Chapter 5:. 1 / 60 Sequential Circuits Combinational Circuit Memory Elements Inputs Outputs ...

Chapter 5:Chapter 5:

22 / 60 / 60

Sequential CircuitsSequential Circuits

CombinationalCircuit

MemoryElements

Inputs Outputs

Asynchronous

Synchronous

CombinationalCircuit

Flip-flops

Inputs Outputs

33 / 60 / 60

LatchesLatches

SR Latch

S R Q0 Q Q’

0 1 Q = Q0

Initial Value

44 / 60 / 60

LatchesLatches

SR Latch

S R Q0 Q Q’

0 0 0 0 10 0 1

1 0 Q = Q0

Q = Q0

55 / 60 / 60

LatchesLatches

SR Latch

S R Q0 Q Q’

0 0 0 0 10 0 1 1 00 1 0 0

1 Q = 0

Q = Q0

66 / 60 / 60

LatchesLatches

SR Latch

S R Q0 Q Q’

0 0 0 0 10 0 1 1 00 1 0 0 10 1 11

0 1Q = 0

Q = Q0

77 / 60 / 60

LatchesLatches

SR Latch

S R Q0 Q Q’

0 0 0 0 10 0 1 1 00 1 0 0 10 1 1 0 11 0 0

Q = Q0

88 / 60 / 60

LatchesLatches

SR Latch

S R Q0 Q Q’

0 0 0 0 10 0 1 1 00 1 0 0 10 1 1 0 11 0 0 1 01 0 1

Q = Q0

99 / 60 / 60

LatchesLatches

SR Latch

S R Q0 Q Q’

0 0 0 0 10 0 1 1 00 1 0 0 10 1 1 0 11 0 0 1 01 0 1 1 01 1 0

Q = Q0

Q = Q’

1010 / 60 / 60

LatchesLatches

SR Latch

S R Q0 Q Q’

0 0 0 0 10 0 1 1 00 1 0 0 10 1 1 0 11 0 0 1 01 0 1 1 01 1 0 0 01 1 1

Q = Q0

Q = Q’

1111 / 60 / 60

LatchesLatches

SR Latch

0 0 Q0

1 1 Q=Q’=0

No change

Invalid

0 0 Q=Q’=1

1 1 Q0

Invalid

No change

1212 / 60 / 60

LatchesLatches

SR Latch

0 0 Q0

1 1 Q=Q’=0

No change

Invalid

S’ R’ Q

0 0 Q=Q’=1

1 1 Q0

Invalid

No change

1313 / 60 / 60

Controlled LatchesControlled Latches

SR Latch with Control Input

C S R Q

0 x x Q0

1 0 0 Q0

1 0 1 0

1 1 0 1

1 1 1 Q=Q’

No change

Invalid

1414 / 60 / 60

D Latch (D = Data)

0 x Q0

No change

Timing Diagram

Output may change

1515 / 60 / 60

D Latch (D = Data)

0 x Q0

No change

Timing Diagram

Output may change

1616 / 60 / 60

Flip-FlopsFlip-Flops

Controlled latches are level-triggered

Flip-Flops are edge-triggered

CLK Positive Edge

CLK Negative Edge

1717 / 60 / 60

Master-Slave D Flip-Flop

D Latch(Master)

QD Latch(Slave)

CLK CLK

QMaster

QSlave

Looks like it is negative edge-triggered

Master Slave

1818 / 60 / 60

Edge-Triggered D Flip-Flop

Positive Edge

Negative Edge

1919 / 60 / 60

JK Flip-Flop

D = JQ’ + K’Q

2020 / 60 / 60

T Flip-Flop

D = TQ’ + T’Q = T Q

D = JQ’ + K’QT Q

2121 / 60 / 60

Flip-Flop Characteristic TablesFlip-Flop Characteristic Tables

D Q(t+1)0 01 1

J K Q(t+1)0 0 Q(t)0 1 01 0 11 1 Q’(t)

No change

Toggle

T Q(t+1)0 Q(t)1 Q’(t)

