sequential logic circuit design. 1 sequential circuits combinational circuit memory elements inputs...
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SEQUENTIAL LOGIC
CIRCUIT DESIGN
22
Sequential CircuitsSequential Circuits
CombinationalCircuit
MemoryElements
Inputs Outputs
Asynchronous
Synchronous
CombinationalCircuit
Flip-flops
Inputs Outputs
Clock
33
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0
0
1
0
0
0 1 Q = Q0
Initial Value
44
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0 0 10 0 1
1
0
0
0
1 0 Q = Q0
Q = Q0
55
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0 0 10 0 1 1 00 1 0 0
0
1
1
0
1 Q = 0
Q = Q0
66
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0 0 10 0 1 1 00 1 0 0 10 1 11
0
1
0
0 1Q = 0
Q = Q0
Q = 0
77
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0 0 10 0 1 1 00 1 0 0 10 1 1 0 11 0 0
0
1
0
1
1 0
Q = 0
Q = Q0
Q = 1
88
LatchesLatches
SR Latch
R
S
Q
Q
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
0
0
1
1 0
Q = 0
Q = Q0
Q = 1
Q = 1
99
LatchesLatches
SR Latch
R
S
Q
Q
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
1
1
1
0 0
Q = 0
Q = Q0
Q = 1
Q = Q’
0
1010
LatchesLatches
SR Latch
R
S
Q
Q
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
1
0
1
1
0 0
Q = 0
Q = Q0
Q = 1
Q = Q’
0
Q = Q’
1111
LatchesLatches
SR Latch
R
S
Q
Q
S R Q
0 0 Q0
0 1 0
1 0 1
1 1 Q=Q’=0
No change
Reset
Set
Invalid
S
R
Q
Q
S R Q
0 0 Q=Q’=1
0 1 1
1 0 0
1 1 Q0
Invalid
Set
Reset
No change
1212
LatchesLatches
SR Latch
R
S
Q
Q
S R Q
0 0 Q0
0 1 0
1 0 1
1 1 Q=Q’=0
No change
Reset
Set
Invalid
S’ R’ Q
0 0 Q=Q’=1
0 1 1
1 0 0
1 1 Q0
Invalid
Set
Reset
No change
S
R
Q
Q
1313
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
No change
Reset
Set
Invalid
S
R
Q
Q
S
R
C
S
RQ
QS
R
C
1414
Controlled LatchesControlled Latches
D Latch (D = Data)
C D Q
0 x Q0
1 0 0
1 1 1
No change
Reset
Set
S
R
Q
Q
D
C
C
Timing Diagram
D
Q
t
Output may change
1515
Controlled LatchesControlled Latches
D Latch (D = Data)
C D Q
0 x Q0
1 0 0
1 1 1
No change
Reset
Set
C
Timing Diagram
D
Q
Output may change
S
R
Q
Q
D
C
1616
Flip-FlopsFlip-Flops
Controlled latches are level-triggered
Flip-Flops are edge-triggered
C
CLK Positive Edge
CLK Negative Edge
1717
Flip-FlopsFlip-Flops
Master-Slave D Flip-Flop
D Latch(Master)
D
C
QD Latch(Slave)
D
C
Q QD
CLK CLK
D
QMaster
QSlave
Looks like it is negative edge-triggered
Master Slave
1818
Flip-FlopsFlip-Flops
Edge-Triggered D Flip-Flop
D
CLK
Q
Q
D Q
Q
D Q
Q
Positive Edge
Negative Edge
1919
Flip-FlopsFlip-Flops
JK Flip-Flop
D Q
Q
Q
QCLK
J
K
J Q
QK
D = JQ’ + K’Q
2020
Flip-FlopsFlip-Flops
T Flip-Flop
D = TQ’ + T’Q = T Q
J Q
QK
T D Q
Q
T
D = JQ’ + K’QT Q
Q
2121
Flip-Flop Characteristic TablesFlip-Flop Characteristic Tables
D Q
Q
D Q(t+1)0 01 1
Reset
Set
J K Q(t+1)0 0 Q(t)0 1 01 0 11 1 Q’(t)
No change
Reset
Set
Toggle
J Q
QK
T Q
Q
T Q(t+1)0 Q(t)1 Q’(t)
No change
Toggle
2222
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
D Q
Q
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
J Q
QK
T Q
Q
T Q(t+1)0 Q(t)1 Q’(t)
Q(t+1) = T Q
2323
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
Analysis / Derivation
J Q
QK
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
Reset
Set
Toggle
2424
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
Analysis / Derivation
J Q
QK
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
Reset
Set
Toggle
2525
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
Analysis / Derivation
J Q
QK
