so what’s going on? i1 is

coming out of it, i4 is coming out of it, i5 is coming

out of it. So the sum of i1, i4,

and i5 has to be 0. Well, KCL says the sum of the

currents coming out of a node must be 0. The super KCL says the sum of

the currents coming out of any closed region is also 0. But the interesting thing about

this closed region is that it encloses all but

one of the nodes. That’s always true. Regardless of the system,

regardless of the circuit, you can always draw a line that will

isolate one node from all the others. So what that proves is that you

can always write KCL for this node in terms of

KCL for those nodes. Ok. So there’s a generalization then

that says that you can always write KCL

for every node. They will always be linearly

dependent. So you can always

throw away one. So in some sense now,

we’re done. We’ve just finished

circuit theory. We talked about how every

element has to have a law. A resistor is Ohm’s Law. A voltage source says that the

voltage across the terminals is always a constant. A current source says that the

current through the current source is always a constant. So every element tells

you one law. We know how to think

about KVL. So we know the rule for how the

across variables behave. What’s the aggregate behavior

of all the across variables? Well, KVL has to be satisfied

for every possible loop. The loops don’t have

to be independent. You have to worry about whether

they’re independent. The only simple rule we came up

with– we’ll come up with another one in a moment. The only simple rule that we

came up with was for planar circuits, where the innermost

loops were linearly independent of each other. And you have to write KCL for

all the nodes, except one. One of them never matters. So in some sense, we’re done. What we would do to

solve the circuit, think about every element. For every element assign a

voltage, a reference voltage. For every element,

assign a current. Make sure they go in the

right direction. We always define currents

to go down the potential gradient. They always go in the directions

through the element from the positive

to the negative. So for every element, assign

a current and a voltage. We have 6 elements, that’s

12 unknowns. Now we dig and we find

12 equations. In this particular circuit, we

found those 12 equations. There were three KCL equations,

one for each of the inner loops. There were three KCL equations,

one for each node except one. There were 5 Ohm’s law

equations, one for each one of the resistors. There was one source equation

for the voltage source. 12 equations, 12 unknowns,

we’re done. The only problem is a

lot of equations. It’s not a very complicated

circuit. We’ve only got 6 elements. I tried to motivate this in

terms of studying networks that had 10 to the 9 elements. This technique is not

particularly great at 10 to the 9. It would probably work. But we would probably

be interested in finding simpler ways. So there are simpler ways you

might imagine, and we’ll discuss two of them

just very briefly. The dumb way that I just talked

about is what we call primitive variables,

element variables. If you write all the element

variables, v1, v2, v3, v4, v5 v6, all of the element currents,

i1, i2, i3, i4, i5, i6, write all the equations,

you can solve it. However, if you’re judicious,

you can figure out a smaller number of unknowns and a

correspondingly smaller number of equations. One method is called

the node method. When we’re thinking about the

individual elements, the thing that matters is the voltage

across the element. However, that’s not

the easy way to write the circuit equations. A much easier way is not to tell

me the voltage across an element, but instead tell me

the voltage associated with each of the nodes. If I tell you the voltage

associated with every node, the important thing about that

way of defining the variables is that you’re guaranteed that

from those variables, you can tell me the voltage

across every part. So for example, in this circuit,

this voltage source– so if I call this one ground,

we’ll always have a magic node called ground. It is not special

in the least. It’s just the reference

voltage. I’ll come back to that. I’ll say words in a minute about

what the reference is. We always get to declare

one node to be ground. We get one free node. It’s a node whose voltage we

don’t care about because it’s the reference for

all voltages. It’s a node whose current we

don’t care about because we get to throw away one node when

we do current equations. So we have one special mode

called ground, about which we don’t care too much. Except that it’s the most

important node in the circuit. Except for that, we don’t

care about it. So this guy’s ground. We think about its

voltage being 0. Then this voltage supply

makes that node be v0. I don’t know what that is,

so I’ll call it e1. And I don’t know what that

is, so I’ll call it e2. So if I tell you the voltage on

all of those nodes, ground voltage is 0, the top voltage is

v0, the left voltage is e1, the right voltage is e2. From those four numbers, 0 and

