Integrated Circuits

Integrated Circuits

IC or Chip

Various gates are interconnected to form a circuit

14 to 64 pins for a typical IC

 

Levels of Integration

Small Scale Integration (SSI)

Less than 10 gates

Medium Scale Integration (MSI)

10 to 100 gates, adders, multiplexers

Large Scale Integration (LSI)

100 to few thousands of gates, processors

Very Large Scale Integration (VLSI)

Thousands of gates, complex computer Chips, memory arrays

Digital Logic Families

TTL (Transistor – Transistor Logic)

Widespread and this is the standard

Bipolar

ECL (Emitter Coupled Logic )

For high speed operation

Bipolar

MOS (Metal Oxide Semiconductors)

High component density

Unipolar

CMOS (Complementary Metal Oxide Semiconductors)

Useful where Low power consumption is needed

unipolar

 

Characteristics of digital logic families

Fan in

Number of inputs of an electronic gate

Physical logic gates with large fan in are slower compared to less fan in

image

The NAND gates have a fan in factor of 4

Fan out

Specifies the number of standard loads that the output of a gate can drive without impairing its normal operation.

Propagation Delay

Average transition delay time for the signal to propagate from input to output

image

A full adder has an overall gate delay of 3 logic gates from the inputs A and B to the carry output Cout shown in red

Power dissipation

Power consumed by the gate

Noise Margin

Minimum external noise voltage that causes an undesirable change in the circuit output

A full adder has an overall gate delay of 3 logic gates from the inputs A and B to the carry output Cout shown in red

3D Glass with 3D–Anaglyph Maker

I recently downloaded a 3D movie through a website. I thought of viewing through 3d glass. So I made a search through www.google.com. I got it through www.ebay.in and ordered 3 pair of 3d glasses for Rs.149. But beforehand I thought of searching that in Chennai, I failed to get it in most of the places in and around Chennai. I got some response from some shop owners that the glasses are available at Ashok Nagar or KK Nagar.

But I want to see the film immediately, so I purchased two glass sheets of paper (Red and Blue). See the pics below. Both together costs just Rs.10 only.

one

Then I took my old sunglass, removed both the glasses and the cuttings from the two glass sheets (Red and Blue) was pasted as per the following arrangement. Now that’s it. I watched the film fantastically. May be you can prepare the mirror on your own.

two

But how to create our own 3D image. Go to this page (http://www.stereoeye.jp/software/index_e.html)and download the software given there. Following is the software window.

Give two images (Left and Right) and decide what type of 3D image you want (Red- Cyan or Gray or ….)

3d2

the superimposing of two images done either U (upper) or R (Right) or  L(Left) or D (Down). Then Save the image and that’s it. Here are few 3d images created with the above software.

demo1

demo2

Canonical Forms and Standard Forms

Standard Forms

  • each term in the function can have any number of literals.

Example, F1 = a +b’c + cde

  • There are 5 variables in F1 (a,b,c,d,e)
  • Canonical form terms should be written as m0, m1… (sum of Products) or M0, M1, M2… (Product of sums).
  • Each term in the canonical form should have all the literals.

Example: F1 = ab’c’d’e’ + abcd’e’ + abcde

More about Canonical Forms

Minterms

  • It is Sum of Products
  • Canonical form is Sum of Minterms
  • three variable minterms are shown below
x a b c minterms
0 0 0 0 m0=a’.b’.c’
1 0 0 1 m1=a’.b’.c
2 0 1 0 m2=a’.b.c’
3 0 1 1 m3=a’.b.c
4 1 0 0 m4=a.b’.c’
5 1 0 1 m5=a.b’.c
6 1 1 0 m6=a.b.c’
7 1 1 1 m7=a.b.c

Maxterms

  • Product of Sum (PoS)
  • Canonical form is Product of MaxTerms
  • three variable maxterms are shown below
x a b c minterms
0 0 0 0 M0=(a+b+c)
1 0 0 1 M1=(a+b+c’)
2 0 1 0 M2=(a+b’+c)
3 0 1 1 M3=(a+b’+c’)
4 1 0 0 M4=(a’+b+c)
5 1 0 1 M5=(a’+b+c’)
6 1 1 0 M6=(a’+b’+c)
7 1 1 1 M7=(a’+b’+c’)

usually

Mi = (mj)’

Express the boolean function F = A + BC in a sum of minterms.

