Number System & Logic Gates

8/14/2019 Number System & Logic Gates

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Departments of Electronics Engg. 2009-2010

__________________________________________________________________________________Prepared by:Mr. A.B. Shinde -1-

Introduction to Number Systems:

There are many ways in which you could represent a numeric value. Each convention for

representing numeric values is called a Number System.The most commonly used system in for day to day applications is the Decimal System. Here,

we will take the overview of four different number systems: decimal, binary, octal, and hexadecimal.

Table 1 illustrates the correspondence in representing numbers between 0 and 16 in each of the

systems.

Decimal

(Base 10)

Binary

(Base 2)

Octal

(Base 8)

Hexadecimal

(Base 16)

0 0000 0 0

1 0001 1 1

2 0010 2 23 0011 3 3

4 0100 4 4

5 0101 5 5

6 0110 6 6

7 0111 7 7

8 1000 10 8

9 1001 11 9

10 1010 12 A

11 1011 13 B12 1100 14 C

13 1101 15 D

14 1110 16 E

15 1111 17 F

16 10000 20 10

Table 1: Number System Correspondence Table

Decimal, Binary, Octal, and Hexadecimal Numbers

The decimal number system uses 10 different digits (0 - 9) to represent numbers. It is alsocalled base 10 for the same reason. In general we call a number system by the number of digits it

uses.

For example:

1) the binary system uses two digits (0 and 1) base 2,

2) octal uses eight digits (0 7) base 8,

3) and hexadecimal uses 16 digits (0 - 9 A B C D E F) base 16.


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Prepared by:Mr. A.B. Shinde -2-

Position #

BasePosition # = Position Value

Decimal

Hexadecimal

3 2 1 0

103 =1000 102 =100 101 =10 100 =1

163 =4096 162 = 256 161 =16 160 =1

In representing numeric values in any number system, the position and base terms are key.

Figure 1 illustrates the definition of position and base.

Fig:1Understanding bases, position numbers, and position values.As shown in Figure 1, we could think of each digit placeholder as a position starting with

position 0 from the right. Each position has a certain multiplier value associated with. This value is

given by the position value formula shown in the figure. How does relate to writing numerical values?

When you write a number each digit will be multiplied by the position value, and the resulting values

are then added to give the total value of the number.

For example:

let's suppose we have three numbers, 3, 30, and 300 in the decimal number system. Applying

the above methodology, we get:

3 = (3 * 100)

30 = (3 * 101) + (0 * 10

0)

300 = (3 * 102) + (0 * 10

1)+ (0 * 10

0)

let's suppose we have another four numbers, 4, 44, 444 and 4444 in the decimal number

system. Applying the above methodology, we get:

4 = (4 * 100)

44 = (4 * 101) + (4 * 10

0)

444 = (4 * 102) + (4 * 10

1)+ (4 * 10

0)

4444 = (4 * 103) + (4 * 10

2) + (4 * 10

1)+ (4 * 10

0)

This is simple enough to understand with the decimal system. The position number we place a digit

in determines what value that digit contributes to the whole number's value.

Conversion from One Number System to Other:


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1) DECIMAL TO BINARY CONVERSION:For example, suppose we want to convert 30 from decimal to binary. We know that the

binary systems base is 2, therefore the divisor should be 2. We divide the initial number by 2, keepthe result for the next division unless it is less than 2, which signifies the end of the iteration, and

note the remainder.

2 30 0

2 15 1 (30)10 = (1 1 1 1 0)2

2 7 1

2 3 1

1 1

Remainders

e.g.2: Convert 14 into binary

2 14 0

2 7 1 (14)10 = (1 1 1 0)2

2 3 1

1 1

Remainders

For example: suppose the given the number is (18.625)10

2 18 0

2 9 1

2 4 0 (18)10 = (1 0 0 1 0)2

2 2 0

1 1

And

0.625 x 2 = 1.25 1

0.25 x 2 = 0.50 0 (0.625)10 = (0.101)2

0.50 x 2 = 1.00 1This method is called as double dable method

(18.625)10 = (1 0 0 1 0 . 1 0 1)2

Problems: Do the following Decimal to Binary conversions

a) 24.365 b) 36.785 c) 55.951 d) 40.240 e) 19.852

f) g) h) i) j)


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2) BINARY TO DECIMAL CONVERSION:In converting any binary number to decimal we use the base and position concept explained

above. We simply multiply the value of the digit by the position value in the source number system

and add all obtained values.

