Lecture 18 M - Copy.pptx

• Two classes of logic circuits:
• Combinational Circuits
• Sequential Circuits
• A Combinational circuit consists of logic
gates
• Output depends only on input
• A Sequential circuit consists of logic gates
and memory
• Output depends on current inputs and previous
ones (stored in memory)
• Memory defines the state of the circuit.

 Output is function of input only
i.e. no feedback
When input changes, output may change (after a delay)
•
•
•
•
•
•
n inputs m outputs
Combinational
Circuits


 Arithmetic & logical functions (adder,
subtractor, comparator)
 Data transmission(decoder, encoder,
multiplexer, de-multiplexer)
 Code converters( BCD, grey code,7-segment)

 Analysis
 Given a circuit, find out its function
 Function may be expressed as:
 Boolean function
 Truth table
 Design
 Given a desired function, determine its circuit
 Function may be expressed as:
 Boolean function
 Truth table
C
B
A
C
B
A
B
A
C
A
C
B
F1
F2
?
?
?

 Boolean Expression Approach
C
B
A
C
B
A
B
A
C
A
C
B
F1
F2
T2=ABC
T1=A+B+C
F2=AB+AC+BC
F’2=(A’+B’)(A’+C’)(B’+C’)
T3=AB'C'+A'BC'+A'B'C
F1=AB'C'+A'BC'+A'B'C+ABC
F2=AB+AC+BC

C
B
A
C
B
A
B
A
C
A
C
B
F1
F2
 Truth Table Approach
A B C F1 F2
0 0 0
= 0
= 0
= 0
= 0
= 0
= 0
= 0
= 0
= 0
= 0
= 0
= 0
0
0
0
0
0
0
1
0
0
0 0

C
B
A
C
B
A
B
A
C
A
C
B
F1
F2
= 0
= 0
= 1
= 0
= 0
= 1
= 0
= 0
= 0
= 1
= 0
= 1
0
1
0
0
0
0
1
1
1
A B C F1 F2
0 0 0 0 0
0 0 1 1 0

C
B
A
C
B
A
B
A
C
A
C
B
F1
F2
= 0
= 1
= 0
= 0
= 1
= 0
= 0
= 1
= 0
= 0
= 1
= 0
0
1
0
0
0
0
1
1
1
A B C F1 F2
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0

C
B
A
C
B
A
B
A
C
A
C
B
F1
F2
= 0
= 1
= 1
= 0
= 1
= 1
= 0
= 1
= 0
= 1
= 1
= 1
0
1
0
0
1
1
0
0
0
A B C F1 F2
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1

C
B
A
C
B
A
B
A
C
A
C
B
F1
F2
= 1
= 0
= 0
= 1
= 0
= 0
= 1
= 0
= 1
= 0
= 0
= 0
0
1
0
0
0
0
1
1
1
A B C F1 F2
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1
1 0 0 1 0

C
B
A
C
B
A
B
A
C
A
C
B
F1
F2
= 1
= 0
= 1
= 1
= 0
= 1
= 1
= 0
= 1
= 1
= 0
= 1
0
1
0
1
0
1
0
0
0
A B C F1 F2
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1

C
B
A
C
B
A
B
A
C
A
C
B
F1
F2
= 1
= 1
= 0
= 1
= 1
= 0
= 1
= 1
= 1
= 0
= 1
= 0
0
1
1
0
0
1
0
0
0
A B C F1 F2
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1
1 1 0 0 1

C
B
A
C
B
A
B
A
C
A
C
B
F1
F2
= 1
= 1
= 1
= 1
= 1
= 1
= 1
= 1
= 1
= 1
= 1
= 1
1
1
1
1
1
1
0
0
1
A B C F1 F2
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1
1 1 0 0 1
1 1 1 1 1
B
0 1 0 1
A 1 0 1 0
C
B
0 0 1 0
A 0 1 1 1
C
F1=AB'C'+A'BC'+A'B'C+ABC F2=AB+AC+BC

 Given a problem statement:
 Determine the number of inputs and outputs
 Derive the truth table
 Simplify the Boolean expression for each output
 Produce the required circuit
Example:
Design a circuit to convert a “BCD” code to “Excess 3”
code
 4-bits
 0-9 values
 4-bits
 Value+3
?