The function has three variables,

so F = A + BC will be

F = A(B + B’) + (A +A’) BC  [since, x + x’ =1]

F = AB + AB’ + ABC + A’BC

F = AB(C+C’) + AB’(C+C’) + ABC + A’BC

F = ABC + ABC’ + AB’C + AB’C’ + ABC + A’BC

F = ABC + ABC’ + AB’C + AB’C’ + A’BC [since x + x = x]

F= m0 + m6 + m5 + m4 + m3

Binary Codes

  • bit is just called as binary digit
  • To represent 2n elements, n bits are needed

Here are the following Binary Codes,

image

For example, the BCD code is otherwise called as 8421 code where the 8,4,2,1 are the weights assigned to the digits

for number 3 (0011), the weightage will be 0 * 8 + 0 * 4 + 1 * 2 + 1 *1

in 84-2-1 code, the weights are arranged like this,

for example, for number 2 (0110),  it is 0 * 8 + 1 * 4 + 1 * –2 + 0 * –1

similarly the same case for 2421 codes.

Excess – 3 is a code which is in excess of 3 in decimal numbers .

Error Detection Codes

  • Binary information is usually transmitted from one place to other through wired medium, due to the electromagnetic radiation or external noise, the information bits can be changed (ie 1 to 0 or 0 to 1), in this scenario, there is a provision to check whether the given word or byte is correct or not. Parity bits are used for that.
  • Odd parity or even parity is adopted based on the application, but mostly even parity is adopted.
  • Parity bit is an extra bit added along with the given byte to make the number of 1’s in the total word is even or odd (based on even or odd parity)
Message odd parity bit Message Even parity bit
0000 1 0000 0
0001 0 0001 1
0010 0 0010 1
0011 0 0011 0
0100 0 0100 1
0101 1 0101 0
0110 1 0110 0
0111 0 0111 1
1000 1 1000 1
1001 1 1001 0
1010 1 1010 0
1011 0 1011 1
1100 1 1100 0
1101 0 1101 1
1110 0 1110 1
1111 1 1111 0

Gray Code

Gray code is different from binary code in which only one bit change is there in the code group. Moving from one number to another number, there is just only one bit change. here is an example of Gray Code

Gray Code Decimal Equivalent
0000 0
0001 1
0011 2
0010 3
0110 4
0111 5
0101 6
0100 7
1100 8
1101 9
1111 10
1110 11
1010 12
1011 13
1001 14
1000 15

 

Other Alpha Numeric Codes

  • ASCII (American Standard Code for Information Interchange)
    • used in Keywords,
    • take 7 bit data
    • can able to address 128 characters

image

  • EBCIC  (Extended Binary Coded Decimal Interchange Code)
    • This code is from IBM
    • uses 8 bits, so addresses up to 256

Radix Complement and Diminished Radix complement

Complements are used in digital computers for simplifying the subtraction and for logical manipulations. There are two types of complement

  • Radix complements ( r complement)
  • diminished radix complements (r-1 complement)

For base 2 or binary number system the r’s complement is 2’s complement and (r-1)’s complement is 1’s complement

For base 10 or decimal system the r’s complement is 10’s complement and (r-1)’s complement is 9’s complement.

10’s Complement

Let us assume the number 3567890, for finding the 10’s complement, there is a rule

3 5 6 7 8 9 0
Most significant Digit           Least Significant Digit

10’s complement can be obtained by

  • leaving all least significant 0’s unchanged
  • subtracting the first non zero least significant digit from 10
  • and subtracting all higher significant digits from 9

For the above example, The 10’s complement is

Given number 3 5 6 7 8 9 0
Process 9-3 9-5 9-6 9-7 9-8 10-9 unchanged
10’s Complement 6 4 3 2 1 1 0

so 10’s complement of 3567890 is 6432110

9’s Complement

9’s complement is a diminished radix complement and can be easily found out by subtracting all the given digits by 9.

For example, the 9’s complement of 3567890 is

9999999 – 3567890 = 6432109

given number 3 5 6 7 8 9 0
Process 9-3 9-5 9-6 9-7 9-8 9-9 9-0
9’s complement 6 4 3 2 1 0 9

In short,

10’s Complement = 9’s Complement + 1

2’s Complement

For binary numbers, there is 2’s complement and 1’s complement

2’s complement can be obtained by

  • leaving the least significant 0’s unchanged and the first 1 unchanged
  • replacing 1’s with 0’s and 0’s with 1’s in all other higher significant digits
given number 1 1 0 1 1 0 0
process 0 0 1 0 unchanged unchanged unchanged
2’s complement 0 0 1 0 1 0 0

ie.

