chapter 6eestaff.kku.ac.th/~sa-nguan/en812200/2021-chap 6...digital logic design @department of...

Chapter 6

Sequential System Design

Computer Engineering KhonKaen University

1

http://www.free-powerpoint-templates-design.com/free-powerpoint-templates-design

Digital Logic Design @Department of Computer Engineering KKU

Topic

❑Shift Registers

❑Synchronous & Asynchronous

Counters

❑Programmable Logic Devices (PLD)

❑Hardware Design Languages

2


Shift Register

▪ data is often stored in registers, rather than

individual flip-flops.

▪ A register is just a collection of flip-flops,

often with a common name (using subscripts

to indicate the individual flip-flops) and

usually with a common clock.

▪ A shift register is a set of flip-flops, such that

the data moves one place to the right on

each clock or shift input.

3


A simple shift register

and its timing trace

x 1 0 1 1 1 0 1 1 1 1 0 0 0

Q10 1 0 1 1 1 0 1 1 1 1 0 0 0

Q2 0 0 1 0 1 1 1 0 1 1 1 1 0 0 0

Q3 0 0 0 1 0 1 1 1 0 1 1 1 1 0 0 0

Q4 0 0 0 0 1 0 1 1 1 0 1 1 1 1 0 0 0

Clock

S

R

q1

q2

S q1

q2R

S q1

q2R

S q1

q2R

X

4


Shift Register (cont.)

▪ Serial-in / Parallel-in Shift Registers

▪ Serial-out / Parallel-out Shift Registers

▪ Right / Left Shift Registers

▪ Leading-edge / Trailing-edge Shift Registers

▪ Enable, Clear, Load pins

5


74164

Serial-in parallel-out shift register

Serial-in

Parallel-out

AQD

SET

CLR

Q

D Q

SET

CLR

Q

SET

D Q

CLR

Q

SET

QD

CLR

Q

QD

SET

CLRQ

SET

QD

CLR

Q

SET

QD

CLR

Q

D

SET

Q

CLR

Q

B

Clock

Clear'

q1 q2 q3 q4 q5 q6 q7 q8

6


Parallel-in Shift Register

loading is donestatically

loading is donesynchronously

Clock

(a) 74165

Q

SET

S

CLR

QREnable'

Load'

q1

q1'

q2

q2'

IN2

Clock

Q

SET

S

CLR

QR

q2

q2'

Enable'

q1

IN2

Load'Clear'

(b) 74166

7


Clear’ S0 S1 q1* q2* q3* q4*

Static clear 0 X X 0 0 0 0

Hold 1 0 0 q1 q2 q3 q4

Shift left 1 0 1 q2 q3 q4 LS

Shift right 1 1 0 RS q1 q2 q3

Load 1 1 1 IN1 IN2 IN3 IN4

Figure 6.7 Right/left shift register.

Table 6.1 Right/left shift register.

Clock

S q2

CLR'

q'2R

q1

IN2

q3

Clear'

S1

S0

8


Clock

Clear'

CJ

CLR'

C'K

Load'

B

A

ENP

ENT

INC z

x

y

w

Figure 6.8 The 74161 counter.

ENT

74161

C B ADENP

CLR'

Load'INAINBINCIND

Clock

OV

9


Figure 6.9 8-bit counter.

8-bit counter from two 74161 counters

can count to 255 (28 - 1)

12-bit counter from three 74161 counterscan count to 4095 (212 - 1)

Clear'

High

D C B A

OV

CLR' LD' ENT ENP

Low

D C B A

OV

CLR' LD' ENT ENP

Clock

1

O7 O6 O5 O4 O3 O2 O1 O0

10


Figure 6.10 Typical bit of the 7419 Down/Up’ counter.

LD’ EN’ D/U’

0 X X Static load

1 1 X Do nothing

1 0 0 Clocked count up

1 0 1 Clocked count down

Figure 6.11 The 74191 Down/Up’ counter.

