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DIGITAL ELECTRONICS
K SATYNARAYANA RAJU
Department of physics and electronics
B V RAJU COLLEGE
TOPICS
Number systems and codes
Boolean algebra and theorems
Combinational digital circuits
Sequential digital circuits
Memory devices
Number systems and codes
Number systems
Decimal Number system
Binary Number system
Octal Number system
Hexa Decimal Number system
Codes
BCD Code
Gray Code
Excess-3 Code
Decimal Number system
Base or radix of a number system
Decimal number system- Base 10
Numerals or symbols- 0,1,2,3,4,5,6,7,8,9
Value- 1976.539(10) - 1×10^3+9×10^2+7×10^1+6×10^0+5×10^-1+3×10^-
2+9×10^-3
Binary Number systems
Binary number system- Base 2
Numerals or Symbols used- 0,1
Value- 11001.101= 1×2^4+1×2^3+0×2^2+0×2^1+1×2^0+1×2^-
1+0×2^-2+1×2^-3
Octal Number system
Octal Number system- Base 8
Numerals or Symbols used- 0,1,2,3,4,5,6,7,
Value- 26345.167= 2×8^4+6×8^3+3×8^2+4×8^1+5×8^0+1×8^-1+6×8^-2+7×8^-3
Hexa Decimal Number system
Hexa decimal Number system- Base 16
Numerals or Symbols used- 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
Value- B98.C6= 11×16^2+9×16^1+8×16^0+12×16^-1+6×16^-2
Decimal to Binary conversion
253.15(10) is a decimal number
Real part- 253
Fractional part- 15
For real part – Successive division by 2
Fractional part – Successive multiplication by 2
Real part- 253÷2=126-->1
126÷2=63 -->0
63÷2=31 -->1
31÷2=15 -->1
15÷2=7 --->1
7÷2=3 -->1
3÷2=1 --->1
1÷2=0 --->1
Collect the reminders from bottom to top
253.15(10) = 11111101.0010 (2)
Fractional part
0.15×2=0.30--->0
0.30×2=0.60--->0
0.60×2=1.20--->1
0.20×2=0.40--->0
Collect the quotients from top to bottom
Decimal to Octal conversion
7546.32(10) is a decimal number
Real part- 754
Fractional part- 32
For real part – Successive division by 8
Fractional part – Successive multiplication by 8
Real part- 7546÷8=943 -->2
943 ÷8=117 -->7
117÷ 8= 14 -->5
14÷8= 1 --> 6
1÷8=0 ---> 1
Collect the reminders from bottom to top
7546.32(10) = 16572.2436 (8)
Fractional part
0.32×8= 2.56---> 2
0.56×8= 4.48 --->4
0.48×8= 3.84--->3
0.84×8= 6.72--->6
Collect the quotients from top to bottom
Decimal to Hexadecimal conversion
6876.94(10) is a decimal number
Real part- 6876
Fractional part- 0.94
For real part – Successive division by 16
Fractional part – Successive multiplication by 16
Real part- 6876÷16=429-->12 (C)
429÷16=26 -->13(D)
26÷16=1 -->10 (A)
1÷16=0 -->1
Collect the reminders from bottom to top
6876.94(10) = 1ADC.F0A3 (16)
Fractional part
0.94×16= 15.04---> 15(F)
0.04×16= 0.64---> 0
0.64×16= 10.24 --> 10(A)
0.24× 16= 3.84 --->3
Collect the quotients from top to bottom
