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Logic Circuits

Gates implement mathematical logic operations applied to binary variables – A, B, X, Y, Z. Logic Circuits. Digital Systems are made out of integrated circuits, transistors and interconnections. Certain well defined – small circuits are called Logic Gates.

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Logic Circuits

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  1. Gates implement mathematical logic operations applied to binary variables – A, B, X, Y, Z Logic Circuits Digital Systems are made out of integrated circuits, transistors and interconnections Certain well defined – small circuits are called Logic Gates Mathematical logic – ideal, that is: gates do not suffer from the physical propagation delays that are inherent in transistors. An example: Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  2. X X Vdd (5V) Pull up o Vin Vout If X = 0 (1) then X (NOT X) is 1 (0) Pull down Vss (0V) Example: inverter circuit Inverter gate Operation: NOT (or complement) Response is instantaneous Not instantaneous Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  3. Or : Z = X Y X Y Z Gates may have more than 2 inputs! 0 0 0 0 1 0 1 0 0 1 1 1 X Y Z W = XYZ Other gates Truth table AND Operation: Z is 1 if and only if x AND y are 1 Note the ordering of the X-Y bits! Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  4. X Z = X + Y Y OR gate Or : Z = X Y X Y Z X 0 0 1 1 0 0 0 0 1 1 1 0 1 1 1 1 0 1 0 1 Y X•Y 0 0 0 1 X+Y 0 1 1 1 X 1 1 0 0 Other gates (cont’d) Truth table OR Operation: Z is 1 if at least one of X and Y is one Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  5. X X buffer Other gates (cont’d) An inverter is often a buffer whose output is the complement of its input. Operation: identity A physical buffer is not an identity. It is a signal restoring devise. The circle is, in fact, the symbol for complementing Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  6. This circuit has a special name and a special symbol (gate) X Y Z = X Y XOR GATE Operation : exclusive OR Z is 1 if either X is 1 or Y is 1 Combinational Circuits X Y X Y Z 0 0 0 0 1 1 1 0 1 1 1 0 Z The physical XOR gate need not be realized this way (see page 82!) Z = XY + XY Other gates ? Yes and no. The gates we have introduced so far form a complete set to build logic circuits without memory, so-called Here is an example: Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  7. Watch the little circles (complementing) What you see is but a symbol: A physical NAND is not an AND + Inverter In fact, NANDs and NORs are simpler than ANDs and ORs! It is the set {NOT, NAND, NOR, exclusive NOR} NAND = F (XY) AND We will see alternative expressions for the outputs of the alternative gates, And also conversions between the two sets. OR NOR = + F (X Y ) XOR Inverter = F X = + = Å F X Y X Y X Y XNOR Buffer = + = Å F XY X Y (X Y ) Other gates ? Yes, but they form an alternative set to {NOT, AND, OR, XOR} It is the set {NOT, NAND, NOR, exclisive NOR} Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  8. How do I know? There must be rules! Z Conversely, given a Boolean function, how do I derive the implementation? There must be rules! = X.Y + X.Y = X  Y ! Z= ((X.Y).X).((X.Y).Y) Boolean Functions and Combinational Circuits This is a four-NAND-gate Combinational Circuit Combinational circuits do not have any memory. They implement Boolean functions: A binary variable (the function), an equal sign, and an algebraic expression formed by using sinary variables, the constants 0 and 1, the logic operation symbols, and parenthesis. Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  9. Hence: X + Y = X.Y Known as DeMorgan’s theorem F = X + Y G = X. Y X Y F 0 0 1 0 1 0 1 0 0 1 1 0 X Y G 0 0 1 0 1 0 1 0 0 1 1 0 Truth tables and Boolean Algebra A Boolean function can be represented by a truth table As we did for the basic gates (which do implement Elementary functions!) A Boolean function can also be manipulated based on Fundamental identities of Boolean Algebra. Examples: Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  10. The two columns are duals of each other: Dual: replace 0 (1) by 1 (0) and OR (AND) by AND (OR). Identity X+0=X X•1=X Consensus theorem (and dual) XY + XZ + YZ = XY + XZ (X+Y)(X+Z)(Y+Z) = (X+Y)(X+Z) Absorption X+1=1 X•0=0 Idempotency X+X=X X•X=X X+X'=1 X•X'=0 Complements Also true in ordinary algebra (X')'=X Involution Not true in ordinary algebra X+Y= Y+X X Commutative • X •Y = Y X+(Y+Z) = (X+Y)+Z X Associative •Z) = (X •Y) • (Y •Z Distributive X(Y+Z)=XY+XZ X+YZ=(X+Y)•(X+Z) De Morgan (X+Y)'=X'•Y' (X•Y)'=X'+Y' Basic Identities of Boolean Algebra Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  11. XY.Y = V XY.X = U UV = U + V = XY.X + XY.Y = (X+Y)X + (X+Y).Y (DeMorgan) (DeMorgan) = YX + XY (XX = YY =0) = X.Y + X.Y = X  Y ! Z= ((X.Y).X).((X.Y).Y) Back to my claim Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  12. NAND Y Y F F NOR X X Y F X F X Y Once more DeMorgan. (X+Y)'=X'•Y' (X•Y)'=X'+Y' Recall: Hence: Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  13. I call this the circle game F = XY’+X’Y+Z Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  14. A term in a Boolean expression is a literal or a group of literals joined by the same operator. Terminology • A literal is any variable in an expression, complemented or not . A variable that appears twice counts for two literals. X'YZ + X'YZ' + XZ 8 literals 3 terms (A+B)(A+C'+D) 5 literals 2 terms Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  15. X'YZ + X'YZ' + XZ two levels: first level AND's, second level OR's (A+B)(A+C'+D) two levels: first level OR's, second level AND's. A + B(C+D) three levels: first level OR, second level AND, third level OR (A+B)C + B(C+D) Three levels Levels of an expression Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  16. F = XY + XZ This function consists of 2 terms XY and XZ, 3 variables X, Y, and Z, and 4 literals, X, X, Y, and Z. It is a 2-level function. F = XY + XZ = XY. XZ =(X + Y).(X+Z) F = (X+Y).(X+Z) Complements of functions Two ways to derive its complement: 1. Use Boolean identities: 2. Take dual of F and complement each literal: (generalization of DeMorgan) Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  17. F = ABC + BCD + ABC + ABCD These are SOPs G = BD + BC + ACD Standard Forms F = G! How do I know We try to express functions in standard form: Either Sum of Products (SOP), and even better Sum of Minterms Or Products of Sums (POS), and even better Products of Maxterms Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  18. For 3 variables X, Y, and Z there are minterms: m0 = XYZ m1 = XYZ m2 = XYZ m3 = XYZ m4 = XYZ m5 = XYZ m6 = XYZ m7 = XYZ product terms of exactly 3 literals m5 m5 X Y Z m0 m1 m2 m3 m4 m6 m7 X Y Z m0 m1 m2 m3 m4 m6 m7 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 If X=1, Y=0, and Z =1 then the function m5 = XYZ is one and only this one. 1 1 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 1 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 Minterms for Three Variables truth table 0 1 2 3 4 5 6 7 Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  19. For 3 variables X, Y, and Z, there are Maxterms: sum terms of exactly 3 literals Mi = mi M0 = X+Y+Z M1 = X+Y+Z M2 = X+Y+Z M3 = X+Y+Z M4 = X+Y+Z M5 = X+Y+Z M6 = X+Y+Z M7 = X+Y+Z X Y Z M0 M1 M2 M3 M4 M5 M6 M7 0 0 0 0 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 1 1 1 0 1 0 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 0 1 1 If X =1 and Y=0, and Z = 0 then M4 is 1 and only this one 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 Maxterms for Three Variables truth table 0 1 2 3 4 5 6 7 Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  20. From Truth Table to Sum of Minterms F(X,Y,Z) m0 = X'Y'Z' m2 = X'YZ' F = X'Y'Z'+X'YZ'+XY'Z+XYZ F = X'Y'Z'+X'YZ'+XY'Z+XYZ F = X'Y'Z'+X'YZ'+XY'Z+XYZ F = X'Y'Z'+X'YZ'+XY'Z+XYZ m5 = XY'Z m7 = XYZ More compact notation: F = m0 + m2 + m5 + m7 = ∑m(0,2,5,7) More compact notation: F = m0 + m2 + m5 + m7 = ∑m(0,2,5,7) Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  21. From Truth Table toProduct of Maxterms M1 = X+Y+Z' M3 = X+Y'+Z' M4 = X'+Y+Z Individual terms only zero for those combinations for which F = 0 M6 = X'+Y'+Z F = (X+Y+Z') F = (X+Y+Z')(X+Y'+Z') F = (X+Y+Z')(X+Y'+Z')(X'+Y+Z) F = (X+Y+Z')(X+Y'+Z')(X'+Y+Z)(X'+Y'+Z) More compact: F = M1•M3•M4•M6 More compact: F = M1•M3•M4•M6 = ∏M(1, 3, 4, 6) Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  22. From E = X' + Y'Z' to Standard E = ∑m(0, 1, 2, 3, 4) = ∏M(5, 6, 7) E' = ∏M(0, 1, 2, 3, 4) = ∑m(5, 6, 7) Look closer: relation mj - Mj and more … Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  23. But clearly: F = X + Y How can I use K-map to find out? Y 0 1 Y 0 1 X X m0 m1 m2 m3 XY XY XY XY 1 1 0 1 0 1 1 What next? Expressions may still be not minimal! If not, circuit is too large. Simplifying expressions is not easy. Some computer programs can go a long way. One way to it is by means of a Karnaugh Map (K-map). It it is a graphical method that does not work for functions with more than 4 (perhaps 5 or 6) variables. An example: Suppose F(X,Y) is 1 for (X,Y)=(0,1), (1,0), (1,1) Thus F = m1+m2+m3 Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  24. Y Y Y 0 1 0 1 0 1 X X X 0 F(0,0) F(0,1) 0 0 1 1 F(1,0) F(1,1) 1 1 1 1 1 Map of F(X,Y) F(X,Y) = X•Y F(X, Y) = X+Y Y Y Y 0 1 0 1 0 1 X X X 0 X'Y' X'Y 0 1 1 0 1 1 1 XY' XY 1 1 1 F(X, Y) = X' + Y Maps of minterms F(X,Y) = X' F(X, Y) = X' +Y F(X, Y) = X' + Y Two variable map Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  25. Three variable map Minterms in three variable map Three variable map 1 1 Map of function m5+m7 = XY'Z+XYZ = XZ(Y+Y') = XZ•1 = XZ Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  26. Simplifying a Function F(X, Y, Z) = ∑m(2,3,4,5) 1 1 1 1 m2+m3=X'YZ'+X'YZ = X'Y(Z+Z') = X'Y m4+m5=XY'Z'+XY'Z=XY'(Z+Z') = XY' F = X'Y+XY' F = X'Y+XY' 2 adjacent 1’s=sum of 2 three-literal minterms=2-literal term Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  27. ZY X' Z' Z Simplifying rules X'Z' Group of two adjacent 1’s gives product of two literals XZ' XZ Group of four Adjacent 1’s gives Single literal Group of 8 Adjacent 1’s gives 1 Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  28. A rectangle of 2 squares represents a product term of three literals 2x1 or 1x2 cells A rectangle of 4 squares represents a product of two literals 2x2, 1x4, or 4x1 cells A rectangle of 8 squares represents one literal 2x4 or 4x2 cells Simplifying rules Four variable map. Variables are W, X, Y, Z Four variable map with formulas for minterms Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  29. F(W,X,Y,Z) = ∑m(0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14) Simplifying 4-variable function W'Z' 1 1 1 1 1 1 1 1 1 1 1 XZ' Y' F = Y'+W'Z'+XZ' Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  30. More simplifying rules To be introduced: implicants, prime implicants, essential prime implicants • THEN: • Given: Map. • Goal: Best SOP expression for this function • 1. Findall prime implicants, identifyessential and nonessentialprime implicants. • 2. Includeall essentialprime implicants in function • 3. Includeenough nonessentialprime implicants to cover function. Minimize overlap among prime implicants. Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  31. Definition: F1 implies F2 if F2 = 1 whenever F1 = 1. Example: F1 = XY, F2 = X+Y F1 = 1 only when both X and Y are equal to 1. When both X and Y are equal to 1, F2 = 1 + 1 = 1 Therefore, F1 implies F2 Does F2 imply F1? Consider X = 1, Y = 0. F1 = 1 + 0 = 1 F2 = 1•0 = 0 Therefore, F2 does not imply F1 Implication Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  32. Aproduct term P is an implicantof function F if P implies F P is an implicant of F(X, Y, Z ...) if F(X, Y, ...) = 1 whenever P = 1 X'Y'Z' 1 1 Z' 1 1 1 XY' XY Implicants F(X,Y,Z) = ∑m(0,2,4,5,6) XY is not an implicant of F. XY', X'YZ', and Z'are implicants of F An implicant is prime (prime implicant) if leaving out any of its literals produces a term that is not an implicant X'Y'Z' is not prime because, e. g. Y'Z' is an implicant XY', Z' are both prime implicants. Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  33. In writing a simple SOP for a map, we only use prime implicants Simplification with Prime Implicants Prime Implicants: A'D and A'B and BD' Why? Good expression: F = A'D + BD' Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  34. All prime implicants may not be needed Which ones should be used? An essential prime implicant (EPI) is a prime implicant that covers a minterm not covered by any otherprime implicant. Essential Prime Implicants With EPIs we try to avoid overlap. Any good SOP expression for F must include the EPIs Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  35. Given: Map. Goal: Best SOP expression for this function 1. Findall prime implicants,identifyessential and nonessential prime implicants. 2. Includeall essential prime implicants in function 3. Includeenough nonessential prime implicants to cover function. Minimize overlap among prime implicants. Selection rules (again) Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  36. All prime implicants: A'B'C'D', ABC', ABD, AB'C, BC'D, ACD Essential prime implicants: A'B'C'D', ABC', AB'C, BC'D Nonessential prime implicants: ABD, ACD Example F = A'B'C'D'+ABC'+AB'C+BC'D+(either ABD or ACD) Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  37. Sometimes the product of sums expression for a function is simpler than the sum of products How do we generate a product of sums? Cover zeros with rectangles! Product of Sum simplification Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  38. F(A,B,C,D) = ∑m(0,1,2,5,8,9,10) 1 1 0 1 0 1 0 0 0 0 0 0 1 0 1 1 POS Simplification F = (A'+B')(C'+D')(B'+D) How did I get it? F' is 1, if either AB or CD or BD' is zero Then invert (take dual and complement literals). Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  39. Sometimes the value of a Boolean function is not specified for some combination of variables A certain combination of variables never occurs The value of the function for this combination of variables may be unimportant This is called a don't carecondition Example: decimal digits are commonly represented by 4-bit binary numbers in the range of 0 to 9 (0000 to 1001). The binary patterns 1010 to 1111 don't occur in this application. For a don't care combination of input variables the value of aBoolean function is represented by an ‘x’. Don't-Care Conditions Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  40. Since the value of a don’t-care function is not specified, we may choose an expression that takes on value 0 or 1 We can use expressions that are not equal for the don’t care combinations. As long as the expression satisfies the specified values, they are acceptable What do we do? We choose values for the don’t-care combinations that produce simple expressions! Expressions for Functions with Don’t-Cares Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  41. Minterm specification: F(A,B,C,D) = ∑m(1, 3, 7, 11, 15) Don’t care combinations: d(A,B,C,D) = ∑m(0,2,5) All other values are 0 x x 1 1 x x 1 1 0 0 x x 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 Example with Don’t-Care Conditions F1 = A’B’ +CD F2 = A’D + CD Notice: F1 is not equal to F2 Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  42. x 1 x 1 0 x 1 0 0 0 0 1 0 0 0 1 POS Implementation Same F and same d F3 = D(A’+C) Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  43. F(A,B,C,D) = ∑m(0,1,2,4,5,10,11,13,15) C 00 01 11 10 1 1 1 00 1 1 01 B 1. A'C' Non overlapping PIs 1 1 11 A 2. BC'D 1 1 10 3. ABD D 4. ACD P1 is the only essential For the cover, we choose1, 3, 5, 7 F = A'C' + ABD + AB'C + A'B'D' 5. AB'C 6. B'CD' 7. A'B'D' Example B’ÇD’ would also do Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

  44. Example: Detailed Specification One bit binary adder (full adder) Fall 2003 Digital Design Fundamentals Ed F. Deprettere Leiden University

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