Circuits & Electronics P4

6.002 Fall 2000 Lecture
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6.002
CIRCUITS
AND
ELECTRONICS
The Digital Abstraction
6.002 Fall 2000 Lecture
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4
Review
z
Discretize matter by agreeing to
observe the lumped matter discipline
zAnalysis tool kit: KVL/KCL, node method,
superposition, Thévenin, Norton
(remember superposition, Thévenin,
Norton apply only for linear circuits)
Lumped Circuit Abstraction
6.002 Fall 2000 Lecture
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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
6.002 Fall 2000 Lecture

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4
Analog signal processing
But first, why digital?
In the past
…
By superposition,
The above is an “adder” circuit.
2
21
1
1
21
2
0
V
RR
R
V
RR
R
V
+
+
+
=
If
,
21
RR =

2
21
0
VV
V
+
=
1
V
1
R
2
R
+
–
2
V
+
–
0
V
and
might represent the
outputs of two
sensors, for example.
1
V
2
V
6.002 Fall 2000 Lecture

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Noise Problem
…
noise hampers our ability to distinguish
between small differences in value —
e.g. between 3.1V and 3.2V.
Receiver:
huh?
add noise on
this wire
t
6.002 Fall 2000 Lecture
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Value Discretization
Why is this discretization useful?
Restrict values to be one of two
HIGH
5V
TRUE
1
LOW
0V
FALSE
0
…
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.)
6.002 Fall 2000 Lecture
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Digital System
sender
receiver
S
V
R
V
noise
S
V
“0” “0”
“1”
0V
2.5V
5V
HIGH
LOW
t
R
V
“0” “0”
“1”
0V
2.5V

5V
t
VV
N
0=
N
V
S
V
“0” “0”
“1”
2.5V
t
With noise
VV
N
2.0=
S
V
“0” “0”
“1”
0V
2.5V
5V
t
0.2V
t
6.002 Fall 2000 Lecture
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Digital System
Better noise immunity
Lots of “noise margin”
For “1”: noise margin
5V to 2.5V = 2.5V
For “0”: noise margin
0V to 2.5V = 2.5V