HALF ADDER & FULL ADDER EXPERIMENT

Students, you have to write this Expt in your RECORD.

Dont write the Eeperiment Number.

EXPERIMENT: 9 HALF ADDER AND FULL ADDER

AIM: To realize Half Adder and Full Adder using Basic gates

LEARNING OBJECTIVE:

_ To realize the half adder circuits using basic gates

_ To realize full adder using two half adders

COMPONENTS REQUIRED:

IC 7408, IC 7486, IC 7432, Patch Cords & IC Trainer Kit.

THEORY:

Adder circuit is a combinational digital circuit that is used for adding two numbers. A typical adder circuit produces a sum bit (denoted by S) and a carry bit (denoted by C) as the output. Typically adders are realized  for adding binary numbers but they can be also realized  for adding other formats like BCD (binary coded decimal, XS-3 etc. Besides addition, adder circuits can be used for a lot of other applications in digital electronics like address decoding, table index calculation etc. Adder circuits are of two types: Half adder ad Full adder.

Half-Adder: A combinational logic circuit that performs the addition of two data bits, A and

B, is called a half-adder. Addition will result in two output bits; one of which is the sum bit, S, and the other is the carry bit, C. The Boolean functions describing the half-adder are: S =A Å B C = A B

Full-Adder: The half-adder does not take the carry bit from its previous stage into account. This carry bit from its previous stage is called carry-in bit. A combinational logic circuit that adds two data bits, A and B, and a carry-in bit, Cin , is called a full-adder. The Boolean functions describing the full-adder are:

S = (x Å y) Å Cin C = xy + Cin (x Å y)

Half adder circuit diagram:

Full adder circuit diagram:

Procedure:

• Check the components for their working.

• Insert the appropriate IC into the IC base.

• Make connections as shown in the circuit diagram.

• Verify the Truth Table and observe the outputs.

Precautions:

·         All IC’s must be checked before start the experiment.

·         All connections should made and tight.

·         While making connections main voltage should be kept switch off.

·         The circuit should be OFF before change the connection.

RESULT:

Attendance on 22-09-2014 for CSE-A & B

Attendance on 22-09-2014 for CSE-A    DLD Lab

Absentees: 542, 548   (531-560  Batch)

Name of the Experiment: Half Adder & Half subtractor


Attendance on 22-09-2014 for CSE-B

Class

Absentees: 65, 71, 78, 79, 91, A1, B0, B2, B9

Topic: Binary Codes

Attendance on 20-09-2014 for CSE-A & B

Attendance on 20-09-2014 for CSE-A

Absentees : 1, 19, 24, 25, 26, 31, 40, 42, 48, 49




Attendance on 20-09-2014 for CSE-B

Absentees : 79, 91, 95, A8, B4, B8, B9

DLD Lab 1st Experiment : Verification of Logic Gates

Students, you have to write this experiment in your  RECORD.

EXPERIMENT: 1 LOGIC GATES

AIM:- To study and verify the truth table of logic gates

LEARNING OBJECTIVE:-

 Identify various ICs and their specification.

COMPONENTS REQUIRED:-

 Logic gates (IC) trainer kit.

 Connecting patch chords.

 IC 7400, IC 7408, IC 7432, IC 7406, IC 7402, IC 7404, IC 7486

THEORY:

The basic logic gates are the building blocks of more complex logic circuits. These logic gates perform the basic Boolean functions, such as AND, OR, NAND, NOR, Inversion, Exclusive-OR, Exclusive-NOR. Fig. below shows the circuit symbol, Boolean function, and truth. It is seen from the Fig that each gate has one or two binary inputs, A and B, and one binary output, C. The small circle on the output of the circuit symbols designates the logic complement. The AND, OR, NAND, and NOR gates can be extended to have more than two inputs. A gate can be extended to have multiple inputs if the binary operation it represents is commutative and associative. These basic logic gates are implemented as small-scale integrated circuits (SSICs) or as part of more complex medium scale (MSI) or very large-scale (VLSI) integrated circuits. Digital IC gates are classified not only by their logic operation, but also the specific logic-circuit family to which they belong. Each logic family has its own basic electronic circuit upon which more complex digital circuits and functions are developed. The following logic families are the most frequently used.

