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6.002 Fall 2000 Lecture 4

6.002 CIRCUITS ANDELECTRONICS

The Digital Abstraction

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6.002 Fall 2000 Lecture 4

ReviewDiscretize matter by agreeing to observe the lumped matter discipline

Analysis tool kit: KVL/KCL, node method, superposition, Thévenin, Norton (remember superposition, Thévenin, Norton apply only for linear circuits)

Lumped Circuit Abstraction

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6.002 Fall 2000 Lecture 4

Discretize value Digital abstraction

Interestingly, we will see shortly that the tools learned in the previous three lectures are sufficient to analyze simple digital circuits

Reading: Chapter 5 of Agarwal & Lang

Today

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6.002 Fall 2000 Lecture 4

Analog signal processing

But first, why digital?In the past …

By superposition,

The above is an “adder” circuit.

221

11

21

20 V

RRRV

RRRV

++

+=

If ,21 RR =

221

0VVV +

=

1V

1R

2R+–

2V +–

0V

andmight represent the outputs of two sensors, for example.

1V 2V

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6.002 Fall 2000 Lecture 4

Noise Problem

… noise hampers our ability to distinguishbetween small differences in value —e.g. between 3.1V and 3.2V.

Receiver:huh?

add noise onthis wire

t

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6.002 Fall 2000 Lecture 4

Value Discretization

Why is this discretization useful?

Restrict values to be one of two

HIGH5V

TRUE1

LOW0V

FALSE0

…like two digits 0 and 1

(Remember, numbers larger than 1 can be represented using multiple binary digits and coding, much like using multiple decimal digits to represent numbers greater than 9. E.g., the binary number 101 has decimal value 5.)

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6.002 Fall 2000 Lecture 4

Digital System

sender receiverSV

RV

noise

SV“0” “0”“1”

0V

2.5V

5V HIGH

LOW

t

RV“0” “0”“1”

0V2.5V

5Vt

VVN 0=

NV

SV“0” “0”“1”

2.5V t

With noiseVVN 2.0=

SV“0” “0”“1”

0V2.5V

5Vt

0.2Vt

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6.002 Fall 2000 Lecture 4

Digital System

Better noise immunityLots of “noise margin”

For “1”: noise margin 5V to 2.5V = 2.5VFor “0”: noise margin 0V to 2.5V = 2.5V

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6.002 Fall 2000 Lecture 4

Voltage Thresholds and Logic Values

1

0

1

0

sender receiver

1

0

0V

2.5V

5V

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6.002 Fall 2000 Lecture 4

forbiddenregion

VH

V L

3V

2V

But, but, but …What about 2.5V?Hmmm… create “no man’s land”or forbidden regionFor example,

sender receiver

0V

5V

1 1

0 0

“1” V 5V

“0” 0V V

H

L

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6.002 Fall 2000 Lecture 4

sender receiver

But, but, but …Where’s the noise margin?What if the sender sent 1: ?VHHold the sender to tougher standards!

5V

0V

11

00

V0H

V0L

VIH

VIL

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6.002 Fall 2000 Lecture 4

sender receiver

5V

0V

VH

But, but, but …Where’s the noise margin?What if the sender sent 1: ?Hold the sender to tougher standards!

“1” noise margin:“0” noise margin:

VIH - V0HVIL - V0L

11

00

V0H

V0L

VIH

VIL

Noise margins

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6.002 Fall 2000 Lecture 4

Digital systems follow static discipline: if inputs to the digital system meet valid input thresholds, then the system guarantees its outputs will meet valid output thresholds.

sender

receiver

0 1 0 1

t

5VV0H

V0L0V

VIHVIL

0 1 0 1

t

5VV0H

V0L0V

VIHVIL

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6.002 Fall 2000 Lecture 4

Processing digital signals

Recall, we have only two values —

Map naturally to logic: T, FCan also represent numbers

1,0

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6.002 Fall 2000 Lecture 4

Processing digital signalsBoolean Logic

If X is true and Y is trueThen Z is true else Z is false.

Z = X AND YX, Y, Z

are digital signals“0” , “1”

Z = X • YBoolean equation

Enumerate all input combinations

Truth table representation:ZX Y

AND gateZXY

0 0 00 1 01 0 01 1 1

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6.002 Fall 2000 Lecture 4

Adheres to static disciplineOutputs are a function ofinputs alone.

Combinational gateabstraction

Digital logic designers do nothave to care about what isinside a gate.

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6.002 Fall 2000 Lecture 4

Demo

Noise

ZXYZ = X • Y

Z

Y

X

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6.002 Fall 2000 Lecture 4

Z = X • Y

Examples for recitation

X

t

Y

t

Z

t

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6.002 Fall 2000 Lecture 4

In recitation…Another example of a gate

If (A is true) OR (B is true)then C is true else C is false

C = A + B Boolean equationOR

OR gateCA

B

ZXY

NANDZ = X • Y

More gatesB B

Inverter

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6.002 Fall 2000 Lecture 4

Boolean Identities

AB + AC = A • (B + C)

X • 1 = XX • 0 = XX + 1 = 1X + 0 = X

1 = 00 = 1

output

BC B • C

A

Digital CircuitsImplement: output = A + B • C