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Chapter 8

Chapter 8. Solving Larger Sequential Problems. Chapter 8 Solving Larger Sequential Problems. 8.1 Shift Registers. A Simple 4-bit Shift Register. Shift registers are most commonly implemented with SR FF. JK FF could be used in place of the SR’s.

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Chapter 8

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  1. Chapter 8 SolvingLarger Sequential Problems

  2. Chapter 8 Solving Larger Sequential Problems 8.1 Shift Registers • A Simple 4-bit Shift Register • Shift registers are most commonly implemented with SR FF. • JK FF could be used in place of the SR’s. • D FF could also be used. Current output q is connected to next D FF input. • At each clock, input x is shifted one place to the right.

  3. Chapter 8 Solving Larger Sequential Problems 8.1 Shift Registers • A Simple 4-bit Shift Register • Circuit • Timing Diagram P. 402

  4. Chapter 8 Solving Larger Sequential Problems 8.1 Shift Registers • Leading-Edge Triggered Shift Register • NOT gate is added at the clock input • The Leading edge of the clock is the trailing edge of the flip-flop clock input. • Clock input signal only goes to NOT gate. • Circuit presents a load of 1 to the clock rather than a load of 4. P. 402

  5. Chapter 8 Solving Larger Sequential Problems 8.1 Shift Registers • Shift Register with Load Reducing NOT gates • When a trailing-edge triggered shift register is desired, a second NOT gate is added. P. 402

  6. Chapter 8 Solving Larger Sequential Problems 8.1 Shift Registers • Shift Register Storage • Load = 0 : circular shift • Load = 1 : store x P. 403

  7. Chapter 8 Solving Larger Sequential Problems 8.1 Shift Registers • Serial-in Serial-out • Only 1 bit may be loaded into the register or read from the register at a time • Large serial-in serial-out shift register is a memory similar to a disk. • When Load = 0, data circulate around the n flip flops. • When Load = 1, new value x is supplied . • To initialize a 4-bit serial-in serial-out shift register to all 0’s, we would have to clock it four times with 0 on input x each time. • To avoid this, most shift registers have an active low clear input.

  8. Chapter 8 Solving Larger Sequential Problems 8.1 Shift Registers • Serial-in Parallel-out • 8-bit Serial-in Parallel-Out shift register using D ff (74164) P. 404

  9. Chapter 8 Solving Larger Sequential Problems 8.1 Shift Registers • Parallel-in • Require an input line for each flip flop. • Static Load(74165) P. 404

  10. Chapter 8 Solving Larger Sequential Problems 8.1 Shift Registers • Parallel-in • Dynamic Load(74166) P. 404

  11. Chapter 8 Solving Larger Sequential Problems 8.1 Shift Registers • Right/Left Shift Register • Truth Table • Circuit P. 405

  12. Chapter 8 Solving Larger Sequential Problems 8.1 Shift Registers • Example of Shift Register • Problem • Output z is 1 if input x has been alternating for 7 clock times • Using 8-bit serial-in parallel-out shift register. P. 406

  13. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • 74161 Counter • Synchronous counting and loading and asynchronous clear. • Load  = 0 : D = IND, • C = INC • B = INB • A = INA • Count (ENP=1, ENT=1) • OV : Overflow P. 408

  14. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • 74161 Counter • Load  = 0 → x=1, y=1, z= INc , w= INc, flip flop= INc • Load  = 1 → x=0, w = z = 1, y = output of red AND gate • y = 1 → J = K = 1 : flip flop change P. 408

  15. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • 8-bit Counter • Low-order counter is enable for the first 15 clocks. • Low-order counter reaches 1111, OV signal enables the second counter. P. 408

  16. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • MOD-120 Counter(static clear) • Use 74161 counter • When reached maximum, count one more and clear the counter. • 01111000(120) → clear the counter P. 409

  17. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • MOD-120 Counter(synchronous clear) • Use 74163 counter • Detect 119(01110111) and reset it on the next clock pulse P. 409

  18. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • Down/Up’ Counter(1) P. 410

  19. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • Down/Up’ Counter(2) • Load’ = 0 → C = Inc (Preset or Clear) • Load’ = 1, D/U’ = 1→ Clocked count down • Load’ = 1, D/U’ = 0→ Clocked count up P. 410

  20. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • Asynchronous Binary Counter • 7493 Asynchronous Binary Counter P. 410

  21. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • Clock Generator for every 9 input clock(1) • Solution 1 • Use 74163(clocked clear) • 0 1 2 3 4 5 6 7 8 0 1 2 … • D is only 1 in state 8 → Clear Counter P. 412