TTL _Transistor-transistor logic

ECL _Emitter-coupled logic

MOS _Metal-oxide semiconductor

CMOS _Complementary metal-oxide semiconductor

TTL and ECL are based upon bipolar transistors. TTL has a well established popularity among logic families. ECL is used only in systems requiring high-speed operation. MOS and CMOS, are based on field effect transistors. They are widely used in large scale integrated circuits because of their high component density and relatively low power consumption. CMOS logic consumes far less power than MOS logic. There are various commercial integrated circuit chips available. TTL ICs are usually distinguished by numerical designation as the 5400 and 7400 series.

AND gate

                       

The AND gate is an electronic circuit that gives a high output (1) only if all its inputs are high.  A dot (.) is used to show the AND operation i.e. A.B.  Bear in mind that this dot is sometimes omitted i.e. AB

OR gate

                      

The OR gate is an electronic circuit that gives a high output (1) if one or more of its inputs are high.  A plus (+) is used to show the OR operation.

NOT gate

                       

The NOT gate is an electronic circuit that produces an inverted version of the input at its output.  It is also known as an inverter. If the input variable is A, the inverted output is known as NOT A.  This is also shown as A', or A with a bar over the top, as shown at the outputs. The diagrams below show two ways that the NAND logic gate can be configured to produce a NOT gate. It can also be done using NOR logic gates in the same way.

NAND gate

                      

This is a NOT-AND gate which is equal to an AND gate followed by a NOT gate.  The outputs of all NAND gates are high if any of the inputs are low. The symbol is an AND gate with a small circle on the output. The small circle represents inversion.

NOR gate

                                        

This is a NOT-OR gate which is equal to an OR gate followed by a NOT gate.  The outputs of all NOR gates are low if any of the inputs are high.

The symbol is an OR gate with a small circle on the output. The small circle represents inversion.

EXOR gate

                                 

The 'Exclusive-OR' gate is a circuit which will give a high output if either, but not both, of its two inputs are high.  An encircled plus sign () is used to show the Ex-OR operation.

EXNOR gate

The 'Exclusive-NOR' gate circuit does the opposite to the Ex-OR gate. It will give a low output if either, but not both, of its two inputs are high. The symbol is an EX-OR gate with a small circle on the output. The small circle represents inversion.

PROCEDURE:

1. Check the components for their working.

2. Insert the appropriate IC into the IC base.

3. Make connections as shown in the circuit diagram.

4. Provide the input data via the input switches and observe the output on output LEDs.

PIN CONFIGURATION OF IC’s :

RESULT: -

Attendance on 19-09-2014 for CSE A & B

Attendance on 19-09-2014 for CSE-A

1st hr  Absentees : 25, 42, 48, 50, 56

2nd hr Absentees : 25, 42, 48, 50, 56



Attendance on 19-09-2014 for CSE-B

1st hr  Absentees :79, 91, A8, B0, B2, B4, B8, B9

2nd hr Absentees : 79, 91, B0, B2, B4, B8, B9

Attendance on 18-09-2014 for CSE - A & B

Attendance on 18-09-2014 for CSE - A  class

Absentees: 2, 3, 9, 11, 13, 15, 16, 21, 25, 27, 33, 34, 40, 42, 43, 44, 47, 48, 50, 52, 56, 57


Attendance on 18-09-2014 for CSE - B  

Class  Absentees :  62, 72, 76, 79, 80, 87, 91, 99, A3, A8, A9, B7


DLD Lab   Absentees:    62, 72, 76, 80, 87        (561-590 batch)


1st batch    - 561 to 564
2nd batch  - 565 to 568
3rd batch  - 569 to 572
4th batch  - 573 to 576
5th batch  - 577 to 581     (579 detained)
6th batch  - 582 to 585
7th batch  - 586 to 590

Expt done: Verification of logic gates.

Attendance on 17-09-2014 (CSE-A & B)

Attendance on 17-09-2014  CSE-A 

DLD lab

Absentees: 1, 13, 15

Expt: Verification of Logic gates

Class:



Attendance on 17-09-2014  CSE-B    class

Absentees: 79, 91, 94

OCTAL ARITHMETIC

OCTAL ARITHMETIC :

(1). Octal Addition:

---> If the sum is less than 8, the addition of two octal digits is same as the equivalent sum.

---> If the sum is greater than or equal to 8, then 8 must be subtracted from the sum to obtain the sum and carry of '1' is produced.

Ex(1):  Add (23)₈ and (67)₈

Sol:       

Column-1: here, 3+7=10,  i.e 10>8, so 8 must be subtracted from 10.

10-8=2 and carry 1 produced.

