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Dr. Wissam Hasan Mahdi Alagele

وزارة التعليم العالي والبحث العلمي جامعة الكوفة - كلية التربية – قسم علوم الحاسوب. Digital Logic Design I I I. Chapter 3 Combinational Design. Dr. Wissam Hasan Mahdi Alagele. e-mail:wisam.alageeli@uokufa.edu.iq. http :// edu-clg.kufauniv.com/staff/Mr.Wesam. Combinational Circuits.

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Dr. Wissam Hasan Mahdi Alagele

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  1. وزارة التعليم العالي والبحث العلمي جامعة الكوفة - كلية التربية – قسم علوم الحاسوب Digital Logic Design III Chapter 3 Combinational Design Dr. WissamHasan Mahdi Alagele e-mail:wisam.alageeli@uokufa.edu.iq http://edu-clg.kufauniv.com/staff/Mr.Wesam

  2. Combinational Circuits • Output is function of input only i.e. no feedback When input changes, output may change (after a delay) Combinational Circuits n inputs m outputs • • • • • • 

  3. ? ? ? Combinational Circuits • Analysis • Given a circuit, find out its function • Function may be expressed as: • Boolean function • Truth table • Design • Given a desired function, determine its circuit • Function may be expressed as: • Boolean function • Truth table

  4. Analysis Procedure • Boolean Expression Approach T2=ABC T1=A+B+C T3=AB'C'+A'BC'+A'B'C F’2=(A’+B’)(A’+C’)(B’+C’) F2=AB+AC+BC F1=AB'C'+A'BC'+A'B'C+ABC F2=AB+AC+BC

  5. Analysis Procedure • Truth Table Approach = 0 = 0 = 0 = 0 = 0 = 0 = 0 = 0 = 0 = 0 = 0 = 0 0 0 0 0 0 0 0 0 0 1 0

  6. Analysis Procedure • Truth Table Approach = 0 = 0 = 1 = 0 = 0 = 1 = 0 = 0 = 0 = 1 = 0 = 1 0 1 0 0 0 1 1 0 1 1 0

  7. Analysis Procedure • Truth Table Approach = 0 = 1 = 0 = 0 = 1 = 0 = 0 = 1 = 0 = 0 = 1 = 0 0 1 0 0 0 1 1 1 0 1 0

  8. Analysis Procedure • Truth Table Approach = 0 = 1 = 1 = 0 = 1 = 1 = 0 = 1 = 0 = 1 = 1 = 1 0 1 0 0 1 0 0 0 1 0 1

  9. Analysis Procedure • Truth Table Approach = 1 = 0 = 0 = 1 = 0 = 0 = 1 = 0 = 1 = 0 = 0 = 0 0 1 0 0 0 1 1 1 1 0 0

  10. Analysis Procedure • Truth Table Approach = 1 = 0 = 1 = 1 = 0 = 1 = 1 = 0 = 1 = 1 = 0 = 1 0 1 0 1 0 0 0 0 0 1 1

  11. Analysis Procedure • Truth Table Approach = 1 = 1 = 0 = 1 = 1 = 0 = 1 = 1 = 1 = 0 = 1 = 0 0 1 1 0 0 0 0 0 1 0 1

  12. Analysis Procedure • Truth Table Approach = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 1 1 1 1 1 1 0 0 1 11 F2=AB+AC+BC F1=AB'C'+A'BC'+A'B'C+ABC

  13. ? Design Procedure • Given a problem statement: • Determine the number of inputs and outputs • Derive the truth table • Simplify the Boolean expression for each output • Produce the required circuit Example: Design a circuit to convert a “BCD” code to “Excess 3” code • 4-bits • 0-9 values • 4-bits • Value+3

  14. Design Procedure • BCD-to-Excess 3 Converter w = A+BC+BD x = B’C+B’D+BC’D’ y = C’D’+CD z = D’

  15. Design Procedure • BCD-to-Excess 3 Converter w = A + B(C+D) y = (C+D)’ + CD x = B’(C+D) + B(C+D)’ z = D’

  16. a b f w x y z a b c d e f g ? g c e d Seven-Segment Decoder BCD code a = w + y + xz+ x’z’ e=x’z’+yz’ b =x’+yz+y’z’ f=w+x+y’z’ c =x+y’+z g=w+x’y+xy’+x’y+xz’ d =x’z’+yz’+x’y+xy’z

  17. Problem Design a circuit which displays the letters A through J on a seven-segment indicator. The circuit has four inputs W, X, Y, Z which represent the last 4 bits of the ASCII code for the letter to be displayed. For example, if WXYZ = 0001, “A” will be displayed. The letters should be displayed in the following form: Design your circuit using only two-, three-, and four-input NOR gates and inverters. Any solution with 22 or fewer gates and inverters (not counting the four inverters for the inputs) is acceptable.