  22. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • Clock Generator for every 9 input clock(2) • Solution 2 • Use 74161(static clear) • Count 9 before clearing counter • State 9 remains for a short time P. 412 - 413

  23. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • Clock Generator for every 9 input clock(3) • Solution 3 • Use 74163(synchronous load) • 8 9 10 11 12 13 14 15 0 8 9 … • Counter reaches 0, load 8 into counter P. 413

  24. Chapter 8 Solving Larger Sequential Problems 8.2 Counters • Clock Generator for every 9 input clock(4) • Solution 4 • Use 74163(Overflow) • 7 8 9 10 11 12 13 14 15 7 8 9 … • When counter is 15, load 7 into counter P. 414

  25. Chapter 8 Solving Larger Sequential Problems 8.3 Programmable Logic Devices(PLDs) • 16R4 PLD • 16 : number of inputs to the and array • 4 : number of flip flops P. 415

  26. Chapter 8 Solving Larger Sequential Problems 8.3 Programmable Logic Devices(PLDs) • Up/Down Counter using PLD • Output F : counter is saturated • Output G : counter is recycling • State Table • Equations • DA = x’A’B + x’AB’ + x’yA + xAB + xy’A’B’ • DB = x’yA + AB’ + x’B’ + y’B’ • F = x’yAB + xyA’B’ • G = x’y’AB + xy’A’B’ P. 416

  27. Chapter 8 Solving Larger Sequential Problems 8.3 Programmable Logic Devices(PLDs) • Up/Down Counter using PLD P. 417

  28. Chapter 8 Solving Larger Sequential Problems 8.3 Programmable Logic Devices(PLDs) • PLDs • Complex Programmable Logic Device(CPLD) • Array of PLD-like blocks and programmable interconnection networks • Field Programmable Logic Device(FPLD) • For larger circuits • Basic Building Block • General purpose logic generator • Multiplexer • Flip flop • These blocks are connected by a programmable routing networks.

  29. Chapter 8 Solving Larger Sequential Problems 8.3 Programmable Logic Devices(PLDs) • PLDs • Field Programmable Logic Device(FPLD) • Lookup Table(LUT) P. 418

  30. Chapter 8 Solving Larger Sequential Problems 8.3 Programmable Logic Devices(PLDs) • PLDs • Field Programmable Logic Device(FPLD) • Lookup Table(LUT) • Example • Cells are programmed ( 0 0 0 1 0 0 1 1) • f = x’yz + xyz’ + xyz = yz + xy

  31. Chapter 8 Solving Larger Sequential Problems 8.3 Programmable Logic Devices(PLDs) • PLDs • Field Programmable Logic Device(FPLD) • f = x1x2’ + x2x3 P. 419

  32. Chapter 8 Solving Larger Sequential Problems 8.4 Design Using ASM Diagram • Basic Blocks • State Box • Moore type output, one entry point, one exit point • Decision Box • Two way branch exit point, one entry point • Mealy Output Box • One entry point, one exit point • Output when state transition takes place. P. 420

  33. Chapter 8 Solving Larger Sequential Problems 8.4 Design Using ASM Diagram • Basic Blocks • State A and input x = 1 • System goes State B and output z • State A and input x = 0 • System goes State A P. 420

  34. Chapter 8 Solving Larger Sequential Problems 8.4 Design Using ASM Diagram • Moore State Diagram • z = 1 iff x has been 1 for 3 consecutive clock times P. 421

  35. Chapter 8 Solving Larger Sequential Problems 8.4 Design Using ASM Diagram • Mealy State Diagram • z = 1 iff x has been 1 for 3 consecutive clock times P. 422

  36. Chapter 8 Solving Larger Sequential Problems 8.4 Design Using ASM Diagram • Serial Adder Controller • 16 bit shift register • Two operands are already loaded into registers P. 422

  37. Chapter 8 Solving Larger Sequential Problems 8.4 Design Using ASM Diagram • Serial Adder Controller • s : start the addition • d : done • Registers shifted right • s = 1 : state 00 → 01 • N = 111 : state 01 → 10 • Next clock : state 10 → 00 P. 423

  38. Chapter 8 Solving Larger Sequential Problems 8.5 One Hot Encoding • One Hot Encoding • Design an ASM diagram • Use one flip flop for each state • That flip flop is 1 and all other are 0 • Example with Figure 8.18 • Four state → four flip flop(A,B,C,D) • A* = x’(A + B + C + D) = x’ • B* = xA • C* = xB • D* = x(C + D) • z = D • This approach is sometime used in designing large controllers.