Binary to Decimal Conversion
11111101.0010 (2) is a binary number
Positional weights
1 1 1 1 1 1 0 1 . 0 0 1 0
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 2^-1 2^-2 2^-3 2^-4
1×2^7+1×2^6+1×2^5+1×2^4+1×2^3+1×2^2+1×2^1+0×2^0+0×2^-1+0×2^-2+1×2^-3+0× 2^-4
128+64+32+16+8+4+1+0.125= 253.125(10)
11111101.0010 (2) = 253.125(10)
Octal to Decimal Conversion
16572.2436(8)
1 6 5 7 2 . 2 4 3 6
8^4 8^3 8^2 8^1 8^0 8^-1 8^-2 8^-3 8^-4
1×8^4+ 6×8^3+5×8^2+7×8^1+2×2^0+2×8^-1+4×8^-2+3×8^-3+6×8^-4
4096+6×512+5×64+7×8+2×1+2×0.125+4×0.01562+3×0.001953+6×0.0002441
4096+3072+320+56+2+0.25+0.06248+0.005859+0.001464
16572.2436(8) = 7546.3198(10)
Hexa Decimal to Decimal conversion
1ADC.F0A3 (16)
1 A D C . F 0 A 3
16^3 16^2 16^1 16^0 16^-1 16^-2 16^-3 16^-4
1×16^3+10×16^2+13×16^1+12×16^0+15×16^-1+0×16^-2+10×16^-3+3×16^-4
= 4096+10×256+13×16+12+15×0.0625+10×0.00024414+3×0.000015258
= 4096+2560+208+12+0.9375+0.0024414+0.00004577
= 6876.9399
1ADC.F0A3 (16) = 6876.9399 (10)
Binary to octal conversion
1011101110.1101(2)
000-0
001-1
010-2
011-3
100-4
101-5
110-6
111-7
1011101110.1101(2) = 1356.64(8)
001 011 101 1 1 0 . 1 1 0 100
1 3 5 6 6 4
Octal to Binary conversion
1356.64(8)
1 3 5 6 . 6 4
001 011 101 110 110 100
1356.64(8) = 001011101110.110100 (2)
Binary to Hexadecimal conversion
110110101110101.1100100110(2)
0000-0
0001-1
0010-2
0011-3
0100-4
0101-5
0110-6
0111-7
1000-8
1001-9
1010-A
1011-B
1100-C
1101-D
1110-E
1111-F
110110101110101.1100100110 (2)= 6D75.C98 (16)
0110 1101 0111 0101. 1100 1001 1000
6 D 7 5 C 9 8
Hexa Decimal to Binary conversion
6D75.C98 (16)
6 D 7 5 . C 9 8
0110 1101 0111 0101 1100 1001 1000
6D75.C98 (16) = 0110110101110101.110010011000(2)
Octal to Hexa Decimal conversion
653.24(8)
6 5 3 . 2 4
110 101 011 010 100
110101011.010100
0001 1010 1011 . 0101 0000
1 A B 5 0
653.24(8) = 1AB.50 (16)
Hexa Decimal to octal conversion
A57. 2C(16)
A 5 7 . 2 C
1010 0101 0111 0010 1100
101001010111.00101100
101 001 010 111 . 001 011 000
5 1 2 7 1 3 0
A57. 2C(16) = 5127.130(2)
Binary Addition
• 0+0=00
• 0+1=01
• 1+0=01
• 1+1=10
• 1+1+1=11
• 1+1+1+1=100
• 1+1+1+1+1=101
Add 1011101.0101 to 1101001.1100
1 0 1 1 1 0 1 . 0 1 0 1
1 1 0 1 0 0 1 . 1 1 0 0
………………………………………………………………………………………………
1 1 0 0 0 1 1 1 0 0 0 1
Binary subtraction
• 0-0=0
• 1-0=1
• 1-1=0
• 0-1=1 with borrow 1
Perform 11001(2) - 10111(2)
1 1 0 0 1
1 0 1 1 1
………………………………………………………
0 0 0 1 0
1’s complement method of subtraction
Perform 11001(2) - 10111(2) using 1’s complement method of subtraction
Minuend - 11001
Subtrahend- 10111
1’s complement of subtrahend - 01000
Add minuend with 1’s complement of subtrahend
11001
01000
……………..