Column-2: here, 1+2+6=9, i.e 9>8, so 8 must be subtracted from 9

9-8=1 and carry 1 is produced.

Column-3: here, we take the carry also.

Ex(2):  Add (456)₈ and (123)₈

Sol:       

Column-1: here, 6+3=9,  i.e 9>8, so 8 must be subtracted from 9.

9-8=1 and carry 1 produced.

Column-2: here, 1+5+2=8, i.e 8=8, so 8 must be subtracted from 8

8-8=0 and carry 1 is produced.

Column-3: here, 1+4+1=6, i.e 6<8. So, we have to take the same equivalent sum, i.e ‘6’.

(2). Octal Subtraction:

 The subtraction of octal numbers follows the same rules as the subtraction of numbers in any other number system. The only variation is in borrowed number. In the decimal system, you borrow a group of 10₁₀. In the binary system, you borrow a group of 2₁₀. In the octal system you borrow a group of 8₁₀.

Ex (1):

Attendance on 16-09-2014 for CSE-A

Attendance on 16-09-2014  for CSE-A

Absentees: 7, 15, 25, 26, 34, 50, 56

Topic: Octal Arithmetic & Hexadecimal Arithmetic



Attendance on 16-09-2014  for CSE-B

DLD LAB : (591-5C0 & LE-Three members)

Absentees: A7

Expt: Verification of Logic gates

BINARY ARITHMETIC

BINARY ARITHMETIC

Binary Addition:

Binary addition is performed in the same manner as decimal addition. However, since binary system has only two digits, the addition table for binary arithmetic is very simple consisting of only four entries. The complete table for binary addition is as follows:

Rules for binary addition ….

0 + 0 = 0,     0+1 =1,     1+0= 1,    1+1 = 10    (here, 0 is result and 1 is carry)

Ex (1). Add the binary numbers  (11011) and (1101)

Ex (2). Add the binary numbers   10111.101 and 110111.010

Binary Subtraction:

 It consists of two steps, which are repeated for each column of the numbers. The first step is to determine if it is necessary to borrow. If the subtrahend (the lower digit) is larger than the minuend (the upper digit), it is necessary to borrow from the column to the left. It is important to note here that the valued borrowed depends upon the base of the number and is always the decimal equivalent of the base. Thus, in decimal, 10 is borrowed; in binary, 2 is borrowed; in octal, 8 is borrowed; in hexadecimal, 16 is borrowed. The second step is to simply to subtract the lower value from the upper value. The complete table for binary subtraction is as follows:

Rules for binary subtraction ….

0 - 0 = 0,     1-0= 1,    1-1 = 0,   0-1 =1  with a barrow of 1.

Ex (1). Subtract the binary numbers  0101 & 0011

 Sol:

                                                                   barrow

Binary Multiplication:

Multiplication in the binary system also follows the same general rules as decimal multiplication. However, learning the binary multiplication is a trivial task become the table for binary multiplication is very short, with only four entries instead of the 100 necessary for decimal multiplication. The complete table for binary multiplication is as follows :

0 x 0 = 0

0 x 1 = 0

1 x 0 = 0

1 x 1 = 1

Ex (1): Multiply the binary numbers 1010 and 1001

       

        1010

     x 1001

        1010

      0000

    0000 

  1010     

1011010

Courtesy : http://bscshortnote.blogspot.in/2013/05/binary-addition-subtraction.html

Attendance on 15-09-2014 (CSE-A & B)

Attendance on 15-09-2014 CSE-B

DLD LAB  (531-560) 

Absentees: 34, 40, 46, 52

Expt. done : Verification of logic gates


Attendance on 15-09-2014 CSE-A

Class

Absentees : 79, 91, 92, B2

Any Base (Radix) & its Characters -Class on 13-09-2014

Any Base (Radix) & its Characters:

Method-1 : Any radix (base) to decimal
Method-2: Decimal to any radix (base)

Attendance on 13-09-2014 (CSE-A & B)

Attendance on 13-09-2014 (CSE-A)   -- 2hrs

Absentees: 6,7,8,23,24,25,26,28,37,38,41,42,50,52,58


Attendance on 13-09-2014 (CSE-B)

Absentees: 65,78,79,91,92,98,A0,A4,B4,B5,B9,C0

Decimal to any base conversion & Hexa to any base conversion (Class on 11-09-14 for CSE-A &B)

Decimal to any base (Radix) conversion:

1.      Decimal to binary number conversion                         * * * * * * * * * * * * * * * * * *

Take the remainders from bottom to top,

The result of 57₁₀ is =(111001)₂

Ex:2 Convert 0.75₁₀to binary

            Multiply the given fraction by 2.