For example, to convert 11110 from binary to decimal we perform the following.

(11110)2 = (1 * 24)+ (1 * 2

3) + (1 * 2

2) + (1 * 2

1)+ (0 * 2

0)

= (1*16) + (1*8) + (1*4) + (1*2) + (0*1)

= 16 + 8 + 4 + 2 + 0

= (30)10

To convert from binary to decimal, we use the same concept as above.

(1011)2 = (1 * 23) + (0 * 2

2) + (1 * 2

1) + (1 * 2

0)

= (1 * 8) + (0 * 4) + (1 * 2) + (1*1)

= 8 + 0 + 2 + 1

= (11)10

But if suppose the given number is (11.101)2(11.101)2 = [(1 * 2

1) + (1 * 2

0)] . [ (1 * 2

-1) + (0 * 2

-2) + (1 * 2

-3)

= [(1 * 2) + (1 * 1)] . [(1 * 0.5) + (0 * 0.25) + (1 * 0.125)]

= [2 + 1] . [0.5 + 0 + 0.125]

= (3.625)10

Problems for Practice:

Convert the following numbers into Decimal

a) (1101.01)2

b)

(110001)2c) (001.101)2

d) (101.1101)2

e) (1111.1110)2

f) (1010.0001)2

g) (1110. 001)2

h) (10111.0101)2

i) (110010.1101)2


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3) DECIMAL TO OCTAL CONVERSION:For example, suppose we want to convert 35 from decimal to octal. We know that for the

octal system is base 8, therefore the divisor should be 8. We divide the initial number by 8, keep theresult for the next division unless it is less than 8, which signifies the end of the iteration, and note

the remainder.

8 35 3

4 4 (35)10 = (43)8

Remainders

Convert 78 to octal

8 78 6

8 9 1 (78)10 = (116)8

1 1

Remainders

The given number is (107.24)10 Convert it to Octal

8 107 3

8 13 51 1 (107)10 = (153)8

And

0.24 x 8 = 1.92 1

0.92 x 8 = 7.36 7 (0.24)10 = (0.17270)8

0.36 x 8 = 2.88 2

0.88 x 8 = 7.04 7

0.04 x 8 = 0.32 0

Therefore (107.24)10 = (153.17270)8

Problems: Do the following Decimal to Octal conversions

a) 215.365

b) 127.321

c) 92.35

d) 27.36

e) 15.15 etc


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4) OCTAL TO DECIMAL CONVERSION:In converting from Octal number system to decimal we use the base and position concept

explained above. We simply multiply the value of the digit by the position value in the source number

system and add all obtained values.

For example, to convert 721 from octal to decimal, we perform the following.

(721)8 = (7 * 82)+ (2 * 8

1) + (1 * 8

0)

= (7*64) + (2*8) + (1*0)

= 448 + 16 + 1

= (465)10

To convert from octal to decimal, we use the same concept as above.

(223)8 = (2 * 82) + (2 * 8

1) + (3 * 8

0)

= (2 * 64) + (2 * 8) + (3 * 1)

= 128 + 16 + 3

= (147)10

But if suppose the given number is (42.36)8(42.36)8 = [(4 * 8

1) + (2 * 8

0)] . [ (3 * 8

-1) + (6 * 8

-2)]

= [(4 * 8) + (2 * 1)] . [(3 * 0.125) + (6 * 0.0156)]

= [32 + 2] . [0.375 + 0.0936]

= (34.4686)10


Convert the following numbers from Octal to Decimal

a) (110.41)8

b)

(234)8c) (654.321)8

d) (12.4563)8

e) (772.114)8

f) (156.56)8

g) (378.26)8

h) (331.454)8

i) (254.247)8


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5) DECIMAL TO HEXADECIMAL CONVERSION:For example, suppose we want to convert 50 from decimal to hexadecimal. We know that for

the hexadecimal system base is 16, therefore the divisor should be 16. We divide the initial numberby 16, keep the result for the next division unless it is less than 16, which signifies the end of the

iteration, and note the remainder.

