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6.002 CIRCUITS ANDELECTRONICS
The Digital Abstraction
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6.002 Fall 2000 Lecture 24
Review
Discretize matter by agreeing toobserve the lumped matter discipline
Analysis tool kit: KVL/KCL, node method,superposition, Thvenin, Norton
(remember superposition, Thvenin,Norton apply only for linear circuits)
Lumped Circuit Abstraction
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Discretize value Digital abstraction
Interestingly, we will see shortly that thetools learned in the previous threelectures are sufficient to analyze simpledigital circuits
Reading: Chapter 5 of Agarwal & Lang
Today
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Analog signal processing
But first, why digital?In the past
By superposition,
The above is an adder circuit.
2
21
11
21
20
VRR
VRR
V+
+
+
=
If ,21 RR =
2
21
0
VVV
+=
1V
1
2+
2V +
0V
and
might represent the
outputs of two
sensors, for example.
1V 2V
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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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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 berepresented using multiple binary digits andcoding, much like using multiple decimal digits torepresent numbers greater than 9. E.g., thebinary number 101 has decimal value 5.)
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Digital System
sender receiverS
VRV
noise
SV
0 01
0V
2.5V
5V HIGH
LOW
t
RV
0 01
0V
2.5V
5V
t
VVN
0=
NV
SV
0 01
2.5V t
With noiseVV
N2.0=
SV
0 01
0V
2.5V
5V
t
0.2V
t
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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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Voltage Thresholdsand Logic Values
1
0
1
0
sender receiver
1
0
0V
2.5V
5V
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6.002 Fall 2000 Lecture 104
forbidden
region
VH
V L
3V
2V
But, but, but What about 2.5V?
Hmmm create no mans landor forbidden region
For example,
senderreceiver
0V
5V
1 1
0 0
1 V 5V
0 0V V
H
L
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6.002 Fall 2000 Lecture 114
sender receiver
But, but, but Wheres the noise margin?
What if the sender sent 1: ?VHHold the sender to tougher standards!
5V
0V
11
00
V
0H
V0L
VIH
VIL
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6.002 Fall 2000 Lecture 124
sender receiver
VH
But, but, but Wheres the noise margin?
What if the sender sent 1: ?Hold the sender to tougher standards!
5V
0V
1 noise margin:
0 noise margin:
VIH
- V0H
VIL
- V0L
11
00
V
0H
V0L
VIH
VIL
Noise margins
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6.002 Fall 2000 Lecture 134
Digital systems follow static discipline: ifinputs to the digital system meet valid inputthresholds, then the system guarantees itsoutputs will meet valid output thresholds.
sender
receiver
0 1 0 1
t
5VV
0H
V0L
0V
VIH
VIL
0 1 0 1
t
5VV
0H
V0L
0V
VIH
VIL
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Processing digital signals
Recall, we have only two values
Map naturally to logic: T, F
Can also represent numbers
1,0
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6.002 Fall 2000 Lecture 154
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 signals0 , 1
Z = X YBoolean equation
Enumerate all input combinations
Truth table representation:
ZX Y
AND gateZX
Y
0 0 0
0 1 01 0 0
1 1 1
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Adheres to static discipline Outputs are a function of
inputs alone.
Combinational gateabstraction
Digital logic designers do not
have to care about what is
inside a gate.
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Demo
Noise
ZXY
Z = X Y
Z
Y
X
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Z = X Y
Examples for recitation
X
t
Y
t
Z
t
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In recitation
Another example of a gateIf (A is true) OR (B is true)
then C is true
else C is false
C = A + B Boolean equation
OR
OR gate
CA
B
ZX
Y NAND
Z = X Y
More gates
B B
Inverter
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Boolean Identities
AB + AC = A (B + C)
X 1 = XX 0 = XX + 1 = 1X + 0 = X
1 = 0
0 = 1
output
BC B C
A
Digital Circuits
Implement: output = A + B C