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?

2

“A circuit whose output depends on current inputs and past outputs”

“A circuit with memory”

Part II. Sequential Networks (Ch. 3)

Memory: Flip flopsSpecification: Finite State MachinesImplementation: Excitation TablesMain Theme: Timing

Present time = t and next time = t+1Timing constraints to separate the present and next times.

Memory / Time steps

Clock

xi yi

si

yi = fi(St,X)si

t+1= gi(St,X)

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

5

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.

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

QQQ

Q

I1

I2

I2 I1

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

QQQ

Q

I1

I2

I2 I1

8

Capacitive Loads

Q

Q

I1

I2

0

1

1

0

• Consider the two possible cases:– Q = 0: then Q’ = 1 and Q = 0 (consistent)

– Q = 1: then Q’ = 0 and Q = 1 (consistent)

– Bistable circuit stores 1 bit of state in the state variable, Q (or Q’ )

• But there are no inputs to control the state

Q

Q

I1

I2

1

0

0

1

9

iClicker

10

Q. Given a memory component made out of a loop of inverters, the number of inverters has to be

A.Even

B.Odd

SR (Set/Reset) Latch

R

S

Q

Q

N1

N2

• SR Latch

• 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:

– S = 0, R = 1:

R

S

Q

Q

N1

N2

0

1

R

S

Q

Q

N1

N2

1

0

12

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

– S = 0, R = 1: then Q = 0 and Q = 1

R

S

Q

Q

N1

N2

0

1

R

S

Q

Q

N1

N2

1

0

13

SR Latch Analysis

–S = 1, R = 1:R

S

Q

Q

N1

N2

1

1

14

SR Latch Analysis– S = 0, R = 0:

15

R

S

Q

Q

N1

N2

SR Latch Analysis– S = 0, R = 0: then Q = Qprev

– S = 1, R = 1: then Q = 0 and Q = 0

R

S

Q

Q

N1

N2

1

1

R

S

Q

Q

N1

N2

0

0

R

S

Q

Q

N1

N2

0

0

0

Qprev = 0 Qprev = 1

16

S

R

y

QQ = (R+y)’

y = (S+Q)’

17

Flip-flop Components

S

R

SR F-F (Set-Reset)

Inputs: S, R State: (Q, y)

y

Q

18

Id Q(t) y(t) S R Q(t1) y(t1) Q(t2)y(t2) Q(t3) y(t3)

0 0 0 0 0 1 1 0 0 1 1

1 0 0 0 1 0 1 0 1 0 12 0 0 1 0 1 0 1 0 1 03 0 0 1 1 0 0 0 0 0 04 0 1 0 0 0 1 0 1 0 15 0 1 0 1 0 1 0 1 0 16 0 1 1 0 0 0 1 0 1 07 0 1 1 1 0 0 0 0 0 08 1 0 0 0 1 0 1 0 1 09 1 0 0 1 0 0 0 1 0 110 1 0 1 0 1 0 1 0 1 011 1 0 1 1 0 0 0 0 0 012 1 1 0 0 0 0 1 1 0 013 1 1 0 1 0 0 0 1 0 114 1 1 1 0 0 0 1 0 1 015 1 1 1 1 0 0 0 0 0 0

Q y

State

TransitionSR

1010

00

11

00

10SR

1110

01 11

0111

01

10 0010

0001

0011

State Diagram

01

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

20

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)

SR Latch Analysis– S = 0, R = 0: then Q = Qprev and Q = Qprev (memory!)

– S = 1, R = 1: then Q = 0 and Q = 0 (invalid state: Q ≠ NOT Q)

R

S

Q

Q

N1

N2

1

1

0

00

0

R

S

Q

Q

N1

N2

0

0

1

01

0

R

S

Q

Q

N1

N2

0

0

0

10

1

Qprev = 0 Qprev = 1

21

CASESSR=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 changingSolutions: Avoid the two cases

1) SR = (0,0), 2) SR = (1,1).