16 50 2

3 3 (50)10 = (23)16

Remainders

Convert 100 to hexadecimal

16 100 4

6 6 (100)10 = (64)16

Remainders

The given number is (260.24)10 Convert it to hexadecimal

16 260 4

16 16 01 1 (260)10 = (104)16

And

0.24 x 16 = 3.84 3

0.84 x 16 = 13.44 D (0.24)10 = (0.3D70)16

0.44 x 16 = 7.04 7

0.04 x 16 = 0.64 0

Therefore (260.24)10 = (104.3D70)16

Problems: Do the following Decimal to Hexadecimal conversions

a) 415.325

b) 307.348

c) 568.315

d) 287.046

e) 115.115

f) 24.15 etc


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6) HEXADECIMAL TO DECIMAL CONVERSION:To convert from hexadecimal to decimal we use the same concept with one minor difference.

In converting from Hexadecimal number system to decimal we use the base and position concept.We simply multiply the value of the digit by the position value in the source number system and add

all obtained values. Whenever we encounter a letter between A and F, the decimal value of the letter

is substituted it in the above formula. Since there are only 6 letters you need to know, it helps to

memorize the values.

A = 10, B = 11, etc. as shown in Table 1.

For example, to convert 3C from hexadecimal to decimal we do the following.

(3C)16 = (3 * 161)+ (12 * 16

0) Note that we converted C to 12 before plugging it into the formula

= (3*16) + (12*1)

= 48 + 12

= (60)10

Convert 104 from hexadecimal to decimal:

(104)16 = (1 * 162)+ (0 * 16

1) + (4 * 16

0)

= (1*256) + (0*16) + (4*1)

= 256 + 0 + 4

= (260)10

To convert from hexadecimal to decimal, we use the same concept as above.

(3A0)16 = (3 * 162) + (10 * 16

1) + (0 * 16

0)

= (3 * 256) + (10 * 16) + (0 * 1)

= 768 + 160 + 0

= (928)10

But if suppose the given number is (9D.2EF)16

(9D.2EF)16 = [(9 * 161) + (13 * 16

0)] . [ (2 * 16

-1) + (14 * 16

-2) + (15 * 16

-3)

= [(9 * 16) + (13 * 1)] . [(2 * 0.0625) + (14 * 0.00391) + (15 * 0.000244)]= [144 + 13] . [0.125 + 0.05474 + 0.00366]

= (157.1834)10


Convert the following hexadecimal numbers to Decimal

a) (C4.52)16 d) (5B.6E)16

b) (247.AC)16 e) (16E.230)16

c) (F5.012)16 f) (5D.ABC)16


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7) OCTAL TO HEXADECIMAL CONVERSION:For converting any given number in Octal number system to the Hexadecimal number system,

we first need to convert the given octal number into its equivalent binary number. Then from thederived binary number prepare the groups of 4 digits (bits) from right to left, because the base of

Hexadecimal number system is 16 i.e. 24.

For Example:

Convert (457)8 to hexadecimal:

First we need to convert the given octal number to binary.

Therefore, 4 5 7

1 0 0 1 0 1 1 1 1

i.e. (457)8 => (100101111)2

Now from the derived binary number form the groups of 4 numbers each fro m right to left.

i.e. 0 0 0 1 0 0 1 0 1 1 1 1 (457)8 => (12F)16

1 2 F

These are the extra added 0s for completing the group of 4 digits (bits)

Convert (4236)8 to hexadecimal

(4 2 3 6)8 => (100 010 011 110)2

1 0 0 0 1 0 0 1 1 1 1 0

8 9 E (4236)8 => (89E)16

Problems: Do the following octal to Hexadecimal conversions

a) 6355

b) 30441c) 5710.152

d) 237.046

e) 416.411

f) 24.57

g) 124.632

h) 447.441

i) 567.765


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8) HEXADECIMAL TO OCTAL CONVERSION:For converting any given number from Hexadecimal number system to Octal the number

system, we first need to convert the given Hexadecimal number into its equivalent binary number.Then from the derived binary number prepare the groups of 3 digits (bits) from right to left, because

the base of Hexadecimal number system is 8 i.e. 23.

