Lecture 7: Sequential Networks CSE 140: Components and Design - PowerPoint PPT Presentation

Lecture 7: Sequential Networks CSE 140: Components and Design Techniques for Digital Systems Fall 2014 CK Cheng Dept. of Computer Science and Engineering University of California, San Diego 1

What is a sequential circuit? “A circuit whose output depends on current inputs and past outputs” “A circuit with memory” 2

Part II. Sequential Networks (Ch. 3) Memory / Time steps s i x i y i y i = f i (S t ,X) Clock t+1 = g i (S t ,X) s i Memory: Flip flops Specification: Finite State Machines Implementation: Excitation Tables Main Theme: Timing Present time = t and next time = t+1 Timing constraints to separate the present and next times. 3

Sequential Networks • Memory Components – Hierarchy of Memory – Basic Mechanism of Memory – Types of Flip-Flops • Implementation – Finite State Machine 4

The usage of a sequential machine iClicker Question: A. Digital systems are implemented using sequential machines. B. Only a small subset of digital systems can be implemented using sequential machines. C. Sequential machines are too simple for complicated digital systems. 5

Hierarchy of Memory Devices • Memory Bank (Farms of memory cells) • Register (A vector of memory cells) • Flip-Flop (One bit memory cell) – SR, D, T, JK flip-flops (Different types of memory cells) – State Tables (Truth table of sequential machine) – Characteristic Expressions (Switching algebraic expression of sequential machine) 6

Fundamental Memory Mechanism I1 Q Q Q I2 I1 I2 Q 7

Memory Mechanism: Capacitive Load • Fundamental building block of sequential circuits • Two outputs: Q , Q • There is a feedback loop! • In a typical combinational logic, there is no feedback loop. • No inputs I1 Q Q Q I2 I1 I2 Q 8

Capacitive Loads • Consider the two possible cases: – Q = 0: then Q’ = 1 and Q = 0 (consistent) 0 I1 Q 1 0 1 I2 Q – Q = 1: then Q ’ = 0 and Q = 1 (consistent) 1 I1 Q 0 1 0 I2 Q – Bistable circuit stores 1 bit of state in the state variable, Q (or Q’ ) • But there are no inputs to control the state 9

iClicker Q. Given a memory component made out of a loop of inverters, the number of inverters has to be A. Even B. Odd 10

SR (Set/Reset) Latch • SR Latch R N1 Q N2 Q S • Consider the four possible cases: – S = 1, R = 0 – S = 0, R = 1 – S = 0, R = 0 – S = 1, R = 1 11

SR Latch Analysis – S = 1, R = 0: 0 R N1 Q 1 N2 Q S 1 – S = 0, R = 1: R N1 Q 0 N2 Q S 12

SR Latch Analysis – S = 1, R = 0: then Q = 1 and Q = 0 0 R N1 Q 1 N2 Q S – S = 0, R = 1: then Q = 0 and Q = 1 1 R N1 Q 0 N2 Q S 13

SR Latch Analysis – S = 1, R = 1: 1 R N1 Q 1 N2 Q S 14

SR Latch Analysis – S = 0, R = 0: R N1 Q N2 Q S 15

SR Latch Analysis – S = 0, R = 0: then Q = Q prev Q prev = 0 Q prev = 1 0 0 R 0 R N1 Q N1 Q 0 N2 Q 0 N2 Q S S – S = 1, R = 1: then Q = 0 and Q = 0 1 R N1 Q 1 N2 Q S 16

y S y = (S+Q)’ Q Q = (R+y)’ R 17

Flip-flop Components SR F-F (Set-Reset) y S R Q Inputs: S, R State: (Q, y) 18

Id Q(t) y(t) S R Q(t 1 ) y(t 1 ) Q(t 2 )y(t 2 ) Q(t 3 ) y(t 3 ) State 0 0 0 0 0 1 1 0 0 1 1 1 0 0 0 1 0 1 0 1 0 1 Q y 2 0 0 1 0 1 0 1 0 1 0 3 0 0 1 1 0 0 0 0 0 0 SR 4 0 1 0 0 0 1 0 1 0 1 Transition 5 0 1 0 1 0 1 0 1 0 1 6 0 1 1 0 0 0 1 0 1 0 7 0 1 1 1 0 0 0 0 0 0 State Diagram 8 1 0 0 0 1 0 1 0 1 0 00 9 1 0 0 1 0 0 0 1 0 1 01 10 00 10 1 0 1 0 1 0 1 0 1 0 10 01 10 10 01 11 1 0 1 1 0 0 0 0 0 0 11 12 1 1 0 0 0 0 1 1 0 0 10 01 11 13 1 1 0 1 0 0 0 1 0 1 10 00 SR 14 1 1 1 0 0 0 1 0 1 0 11 01 15 1 1 1 1 0 0 0 0 0 0 00 00 11 11 19

CASES: SR=01, (Q,y) = (0,1) SR=10, (Q,y) = (1,0) SR=11, (Q,y) = (0,0) SR = 00 => if (Q,y) = (0,0) or (1,1), the output keeps changing Q. To avoid the SR latch output from toggling or behaving in an undefined way which input combinations should be avoided: A. (S, R) = (0, 0) B. (S, R) = (1, 1) 20