3 nontrivial numbers, you can find all of the component

voltages. So for example, the voltage v6,

the voltage across R6 is e2 minus e1. The voltage v4, the voltage

across the R4 resistor is e1 minus 0. So if I tell you all the node

voltages, you can tell me all of the element voltages. And in general, there’s fewer

nodes than there are components. OK, that’s great. So instead of naming the volts

across the elements, we’ll name the voltages at the nodes

because there’s fewer of them. Then, all we need to do in the

node method is write the minimum number of

KCL equations. We know we only have two

unknowns, e1 and e2. And it turns out– and you

can prove this, but I won’t prove it today. It turns out that you need

two KCL equations. Two unknowns, e1, e2,

two KCL equations. And it turns out those two KCL

equations are exactly the KCL equations associated

with the two nodes. So the current leaving

e1, so KCL at e1 — well, there’s a current

that goes that way. Well, that’s the voltage drop

in going from e1 to v0, e1 minus v0, divided by R2. That’s Ohm’s law. So this term represents the

current going up that leg plus the current that goes through

this leg, which is e1 minus e2 over R6. Plus the current going in

that leg, which is e1 minus 0 over R4. The sum of those three

currents better be 0. Analogously, the sum of the

currents at this node must be 0, and the equation looks

virtually the same. Because v0 is known, so it

didn’t add an unknown. v0 was set by the voltage,

by the voltage source. So I have two equations,

two unknowns, solved. Done. So rather than solving 12

equations and 12 unknowns, I can do it with two equations

and two unknowns. That’s called the node method. One of the most interesting

theories about circuits is that every simplification that

you can think about for voltage has an analogous

simplification that you can think about in current. That’s called duality. We won’t do that because it’s

kind of complicated. But it’s kind of

a cute result. If you can think of a

simplification that works in voltage, then there is

an analogous one, and you can prove it. In fact, you can formally derive

what it must have been. This is a rule for how you can

simplify things by thinking about voltages in aggregate. Rather than thinking about the

element voltages, think about the node voltages. The analogous current law is

rather than thinking about the currents through the elements,

the element currents, think about loop currents. OK, that’s a little bizarre. So we name the loop, the current

that flows in this loop, IA, the current that flows

in this loop, IB, and the current hat flows

in this loop, IC. What on earth is he doing? Well, the element voltages are

some linear combination of those loop currents. And in fact, the coefficients in

the linear combination are one and minus one. So the element current I4, the

current that flows through the R4 resistor is the sum of

IA coming down minus IC, which is going up. So there’s a way of thinking

about each element current as a sum or difference of

the loop currents. Everybody get that? So instead of thinking about

the individual element currents, I think about

the loop currents. And now, I need to write

three KVL equations. So in the node method, I named

the nodes and had to write two KCL equations. Here, I named the loop currents

and I have to write three KVL equations,

one for each loop. It’s completely analogous. If you write out a sentence,

what did you do? I assigned a voltage to every

node, and I wrote KCL of all the nodes. Then if you turn the word

“current” into the word “voltage,” the word “node”

into the word “loop,” you derive this new method. So this says that if I write KVL

at the A loop, think about spinning around this loop, as

I go up through the voltage source, so I go in the negative

terminal here. So that’s minus v0. As I go down through this

resistor, I have to use Ohm’s law, so that’s R2 times

the down current. Well, the down current is

IA down minus IB up. So I went up through here,

down through here. Now I go down through

this one. When I go down through that one,

according to Ohm’s law, that’s R4 times the current

through that element. That current– well, it’s

IA down and it’s IC up. So this is the KVL equation

for that loop. I write two more of them, and I

end up with three equations and three unknowns. Both the node method and the

loop method resulted in a lot fewer equations than

the primitives did. I had 12 primitive unknowns,

6 voltages and 6 currents. In the node method, I get the

number of independent nodes as the number of equations and

unknowns, which is less than the number of primitive

variables. In the loop method, I have the

number of independent loops. Which is again, smaller. So the idea then is that we have

a couple of ways to think about solving circuits. Fundamentally, all we have are

the element relationships and the rules for combination. Oh, this is starting to sound

like PCAP, primitives and combinations. So the primitives are, how does

the element constrain the voltages and currents? We know three of those,

Ohm’s law, voltage source, current source. And what are the rules

for combination? Well, the currents add to the

node, and the voltages add around loops. OK, just to make sure you’ve

absorb all that, figure out the current I for

this circuit. OK, what’s a good

way to start? What should I do to start

thinking about calculating I? OK, bad way. Assign voltages and currents

to everything. 4 elements, that’s 4 voltages,

4 currents. That’s 8 unknowns. Find 8 equations, solve. That’ll work. Bad way. What’s a better way? OK, [UNINTELLIGIBLE PHRASE]. AUDIENCE:

[UNINTELLIGIBLE PHRASE]. PROFESSOR: It was on

the previous sheet. [UNINTELLIGIBLE PHRASE]. AUDIENCE: KVL [UNINTELLIGIBLE]. PROFESSOR: KVL for where? AUDIENCE: Loops. PROFESSOR: Which loop? AUDIENCE: Left loop. PROFESSOR: So do KVL

on the left loop? AUDIENCE: Yes. PROFESSOR: OK, that’s good. But you have to tell me how

to assign variables. Do you want 8 primitive

variables? 8 primitive variables are v1,

i1, v2, i2, v3, i3, v4, i4. So that’s what I mean by

primitive variables. Or element variables is

another word for it. What’s a better way than using

element variables? Yeah. AUDIENCE: Create 2

loop equations. PROFESSOR: Create 2 loop

equations, that’s fantastic. AUDIENCE: I1 for the first loop,

I2 for the second loop. PROFESSOR: So if you do I1 going

around here, then I1 is actually I. And if you do I2

going around here, what’s I2? AUDIENCE: [INAUDIBLE]. PROFESSOR: So if I think about

I2 spinning around this loop, so the sum of I1 and I2

goes through that box. But the only current that goes

through this box is? AUDIENCE: [INAUDIBLE]. PROFESSOR: So the suggestion

is that I think about– so if I have I1 here, but I know

that’s I. Then I can see immediately that since

the only current that goes through here– so if I have I1 and I2. That was a very clever idea. If you have I1 and I2, the only

current that goes through here is I. So I1

must’ve been I. The only current that

goes over here must’ve been this guy. So this must be minus 10. So I could redo that this way. I could say I’ve got

10 going that way. That make sense? So now I only have one unknown

which is I. So that’s a very clever way of doing it. So what I could do

is showed here. I have I going around one

loop and I have 10 going around that loop. That completely specifies

all the currents. So now all I need to

do is write KVL for these different cases. So if I write KVL for the left

loop, then I get going up through here, that’s minus 15,

and going down through here, going to the right through

this guy is 3I. Going down through this guy

is 2 times I plus 10. Both of these are going down,

so you have to add them. So I get one equation

and one unknown. And when I solve it,

I get minus one. That make sense? There’s an analogous

way you could have done it with one node. You could have said that the

circuit has a single node and figured out KCL for

that one node. KCL would be the sum of

the currents here. There’s a current that

goes that way, that way, and that way. And again, you end up with

1 equation and 1 unknown. Yes? AUDIENCE:

[UNINTELLIGIBLE PHRASE]. PROFESSOR: Correct. If I thought about this current

going this way, it would be minus 10. If I flipped the direction,

then it’s plus 10. So the loop current has the

property that it’s the only current through this element. So that has to match. It’s one of two currents that

go through this element. AUDIENCE: You said that

everything that [UNINTELLIGIBLE PHRASE]. PROFESSOR: This loop is

[UNINTELLIGIBLE]. Yes. AUDIENCE: So why is the

[UNINTELLIGIBLE PHRASE]. PROFESSOR: Correct. I want to have this

picture now. So if I’m doing it with loops,

I have two loops. The current through this element

is just I. The current through this element is just

I. The current through this element is just 10. The current through this

element– well, the sum of these two currents go through

that element. Does that make sense? AUDIENCE: [INAUDIBLE] PROFESSOR: This loop current

is just a fraction of the current in the whole system. So this loop current goes

through this element and contributes to this element. But so does that one. OK, if you’re still confused,

you should try to get it straightened out in one of the

software labs or the hardware lab, or talk to me

after lecture. But the idea is to

decompose in the case of the node voltages. Think about the element

voltages in terms of differences in the

node voltages. In the case of the loop

currents, think about the element currents in terms of

a sum of loop currents. OK, so the answer is minus 1

regardless of how you do it. Ok. The remaining thing I want to

do today is think about abstraction. We’ve talked about the

primitives, which are things like resistors, voltage sources,

and current sources. Means of combinations,

that’s KVL and KCL. Now we want to think

about abstraction. And the first abstraction that

we’ll talk about is, how do you think about one element

that represents more than one element? This is the same thing that we

did when we thought about linear systems, when we did

signals and systems. We started with R’s and K’s and

pluses, and we made single boxes that had lots of R’s and

pluses and gains in them. What was the name of the thing

that was inside the box? If we combined lots of R’s,

gains, and pluses into a single box, what would we call

the thing that’s in the box? AUDIENCE: [INAUDIBLE]. PROFESSOR: Shout again. AUDIENCE: System function. PROFESSOR: System function. Right? So we started with boxes