For Example:

Convert (489A)16 to octal:

First we need to convert the given Hexadecimal number to binary.

Therefore, 4 8 9 A

0 1 0 0 1 0 0 0 1 0 0 1 1 0 1 0

i.e. (489A)16 => (0100100010011010)2

Now from the derived binary number form the groups of 3 numbers each from right to left.

i.e. 0 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 1 0

(489A)16 => (044232)8

0 4 4 2 3 2

These are the extra added 0s for completing the group of 3 digits (bits)

Convert (20)16 to octal

(2 0)8 => (0010 0000)2

0 0 0 1 0 0 0 0 0

0 4 0 (20)16 => (40)8

Problems: Do the following Hexadecimal to octal conversions

a) F5

b) 30C1c) 102.AB

d) A9C.6E2

e) 958D.223

f) CD.10F

g) 68EF.9

h) 4F7.4E1


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Digital Logic Gates:

Standard commercially available Digital Logic Gates are available in two basic forms, TTL

which stands for Transistor-Transistor Logic such as the 7400 series, and CMOS which stands for

Complementary Metal-Oxide-Silicon which is the 4000 series of chips.

Generally speaking, this refers to the logic technology used to manufacture the Integrated

Circuit, (IC) or "chip" as it is commonly called. TTL IC's use NPN type Bipolar Junction Transistors

while CMOS IC's use Field Effect Transistors or FET's for both their input and output circuitry. By

using TTL and CMOS technology, simple digital logic gates can also be made by connecting together

diodes and resistors to produce RTL, Resistor-Transistor Logic circuits but these are now less

common.

Integrated Circuits or IC's as they are more commonly called, can be grouped together into

families according to the number of transistors or "gates" that they contain. For example, a simple

AND gate my contain only a few individual transistors, where as a more complex microprocessor may

contain many thousands of individual transistor gates. A general classification for the number of

gates within a single chip is given as:

Classification of Integrated Circuits

Small Scale Integration or SSI - up to 10 transistors or gates

Medium Scale Integration or MSI - in between 10 and 100 transistors or gates

Large Scale Integration or LSI - in between 100 and 1,000 transistors or gates

Very-Large Scale Integration or VLSI - between 1,000 and 10,000 transistors or gates

Super-Large Scale Integration or SLSI - between 10,000 and 100,000 transistors or gates

Ultra-Large Scale Integration or ULSI - more than 1 million transistors or gates

While the ultra large scale ULSI classification is less well used, another level of integration

which represents the complexity of the Integrated Circuit is known as the System-on-Chip or (SOC).Here individual components such as the microcontroller, memory, peripherals, I/O logic etc, are all

produced on a single piece of silicon and which represents a whole electronic system within one

single chip. These chips are generally used in Mobile Phones, Digital Cameras, Microcontrollers and

robotic applications etc, and can contain up to 100 million individual Silicon-CMOS transistor gates

within a single chip.


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The Digital Logic "AND" Gate

2-input AND Gate

Symbol Truth Table

B A Q

0 0 0

0 1 0

1 0 0

1 1 1

Boolean Expression Q = A.B Read as A AND B gives Q

3-input AND Gate

Symbol Truth Table

C B A Q

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 0

1 0 0 0

1 0 1 0

1 1 0 0

1 1 1 1

Boolean Expression Q = A.B.C Read as A AND B AND C gives Q

The logic AND function can have any number of individual inputs. However, commercial

available AND Gate IC's are available in standard 2, 3, or 4 input types. If additional inputs are

required, then the standard AND gates will need to be cascaded together for example.

The Boolean Expression for this 6-input AND gate will therefore be: Q = (A.B).(C.D).(E.F)


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Transistor AND Gate

A simple 2-input logic AND gate can be constructed using transistor switches connected together as

shown below with the inputs connected directly to the transistor bases.