  18. x y S C HA x y S C Binary Adder • Half Adder • Adds 1-bit plus 1-bit • Produces Sum and Carry x + y ─── CS

  19. S C x y z FA Binary Adder • Full Adder • Adds 1-bit plus 1-bit plus 1-bit • Produces Sum and Carry x + y + z ─── CS S = xy'z'+x'yz'+x'y'z+xyz= z’(xy’+x’y)+z(x’y’+xy) =z’(x  y) + z(x  y)’ = x y z C = xy+ xz+ yz

  20. Binary Adder S = xy'z'+x'yz'+x'y'z+xyz=x y z • Full Adder C = xy+ xz+ yz x y z S C S C x y z

  21. Binary Adder • Full Adder HA HA x y z S C x y z S C

  22. ? ? QUIZ Q#1- This combinational logic circuit is described as a(n) ___. Q#2- What are the sum and carry out outputs of this full-adder circuit? Q#3- What are the sum and carry out outputs of this full-adder circuit? Q#4- What are the sum and carry out outputs of this full-adder circuit? Q#5- What are the sum and carry out outputs of this full-adder circuit? Q#6- What are the sum and carry out outputs of this full-adder circuit? ANS: full-adder ANS: Sum=0, Carry out=0 ANS: Sum=1, Carry out=0 ANS: Sum=0, Carry out=1 ANS: Sum=1, Carry out=1 ANS: Sum=0, Carry out=1 Cin = 0 A = 0 B = 0 Cin = 0 A = 0 B = 1 Cin = 1 A = 0 B = 1 Cin = 1 A = 1 B = 1 Cin = 0 A = 1 B = 1

  23. x3x2x1x0 y3y2y1y0 Binary Adder C0 Cy S3S2S1S0 Binary Adder c3 c2 c1 . + x3 x2 x1 x0 + y3 y2 y1 y0 ──────── Cy S3 S2 S1 S0 Carry Propagate Addition x3 x2 x1 x0 y3 y2 y1 y0 0 FA FA FA FA C4 C3 C2 C1 S3 S2 S1 S0

  24. 74x2834-bit adder • Uses carry look-ahead internally

  25. CPA A3 A2 A1 A0 B3 B2 B1 B0 Cy C0 S3 S2 S1 S0 Binary Adder c3c2 c1 . x3x2 x1 x0 y3y2 y1 y0 ──────── Cy S3 S2 S1 S0 c7 c6 c5Cy. + x7 x6 x5 x4 + y7 y6 y5 y4 ──────── CyS7 S6 S5 S4 • Carry Propagate Adder x3 x2 x1 x0 x7 x6 x5 x4 y3 y2 y1 y0 y7 y6 y5 y4 CPA A3 A2 A1 A0 B3 B2 B1 B0 0 Cy C0 S3 S2 S1 S0 S7 S6 S5 S4 S3 S2 S1 S0

  26. Subtractors Half Subtractor Full Subtractor Adder/Subtractor - 1 Adder/Subtractor - 2

  27. A Di B Bo Half Subtractor Input Output Logic Symbol: A Di (difference) Half Subtractor B B0 (borrow out) A B Di Bo 0 0 00 0 1 11 1 0 10 1 1 00 2 1 0 -1 1 Logic Diagram:

  28. ? ? QUIZ Q#1- What is the difference and borrow outputs from This half-subtractor circuit? Q#2- What is the difference and borrow outputs from this half-subtractor circuit? Q#3- What is the difference and borrow outputs from this half-subtractor circuit? Q#4- What is the difference and borrow outputs from this half-subtractor circuit? ANS: Di= 0, Bo= 0 ANS: Di= 1, Bo= 0 ANS: Di= 0, Bo= 0 ANS: Di= 1, Bo= 1 (A – B) A = 0 B = 0 (A – B) A = 1 B = 0 (A – B) A = 1 B = 1 (A – B) A = 0 B = 1

  29. Input Output Bin Di (difference) Full Subtractor A B B0 (borrow out) Bin Di H. S. half subtractor A B0 H. S. B half subtractor Full Subtractor Logic Symbol: Logic Diagram: Bo

  30. AB 00 01 11 10 Ci 0 1 Full Subtractor CiA B Di Bo 0 0 0 0 0 0 0 1 1 1 0 1 0 1 0 0 1 1 0 0 1 0 0 1 1 1 0 1 0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 Di Di Same as S in full adder 00 01 11 10 0 1 Bo

  31. ? ? QUIZ Q#1- What are the Difference and Borrow out output from this full-subtractor circuit? Q#2- What are the Difference and Borrow out output from this full-subtractor circuit? Q#3- What are the Difference and Borrow out output from this full-subtractor circuit? Q#4- What are the Difference and Borrow out output from this full-subtractor circuit? Q#5- What are the Difference and Borrow out output from this full-subtractor circuit? Q#6- What are the Difference and Borrow out output from this full-subtractor circuit? HINT: truth table from textbook (Fig. 10-10) is helpful ANSWER: Di = 0, Bo = 0 ANSWER: Di = 1, Bo = 1 ANSWER: Di = 1, Bo = 1 ANSWER: Di = 0, Bo = 1 ANSWER: Di = 1, Bo = 0 ANSWER: Di = 0, Bo = 0 (A – B - Bin) Bin = 0 A = 0 B = 0 (A – B - Bin) Bin = 1 A = 0 B = 0 (A – B - Bin) Bin = 1 A = 1 B = 1 (A – B - Bin) Bin = 1 A = 0 B = 1 (A – B - Bin) Bin = 0 A = 1 B = 0 (A – B - Bin) Bin = 0 A = 1 B = 1

  32. A S B C A Di B Bo Adder/Subtractor - 1 Half adder Half subtractor i E = 0: Half adder o E = 1: Half subtractor

  33. C i A Di B Bo E Adder/Subtractor-1 E = 0: Full adder E = 1: Full subtractor

  34. Adder/Subtractor-2 E = 0: 4-bit adder E = 1: 4-bit subtractor

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