Load'

Clock

CJ

CLR'

C'K

PRE'

Enable'

D/U'

INC

A'B'

AB

AD C B A

OV

CLR' LD' EN' D/U'

B

C

D

IN

11


Flip-flop Design Techniques

▪ D Flip-flop

▪ SR Flip-flop

▪ T Flip-flop

▪ JK Flip-flop

q*=D

qKqJq* +=

qRSq* +=

q*=Tq

12


Flip-flop Design Techniques (cont.)

From the truth table, it’s

clear that

We need to create the

appropriate flip-flop

design table to obtain a

truth table for the flip-flop inputs.

z=q1q2

q q* Input(s)

0 0

0 1

1 0

1 1

• D flip-flops

• SR flip-flops

• T flip-flops• JK flip-flops

x q1 q2 q1* q2* z

0 0 0 0 0 0

0 0 1 0 0 0

0 1 0 0 0 0

0 1 1 0 0 1

1 0 0 0 1 0

1 0 1 1 0 0

1 1 0 1 1 0

1 1 1 1 1 1

Main truth table

13


Design with D Flip-flop

D q*

0 0

1 1

q q* D

0 0 0

0 1 1

1 0 0

1 1 1

x q1 q2 D1=q1* D2=q2*

0 0 0 0 0

0 0 1 0 0

0 1 0 0 0

0 1 1 0 0

1 0 0 0 1

1 0 1 1 0

1 1 0 1 1

1 1 1 1 1

From the main truth table

q*=D

14


0

1

1

1

1

00

01

11

10

Xq1q2

0

1

1

1

1

00

01

11

10

Xq1q2

Design with D Flip-flop (cont.)

122 121 xqqxDxqxqD +=+=

15


Implementation using D Flip-flops

back

Clock

ZD q1

q'1

D q2

q'2

X

16


Design with SR Flip-flop

S R q*

0 0 q

0 1 0

1 0 1

1 1 -

q q* S R

0 0 0 X

0 1 1 0

1 0 0 1

1 1 X 0

x q1 q2 q1* q2* S1 R1 S2 R2

0 0 0 0 0 0 X 0 X

0 0 1 0 0 0 X 0 1

0 1 0 0 0 0 1 0 X

0 1 1 0 0 0 1 0 1

1 0 0 0 1 0 X 1 0

1 0 1 1 0 1 0 0 1

1 1 0 1 1 X 0 1 0

1 1 1 1 1 X 0 X 0


qRSq* +=

17


J1

0

1

X

X

1

00

01

11

10

Xq1q2 0

X X

X

1

1

1

00

01

11

10

Xq1q2 0

1

X

1

1

00

01

11

10

Xq1q2 0

X

1 1

1

X

1

00

01

11

10

Xq1q2

K1 J2 K2

Design with SR Flip-flop (cont.)

21222

121

qqxR ; qxS

xR ; xqS

+==

==

18


Design with T Flip-flop

T q*

0 q

1

q q* T

0 0 0

0 1 1

1 0 1

1 1 0

x q1 q2 q1* q2* T1 T2

0 0 0 0 0 0 0

0 0 1 0 0 0 1

0 1 0 0 0 1 0

0 1 1 0 0 1 1

1 0 0 0 1 0 1

1 0 1 1 0 1 1

1 1 0 1 1 0 1

1 1 1 1 1 0 0


q*=Tq

q

19


0

1

1

1

1

00

01

11

10

Xq1q2

0

1

1 1

1

1

1

00

01

11

10

Xq1q2

Design with T Flip-flop (cont.)

21222 2111 qqxqxqxTqqxqxT +=+= +

20


Design with JK Flip-flop

J K q*

0 0 q

0 1 0

1 0 1

1 1

q q* J K

0 0 0 X

0 1 1 X

1 0 X 1

1 1 X 0

x q1 q2 q1* q2* J1 K1 J2 K2

0 0 0 0 0 0 X 0 X

0 0 1 0 0 0 X X 1

0 1 0 0 0 X 1 0 X

0 1 1 0 0 X 1 X 1

1 0 0 0 1 0 X 1 X

1 0 1 1 0 1 X X 1

1 1 0 1 1 X 0 1 X

1 1 1 1 1 X 0 X 0

q


qKqJq* +=

21


Design with JK Flip-flop (cont.)

