CENG 241 Digital Design 1 Lecture 7 - PowerPoint PPT Presentation

About This Presentation
Title:

CENG 241 Digital Design 1 Lecture 7

Description:

CENG 241 Digital Design 1 Lecture 7 Amirali Baniasadi amirali_at_ece.uvic.ca – PowerPoint PPT presentation

Number of Views:84
Avg rating:3.0/5.0
Slides: 29
Provided by: Origi5
Category:
Tags: ceng | adder | design | digital | full | lecture

less

Transcript and Presenter's Notes

Title: CENG 241 Digital Design 1 Lecture 7


1
CENG 241Digital Design 1Lecture 7
  • Amirali Baniasadi
  • amirali_at_ece.uvic.ca

2
4-bit by 3-bit Binary Multiplier
B3 B2 B1 B0
A2 A1 A0 A0B3
A0B2 A0B1 A0B0 A1B3 A1B2 A1B1
A1B0 A2B3 A2B2 A2B1 A2B0
3
Decimal adder
  • When dealing with decimal numbers BCD code is
    used.
  • A decimal adders requires at least 9 inputs and 5
    outputs.
  • BCD adder each input does not exceed 9, the
    output can not exceed 19
  • How are decimal numbers presented in BCD?
  • Decimal Binary BCD
  • 9 1001 1001
  • 19 10011
    (0001)(1001)

  • 1 9

4
Decimal Adder
  • Decimal numbers should be represented in binary
    code number.
  • Example BCD adder
  • Suppose we apply two BCD numbers to a binary
    adder then
  • The result will be in binary and ranges from 0
    through 19.
  • Binary sum K(carry) Z8 Z4 Z2 Z1
  • BCD sum C(carry) S8 S4 S2 S1
  • For numbers equal or less than 1001 binary and
    BCD are identical.
  • For numbers more than 1001, we should add 6(0110)
    to binary to get BCD.
  • example 10011(binary) 11001(BCD) 19
  • ADD 6 to correct.

5
BCD adder
Numbers that need correction (add 6) are 01010
(10) 01011 (11) 01100 (12) 01101 (13) 01110
(14) 01111 (15) 10000 (16) 10001 (17) 10010
(18) 10011 (19)
Decides to add 6?
Adds 6
6
BCD adder
Numbers that need correction (add 6) are K Z8 Z4
Z2 Z1 0 1 0 1 0 (10) 0 1 0 1 1
(11) 0 1 1 0 0 (12) 0 1 1 0 1
(13) 0 1 1 1 0 (14) 0 1 1 1 1
(15) 1 0 0 0 0 (16) 1 0 0 0 1
(17) 1 0 0 1 0 (18) 1 0 0 1 1
(19)
C K Z8Z4 Z8Z2
7
Magnitude Comparators
  • Compares two numbers, determines their relative
    magnitude.
  • We look at a 4-bit magnitude comparator
  • AA3A2A1A0, BB3B2B1B0
  • Two numbers are equal if all bits are equal.
  • AB if A3B3 AND A2B2 AND A1B1 AND A0B0
  • Xi AiBi AiBi AiBi Xi1
    (remember exclusive NOR?)

8
Magnitude Comparators
  • How do we know if AgtB?
  • 1.Compare bits starting from the most significant
    pair of digits
  • 2.If the two are equal, compare the next lower
    significant bits
  • 3.Continue until a pair of unequal digits are
    reached
  • 4.Once the unequal digits are reached, AgtB if
    Ai1 and Bi0, AltB if Ai0 and Bi 1
  • AgtB A3B3X3A2B2X3X2A1B1X3X2X1A0B0
  • AltB A3B3X3A2B2X3X2A1B1X3X2X1A0B0
  • Xi1 if AiBi

9
Magnitude Comparators
A3B3 ?
X3A2B2
10
Decoders
  • A decoder converts binary information from n
    input lines to a maximum of 2n output lines
  • Also known as n-to-m line decoders where mlt 2n
  • Example 3-to-8 decoders.

11
Decoders Truth Table
  • X Y Z D0 D1 D2
    D3 D4 D5 D6 D7
  • 0 0 0 1 0 0
    0 0 0 0 0
  • 0 0 1 0 1 0
    0 0 0 0 0
  • 0 1 0 0 0 1
    0 0 0 0 0
  • 0 1 1 0 0 0
    1 0 0 0 0
  • 1 0 0 0 0 0
    0 1 0 0 0
  • 1 0 1 0 0 0
    0 0 1 0 0
  • 1 1 0 0 0 0
    0 0 0 1 0
  • 1 1 1 0 0 0
    0 0 0 0 1

12
Decoders AND implementation
13
2-to-4 Decoder NAND implementation
Decoder is enabled when E0
14
How to build bigger decoders?
We can combine two 3-to-8 decoders to build a
4-to-16 decoder.
Generates from 0000 to 0111
Generates from 1000 to 1111
15
Combinational Logic implementation
  • A decoder provides the 2n minterms of n input
    variables.
  • Any function is can be expressed in sum of
    minterms.
  • Use a decoder to make the minterms and an
    external OR gate to make the sum.
  • Example consider a full adder.
  • S(x,y,z) S(1,2,4,7)
  • C(x,y,z) S (3,5,6,7)

16
Combinational Logic implementation
17
Encoders
  • Encoders perform the inverse operation of a
    decoder
  • Encoders have 2n input lines and n output line.
  • Output lines generate the binary code
    corresponding to the input value.

18
Encoders Truth Table
  • Outputs
    Inputs
  • X Y Z D0 D1 D2
    D3 D4 D5 D6 D7
  • 0 0 0 1 0 0
    0 0 0 0 0
  • 0 0 1 0 1 0
    0 0 0 0 0
  • 0 1 0 0 0 1
    0 0 0 0 0
  • 0 1 1 0 0 0
    1 0 0 0 0
  • 1 0 0 0 0 0
    0 1 0 0 0
  • 1 0 1 0 0 0
    0 0 1 0 0
  • 1 1 0 0 0 0
    0 0 0 1 0
  • 1 1 1 0 0 0
    0 0 0 0 1
  • zD1D3D5D7 yD2D3D6D7
    xD4D5D6D7

19
Priority Encoders
  • Encoder limitations
  • If two inputs are active, the output is
    undefined.
  • Solution we need to take into account priority.
  • What if all inputs are 0?
  • Solution we need a valid bit
  • Input
    Output
  • D0 D1 D2 D3 x
    y v
  • 0 0 0 0 X
    X 0
  • 1 0 0 0 0
    0 1
  • X 1 0 0 0
    1 1
  • X X 1 0 1
    0 1
  • X X X 1 1
    1 1

20
Priority Encoders Map
21
Priority Encoders Circuit
22
Multiplexers
  • Multiplexer selects one binary input from many
    selections
  • example 2-to-1 MUX

23
4-to-1 MUX
Directs 1 of the 4 inputs to the output
24
Multi-bit selection logic
  • Multiplexers can be combined with common
    selection inputs to support multi-bit selection
    logic

25
Implementing Boolean functions w/ MUX
  • General rules for implementing any Boolean
    function with n variables
  • Use a multiplexer with n-1 selection inputs and 2
    n-1 data inputs
  • List the truth tabel
  • Apply the first n-1 variables to the selection
    inputs of multiplexer
  • For each combination evaluate the output as a
    function of the last variable.
  • The function can be 0, 1 the variable or the
    complement of the variable.

26
Implementing Boolean functions w/ MUX
27
Implementing Boolean functions w/ MUX
28
Summary
  • Reading up to page 156
Write a Comment
User Comments (0)
About PowerShow.com