[1]00001
Add carry 1 to 00001 to get the final result
00010
2’s complement method of subtraction
Perform 11001(2) - 10111(2) using 2’s complement
method of subtraction
Minund - 11001
Subtrahend- 10111
1’s complement of subtrahend - 01000
2’s complement of subtrahend- 01000+1 = 01001
Add minuend with 1’s complement of subtrahend
11001
01001
• ……………..
• [1]00010
Leave the carry to get the final result
00010
Note:- if carry is 0 then find 2’s complement to
Result obtained after addition to get the final result.
•
•
•
• ```
Binary multiplication
0×0=0
0×1=0
1×0=0
1×1=1
Multiply 10110(2) with 110 (2)
1 0 1 1 0
× 1 1 0
……………………….
0 0 0 0 0
1 0 1 1 0
1 0 1 1 0
…… ..……………………………
1 0 0 0 0 1 0 0
Binary Division
0÷0 is undefined
0÷1=0
1÷0 is undefined
1÷1=1
Divide 11001(2) with 101 (2)
101 ) 11001 ( 101
101
.................
101
101
…………..
000
BCD Code
Binary coded decimal
Input and output of digital system
8421 code
Convert decimal number 4782 in to BCD code
4 7 8 2
0100 0111 1000 0010
BCD code- 0100011110000010
Gray code
Cyclic code or reflection code or unit distance code
Binary code Gray code
0000 0000
0001 0001
0010 0011
0011 0010
0100 0110
0101 0111
0110 0101
0111 0100
1000 1100
1001 1101
Excess 3 code
Decimal Binary code Excess 3 code
0 0000 0011
1 0001 0100
2 0010 0101
3 0011 0110
4 0100 0111
5 0101 1000
6 0110 1001
7 0111 1010
8 1000 1011
9 1001 1100

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DIGITAL ELECTRONICS GUIDE

  • 1. DIGITAL ELECTRONICS K SATYNARAYANA RAJU Department of physics and electronics B V RAJU COLLEGE
  • 2. TOPICS Number systems and codes Boolean algebra and theorems Combinational digital circuits Sequential digital circuits Memory devices
  • 3. Number systems and codes Number systems Decimal Number system Binary Number system Octal Number system Hexa Decimal Number system Codes BCD Code Gray Code Excess-3 Code
  • 4. Decimal Number system Base or radix of a number system Decimal number system- Base 10 Numerals or symbols- 0,1,2,3,4,5,6,7,8,9 Value- 1976.539(10) - 1×10^3+9×10^2+7×10^1+6×10^0+5×10^-1+3×10^- 2+9×10^-3
  • 5. Binary Number systems Binary number system- Base 2 Numerals or Symbols used- 0,1 Value- 11001.101= 1×2^4+1×2^3+0×2^2+0×2^1+1×2^0+1×2^- 1+0×2^-2+1×2^-3
  • 6. Octal Number system Octal Number system- Base 8 Numerals or Symbols used- 0,1,2,3,4,5,6,7, Value- 26345.167= 2×8^4+6×8^3+3×8^2+4×8^1+5×8^0+1×8^-1+6×8^-2+7×8^-3
  • 7. Hexa Decimal Number system Hexa decimal Number system- Base 16 Numerals or Symbols used- 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F Value- B98.C6= 11×16^2+9×16^1+8×16^0+12×16^-1+6×16^-2