  39. Chapter 8 Solving Larger Sequential Problems 8.6 Verilog for Sequential Systems • Structural Model of a D flip flop module D_ff ( q, clk, D, CLR); input clk, D, CLR; output q; reg q; always @ (negedge clk or negedge CLR) begin if (!CLR) q <= 0; else q <= D; end end module

  40. Chapter 8 Solving Larger Sequential Problems 8.6 Verilog for Sequential Systems • Shift Register using flip flop module module shift ( Q, x, clk, CLR); input x, clk, CLR; output [7:0]Q; wire [7:0]Q; D_ff Stage 7 (Q[7], x, clk, CLR); D_ff Stage 6 (Q[6], Q[7], clk, CLR); D_ff Stage 5 (Q[5], Q[6], clk, CLR); D_ff Stage 4 (Q[4], Q[5], clk, CLR); D_ff Stage 3 (Q[3], Q[4], clk, CLR); D_ff Stage 2 (Q[2], Q[3], clk, CLR); D_ff Stage 1 (Q[1], Q[2], clk, CLR); D_ff Stage 0 (Q[0], Q[1], clk, CLR); end module

  41. Chapter 8 Solving Larger Sequential Problems 8.6 Verilog for Sequential Systems • Single Module Shifter module shift ( Q, x, clk, CLR); input x, clk, CLR; output [7:0]Q; reg [7:0]Q; always (@ negedge clk) begin Q[0] <= Q[1]; Q[1] <= Q[2]; Q[2] <= Q[3]; Q[3] <= Q[4]; Q[4] <= Q[5]; Q[5] <= Q[6]; Q[6] <= Q[7]; Q[7] <= x; end module

  42. Chapter 8 Solving Larger Sequential Problems 8.7 More Complex Examples • First Example • The system keeps track of how many consecutive 1 inputs occur on input line x and then, starting at the first time that the input x is 0, it outputs on line z that same number of 1’s at consecutive clocks(z is 0 at all other time) • Simple timing trace P. 426

  43. Chapter 8 Solving Larger Sequential Problems 8.7 More Complex Examples • First Example • Solution 1 • x = 1 → counts up • x = 0 → counts down, output 1 • Max counting range : 14 P. 426

  44. Chapter 8 Solving Larger Sequential Problems 8.7 More Complex Examples • First Example • Solution 2 • Max counting range : 15 • Large number of 1’s • Count 15 and output 1 P. 427

  45. Chapter 8 Solving Larger Sequential Problems 8.7 More Complex Examples • First Example • Solution 3(ignore inputs until 1 outputs are completed) • x = 0 Q = 0 count = 0 EN = 0 z = 0 D/U = X • x = 0 Q = 0 count = 1 EN = 1 z = 1 D/U = 1 • x = 0 Q = 0 count > 1 EN = 1 z = 1 D/U = 1 Q <- 1 • x = 1 Q = 0 count ≠ 15 EN = 1 z = 0 D/U = 0 • x = 1 Q = 0 count = 15 EN = 0 z = 0 D/U = X • x = X Q = 1 count > 1 EN = 1 z = 1 D/U = 1 • x = X Q = 1 count = 1 EN = 1 z = 1 D/U = 1 Q <- 0

  46. Chapter 8 Solving Larger Sequential Problems 8.7 More Complex Examples • First Example • Solution 3(ignore inputs until 1 outputs are completed) • J = x (D + C + B) K = D C B A • z = Q + x Q (D + C + B + A) = Q + x (D + C + B + A) • EN = x (A B C D) + z • D/U = Q + x (D + C + B + A)

  47. Chapter 8 Solving Larger Sequential Problems 8.7 More Complex Examples • First Example • Solution 4 (using shift register) • 12 bit shift register • X = 1 → S0 = 1, S1 = 0, shift right, leftmost bit = 1 • X = 0 → shift left, loading 0 from right P. 429

  48. Chapter 8 Solving Larger Sequential Problems 8.7 More Complex Examples • Second Example • Count 16 state (1 2 4 7 11 0 6 13 5 14 8 3 15 12 10 9) • State Table P. 430

  49. Chapter 8 Solving Larger Sequential Problems 8.7 More Complex Examples • Second Example • Solution 1 (4 JK ff) • JD = CA + CB + BA, KD = C’B’ + C’A + B’A • JC = D’A’ + D’B, KC = DA’ + D’BA • JB = D’ + A’, KB = D + A’ • JA = D’C + DC’, KA = D’B’ + CB’ + DC’B P. 430

  50. Chapter 8 Solving Larger Sequential Problems 8.7 More Complex Examples • Second Example • Solution 2 (Decoder Block) • Truth Table P. 431

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