122

121

qxK ;x J

xK ; xqJ

+==

==

J1

0

1

X X

X X

1

00

01

11

10

Xq1q2 0

X X

X X

1

1

1

00

01

11

10

Xq1q2 0

1

X X

X X

1

1

00

01

11

10

Xq1q2 0

X X

1 1

1

X X

1

00

01

11

10

Xq1q2

K1 J2 K2

22


Quick method for JK

*qJ

J0K1Jq*

=

=+=

*qK

K1K0Jq*

=

=+=

Because

Notice that when q=0

And when q=1

qKqJq* +=

23


Quick method for JK (cont.)

state table to maps

q1q2

q1*q2*

ZX = 0 X = 1

00 0 0 0 1 0

01 0 0 1 0 0

10 0 0 1 1 0

11 0 0 1 1 1 q1*

0

1

1

1

1

00

01

11

10

Xq1q2 0

1

1

1

1

00

01

11

10

Xq1q2

q2*

24


Quick method for JK (cont.)

Conventional q q* J K

0 0 0 X

0 1 1 X

1 0 X 1

1 1 X 0

First column

of J1 and K1

Second column of J1 and K1

J1

0

1

1

1

1

0

1

1

0

Xq1q2 0

1

x x

x x

1

00

01

11

10

Xq1q2 0

x x

x x

1

1

1

00

01

11

10

Xq1q2

K1q1*

0

0

1

1

J1

0

1

1

1

1

0

1

1

0

Xq1q2 0

1

x x

x x

1

00

01

11

10

Xq1q2 0

x x

x x

1

1

1

00

01

11

10

Xq1q2

K1q1*

0

0

1

1

25


Quick method for JK (cont.)Computation of J2 and K2

Conventional

J2

0

1

1

1

1

0

0

1

1

Xq1q2 0

1

x x

x x

1

1

00

01

11

10

Xq1q2 0

x x

1 1

1

x x

1

00

01

11

10

Xq1q2

K2q2*

0

1

1

0

26


Quick method for JK (cont.)Computation of J1 and K1

Quick method0

1

1

1

1

0

1

1

0

Xq1q2

q1*

1

1

1

0 1 0 1

0

1

1

0

J1

Xq2Xq2

0

0

1

1

27


Example 1

For each of the following state tables and state

assignments, find the flip flop input equations and

the system output equation for an implementation

using

a. D flip flopb. JK flip flop

q q* z

X = 0 X = 1 X = 0 X = 1

A B C 1 1

B A B 1 0

C B A 1 0

q q1 q2

A 1 1

B 1 0

C 0 1

28


Example 1 (cont.)

x q1 q2 z q*1 q*2

0 0 0 X X X

0 0 1 1 1 0

0 1 0 1 1 1

0 1 1 1 1 0

1 0 0 X X X

1 0 1 0 1 1

1 1 0 0 1 0

1 1 1 1 0 1

We will first produce the truth table.

q

C

B

A

C

B

A

29


Z

0

X X

1

1 1

1

1

00

01

11

10

Xq1q2

0

X X

1 1

1

1 1

1

00

01

11

10

Xq1q2

0

X X

1

1

1

1

00

01

11

10

Xq1q2

D1 D2

Example 1 (cont.)

a. We can now map z, D1(q*1) and D2(q*2)

z = x’ + q1q2 D1 = x’ + q’1 + q’2 D2 = x’q’2 + q2

30


b. Using the quick method, we look at the shaded

part of the maps for J and the unshaded parts for K. (z is unchanged.)

0

X X

1

1 1

1

1

00

01

11

10

Xq1q2

0

X X

1

1

1

1

00

01

11

10

Xq1q2

J1 = 1 K1 = x q2 J2 = x’ K2 = x’

31


Example 2

For the state table and each of the state

assignments shown, design a system using D flip flops.

q q* z

X = 0 X = 1 X = 0 X = 1

A B C 1 0

B D A 0 0

C B C 1 1

D D A 1 0

q q1 q2

A 0 0

B 0 1

C 1 1

D 1 0

y q1 q2

A 0 0

B 1 1

C 1 0

D 0 1

a. b.