  • 8. Decimal to Binary conversion 253.15(10) is a decimal number Real part- 253 Fractional part- 15 For real part – Successive division by 2 Fractional part – Successive multiplication by 2 Real part- 253÷2=126-->1 126÷2=63 -->0 63÷2=31 -->1 31÷2=15 -->1 15÷2=7 --->1 7÷2=3 -->1 3÷2=1 --->1 1÷2=0 --->1 Collect the reminders from bottom to top 253.15(10) = 11111101.0010 (2) Fractional part 0.15×2=0.30--->0 0.30×2=0.60--->0 0.60×2=1.20--->1 0.20×2=0.40--->0 Collect the quotients from top to bottom
  • 9. Decimal to Octal conversion 7546.32(10) is a decimal number Real part- 754 Fractional part- 32 For real part – Successive division by 8 Fractional part – Successive multiplication by 8 Real part- 7546÷8=943 -->2 943 ÷8=117 -->7 117÷ 8= 14 -->5 14÷8= 1 --> 6 1÷8=0 ---> 1 Collect the reminders from bottom to top 7546.32(10) = 16572.2436 (8) Fractional part 0.32×8= 2.56---> 2 0.56×8= 4.48 --->4 0.48×8= 3.84--->3 0.84×8= 6.72--->6 Collect the quotients from top to bottom
  • 10. Decimal to Hexadecimal conversion 6876.94(10) is a decimal number Real part- 6876 Fractional part- 0.94 For real part – Successive division by 16 Fractional part – Successive multiplication by 16 Real part- 6876÷16=429-->12 (C) 429÷16=26 -->13(D) 26÷16=1 -->10 (A) 1÷16=0 -->1 Collect the reminders from bottom to top 6876.94(10) = 1ADC.F0A3 (16) Fractional part 0.94×16= 15.04---> 15(F) 0.04×16= 0.64---> 0 0.64×16= 10.24 --> 10(A) 0.24× 16= 3.84 --->3 Collect the quotients from top to bottom
  • 11. Binary to Decimal Conversion 11111101.0010 (2) is a binary number Positional weights 1 1 1 1 1 1 0 1 . 0 0 1 0 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 2^-1 2^-2 2^-3 2^-4 1×2^7+1×2^6+1×2^5+1×2^4+1×2^3+1×2^2+1×2^1+0×2^0+0×2^-1+0×2^-2+1×2^-3+0× 2^-4 128+64+32+16+8+4+1+0.125= 253.125(10) 11111101.0010 (2) = 253.125(10)
  • 12. Octal to Decimal Conversion 16572.2436(8) 1 6 5 7 2 . 2 4 3 6 8^4 8^3 8^2 8^1 8^0 8^-1 8^-2 8^-3 8^-4 1×8^4+ 6×8^3+5×8^2+7×8^1+2×2^0+2×8^-1+4×8^-2+3×8^-3+6×8^-4 4096+6×512+5×64+7×8+2×1+2×0.125+4×0.01562+3×0.001953+6×0.0002441 4096+3072+320+56+2+0.25+0.06248+0.005859+0.001464 16572.2436(8) = 7546.3198(10)
  • 13. Hexa Decimal to Decimal conversion 1ADC.F0A3 (16) 1 A D C . F 0 A 3 16^3 16^2 16^1 16^0 16^-1 16^-2 16^-3 16^-4 1×16^3+10×16^2+13×16^1+12×16^0+15×16^-1+0×16^-2+10×16^-3+3×16^-4 = 4096+10×256+13×16+12+15×0.0625+10×0.00024414+3×0.000015258 = 4096+2560+208+12+0.9375+0.0024414+0.00004577 = 6876.9399 1ADC.F0A3 (16) = 6876.9399 (10)
  • 14. Binary to octal conversion 1011101110.1101(2) 000-0 001-1 010-2 011-3 100-4 101-5 110-6 111-7 1011101110.1101(2) = 1356.64(8) 001 011 101 1 1 0 . 1 1 0 100 1 3 5 6 6 4
  • 15. Octal to Binary conversion 1356.64(8) 1 3 5 6 . 6 4 001 011 101 110 110 100 1356.64(8) = 001011101110.110100 (2)
