1. ATS’s COLLEGE OF ENGINEERING, MIRAJ Subject:- Basic Electronics Lecturer:- Amit Totade

2. CONTENT GRADATION OF CHAPTERS CHAPTER – 5 Number System and Logic gates Bit Description:- Bit 1 – Number System Bit 2 – Logic Gates

3. Number System: The system in which we are able to count the Quantity is called Number System. We always used different number for counting the entity. But that number system is divided into two types –  Digital Number System,  Analog Number System.

4. Digital Number System :- It is a system which involves a discrete unit.i.e ( accurate units). this number system is used in digital electronics.

5. In digital electronics, there are four number systems, they are – Decimal Number System (X) 10 Binary Number System (X) 2 Hex – Decimal Number System (X) 16 Octal Number System (X) 8

6. Radix Every Number System is having different radix or base. Radix: It is nothing but the number of digits used in number system.

7. Decimal Number System (X) 10 It is very old number system. In this number system we used number from 0,1,2,3,4,5,6,7,8,9.so the radix is 10. For finding the value, W= ∑ D k * 10 k Where, D k =>digit at K th Position having weight 10 ^k

8. Decimal Number System (X) 10 Example: (752.12) 10 = 7 * 10^2 +2 * 10^1 +5 * 10^0 +1 * 10^-1+ 2 * 10^-2 D k =>digit at K th Position having weight 10 ^k

9. Binary Number System (X) 2 In this number system we used number from 0,1 only, so the radix is 2. For finding the value, W= ∑ D k * 2 k Where, D k =>digit at K th Position having weight 2 ^k

10. Binary Number System (X) 2 Example: (110.01) 2 = 1 * 2^2 +0 * 2^1 +0 * 2^0 +0 * 2^-1+ 1 * 2^-2 D k =>digit at K th Position having weight 2 ^k

11. Hex-Decimal Number System (X) 16 In this number system we used number from 0 to 9 and Alpha bate from A to F, so the radix is 16. For finding the value, W= ∑ D k * 16 k Where, D k =>digit at K th Position having weight 16 ^k

12. Hex-Decimal Number System (X) 16 Example: (1A8.02) 2 = 1 * 16^2 +A* 16^1 +8* 16^0 +0 * 16^-1+ 2 * 16^-2 D k =>digit at K th Position having weight 16 ^k

13. DIGITValue 00 11 22 33 44 55 66 77 88 99 BINARY 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001

14. DIGITValue A10 B11 C12 D13 E14 F15 BINARY 1010 1011 1100 1101 1110 1111

15. Octal Number System (X) 8 In this number system we used number from 0 to 7, so the radix is 8. For finding the value, W= ∑ D k * 8 k Where, D k =>digit at K th Position having weight 8 ^k

16. Octal Number System (X) 16 Example: (156.02) 8 = 1 * 8^2 +5* 8^1 +6* 8^0 +0 * 8^-1+ 2 * 8^-2 D k =>digit at K th Position having weight 8^k

17. Solve the following problem: (562.369) 10 =? (1001.01) 2 =? (56C.A69) 16 = ? (123.56) 8 =?

18. Solve the following problem: (562.369)10 =5*10^2+6*10^1+2*10^0+3*10^- 2+3*10^-2+3*10^-2 (101.01)2 = (101.01)2 =1 * 2^2 +0* 2^1 +1* 2^0+0 * 2^-1+ 1 *2^-2 (56C.A69)16 = (56C.A69)16 = 5 * 16^2 +6* 16^1 +C* 16^0+ A*16^- 1+ 6*16^-2+ 9 * 16^-3 (123.56)8 = (123.56)8 =1 * 8^2 +2* 8^1 +3* 8^0 +5 * 8^-1+ 6 * 8^-2

19. CONVERSION OF DECIMAL TO BINARY : Go on dividing number by 2 until the quotation in zero and write down the remainder so that the last remainder forms the most significant bit (MSB).

20. Example: Convert (37.15) 10 = (?) 2 2 18 9 4 2 22 2 1 37 2 22 2 2 2 2 22 2 2 22 2 0 00 0 Reminder 1 0 1 0 1 MSB :. (37) 10 =(100101) 2

21. For Conversion Of fraction into Binary Go on multiplying the decimal fraction by 2 and write down all digits on left hand side of decimal of product. take reading up to 3 pints.

22. 0.15*2=0.30 0 0.30*2=0.60 0 0.60*2=1.20 1 :.(0.15) 10 =(001) 2 :.(32.15) 10 =(100101.001) 2

23. CONVERSION OF DECIMAL TO Hex-Decimal : Go on dividing number by 16 until the quotation in zero and write down the remainder so that the last remainder forms the most significant bit (MSB).

24. Example: Convert (532.25) 10 = (?) 16 16 16 33 2 532 16 0 00 0 Reminder 4 1 2 MSB :. (532) 10 =(214) 2

25. For Conversion Of fraction into Hex-Decimal Go on multiplying the decimal fraction by 16 and write down all digits on left hand side of decimal of product. take reading up to 3 pints.

26. 0.25*16=4.16 4 0.416*16=6.65 6 :.(0.25) 10 =(46) 2 :.(32.15) 10 =(214.46) 16

27. CONVERSION OF DECIMAL TO OCTAL : Go on dividing number by 8 until the quotation in zero and write down the remainder so that the last remainder forms the most significant bit (MSB).

28. Example: Convert (2080.55) 10 = (?) 8 8 260 32 4 2080 8 88 8 8 8 0 00 0 Reminder 0 4 0 4 MSB :. (2080) 10 =(4040) 8

29. For Conversion Of fraction into Octal Go on multiplying the decimal fraction by 8 and write down all digits on left hand side of decimal of product. take reading up to 3 pints.