32


Example 2 (cont.)

a. This assignment is in map order and once

again we will go directly to the maps without first producing a truth table.

0

1

1

1

1

1

00

01

11

10

Xq1q2

0

1 1

1 1

1

00

01

11

10

Xq1q2

0

1

1 1

1

1

00

01

11

10

Xq1q2

D1 D2 z

D1 = x’q’1q2 + x’q1q’2 + xq’1q’2 + xq1q2

D2 = q’1q’2 + q1q2 z = x’q’2 + q1q2

33


Example 2 (cont.)

b. For this part, we will first construct the truth table and then go to the maps.

x q1 q2 z q*1 q*2

0 0 0 1 1 1

0 0 1 1 0 1

0 1 0 1 1 1

0 1 1 0 0 1

1 0 0 0 1 0

1 0 1 0 0 0

1 1 0 1 1 0

1 1 1 0 0 0

q

A

D

C

B

A

D

C

B

34


Example 2 (cont.)

0

1 1

1 1

1

00

01

11

10

Xq1q2

0

1

1

1

1

1

00

01

11

10

Xq1q2

0

1

1

1 1

1

00

01

11

10

Xq1q2

D1 D2 z

D1 = q’2 D1 = x’ z = x’ q’1 + q1q’2

We really did not need to map D1 and D2; we should be able to see those from the truth table.

35


Design of Synchronous

Counters

D C B A D* C* B* A*

0 0 0 0 0 0 0 1

0 0 0 1 0 0 1 0

0 0 1 0 0 0 1 1

0 0 1 1 0 1 0 0

0 1 0 0 0 1 0 1

0 1 0 1 0 1 1 0

0 1 1 0 0 1 1 1

0 1 1 1 1 0 0 0

1 0 0 0 1 0 0 1

1 0 0 1 0 0 0 0

1 0 1 0 x x x x

… … … … … … … …

1 1 1 1 X x x x

▪ Design a decimal or

decade counter

using JK flip-flops:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, …

36


D C B A D* C* B* A* TD TC TB TA

0 0 0 0 0 0 0 1 0 0

0 0 0 1 0 0 1 0 0 0

0 0 1 0 0 0 1 1 0 0

0 0 1 1 0 1 0 0 0 1

0 1 0 0 0 1 0 1 0 0

0 1 0 1 0 1 1 0 0 0

0 1 1 0 0 1 1 1 0 0

0 1 1 1 1 0 0 0 1 1

1 0 0 0 1 0 0 1 0 0

1 0 0 1 0 0 0 0 1 0

1 0 1 0 x x x x X X X X

… … … … … … … …

1 1 1 1 X x x x X X X X

37


Design of Synchronous Counters (cont.)