that only had things like R’s in them. But eventually, we got boxes

that looked like much more complicated things like that. We thought about a system

function which was a generalized box, that could have

lots of R’s, or lots of gains, or lots of

pluses in it. And that was a way of

abstracting complicated systems so they looked

like simple systems. What we want to do here is the

same thing for circuits. We want to have a single

element, a single circuit element, that represents

many circuit elements. And the simplest case of that

is for series in parallel combinations of resistors. It’s very simple to think about

how if you had two Ohm’s law devices connected in series,

you could replace those two with a single

resistor. And the voltage-current

relationships measured at the outside of the box would

be the same. That’s how we think about an

abstraction in circuits. When is it that you can draw

a box around a piece of a circuit and think about

that as one element? The very simplest cases, the

series combination of two resistors, same sort of thing

happens for the parallel combination. And that simple abstraction

makes some things very easy. What would be the equivalent

resistance for a complicated system like that? Well, that’s easy. All you need to do is think

about successively reducing the pieces. Here I’m thinking about that

having four resistors. I can just successively apply

series and parallel in order to reduce that, make it

less complicated. So I can think about combining

these two in series to get, instead of two 1 Ohm resistors,

one 2 Ohm resistor. Then I can think about these

two 2 Ohm resistors being equivalently one parallel

1 Ohm resistor. And so this whole thing looks

as though it’s just 2 Ohms from the outside world. That’s what we mean

by an abstraction. What we’re trying to do and what

we will do over the next two weeks, is we’ll think

about ways of combining circuits so that we can reduce

the complexity this way. Another convenient way of

thinking about reducing the work that you need to do is to

think about common patterns that result. PCAP, Primitives, Combinations,

Abstractions. So the series of parallel

idea was an abstraction. A pattern, here’s a

common pattern. If you’ve got two resistors in

series, if the same current flows through two resistors,

then there’s a way of very simply calculating the voltage

that falls across each. So you can think about the sum

resistor, R1 plus R2 since they’re in series. So that allows you then

to compute the current from the voltage. Then the voltage that falls

across this guy is by Ohm’s law, just the current

times its resistor, which is like that. And similarly with this one. So you can see that some

fraction of this voltage v occurs across the v1 terminal. And some different fraction

appears across the v2 terminal, such that the sum of

the fractions is, of course, v. That’s what has to

happen for the two. And there’s a proportional

drop. The bigger R1, the bigger is the

proportion of the voltage that falls across R1. So it’s a simple way of thinking

about how voltage drops across two resistors. There’s a completely analogous

way of thinking about how current splits between

two resistors. Here the result looks virtually

the same, except it has kind of the unintuitive

property that most of the current goes through the

resistor that is the smallest. So you get a bigger current in

i1 in proportion to the R2. So it works very much like the

voltage case, except that it has this inversion in it, that

the current likes to go through the smaller resistor. OK, so last problem. Using those kinds of ideas,

think about how you could compute the voltage v0 and

determine what’s the answer. So what’s the easy way to

think about this answer? What do I do first? AUDIENCE: Superposition. PROFESSOR: So superposition? That’s one thing. AUDIENCE: Simplify

[UNINTELLIGIBLE]. PROFESSOR: Simplify. So what’s a good

simplification? Collapse? AUDIENCE: You can put the two

[UNINTELLIGIBLE] in series and treat them as one. PROFESSOR: So you can treat this

as a series combination, and you can replace the series

of 1 and 3 with a? AUDIENCE:

[UNINTELLIGIBLE PHRASE]. AUDIENCE: 4.. PROFESSOR: 4. So this can be replaced by

four [UNINTELLIGIBLE] resistor. Now what? AUDIENCE: You can do the same

on the parallel one. PROFESSOR: So you can replace

the parallel of the 6 and a 12 with a? AUDIENCE: 4. PROFESSOR: Amazing

— with a 4. So there’s a 4 there and

there’s a 4 there. And the answer is? AUDIENCE: It’s half. PROFESSOR: Half of whatever

it was by voltage [UNINTELLIGIBLE] relationship. So you think about this

becoming that. You think about the parallel

becoming that. You get a simple divide

by 2 voltage divider. So the answer is 7 and 1/2,

which was the middle answer. And so what we did today was

basically a whirlwind tour of the theory of circuits. And the goal for the rest of the

week is to go to the lab and do the same sort of thing

with practical where you build a circuit, and try to use some

of these ideas to understand what it does.

ok mit kids, can you demodulate this am radio signal?

https://www.youtube.com/watch?v=RxU9-VYta0w