 Excess 3 code It is a basically a binary code which is
made by adding 3 with the equivalent decimal of a
binary number and again converting it into binary
number. So if we consider any binary number we have to
first convert it into decimal number then add 3 with it
and then convert it into binary and we will get the
excess 3 equivalent of that number.

 BCD-to-Excess 3 Converter
A B C D w x y z
0 0 0 0 0 0 1 1
0 0 0 1 0 1 0 0
0 0 1 0 0 1 0 1
0 0 1 1 0 1 1 0
0 1 0 0 0 1 1 1
0 1 0 1 1 0 0 0
0 1 1 0 1 0 0 1
0 1 1 1 1 0 1 0
1 0 0 0 1 0 1 1
1 0 0 1 1 1 0 0
1 0 1 0 x x x x
1 0 1 1 x x x x
1 1 0 0 x x x x
1 1 0 1 x x x x
1 1 1 0 x x x x
1 1 1 1 x x x x
C
1 1 1
B
A
x x x x
1 1 x x
D
C
1 1 1
1
B
A
x x x x
1 x x
D
C
1 1
1 1
B
A
x x x x
1 x x
D
C
1 1
1 1
B
A
x x x x
1 x x
D
w = A+BC+BD x = B’C+B’D+BC’D’
y = C’D’+CD z = D’

 BCD-to-Excess 3 Converter
A B C D w x y z
0 0 0 0 0 0 1 1
0 0 0 1 0 1 0 0
0 0 1 0 0 1 0 1
0 0 1 1 0 1 1 0
0 1 0 0 0 1 1 1
0 1 0 1 1 0 0 0
0 1 1 0 1 0 0 1
0 1 1 1 1 0 1 0
1 0 0 0 1 0 1 1
1 0 0 1 1 1 0 0
1 0 1 0 x x x x
1 0 1 1 x x x x
1 1 0 0 x x x x
1 1 0 1 x x x x
1 1 1 0 x x x x
1 1 1 1 x x x x
w
x
D
C
z
y
B
A
w = A + B(C+D)
x = B’(C+D) + B(C+D)’
y = (C+D)’ + CD
z = D’

 Seven-Segment Display
• A seven-segment display is digital readout
found in electronic devices like clocks, TVs,
etc.
• Made of seven light-emitting diodes (LED)
segments; each segment is controlled separately.

a
b
c
g
e
d
f
?
w
x
y
z
a
b
c
d
e
f
g
w x y z a b c d e f g
0 0 0 0 1 1 1 1 1 1 0
0 0 0 1 0 1 1 0 0 0 0
0 0 1 0 1 1 0 1 1 0 1
0 0 1 1 1 1 1 1 0 0 1
0 1 0 0 0 1 1 0 0 1 1
0 1 0 1 1 0 1 1 0 1 1
0 1 1 0 1 0 1 1 1 1 1
0 1 1 1 1 1 1 0 0 0 0
1 0 0 0 1 1 1 1 1 1 1
1 0 0 1 1 1 1 1 0 1 1
1 0 1 0 x x x x x x x
1 0 1 1 x x x x x x x
1 1 0 0 x x x x x x x
1 1 0 1 x x x x x x x
1 1 1 0 x x x x x x x
1 1 1 1 x x x x x x x
y
1 1 1
1 1 1
x
w
x x x x
1 1 x x
z
BCD code
a = w + y + xz + x’z’ b = . . .
c = . . .
d = . . .

 Design Problem:01
Write the Boolean logic Equation and draw
the logic circuit that represents the following
statement:
“A bank burglar alarm (A) is to activate if it
is after bank hours (H) and the front door (F)
is opened or if it is after banking hours (H)
and the vault door is opened (V)”.

Design a Majority function generator for 3-bit
input.

Design a system called a parallel binary
comparator, that compares the 4 – bit binary
string A to the 4 – bit binary string B. if the
strings are exactly equal, provide a HIGH –
level output to drive a warning buzzer.