  • 16. Binary to Hexadecimal conversion 110110101110101.1100100110(2) 0000-0 0001-1 0010-2 0011-3 0100-4 0101-5 0110-6 0111-7 1000-8 1001-9 1010-A 1011-B 1100-C 1101-D 1110-E 1111-F 110110101110101.1100100110 (2)= 6D75.C98 (16) 0110 1101 0111 0101. 1100 1001 1000 6 D 7 5 C 9 8
  • 17. Hexa Decimal to Binary conversion 6D75.C98 (16) 6 D 7 5 . C 9 8 0110 1101 0111 0101 1100 1001 1000 6D75.C98 (16) = 0110110101110101.110010011000(2)
  • 18. Octal to Hexa Decimal conversion 653.24(8) 6 5 3 . 2 4 110 101 011 010 100 110101011.010100 0001 1010 1011 . 0101 0000 1 A B 5 0 653.24(8) = 1AB.50 (16)
  • 19. Hexa Decimal to octal conversion A57. 2C(16) A 5 7 . 2 C 1010 0101 0111 0010 1100 101001010111.00101100 101 001 010 111 . 001 011 000 5 1 2 7 1 3 0 A57. 2C(16) = 5127.130(2)
  • 20. Binary Addition • 0+0=00 • 0+1=01 • 1+0=01 • 1+1=10 • 1+1+1=11 • 1+1+1+1=100 • 1+1+1+1+1=101 Add 1011101.0101 to 1101001.1100 1 0 1 1 1 0 1 . 0 1 0 1 1 1 0 1 0 0 1 . 1 1 0 0 ……………………………………………………………………………………………… 1 1 0 0 0 1 1 1 0 0 0 1
  • 21. Binary subtraction • 0-0=0 • 1-0=1 • 1-1=0 • 0-1=1 with borrow 1 Perform 11001(2) - 10111(2) 1 1 0 0 1 1 0 1 1 1 ……………………………………………………… 0 0 0 1 0
  • 22. 1’s complement method of subtraction Perform 11001(2) - 10111(2) using 1’s complement method of subtraction Minuend - 11001 Subtrahend- 10111 1’s complement of subtrahend - 01000 Add minuend with 1’s complement of subtrahend 11001 01000 …………….. [1]00001 Add carry 1 to 00001 to get the final result 00010
  • 23. 2’s complement method of subtraction Perform 11001(2) - 10111(2) using 2’s complement method of subtraction Minund - 11001 Subtrahend- 10111 1’s complement of subtrahend - 01000 2’s complement of subtrahend- 01000+1 = 01001 Add minuend with 1’s complement of subtrahend 11001 01001 • …………….. • [1]00010 Leave the carry to get the final result 00010 Note:- if carry is 0 then find 2’s complement to Result obtained after addition to get the final result. • • • • ```
  • 24. Binary multiplication 0×0=0 0×1=0 1×0=0 1×1=1 Multiply 10110(2) with 110 (2) 1 0 1 1 0 × 1 1 0 ………………………. 0 0 0 0 0 1 0 1 1 0 1 0 1 1 0 …… ..…………………………… 1 0 0 0 0 1 0 0
  • 25. Binary Division 0÷0 is undefined 0÷1=0 1÷0 is undefined 1÷1=1 Divide 11001(2) with 101 (2) 101 ) 11001 ( 101 101 ................. 101 101 ………….. 000
  • 26. BCD Code Binary coded decimal Input and output of digital system 8421 code Convert decimal number 4782 in to BCD code 4 7 8 2 0100 0111 1000 0010 BCD code- 0100011110000010
  • 27. Gray code Cyclic code or reflection code or unit distance code Binary code Gray code 0000 0000 0001 0001 0010 0011 0011 0010 0100 0110 0101 0111 0110 0101 0111 0100 1000 1100 1001 1101
  • 28. Excess 3 code Decimal Binary code Excess 3 code 0 0000 0011 1 0001 0100 2 0010 0101 3 0011 0110 4 0100 0111 5 0101 1000 6 0110 1001 7 0111 1010 8 1000 1011 9 1001 1100