30. 0.55*8=4.40 4 0.40*8=3.20 3 0.20*8=1.60 1 :.(0.55) 10 =(431) 8 :.(2080.55) 10 =(4040.431) 2

31. Conversion of Binary number into Decimal Number Multiply the bit at K th position and sum up all product.The number which we get that will be the decimal number.

32. Example: Convert (1011.11) 2 =(?) 10 ( 1011.11) 2 = 1 * 2^3 + 0 * 2^2 + 1* 2^1 + 1 * 2^0 + 1 * 2^2 (-1) + 1 * 2 ^2 (-2) = 8 + 0 + 2 + 1 + 0.5 + 0.25 = (11.75) 10

33. Conversion of Binary number into Hex-Decimal Number Starting from Binary Number, make group of 4 bits then convert each group of 4 bit into its hex decimal value. To make group you can put zero on right end of fraction part and zero on left end of integer part.

34. Example: Convert (1100101011.01) 2 =(?) 16 0011 0010 1011. 0100 3 2 B. 4 :. (1100101011.01) 2 = (32B.4) 16

35. Conversion of Binary number into Octal number Starting from Binary Number, make group of 3 bits then convert each group of 3 bit into its Octal value. To make group you can put zero on right end of fraction part and zero on left end of integer part.

36. Example: Convert (1100111.11) 2 =(?) 16 001 100 111. 110 1 4 7. 6 :. ( 1100111.11 ) 2 = (147.6) 16

37. Conversion of Hex-Decimal number into Decimal number Multiply the bit at K th position and sum up all product.The number which we get that will be the decimal number.

38. Example: Convert (1011.11) 2 =(?) 10 ( 1D9.48) 2 = 1 * 16^2 + 13* 16^1 + 9 * 16^0 + 4 * 2^16 (-1) +8 * 2 ^16 (-2) = 256+208+9+0.25+0.03125 = (473.28125) 10

39. Conversion of Hex-Decimal number into Binary number Convert each Hex digit into Binary group of 4 bit.

40. Example: Convert (3BC4) 16 =(?) 2 3 B C 4 0011 1011 1100 0100 :.(3BC4) 16 =(0011101111000100) 2

41. Conversion of Hex-Decimal number into Octal number Convert each Hex digit into Binary and then convert binary number into Octal.

42. Example: Convert (1A5.4) 16 =(?) 8 1 A 5. 4 000 1 1010 0101. 010 0 :.(1A5.4) 16 = (645.2) 8 6 4 5. 2

43. Conversion of Octal number into Decimal Number Multiply each digit at K th position and sum up all product.The number which we get that will be the decimal number.

44. Example: Convert (7521.6) 8 =(?) 10 (7521.6) 8 = 7 * 8^3 + 5 * 8^2 + 2* 8^1 + 1 * 8^0 + 6 * 8^(-1) = 3584+320+16+1+0.75 = (3921.75) 10

45. Conversion of Octal number into Binary Number Convert Each octal number into group of 3 binary digits.

46. Example: Convert (123.7) 8 =(?) 2 1 2 3. 7 001 010 011. 111 :.(1A5.4) 8 = (001010011.111) 2

47. Conversion of Octal number into Hex Decimal Number Convert Octal number into binary First and then convert binary into Hex Decimal Number.

48. Example: Convert (225.5) 8 =(?) 16 2 2 5. 5 0 10 0 10 101. 1010 :.(225.5) 8 = (95.A) 16 9 5. A

49. Solve the following problem: (562.369) 10 =( ? ) 2 ( ? ) 8 ( ? ) 16 (1001.01) 2 = ( ? ) 10 ( ? ) 8 ( ? ) 16 (56C.A69) 16 = ( ? ) 2 ( ? ) 8 ( ? ) 10 (123.56) 8 = ( ? ) 2 ( ? ) 10 ( ? ) 16

50. AND gate

51. 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. AND gate

52. OR gate

53. The OR gate is an electronic circuit that gives a high output (1) if one or more of its inputs are high. (2)A plus (+) is used

54. 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.

55. NOT gate 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

56. NAND gate The diagrams below show two ways that the NAND logic gate can be configured. It can also be done using NOR logic gates in the same way.

57. NAND gate

58. This is a NOT-AND gate which is equal to an AND gate followed by a NOT gate.

59. . 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. NAND gate

60. NOR gate

61. 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.

62. NOR gate The symbol is an OR gate with a small circle on the output. The small circle represents inversion.

63. EX-OR gate

64. 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 EOR operation.

65. EX-NOR gate

66. 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 EXOR gate with a small circle on the output. The small circle represents inversion.

67. The NAND and NOR gates are called universal functions since with either one the AND and OR functions and NOT can be generated.

69. Truth Table

70.  (Assignment) Solve the following problem: (36.89) 10 =( ? ) 2 ( ? ) 8 ( ? ) 16 (1011.11) 2 = ( ? )2 ( ? ) 8 ( ? ) 16 (5A.BC) 16 = ( ? )2 ( ? ) 8 ( ? ) 16 (16.23) 8 = ( ? )2 ( ? ) 8 ( ? ) 16

71.  (Assignment) Solve the following problem: Q.Draw and Explain following logic Gate with T.T and Symbol 1.AND 2.OR 3.NAND Q. Explain NAND & NOR as universal gates

72. NOW, TEST IS IN NEXT LECTURE BEST OF LUCK!!