The Digital Logic "OR" Gate

2-input OR Gate

Symbol Truth Table

B A Q

0 0 0

0 1 11 0 1

1 1 1

Boolean Expression Q = A+B Read as A OR B gives Q

3-input OR Gate

Symbol Truth Table

C B A Q

0 0 0 0

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1

Boolean Expression Q = A+B+C Read as A OR B OR C gives Q


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The OR function can have any number of individual inputs. However, commercial available OR gates

are available in 2, 3, or 4 inputs types. Additional inputs will require gates to be cascaded together for

example.

The Boolean Expression for this 6-input OR gate will therefore be: Q = (A+B)+(C+D)+(E+F)

Transistor OR Gate

A simple 2-input logic OR gate can be constructed using transistor switches connected together as

shown below with the inputs connected directly to the transistor bases.

The Digital Inverter or NOT gate

Symbol Truth Table

A Q

0 1

1 0

Boolean Expression Q = not A or A Read as inverse ofA gives Q

Logic NOT gates provide the complement of their input signal and are so called because when

their input signal is "HIGH" their output state will NOT be "HIGH". Likewise, when their input signal is

"LOW" their output state will NOT be "LOW". As they are single input devices, logic NOT gates are

not normally classed as "decision" making devices or even as a gate, such as the AND or OR gates


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Then we can define the operation of a 2-input logic NAND gate as being:

"If both A and B are true, then Q is NOT true"

The Digital Logic "NAND" Gate

2-input NAND Gate

Symbol Truth Table

B A Q

0 0 1

0 1 1

1 0 1

1 1 0

Boolean Expression Q = A.B Read as A AND B gives NOT Q

3-input NAND Gate

Symbol Truth Table

C B A Q

0 0 0 1

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 0

Boolean Expression Q = A.B.C Read as A AND B AND C gives NOT Q

As with the AND function seen previously, the NAND function can also have any number of

individual inputs and commercial available NAND Gate i.c.s are available in standard 2, 3, or 4 input

types. If additional inputs are required, then the standard NAND gates can be cascaded together to

provide more inputs for example.

The Boolean Expression for this 4-input logic NAND gate will therefore be: Q = A.B.C.D


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The Logic "NOR" Gate

Definition

The Logic NOR Gate or Inclusive-NOR gate is a combination of the digital logic OR gate with

that of an inverter or NOT gate connected together in series. The NOR (Not - OR) gate has an output

that is normally at logic level "1" and only goes "LOW" to logic level "0" when ANY of its inputs are at

logic level "1". The Logic NOR Gate is the reverse or "Complementary" form of the OR gate we have

seen previously.

NOR Gate Equivalent

The logic or Boolean expression given for a logic NOR gate is that for Logical Multiplication

which it performs on the complements of the inputs. The Boolean expression for a logic NOR gate is

denoted by a plus sign, (+) with a line or Overline, () over the expression to signify the NOT or

logical negation of the NOR gate giving us the Boolean expression of: A+B = Q.

Then we can define the operation of a 2-input logic NOR gate as being:

"If both A and B are NOT true, then Q is true"

The Digital Logic "NOR" Gate

2-input NOR Gate

Symbol Truth Table

B A Q

0 0 1

0 1 0

1 0 0

1 1 0

Boolean Expression Q = A+B Read as A OR B gives NOT Q


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3-input NOR Gate

Symbol Truth Table

C B A Q

0 0 0 1

0 0 1 0

0 1 0 0

0 1 1 0

1 0 0 0

1 0 1 0

1 1 0 0

1 1 1 0

Boolean Expression Q = A+B+C Read as A OR B OR C gives NOT Q

As with the OR function, the NOR function can also have any number of individual inputs and

commercial available NOR Gate IC's are available in standard 2, 3, or 4 input types. If additional

inputs are required, then the standard NOR gates can be cascaded together to provide more inputs

for example.