No change

Toggle

2222 / 60 / 60

Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations

D Q(t+1)0 01 1

Q(t+1) = D

J K Q(t+1)0 0 Q(t)0 1 01 0 11 1 Q’(t)

Q(t+1) = JQ’ + K’Q

T Q(t+1)0 Q(t)1 Q’(t)

Q(t+1) = T Q

2323 / 60 / 60

Analysis / Derivation

J K Q(t) Q(t+1)0 0 0 00 0 1 10 1 00 1 11 0 01 0 11 1 01 1 1

No change

Toggle

2424 / 60 / 60

J K Q(t) Q(t+1)0 0 0 00 0 1 10 1 0 00 1 1 01 0 01 0 11 1 01 1 1

No change

Toggle

2525 / 60 / 60

J K Q(t) Q(t+1)0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 11 1 01 1 1

No change

Toggle

2626 / 60 / 60

J K Q(t) Q(t+1)0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 11 1 0 11 1 1 0

No change

Toggle

2727 / 60 / 60

J K Q(t) Q(t+1)0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 11 1 0 11 1 1 0

0 1 0 0J 1 1 0 1

Q(t+1) = JQ’ + K’Q

2828 / 60 / 60

Flip-Flops with Direct InputsFlip-Flops with Direct Inputs

Asynchronous Reset

R’ D CLK Q(t+1)

0 x x 0

2929 / 60 / 60

Asynchronous Reset

R’ D CLK Q(t+1)

0 x x 0

1 0 ↑ 0

1 1 ↑ 1

3030 / 60 / 60

Asynchronous Preset and Clear

PR’ CLR’ D CLK Q(t+1)

1 0 x x 0D Q

Preset

3131 / 60 / 60

1 0 x x 0

0 1 x x 1

Preset

3232 / 60 / 60

1 0 x x 0

0 1 x x 1

1 1 0 ↑ 0

1 1 1 ↑ 1

Preset

3333 / 60 / 60

Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits

The State

● State = Values of all Flip-Flops

Example

A B = 0 0

3434 / 60 / 60

State Equations

A(t+1) = DA

= A(t) x(t)+B(t) x(t)

= A x + B x

B(t+1) = DB

= A’(t) x(t)

= A’ x

y(t) = [A(t)+ B(t)] x’(t)

= (A + B) x’

3535 / 60 / 60

State Table (Transition Table)

A(t+1) = A x + B x

B(t+1) = A’ x

y(t) = (A + B) x’

Present State

InputNext State

Output

A B x A B y0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1

t+1 tt

0 0 00 1 00 0 11 1 00 0 11 0 00 0 11 0 0

3636 / 60 / 60

State Table (Transition Table)

A(t+1) = A x + B x

B(t+1) = A’ x

y(t) = (A + B) x’

Present State

Next State Output

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

A B A B A B y y0 0 0 0 0 1 0 00 1 0 0 1 1 1 01 0 0 0 1 0 1 01 1 0 0 1 0 1 0

t+1 tt

3737 / 60 / 60

State Diagram

0 0 1 0

0 1 1 1

1/0 0/10/1

AB input/output

Present State

Next State Output

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

A B A B A B y y

0 0 0 0 0 1 0 0

0 1 0 0 1 1 1 0

1 0 0 0 1 0 1 0

1 1 0 0 1 0 1 0

3838 / 60 / 60

D Flip-Flops

Example: D Q

Present State

InputNext State

A x y A0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1

01101001

0 100,11 00,11

A(t+1) = DA = A x y

3939 / 60 / 60

JK Flip-Flops

Example:

JA = B KA = B x’

JB = x’ KB = A x

A(t+1) = JA Q’A + K’A QA

= A’B + AB’ + AxB(t+1) = JB Q’B + K’B QB

= B’x’ + ABx + A’Bx’

Present State

I/PNext State

Flip-FlopInputs

A B x A B JA KA JB KB

0 0 1 0

0 0 0 1

1 1 1 0

1 0 0 1

0 0 1 1

0 0 0 0

1 1 1 1

1 0 0 0

4040 / 60 / 60

JK Flip-Flops

Example:

BPresent State

I/PNext State

Flip-FlopInputs

0 0 1 0

0 0 0 1

1 1 1 0

1 0 0 1

0 0 1 1

0 0 0 0

1 1 1 1

1 0 0 0

0 0 1 1

0 1 1 0

4141 / 60 / 60

T Flip-Flops

Example:

TA = B x TB = xy = A B

A(t+1) = TA Q’A + T’A QA

= AB’ + Ax’ + A’BxB(t+1) = TB Q’B + T’B QB

CLK Reset

Present State

I/PNext State

F.FInputs

A B x A B TA TB y

4242 / 60 / 60

T Flip-Flops

Example:

CLK Reset

Present State

I/PNext State

F.FInputs

A B x A B TA TB y

0 0 0 1

1 1 1 0

0/01/0

0/00/1

4343 / 60 / 60

Mealy and Moore ModelsMealy and Moore Models

Present State

I/PNext State

A B x A B y0 0 0 0 0 00 0 1 0 1 00 1 0 0 0 10 1 1 1 1 01 0 0 0 0 11 0 1 1 0 01 1 0 0 0 11 1 1 1 0 0

MealyMealy

For the same statestate,the outputoutput changes with the inputinput

Present State

I/PNext State

A B x A B y0 0 0 0 0 00 0 1 0 1 00 1 0 0 1 00 1 1 1 0 01 0 0 1 0 01 0 1 1 1 01 1 0 1 1 11 1 1 0 0 1

MooreMoore

For the same statestate,the outputoutput does not change with the inputinput

4444 / 60 / 60

Moore State DiagramMoore State Diagram

State / Output

0 0 / 0 0 1 / 0

1 1 / 1 1 0 / 0

4545 / 60 / 60

Timing DiagramTiming Diagram

0 0 / 0 0 1 / 0

1 1 / 1 1 0 / 0

StateA

No effect

4646 / 60 / 60

Timing DiagramTiming Diagram

0 0 0 1

1 1 1 0

0/0 0/0

StateA

4747 / 60 / 60

Design of Clocked Sequential CircuitsDesign of Clocked Sequential Circuits

Example:

Detect 3 or more consecutive 1’s

S0 / 0 S1 / 0

S3 / 1 S2 / 0

State A BS0 0 0

S1 0 1S2 1 0

S3 1 1

4848 / 60 / 60

Example:

Present State

InputNext State

Output

A B x A B y0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1

0 0 00 1 00 0 01 0 00 0 01 1 00 0 11 1 1

S0 / 0 S1 / 0

S3 / 1 S2 / 0

4949 / 60 / 60

Example:

Present State

InputNext State

Output

A B x A B y0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1

0 0 00 1 00 0 01 0 00 0 01 1 00 0 11 1 1

A(t+1) = DA (A, B, x)

= ∑ (3, 5, 7)

B(t+1) = DB (A, B, x)

= ∑ (1, 5, 7)

y (A, B, x) = ∑ (6, 7)

Synthesis using DD Flip-Flops

5050 / 60 / 60

Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with DD F.F. F.F.

Example:

DA (A, B, x) = ∑ (3, 5, 7)

= A x + B x

DB (A, B, x) = ∑ (1, 5, 7)

= A x + B’ x

y (A, B, x) = ∑ (6, 7)

0 0 1 0

A 0 1 1 0x B

0 1 0 0

A 0 1 1 0xB

0 0 0 0

A 0 0 1 1x

5151 / 60 / 60

Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with DD F.F. F.F.

Example:

DA = A x + B x

DB = A x + B’ x

y = A B

5252 / 60 / 60

Flip-Flop Excitation TablesFlip-Flop Excitation Tables

Present State

Next State

F.F.Input

Q(t) Q(t+1) D0 00 11 01 1

Present State

Next State

F.F.Input

Q(t) Q(t+1) J K0 00 11 01 1

0 0 (No change)0 1 (Reset)

1 0 (Set)1 1 (Toggle)0 1 (Reset)1 1 (Toggle)0 0 (No change)1 0 (Set)

Q(t) Q(t+1) T0 00 11 01 1

5353 / 60 / 60

Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with JKJK F.F. F.F.