SR Latch Analysis – S = 0, R = 0: then Q = Q prev and Q = Q prev (memory!) Q prev = 0 Q prev = 1 0 0 R 0 R 1 N1 Q N1 Q 1 0 0 1 1 0 0 N2 Q 0 N2 Q S S – S = 1, R = 1: then Q = 0 and Q = 0 (invalid state: Q ≠ NOT Q ) 1 R 0 N1 Q 0 0 0 1 N2 Q S 21

CASES SR=01: (Q,y) = (0,1) SR=10: (Q,y) = (1,0) SR=11: (Q,y) = (0,0) SR = 00: if (Q,y) = (0,0) or (1,1), the output keeps changing Solutions: Avoid the two cases 1) SR = (0,0), 2) SR = (1,1). State table SR inputs 00 01 10 11 PS 0 0 0 1 - Characteristic Expression Q(t) 1 1 0 1 - Q(t+1) = S(t)+R’(t)Q(t) Q(t+1) NS (next state) 22

SR Latch Symbol • SR stands for Set/Reset Latch – Stores one bit of state ( Q ) • Control what value is being stored with S , R inputs – Set: Make the output 1 ( S = 1, R = 0, Q = 1 ) – Reset: Make the output 0 ( S = 0, R = 1, Q = 0 ) SR Latch • Must do something to avoid Symbol invalid state (when S = R = 1) R Q S Q 23

D Latch • Two inputs: CLK , D – CLK : controls when the output changes – D (the data input): controls what the D Latch output changes to Symbol • Function CLK – When CLK = 1, D passes through to Q D Q (the latch is transparent ) – When CLK = 0, Q holds its previous Q value (the latch is opaque ) • Avoids invalid case when Q ≠ NOT Q 24

D Latch Internal Circuit SR Latch Symbol CLK R Q D Q S Q Q 25

D Latch Internal Circuit CLK CLK R R Q Q D D Q S S Q Q D Q CLK D D S R Q Q 0 X 1 0 1 1 26

D Latch Internal Circuit CLK CLK R R Q Q D D Q S S Q Q Q D CLK D D S R Q Q 0 X X 0 0 Q prev Q prev 1 0 1 0 1 0 1 1 1 0 1 0 1 0 27

D Flip-Flop • Two inputs: CLK , D • Function – The flip-flop “samples” D on the rising D Flip-Flop edge of CLK Symbols • When CLK rises from 0 to 1, D passes through to Q • Otherwise, Q holds its previous value D Q – Q changes only on the rising edge of CLK Q • A flip-flop is called an edge-triggered device because it is activated on the clock edge 28

D Flip-Flop Internal Circuit CLK CLK CLK N1 D D Q D Q Q L1 Q L2 Q Q 29

D Flip-Flop Internal Circuit • Two back-to-back latches (L1 and L2) controlled by complementary clocks • When CLK = 0 CLK CLK – L1 is transparent, L2 is opaque – D passes through to N1 CLK CLK CLK CLK N1 N1 D D D D Q Q D D Q Q Q Q • When CLK = 1 L1 L1 Q Q L2 L2 Q Q Q Q – L2 is transparent, L1 is opaque – N1 passes through to Q • Thus, on the edge of the clock (when CLK rises from 0 1) – D passes through to Q 30

D Flip-Flop vs. D Latch CLK D Q D Q Q Q CLK D Q (latch) Q (flop) 31

D Flip-Flop vs. D Latch CLK D Q D Q Q Q CLK D Q (latch) Q (flop) 32

Latch and Flip-flop (two latches) A latch can be considered as a door CLK = 0, door is shut CLK = 1, door is unlocked A flip-flop is a two door entrance CLK = 1 CLK = 0 CLK = 1 33

D Flip-Flop (Delay) CLK Q D CLK CLK N1 D D Q D Q Q CLK L1 Q L2 Q Q Q’ Id D Q(t) Q(t+1) State table 0 0 0 0 PS D 0 1 1 0 1 0 0 0 1 2 1 0 1 1 0 1 3 1 1 1 NS= Q(t+1) Characteristic Expression: Q(t+1) = D(t) 34

iClicker Can D flip-flip serve as a memory component? A.Yes B.No 35

JK F-F State table Q J JK 00 01 10 11 PS CLK 0 0 0 1 ? Q’ K 1 1 0 1 ? Q(t+1) 36

JK F-F State table Q J JK 00 01 10 11 PS CLK 0 0 0 1 1 Q’ K 1 1 0 1 0 Q(t+1) Characteristic Expression Q(t+1) = Q(t)K’(t)+Q’(t)J(t) 37

T Flip-Flop (Toggle) State table Q T PS T 0 1 CLK 0 0 1 Q’ 1 1 0 Q(t+1) Characteristic Expression Q(t+1) = Q’(t)T(t) + Q(t)T’(t) 38

Using a JK F-F to implement a D and T F-F Q x J CLK Q’ K iClicker What is the function of the above circuit? A. D F-F B. T F-F C. None of the above 39

Using a JK F-F to implement a D and T F-F Q T J CLK Q’ K T flip flop 40

Reading [Harris] Chapter 3: 3.3, 3.4.1, 3.4.2 41