The Boolean Expression for this 4-input NOR gate will therefore be: Q = A+B+C+D

The Exclusive-OR Gate

Definition

Previously, we have seen that for a 2-input OR gate, if A = "1", OR B = "1", OR BOTH A + B ="1" then the output from the gate is also at logic level "1" and this is known as an Inclusive-OR

function because it includes the case of Q = "1" when both A and B = "1". If however, an output "1" is

obtained ONLY when A = "1" or when B = "1" but NOT both together at the same time, then this type

of gate is known as an Exclusive-OR function or an Ex-Or function for short because it excludes the

"OR BOTH" case of Q = "1" when both A and B = "1".

In other words the output of an Exclusive-OR gate ONLY goes "HIGH" when its two input

terminals are at "DIFFERENT" logic levels with respect to each other and they can both be at logic

level "1" or both at logic level "0" giving us the Boolean expression of: Q = (A B) = A.B + A.B


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The Exclusive-OR Gate function is achieved is achieved by combining standard gates together to

form more complex gate functions. An example of a 2-input Exclusive-OR gate is given below.

The Digital Logic "Ex-OR" Gate

2-input Ex-OR Gate

Symbol Truth Table

B A Q

0 0 0

0 1 1

1 0 1

1 1 0

Boolean Expression Q = A B Read as A OR B but NOT BOTH gives Q

Then, the logic function implemented by a 2-input Ex-OR is given as "either A OR B but NOT

both" will give an output at Q. In general, an Ex-OR gate will give an output value of logic "1" ONLY

when there are an ODD number of 1's on the inputs to the gate and this description can be expanded

to apply to any number of individual inputs as shown below for a 3-input Ex-OR gate.

3-input Ex-OR Gate

Symbol Truth Table

C B A Q

0 0 0 0

0 0 1 1

0 1 0 1

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 01 1 1 1

Boolean Expression Q = A B C Read as "any ODD number of Inputs" gives Q

The symbol used to denote and Inclusive-OR function is slightly different to that for the

standard inclusive OR gate and is given a plus sign within a circle which is the mathematical "direct

sum" symbol, and this is given as.


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We said previously that the Ex-OR function is a combination of different basic logic gates and

using the 2-input truth table above, we can expand the Ex-OR function to: Q = (A B) = (A+B).(A.B)

which means we can realise this new expression using the following individual gates.

The Exclusive-NOR Gate

Definition

The Exclusive-NOR Gate function or Ex-NOR for short, is a digital logic gate that is the reverse

or complementary form of the Exclusive-OR function we look at in the previous section. It is a

combination of the Exclusive-OR gate and the NOT gate but has a truth table similar to the standard

NOR gate in that it has an output that is normally at logic level "1" and goes "LOW" to logic level "0"when ANY of its inputs are at logic level "1". However, an output "1" is also obtained if BOTH of its

inputs are at logic level "1". For example, A = "1" and B = "1" at the same time giving us the Boolean

expression of: Q = (A B) = A.B + A.B

In other words, the output of an Exclusive-NOR gate ONLY goes "HIGH" when its two input

terminals, A and B are at the "SAME" logic level which can be either at a logic level "1" or at a logic

level "0". Then this type of gate gives and output "1" when its inputs are " logically equal" or

"equivalent" to each other, which is why an Exclusive-NOR gate is sometimes called an Equivalence

Gate. The logic symbol for an Exclusive-NOR gate is simply an Exclusive-OR gate with a circle or

"inversion bubble" at its output to represent the NOT function.

The Exclusive-NOR Gate function is achieved by combining standard gates together to form

more complex gate functions and an example of a 2-input Exclusive-NOR gate is given below.

The Digital Logic "Ex-NOR" Gate

2-input Ex-NOR Gate

Symbol Truth Table

B A Q

0 0 1

0 1 0

1 0 0

1 1 1

Boolean Expression Q = A B Read if A AND B the SAME gives Q

Then, the logic function implemented by a 2-input Ex-NOR gate is given as "when both A AND

B are the SAME" will give an output at Q. In general, an Exclusive-NOR gate will give an output value


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of logic "1" ONLY when there are an EVEN number of 1's on the inputs to the gate (the inverse of the

Ex-OR gate) except when all its inputs are "LOW", and this description can be expanded to apply to

any number of individual inputs as shown below for a 3-input Exclusive-NOR gate.

