6002L1

6.002 Fall 2000 Lecture 1

1󰂀

ADMINISTRIVIA󰂀

󰀱 Lecturer: Prof. Anant Agarwal

󰂄 Textbook: Agarwal and Lang (A&L)󰂀󰂄 󰂄󰀁Handout no. 3

Assignments —

Homework exercises Labs Quizzes Final exam

Fall 2000 Lecture 1

2󰂀

Readings are important!

6.002 󰂄Two homework assignments can be missed (except HW11). 󰂄 Collaboration policy Homework

You may collaborate with others, but do your own write-up. Lab

You may work in a team of two, but do you own write-up.

󰂄 Info handout 󰂄󰀁Reading for today —

Chapter 1 of the book

6.002 Fall 2000 Lecture 1

3󰂀

Purposeful use of science Gainful employment of Maxwell’s equations

From electrons to digital gates and op-amps

6.002 Fall 2000 Lecture 1

4󰂀

Nature as observed in experiments V I 30.160.290.3120.4……Physics laws or “abstractions” 󰁺 Maxwell’s abstraction for 󰁺 Ohm’s tables of data V = R I Lumped circuit abstraction + – V C L R Simple amplifier abstraction Digital abstraction Operational amplifier abstraction abstraction Combinational logic + 6002M S -f Filters Analog system components: Modulators, oscillators, RF amps, power supplies 6.061 Clocked digital abstraction Instruction set abstraction Pentium, MIPS 6.004 Programming languages Java, C++, Matlab 6.001 Software systems 6.033 Operating systems, Browsers Mice, toasters, sonar, stereos, doom, space shuttle6.455 6.1705󰂀6.002 Fall 2000 Lecture 1 Lumped Circuit Abstraction󰂀

The Big Jumpfrom physicsto EECS I Consider

+ V󰂀

-

Suppose we wish to answer this question: What is the current through the bulb?

6.002 Fall 2000 Lecture 1

6󰂀

We could do it the Hard Way… Apply Maxwell’s

Differential form ∇× E =−∂B∂t Continuity ∇⋅ J = − ∂ρ ∂tOthers

∇⋅ E =

ρ 󰁺ε0

󰁺󰁺

6.002 Fall 2000 Lecture 1

Integral form

∫ E ⋅ dl = −∂φB

∂t

∫ J ⋅ dS = −∂q

∂t

∫ E ⋅ dS = q

ε0

󰁺 󰁺 󰁺

7Faraday’s 󰂀

Instead, there is an Easy Way…󰂀First, let us build some insight: Analogy F

a ?

I ask you: What is the acceleration? You quickly ask me: What is the mass? I tell you:

m

You respond: a =

F

m

Done

Fall 2000 Lecture 1

86.002 󰂀

Instead, there is an Easy Way…󰂀First, let us build some insight: Analogy F

a ?

In doing so, you ignored

󰁺 the object’s shape 󰁺 its temperature 󰁺 its color

󰁺 point of force application

Point-mass discretization󰂀

6.002 Fall 2000 Lecture 1

9󰂀

The Easy Way…󰂀Consider the filament of the light bulb.

A B

We do not care about󰂀

󰁺 how current flows inside the filament󰂀󰁺 its temperature, shape, orientation, etc.󰂀Then, we can replace the bulb with a for the purpose of calculating the current.

discrete resistor 6.002 Fall 2000 Lecture 1

10󰂀

The Easy Way…󰂀A B

Replace the bulb with a

for the purpose of calculating the current.

A+ V – B I R

discrete resistor V

and I = R

In EE, we do thingsthe easy way… R represents the only property of interest! Like with point-mass: replace objects

F

with their mass m to find a =

m󰂀

6.002 Fall 2000 Lecture 1

11󰂀

The Easy Way…󰂀A+ V – B I󰂀R

V󰂀

and I = R󰂀

In EE, we do thingsthe easy way… R represents the only property of interest!󰂀R relates element v and i󰂀

V󰂀I =

R

called element v-i relationship󰂀

6.002 Fall 2000 Lecture 1

12󰂀

R is a lumped element abstraction for the bulb. 6.002 Fall 2000 Lecture 1

13󰂀

R is a lumped element abstraction for the bulb. Not so fast, though …

A + I S A V B – SB black box

Although we will take the easy way using lumped abstractions for the rest of this course, we must make sure (at least the first time) that our

abstraction is reasonable. In this case, ensuring that VI are defined for the element

Fall 2000 Lecture 1

6.002 14󰂀

A󰂀+ I S A V B – SB black box

Fall 2000 Lecture 1

VI󰂀must be defined for the element

156.002 󰂀

I

must be defined. True when I into SA

= I out of SB True only when󰀁∂q

∂t

= 0 in the filament!

S∫ J ⋅ dS A

S∫ J ⋅ dS

B

∫ J ⋅ dS −∫ J ⋅ dS =∂q

SA SB

∂t fIA

IB

Mroamxwell II∂q A = B only if = 0

So let’s assume this ∂t

Fall 2000 Lecture 1

16

6.002 V

Must also be defined.󰂀

sAee&L So let’s assume this too V∂φAB defined when B

= So ∂t

0

VAB =∫AB

E ⋅ dl outside elements Fall 2000 Lecture 1

6.002 17󰂀

Lumped Matter Discipline (LMD) Or self imposed constraints: More in Chapter 1of A & L ∂φ B= 0outside ∂t ∂q = 0inside elements ∂t bulb, wire, battery Lumped circuit abstraction applies when

elements adhere to the lumped matter discipline.

6.002 Fall 2000 Lecture 1

18󰂀

Demo only for the sorts of questions we as EEs would like to ask! Lumped element examples whose behavior is completely captured by their V–I relationship. Demo Exploding resistor demo

can’t predict that! Pickle demo

can’t predict light, smell

6.002 Fall 2000 Lecture 1

19󰂀

So, what does this buy us?󰂀

Replace the differential equations with simple algebra using lumped circuit abstraction (LCA). For example —

a R1 b R2 V󰂀+ – R3R4󰂀

d R5󰂀

c󰂀What can we say about voltages in a loop under the lumped matter discipline?

6.002 Fall 2000 Lecture 1

20󰂀

What can we say about voltages in a loop under LMD?

a R1 b R2 V󰂀+ – R3 R4󰂀

d R5󰂀

c ∂φB under DMD ∫ E ⋅ dl = − ∂t 0

∫ E ⋅ dl +∫ E ⋅ dl +∫ E ⋅ dl = 0

ab

bc

ca

+ Vca + Vab + Vbc = 0

Kirchhoff’s Voltage Law (KVL): The sum of the voltages in a loop is 0. 6.002 Fall 2000 Lecture 1

21󰂀

What can we say about currents?󰂀

Consider

ISca a Ida I baLecture 1

226.002 Fall 2000 󰂀

What can we say about currents?󰂀

I ca Sa I󰂀daIba ∂q ∫S J ⋅ dS = − ∂t Ica + Ida + Iba = 0

under LMD 0

Kirchhoff’s Current Law (KCL): The sum of the currents into a node is 0. simply conservation of charge󰂀

6.002 Fall 2000 Lecture 1

23󰂀

KVL and KCL Summary󰂀

KVL:

KCL:

∑ jν j = 0

loop

∑ jij = 0

node

Lecture 1

246.002 Fall 2000 󰂀