Keep the integer in the product as it is and multiply the new fraction in the product by 2. Continue this process & read the integers in the products from the top to bottom.

             

                 Given fraction                    0.75

            Multiply 0.75 by 2                  1.50

            Multiply 0.50 by 2                  1.00

Reducing the integers from top to bottom (0.75)₁₀=(0.11)₂

Ex (3): Convert (105.15)₁₀ to binary.

Sol:      Conversion of integer 105

Remainder from bottom to Top = 1101001   

This particular fraction can never be expressed exactly in binary. This process may be terminated after few steps.

                        Therefore (105.15)₁₀= (1101001.001001)₂

Ex (4): Convert (13)₁₀ to an equivalent base-2 number.

Ex (5): Convert (0.65625)₁₀ to an equivalent base-2 number.

1.      Decimal to Octal number conversion                         * * * * * * * * * * * * * * * * * * *

  Ex (1): (574)₁₀  

                                            Thus  (574)₁₀  = (1076)₈

Ex (2): (247)₁₀  

Ex (3): Convert (0.6875)₁₀ into Octal number.

Ex (4): Convert (3287.5100098)₁₀  into Octal number.

Fractional Part 0.5100098

1.      Decimal to Hexadecimal number conversion                         * * * * * * * * * * * * * * *

Ex (1):    (95.5)₁₀

Ex (2):    (675.625)₁₀

Hexa decimal to any base (Radix) conversion:

1.      Hexadecimal to binary number conversion                                * * * * * * * * * * * * *

1.      Hexadecimal to Octal number conversion                                * * * * * * * * * * * * *

Ex 1:convert (A72E)₁₆ to octal number

                    A      7       2       E

   

     1010  0111 0010 1110

  = 001 010 011 100 101 110

        1     2     3    4     5    6

   = (123456)₈

Ex 2: Convert (0.BF85)₁₆ to octal number

            0.1011111110000101

           

            0.101|111|111|000|010|100

           

           0.   5    7     7    0      2    4

Thus (0.BF85)₁₆=(0.577024)₈

Ex 3: Convert B9F.AE₁₆ to octal

Given hex number is              B        9        F       .    A           E

Convert hex digit to binary  1011  1001   1111   .  1010     1110

Groups of three bits are                   101 110 011 111. 101 011 100

Convert each 3 bit group to octal      5     6      3     7   .   5      3      4

           

The result is (5637.534)₈

Ex 4: (2AB)₁₆ -------(  )₈

             

Ex 5: (42FD)₁₆--------(   )₈

                

          =   4        2         F        D

            0100   0010  1111  1101

            

          = 000|100|001|011|111|101

            

              0      4     1    3      7    5

Thus (42FD)₁₆=(41375)₈

2.      Hexadecimal to decimal number conversion                                * * * * * * * * * * * * *

Ex 1: convert 5C7₁₆ to decimal

            =(5x16²)+(12x16¹)+(7x16⁰)

            =1280+192+7             

            =(1479)₁₀

Ex 2: convert A0FP.0EB₁₆ to decimal

            =(10x16³)+(0+16²)+(15x16¹)+(9x16⁰).(0x16^-1)+(14x16^-2)+(11x16^-3)

            =(40960+0+240+9)+(0+0.0546+0.0026)

            =(41209.0572)₁₀

Ex 3: convert AB6 to decimal

            =10x16²+11x16¹+6x16⁰

             =(2742)₁₀

Ex 4: convert 2EB7 to decimal

            =2x16³+14x16²+11x16¹+7x16⁰            

            =(11959)₁₀

Ex 5: convert A08F.EA to decimal

            =(10x16³+0x16²+8x16¹+15x16^{0 )} + (14x16^-1+11x16^-2)

            =(41103.0.917)₁₀

Ex 6: (8E47.AB)₁₆------(   )₁₀

            =(8x16³+14x16²+4x16¹+7x16^{0 )} + (10x16^-1+11x16^-2)

            =(36423.667)₁₀

Ex 7: obtain decimal equivalent of hexadecimal(3A.2F)₁₆

            =(3x16¹+10x16⁰⁾  + (2x16^-1+15x16^-2)

            =48+10+(2/16)+(15/16²)

            =(58.1836)₁₀