3-input Ex-NOR Gate

Symbol Truth Table

C B A Q

0 0 0 1

0 0 1 0

0 1 0 0

0 1 1 11 0 0 0

1 0 1 1

1 1 0 1

1 1 1 0

Boolean Expression Q = A B C Read as "any EVEN number of Inputs" gives Q

We said previously that the Ex-NOR function is a combination of different basic logic gates Ex-

OR and a NOT gate, and by using the 2-input truth table above, we can expand the Ex-NOR function

to: Q = A B = (A.B) + (A.B) which means we can realise this new expression using the followingindividual gates.

Ex-NOR Gate Equivalent Circuit

One of the main disadvantages of implementing the Ex-NOR function above is that it contains

three different types logic gates the AND, NOT and finally an OR gate within its basic design. One

easier way of producing the Ex-NOR function from a single gate type is to use NAND gates as shown

below.


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The "Universal" NAND Gate:

The Logic NAND Gate is generally classed as a "Universal" gate because it is one of the mostcommonly used logic gate types. NAND gates can also be used to produce any other type of logic

gate function, and in practice the NAND gate forms the basis of most practical logic circuits. By

connecting them together in various combinations the three basic gate types of AND, OR and NOT

function can be formed using only NAND's, for example.

Various Logic Gates using only NAND Gates

As well as the three common types above, Ex-Or, Ex-Nor and standard NOR gates can be

formed using just individual NAND gates.

Ex-OR Function Realisation using NAND gates

Exclusive-OR Gates are used mainly to build circuits that perform arithmetic operations and

calculations especially Adders and Half-Adders as they can provide a "carry-bit" function or as a

controlled inverter, where one input passes the binary data and the other input is supplied with a

control signal.


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Ex-NOR Function Realisation using NAND gates

The "Universal" NOR Gate:

Like the NAND gate seen in the last section, the NOR gate can also be classed as a "Universal"

type gate. NOR gates can be used to produce any other type of logic gate function just like the NAND

gate and by connecting them together in various combinations the three basic gate types of AND, OR

and NOT function can be formed using only NOR's, for example.

Various Logic Gates using only NOR Gates

As well as the three common types above, Ex-Or, Ex-Nor and standard NOR gates can also be

formed using just individual NOR gates.


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Digital Logic Gates Summary

We have seen that there are 3 main basic types of digital logic gates, the AND gate, the OR

gate and the NOT gate. We also saw that each gate has an opposite or complementary form of itselfin the form of the NAND gate, the NOR gate and the Buffer respectively, and that any of these

individual gates can be connected together to form more complex Combinational Logic circuits.

We also saw that both the NAND gate and the NOR gate can both be classed as "Universal"

gates as they can be used to construct any other gate type. In fact, any combinational circuit can be

constructed using only 2 or 3-input NAND or NOR gates. We also saw that NOT gates and Buffers are

single input devices that can also have a 3-state High-impedance output which can be used to control

the flow of data onto a common Data Bus wire.

Logic Gates can be made from discrete components such as Resistors, Transistors and Diodes

to form RTL (resistor-transistor logic) or DTL (diode-transistor logic) circuits, but today's moderndigital 74xxx series integrated circuits are manufactured using TTL (transistor-transistor logic) based

on NPN bipolar transistors or the faster CMOS MOSFET transistor logic used in 74Cxx and 4000 series

logic chips.

The 8 individual "standard" Digital Logic Gates are summarized below along with their corresponding

truth tables.

Summary of all the Digital Logic Gates:

Inputs Truth Table Outputs for 2-input Logic Gates

B A AND NAND OR NOR EX-OR EX-NOR

0 0 0 1 0 1 0 1

0 1 0 1 1 0 1 0

1 0 0 1 1 0 1 0

1 1 1 0 1 0 0 1

Truth Table Output for Single-input GatesA NOT Buffer

0 1 0

1 0 1

Number System & Logic Gates

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