D* C*

B* A*

D

1

X 1

X

X X

X X

DDC

B

B

A

A

A

CCC

BA D

1

1

1

1

X

X

X X

X X

DDC

B

B

A

A

A

CCC

BA

D

1 1

1 1

X

X

X X

X X

DDC

B

B

A

A

A

CCC

BA D

1 1

1 1

X 1

X

X X

X X

DDC

B

B

A

A

A

CCC

BA

38



D* JD

KDAK ;CBA J DD ==

From the main truth table of the D*

D

1

X 1

X

X X

X X

DDC

B

B

A

A

A

CCC

BA D

1

X X

X X

X X

X X

DDC

B

B

A

A

A

CCC

BA

D

X X

X X

X X

X X

X

X 1

X X

X X

DDC

B

B

A

A

A

CCC

BA

39



C* JC

KC

BAKJ CC ==

From the main truth table of the C*

D

1

1

1

1

X

X

X X

X X

DDC

B

B

A

A

A

CCC

BA D

X

X

1 X

X

X

X

X X

X X

DDC

B

B

A

A

A

CCC

BA

D

X

X

X 1

X

X X

X X

X X

X X

DDC

B

B

A

A

A

CCC

BA

40



B* JB

KBAK ;A DJ BB ==

From the main truth table of the C*

D

1 1

1 1

X

X

X X

X X

DDC

B

B

A

A

A

CCC

BA D

1 1

X X

X X

X

X

X X

X X

DDC

B

B

A

A

A

CCC

BA

D

X X

X X

1 1

X X

X X

X X

X X

DDC

B

B

A

A

A

CCC

BA

41



A* JA

KA

1K J AA ==

From the main truth table of the A*

D

1 1

1 1

X 1

X

X X

X X

DDC

B

B

A

A

A

CCC

BA D

1 1

X X

X X

1 1

X 1

X X

X X

X X

DDC

B

B

A

A

A

CCC

BA

D

X X

1 1

1 1

X X

X X

X 1

X X

X X

DDC

B

B

A

A

A

CCC

BA

42


Example: A Base-16 CounterD C B A D* C* B* A* remark

0 0 0 0 0 0 0 1 0→1

0 0 0 1 0 0 1 0 1→2

0 0 1 0 0 0 1 1 2→3

0 0 1 1 0 1 0 0 3→4

0 1 0 0 0 1 0 1 4→5

0 1 0 1 0 1 1 0 5→6

0 1 1 0 0 1 1 1 6→7

0 1 1 1 1 0 0 0 7→8

1 0 0 0 1 0 0 1 8→9

1 0 0 1 1 0 1 0 9→A

1 0 1 0 1 0 1 1 A→B

1 0 1 1 1 1 0 0 B→C

1 1 0 0 1 1 0 1 C→D

1 1 0 1 1 1 1 0 D→E

1 1 1 0 1 1 1 1 E→F

1 1 1 1 0 0 0 0 F→0

43


Example: A Base-16 Counter (cont.)

JD = KD = CBA JC = KC = BA

D

1

1 1

1 1

1

1 1

DDC

B

B

A

A

A

CCC

BA D

1

1

1

1

1

1

1

1

DDC

B

B

A

A

A

CCC

BA

44


Example: A Base-16 Counter (cont.)

JB = KB = A JA = KA = 1

D

1 1

1 1

1 1

1 1

DDC

B

B

A

A

A

CCC

BA D

1 1

1 1

1 1

1 1

DDC

B

B

A

A

A

CCC

BA

45


Example:

An Up/Down CounterX C B A C* B* A* remark

0 0 0 0 0 0 1 0→1

0 0 0 1 0 1 0 1→2

0 0 1 0 0 1 1 2→3

0 0 1 1 1 0 0 3→4

0 1 0 0 1 0 1 4→5

0 1 0 1 1 1 0 5→6

0 1 1 0 1 1 1 6→7

0 1 1 1 0 0 0 7→0

1 0 0 0 1 1 1 0→7

1 0 0 1 0 0 0 1→0

1 0 1 0 0 0 1 2→1

1 0 1 1 0 1 0 3→2

1 1 0 0 0 1 1 4→3

1 1 0 1 1 0 0 5→4

1 1 1 0 1 0 1 6→5

1 1 1 1 1 1 0 7→6

JA = KA = 1

JB = KB = xA + xA

JC = KC = xBA + xBA

46


Example: 0, 3, 2, 4, 1, 5, 7, and repeat

q1 q2 q3 q1* q2* q3* remark

0 0 0 0 1 1 0→3

0 0 1 1 0 1 1→5

0 1 0 1 0 0 2→4

0 1 1 0 1 0 3→2

1 0 0 0 0 1 4→1

1 0 1 1 1 1 5→7

1 1 0 X X X 6→X

1 1 1 0 0 0 7→0

0

7 3

5 2

1 4 6

47


Example: 0, 3, 2, 4, 1, 5, 7, and

repeat (cont.)

D1 = q’2q3 + q2q’3

D1 = q’1q’2q’3 + q’1q2q3 + q1q’2q3

D1 = q’2

D1

0

1 1

1 X

1

00

01

11

10

q1q2q3 0

1

1

1

X

1

00

01

11

10

0

1 1

1 1

X

1

00

01

11

10

D2 D3

q1q2q3q1q2q3

48


Example: 0, 3, 2, 4, 1, 5, 7, and

repeat (cont.)