 Half Adder
The half adder adds two one-bit binary numbers
(AB). The output is the sum of the two bits (S)
and the carry (C)
 Adds 1-bit plus 1-bit
 Produces Sum and Carry
HA
x
y
S
C
x y C S
0 0 0 0
0 1 0 1
1 0 0 1
1 1 1 0
x
+ y
───
C S
x
y
S
C

 Full Adder
The full-adder circuit adds three one-bit binary
numbers and outputs two one-bit binary
numbers, a sum (S) and a carry (C1)
 Adds 1-bit plus 1-bit plus 1-bit
 Produces Sum and Carry
x y z C S
0 0 0 0 0
0 0 1 0 1
0 1 0 0 1
0 1 1 1 0
1 0 0 0 1
1 0 1 1 0
1 1 0 1 0
1 1 1 1 1
x
+ y
+ z
───
C S
FA
x
y
z
S
C
y
0 1 0 1
x 1 0 1 0
z
y
0 0 1 0
x 0 1 1 1
z
S = xy'z'+x'yz'+x'y'z+xyz = x  y  z
C = xy + xz + yz

 Full Adder
x
y
z
S
C
x
y
x
z
y
z
x
y
z
x
y
z
x
y
z
x
y
z
x
y
z
x
y
z
x
y
x
z
y
z
S
C
S = xy'z'+x'yz'+x'y'z+xyz = x  y  z
C = xy + xz + yz

Manipulating the Equations:
S = X  Y  Z
C = XY + XZ + YZ

S = ( X  Y )  Z
C = XY + XZ + YZ
= XY + XYZ + XY’Z + X’YZ + XYZ
= XY( 1 + Z) + Z(XY’ + X’Y)
= XY + Z(X  Y )

S = ( X  Y )  Z
C = XY + XZ + YZ = XY + Z(X  Y )
Think
of Z as
a carry
in

A digital circuit that produces the arithmetic
sum of two binary numbers
• How to build an adder for n-bit numbers?
• Example: 4-Bit Adder
• Inputs ?
• Outputs ?
• What is the size of the truth table?
• How many functions to optimize?

• How to build an adder for n-bit numbers?
• Example: 4-Bit Adder
• Inputs ? 9 inputs
• Outputs ? 5 outputs
• What is the size of the truth table? 512 rows!
• How many functions to optimize? 5 functions

1 0 0 0
0 1 0 1
+ 0 1 1 0
1 0 1 1
 To add n-bit numbers:
• Use n Full-Adders in Cascade.
• The carries propagates as in addition by hand.
Carry in
This adder is called ripple carry adder

c3 c2 c1 .
+ x3 x2 x1 x0
+ y3 y2 y1 y0
────────
Cy S3 S2 S1 S0
FA
x3 x2 x1 x0
FA
FA
FA
y3 y2 y1 y0
S3 S2 S1 S0
C4 C3 C2 C1
0
Binary Adder
x3x2x1x0 y3y2y1y0
S3S2S1S0
C0
Cy
Carry
Propagate
Addition

 Half Subtractor
A logic circuit which is used for subtracting one
single bit binary number from another single
bit binary number is called half subtractor.

S.No
INPUT OUTPUT
A B DIFF BORR
1. 0 0 0 0
2. 0 1 1 1
3. 1 0 1 0
4. 1 1 0 0
 Half Subtractor

 Full Subtractor
The Full subtractor is a combinational circuit
which is used to perform subtraction of three
bits.

 Full Subtractor
S.
No
INPUT OUTPUT
A B C DIFF BORR
1. 0 0 0 0 0
2. 0 0 1 1 1
3. 0 1 0 1 1
4. 0 1 1 0 1
5. 1 0 0 1 0
6. 1 0 1 0 0
7. 1 1 0 0 0
8. 1 1 1 1 1

 Derive simplified Boolean expressions for
Half Subtractor and Full Subtractor.

 How to build a subtractor using 2’s
complement?

 How to build a subtractor using 2’s
complement?
1
S = A + ( -B)

 How to build a circuit that performs both
addition and subtraction?

Using full adders and XOR we can build an Adder/Subtractor!
0 : Add
1: subtract

Lecture 18 M - Copy.pptx

Recomendados

Recomendados

Mais conteúdo relacionado

Semelhante a Lecture 18 M - Copy.pptx

Semelhante a Lecture 18 M - Copy.pptx (20)

Último

Último (20)

Lecture 18 M - Copy.pptx