0 0 0 1 -1 1 0 1 -

PSinputs

00 01 10 11

State table

Q(t+1)

SR

Characteristic ExpressionQ(t+1) = S(t)+R’(t)Q(t)

NS (next state)

Q(t)

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)

• Must do something to avoid

invalid state (when S = R = 1)

S

R Q

Q

SR LatchSymbol

23

D Latch

D LatchSymbol

CLK

D Q

Q

• Two inputs: CLK, D– CLK: controls when the output changes

– D (the data input): controls what the output changes to

• Function– When CLK = 1, D passes through to Q

(the latch is transparent)

– When CLK = 0, Q holds its previous value (the latch is opaque)

• Avoids invalid case when Q ≠ NOT Q

24

D Latch Internal Circuit

CLK

D Q

Q

25

S

R Q

Q

SR LatchSymbol

D Latch Internal Circuit

S

R Q

Q

Q

QD

CLKD

R

S

CLK

D Q

Q

S R Q QCLK D

0 X1 01 1

D

26

D Latch Internal Circuit

S

R Q

Q

Q

QD

CLKD

R

S

CLK

D Q

Q

S R Q

0 0 Qprev0 1 01 0 1

Q

10

CLK D

0 X1 01 1

D

X10

Qprev

27

D Flip-Flop• Two inputs: CLK, D

• Function

– The flip-flop “samples” D on the rising edge of CLK

• When CLK rises from 0 to 1, D passes through to Q

• Otherwise, Q holds its previous value

– Q changes only on the rising edge of CLK

• A flip-flop is called an edge-triggered device because it is activated on the clock edge

D Flip-FlopSymbols

D Q

Q

28

D Flip-Flop Internal Circuit

CLK

D Q

Q

CLK

D Q

Q

Q

Q

DN1

CLK

L1 L2

29

D Flip-Flop Internal Circuit

CLK

D Q

Q

CLK

D Q

Q

Q

Q

DN1

CLK

L1 L2

• Two back-to-back latches (L1 and L2) controlled by complementary clocks

• When CLK = 0– L1 is transparent, L2 is opaque

– D passes through to N1

• When CLK = 1– 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

CLK

D Q

Q

CLK

D Q

Q

Q

Q

DN1

CLK

L1 L2

30

D Flip-Flop vs. D LatchCLK

D Q

Q

D Q

Q

CLK

D

Q (latch)

Q (flop)

31

D Flip-Flop vs. D LatchCLK

D Q

Q

D 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)

D

CLK

Q

Q’

Id D Q(t) Q(t+1)

0 0 0 0

1 0 1 0

2 1 0 1

3 1 1 1

Characteristic Expression: Q(t+1) = D(t)

0 0 11 0 1

PSD

0 1

State table

NS= Q(t+1)

34

CLK

D Q

Q

CLK

D Q

Q

Q

Q

DN1

CLK

L1 L2

iClicker

35

Can D flip-flip serve as a memory component?A.YesB.No

JK F-F

J

CLK

Q

Q’ 0 0 0 1 ?1 1 0 1 ?

PSJK

00 01 10 11

State table

Q(t+1)

K

36

JK F-F

J

CLK

Q

Q’

Characteristic ExpressionQ(t+1) = Q(t)K’(t)+Q’(t)J(t)

0 0 0 1 11 1 0 1 0

PSJK

00 01 10 11

State table

Q(t+1)

K

37

T

CLK

Q

Q’

Characteristic Expression Q(t+1) = Q’(t)T(t) + Q(t)T’(t)

0 0 11 1 0

PS T 0 1

State table

Q(t+1)

T Flip-Flop (Toggle)

38

Using a JK F-F to implement a D and T F-F

J

K

Q

Q’

xCLK

39

iClickerWhat is the function of the above circuit?A.D F-FB.T F-FC.None of the above

Using a JK F-F to implement a D and T F-F

J

K

Q

Q’

TCLK

T flip flop

40

Reading

41

[Harris] Chapter 3: 3.3, 3.4.1, 3.4.2