= + = + =

= +

= + = +

= =

= + + +

= +

= +

= + = +

= + = +

= =

1 2 3 2 3 1 2 3 2 3 1

2 3

2 1 2 3 1 2 3 2 2 3

3 2 3 2

1 1 2 3 2 3 1 2 1

1 2

1

3

2 1 3 1 3

3 2 3 2 3

1 2 3 2

2

3 1 3 2

2 1 3 1 3 2 1 3

3 2 3

S q q q q R q q q q S

q q

S q q q q q q R q q

S q R q

T q q q q q q q q q

T q q q q

T q q q q

J q q q q K q q

J q

q q

q q

q q q K q q

J q K q2

49


Design a synchronous counter that goes through the

sequence

2 6 1 7 5 and repeat

Using JK flip flops

Show a state diagram, indicating what happens if it initially is in one of the unused states(0,3,4) for each of the design.

Example 3

A B C A* B* C*

0 0 0 X X X

0 0 1 1 1 1

0 1 0 1 1 0

0 1 1 X X X

1 0 0 X X X

1 0 1 0 1 0

1 1 0 0 0 1

1 1 1 1 0 1

50


The maps for the JK solution, with the J section shaded, and the equations are shown below.

0

X X

1

X 1

1

1

00

01

11

10

AB C

0

X X

1 1

X

1

1

00

01

11

10

AB C

0

X X

1

X 1

1

1

00

01

11

10

AB C

A* B* C*

JA = 1 KA = B’ + C’

JB = 1 KB = A JC = A KC = AB’

Substituting the values for the three states not

in the cycle :

0(000): JA = KA = 1, JB = 1, KB = 0,

JC = KC = 0 → 110

3(011): JA = 1, KA = 0, JB = 1, KB = 0,

JC = KC = 0 → 111

4(100): JA = KA = 1, JB = KB = 1,JC = KC = 1 → 011

Example 3 (cont.)

51


This is still a different behavior for the unused states. Now

state 4 goes to 3 and then, on the nect clock, it will get back into the cycle. The state diagram is shown below.

2

5 6

0

7 14 3

Example 3 (cont.)

52


Design of Asynchronous Counters

▪ Page 350….

▪ Figure 5.31(2-bit counter)&5.32(timing delay)

▪ Advantage:

❑Simplicity of the hardware no combinational

logic required

▪ Disadvantage:

❑Speed

53


Derivation of State tables and

State Diagrams

▪ Consider the problem:

▪ Another way of wording this same problem is

A Mealy system with one input x and one output z

such that z = 1 iff x has been 1 for three consecutive

clock times.

A system with one input x and one output z such that z = 1 at a clock time iff x is currently 1 and was also 1 at the previous two clock times.

54


▪ A sample input/output trace is

x 0 1 1 1 1 1 1 1 0 1 1 0 1 1 1 0 1

z 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 0 0 0

q1*q2* z

q1 q2 x=0 x=1 x=0 x=1

0 0 0 0 0 1 0 0

0 1 1 0 1 1 0 0

1 0 0 0 0 1 0 0

1 1 1 0 1 1 0 1

◼ First approach:

save the previous

2 inputs.

◼ Knowing them

and the present

input → output.

◼ Just discard the older input stored in memory

and store the newer one plus the current input.55


▪ Second approach: store in memory the

number of consecutive 1’s as follows:

❑A none, that is, the last input was 0

❑B one

❑C two or more

q* z

q x=0 x=1 x=0 x=1

A

B

C

A

A

A

B

C

C

0 0

0 0

0 1

0/0

no 1'sA

B C

0/0 0/0

1/0

1/1

1/0

one 1

two or more 1's

56


▪ The second approach requires only 3 states, whereas the first requires 4. Both, however, use 2 flip-flops.

▪ Consider: the system produces a 1 if the input has been 1 for 25 consecutive clock times.

▪ Now the first approach requires to store the last 24 inputs and a state table of 224

rows.