Example:

Present State

InputNext State

Flip-FlopInputs

0 0 0 0 00 0 1 0 10 1 0 0 00 1 1 1 01 0 0 0 01 0 1 1 11 1 0 0 01 1 1 1 1

0 x0 x0 x1 xx 1x 0x 1x 0

JA (A, B, x) = ∑ (3)dJA (A, B, x) = ∑ (4,5,6,7)KA (A, B, x) = ∑ (4, 6)dKA (A, B, x) = ∑ (0,1,2,3)JB (A, B, x) = ∑ (1, 5)dJB (A, B, x) = ∑ (2,3,6,7)KB (A, B, x) = ∑ (2, 3, 6)dKB (A, B, x) = ∑ (0,1,4,5)

Synthesis using JKJK F.F.

0 x1 xx 1x 10 x1 xx 1x 0

5454 / 60 / 60

Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with JKJK F.F. F.F.

Example:

JA = B x KA = x’

JB = x KB = A’ + x’

Synthesis using JKJK Flip-Flops

0 0 1 0

A x x x xx

x x x x

A 1 0 0 1x

0 1 x x

A 0 1 x xx

x x 1 1

A x x 0 1x

5555 / 60 / 60

Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with TT F.F. F.F.

Example:

Present State

InputNext State

F.F.Input

A B x A B TA TB

0 0 0 0 00 0 1 0 10 1 0 0 00 1 1 1 01 0 0 0 01 0 1 1 11 1 0 0 01 1 1 1 1

00011010

Synthesis using TT Flip-Flops01110110

TA (A, B, x) = ∑ (3, 4, 6)TB (A, B, x) = ∑ (1, 2, 3, 5, 6)

5656 / 60 / 60

Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with TT F.F. F.F.

Example:

TA = A x’ + A’ B x

TB = A’ B + B x

Synthesis using TT Flip-Flops

0 0 1 0

A 1 0 0 1x

0 1 1 1

A 0 1 0 1x

5757 / 60 / 60

HomeworkHomework

● Chapter 5

♦ 5-1

♦ 5-3

♦ 5-6

♦ 5-8

♦ 5-9

5858 / 60 / 60

HomeworkHomework

5-1 The D latch is constructed with four NAND gates and an inverter. Consider the following three other ways for obtaining a D latch. In each case, draw the logic diagram and verify the circuit operation.

(a) Use NOR gates for the SR latch part and AND gates for the other two. An inverter may be needed.

(b) Use NOR gates for all four gates. Inverters may be needed.

(c) Use four NAND gates only (without an inverter). This can be done by connecting the output of the upper gate that goes to the SR latch to the input of the lower gate instead of the inverter output.

5959 / 60 / 60

HomeworkHomework

5-3 Show that the characteristic equation for the complement output of a JK flip-flop is

Q’(t+1) = J’Q + K Q

5-6 A sequential circuit with two D flip-flops, A and B; two inputs, x and y; and one output, z, is specified by the following next-state and output equations:

A(t+1) = x’ y + x A B(t+1) = x’ B + x A z = B

(a) Draw the logic diagram of the circuit.

(b) List the state table for the sequential circuit.

(c) Draw the corresponding state diagram.

6060 / 60 / 60

HomeworkHomework

5-8 Derive the state table and the state diagram of the sequential shown circuit. Explain the function that the circuit performs.

6161 / 60 / 60

HomeworkHomework

5-9 A sequential circuit has two JK flip-flops A and B and one input x. The circuit is described by the following flip-flop input equations:

JA = x KA = B’

JB = x KB = A

(a) Derive the state equations A(t+1) and B(t+1) by substituting the input equations for the J and K variables.

(b) Draw the state diagram of the circuit.

Chapter 5:. 1 / 60 Sequential Circuits Combinational Circuit Memory Elements Inputs Outputs ...

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