2’s complement of 1101100 is 0010100

1’s complement

Finding 1’s complement is just to replace all 1’s by 0’s and all 0’s by 1’s

1’s complement of 1101100 is 0010011

in short

2’s complement = 1’s complement + 1

Number Conversion

Number conversion is the fundamental operation of any digital systems.

There are different bases like base2, base8, base10 and base 16

Base – 10 Base 2
(Binary)
Base 8
(Octal)
Base 16
(Hexadecimal)
0 0000 00 0
1 0001 01 1
2 0010 02 2
3 0011 03 3
4 0100 04 4
5 0101 05 5
6 0110 06 6
7 0111 07 7
8 1000 10 8
9 1001 11 9
10 1010 12 A
11 1011 13 B
12 1100 14 C
13 1101 15 D
14 1110 16 E
15 1111 17 F

The above table shows the various base systems.

Converting between one base to another base is of importance.  Usually all the numbering systems of day to day life is done using Base 10 or Decimal system. So it is necessary to convert

  • Other Base to Decimal
  • Decimal to other base

Any number represented as

anan-1an-2an-3…..a1a0 . a-1a-2……..a-n

We will see one by one

Decimal to Binary (Base 10 to Base 2)

(19.456)10 – (?.?)2

In the above, the number 19.456 has to be converted to binary or base2 which will have only 0s and 1s.

First we will take 19

Division Quotient Remainder Remarks
19/2 9 1 a0=1
9/2 4 1 a1=1
4/2 2 0 a2=0
2/2 1 0 a3=0
1/2 0 1 a4=1

so the conversion is 10011

Secondly we will take (0.456)10

Multiplication Whole number decimal Remarks
0.456 * 2 0 0.912 a-1=0
0.912 * 2 1 0.824 a-2=1
0.824 * 2 1 0.648 a-3=1
0.648 * 2 1 0.296 a-4=1

Therefore (0.456)10 is (0.0111)2

Finally (19.456)10 = (10011.0111)2

Binary (Base 2) to Decimal (Base 10) Conversion

(11110.0111)2 = (?.?)10

Let us take the whole portion 11110

1 1 1 1 0 . 0 1 1 1
a4 a3 a2 a1 a0 . a-1 a-2 a-3 a-4
1 * 24 1 * 23 1 * 22 1 * 21 0 * 20 . 0 * 2-1 1 * 2-2 1 * 2-3 1 * 2-4
16 8 4 2 0 . 0.5 0.25 0.125 0.0625

Total is 30.9375

Decimal to Octal (Base 10 to Base 8)

(19.456)10 – (?.?)8

In the above, the number 19.456 has to be converted to binary or base2 which will have only 0s and 1s.

First we will take 19

Division Quotient Remainder Remarks
19/8 2 3 a0=3
2/8 0 2 a1=2

so the conversion is 23

Secondly we will take (0.456)10

Multiplication Whole number decimal Remarks
0.456 * 8 3 0.648 a-1=3
0.648 * 8 5 0.184 a-2=5
0.824 * 8 1 0.472 a-3=1

Therefore (0.456)10 is (0.351)8

Finally (19.456)10 = (23.351)8

Octal (Base 8) to Decimal (Base 10) Conversion

(337.64)8 = (?.?)10

Let us take the whole portion 11110

3 3 7 . 6 4
a2 a1 a0 . a-1 a-2
3 * 82 3 * 81 7 * 80 . 6 * 8-1 4 * 8-2
192 24 7 . 0.75 0.0625

Total is 267.8125

Decimal to Hexadecimal (Base 10 to Base 16)

(19.456)10 – (?.?)16

In the above, the number 19.456 has to be converted to binary or base2 which will have only 0s and 1s.

First we will take 19

Division Quotient Remainder Remarks
19/16 1 3 a0=3
1/16 0 1 a1=1

so the conversion is (13)16

Secondly we will take (0.456)10

Multiplication Whole number decimal Remarks
0.456 * 16 7 0.296 a-1=7
0.296 * 16 4 0.736 a-2=4
0.736 * 16 B 0.776 a-3=B

Therefore (0.456)10 is (0.74B)16

Finally (19.456)10 = (13.74B)16

Octal (Base 8) to Decimal (Base 10) Conversion

(1AB.62)16 = (?.?)10

Let us take the whole portion 11110

1 A B . 6 2
a2 a1 a0 . a-1 a-2
1 * 162 10 * 161 11 * 160 . 6 * 16-1 2 * 16-2
256 160 11 . 0.375 0.007

Total is (427.382)10