▪ The second approach requires 25 states which can be coded with 5 flip-flops.

57


Example 4

For each of the following problems show a state table or a

state diagram. (A sample input/output trace is shown for

each.)

A Meaaly system that produces a 1 output iff there

have been four or more consecutive 1 inputs or two or

more consecutive 0 input.

x 0 1 1 0 0 1 0 0 1 1 1 1 1 0 0 0 1z ? 0 0 0 1 0 0 1 0 0 0 1 1 0 1 1 0 0

58


Example 4 (cont.)

q q* z

X = 0 X = 1 X = 0 X = 1

A A B 1 0

B A C 0 0

C A D 0 0

D A D 0 1

0/1

A D

B C

0/0

0/0

1/0

1/0

1/1

0/0 1/0

We start with a state for which the last input is 0. From there, we need

four consecutive 1’s to get a 1 output or another 0. Thus, on additional

0’s, we loop back to state A. On a 1, we go to B; on a second 1, we go

to C; and on a third 1, we go to D In D, additional 1’s produce a 1 output; 0’s return the system to state A.

59


Programmable Logic Devices

(PLD)

▪ Since sequential system is a combination of

memory and combinational logic, it can be

implemented using any of the programmable logic

devices of Chapter 4 (for the combinational logic)

and some flip-flops (for the memory).

▪ There are a variety of devices commercially

available that combine primarily a PAL and some D

flip-flops. There are also some very complex

programmable devices available.

60


Figure 6.13 One output of a 16R8 PLD.

OE'

Clock

D

Q

Q'

Out'AND array

61


Design using ASM diagrams

▪ ASM (algorithmic state machine) is the

same as sequential system.

▪ A tool is cross between a state diagram

and a flow chart is the ASM diagram.

62


B

A

X

Z

10

Figure 6.16a State box. Figure 6.16b Decision box. Figure 6.16c Mealy output box.

Figure 6.17 An ASM block.

name

output condition0 1 output

63


Figure 6.18 Moore state diagram and ASM diagram.

1one 1

A0

B0

C0

D1

two 1's

no 1's

three or more 1's

11

1 0 0

0

0

A

X

X

X

Z

X

1

1

1

1

0

0

0

0

B

C

D

64


Figure 6.19 Mealy state diagram and ASM diagram.

one 1

A

B C

two or more 1's

no 1's

0/0

0/0 0/0

1/0

1/0

1/1

10

A

X

X

X

1

1

B

C

Z

1

01

00

0

0

0

65


Hardware Design Languages (HDLs)

module full_adder (c_out, s, a, b, c);

input a, b, c;

wire a, b, c;

output c_out, s;

wire c_out, s;

wire w1, w2, w3;

xor x1 (w1, a, b);

xor x2 (s, w1, c);

nand n1 (w2, a, b);

nand n2 (w3, w1, c);

nand n3 (c_out, w3, w2);

endmodule

Verilog structural description of a full adder

Figure 6.20A full adder.

a

X2

X1

n 1

n 2

n 3

b

c

w 1

w 2

w 3

c_out

S

66


Behavioural Verilog

for the full adder

module full_adder (c_out, s, a, b c);

input a, b, c;

wire a, b, c;

output c_out, s;

reg c_out, s;

alwaysbegin

s = a^b^c;

c_out = (a&b) | (a&c_in) | (b&c_in);

end

endmodule

module full_adder (c_out, s, a, b c);

input a, b, c;

wire a, b, c;

output c_out, s;

reg c_out, s;

alwaysbegin

s = a^b^c;

{c_out, s} = a + b + c;

end

endmodule

(a) With logic equations

(b) With algebraic equations

67


module D_ff (q, ck, D, CLR);

input ck, D, CLR;

output q;

reg q;

always @ (negedge ck ro negedge CLR)begin

if (!CLR)

q <= 0;

else

q <= D;

end

endmodule

Structural model od a D flip flop.

68

chapter 6eestaff.kku.ac.th/~sa-nguan/en812200/2021-chap 6...digital logic design @department of...

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