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
Reset
Set
Toggle
2626
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
Analysis / Derivation
J Q
QK
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
Reset
Set
Toggle
2727
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
Analysis / Derivation
J Q
QK
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
K
0 1 0 0J 1 1 0 1
Q
Q(t+1) = JQ’ + K’Q
2828
Flip-Flops with Direct InputsFlip-Flops with Direct Inputs
Asynchronous Reset
D Q
Q
R
Reset
R’ D CLK Q(t+1)
0 x x 0
2929
Flip-Flops with Direct InputsFlip-Flops with Direct Inputs
Asynchronous Reset
D Q
Q
R
Reset
R’ D CLK Q(t+1)
0 x x 0
1 0 ↑ 0
1 1 ↑ 1
3030
Flip-Flops with Direct InputsFlip-Flops with Direct Inputs
Asynchronous Preset and Clear
PR’ CLR’ D CLK Q(t+1)
1 0 x x 0D Q
Q
CLR
Reset
PR
Preset
3131
Flip-Flops with Direct InputsFlip-Flops with Direct Inputs
Asynchronous Preset and Clear
PR’ CLR’ D CLK Q(t+1)
1 0 x x 0
0 1 x x 1
D Q
Q
CLR
Reset
PR
Preset
3232
Flip-Flops with Direct InputsFlip-Flops with Direct Inputs
Asynchronous Preset and Clear
PR’ CLR’ D CLK Q(t+1)
1 0 x x 0
0 1 x x 1
1 1 0 ↑ 0
1 1 1 ↑ 1
D Q
Q
CLR
Reset
PR
Preset
3333
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
The State
● State = Values of all Flip-Flops
Example
A B = 0 0
D Q
Q
CLK
D Q
Q
A
B
y
x
3434
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
State Equations
D Q
Q
CLK
D Q
Q
A
B
y
x
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
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
State Table (Transition Table)
D Q
Q
CLK
D Q
Q
A
B
y
x
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
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
State Table (Transition Table)
D Q
Q
CLK
D Q
Q
A
B
y
x
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
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
State Diagram
D Q
Q
CLK
D Q
Q
A
B
y
x
0 0 1 0
0 1 1 1
0/0
0/1
1/0
1/0
1/0
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
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
D Flip-Flops
Example: D Q
Q
x
CLK
yA
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
01,10
01,10
A(t+1) = DA = A x y
3939
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
JK Flip-Flops
Example:
J Q
QK
CLK
J Q
QK
x
A
B
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 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
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 1
0 0
1 1
1 0
1 1
1 0
0 0
1 1
4040
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
JK Flip-Flops
Example:
J Q
QK
CLK
J Q
QK
x
A
BPresent State
I/PNext State
Flip-FlopInputs
A B x A B JA KA JB KB
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
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 1
0 0
1 1
1 0
1 1
1 0
0 0
1 1
0 0 1 1
0 1 1 0
1 0 1
0
1
00
1
4141
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
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
= x B
A
B
T Q
QR
T Q
QR
CLK Reset
xy
Present State
I/PNext State
F.FInputs
O/P
A B x A B TA TB y
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0
0 1
0 0
1 1
0 0
0 1
0 0
1 1
0 0
0 1
0 1
1 0
1 0
1 1
1 1
0 0
0
0
0
0
0
0
1
1
4242
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
T Flip-Flops
Example:
A
B
T Q
QR
T Q
QR
CLK Reset
xy
Present State
I/PNext State
F.FInputs
O/P
A B x A B TA TB y
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0
0 1
0 0
1 1
0 0
0 1
0 0
1 1
0 0
0 1
0 1
1 0
1 0
1 1
1 1
0 0
0
0
0
0
0
0
1
1
0 0 0 1
1 1 1 0
0/01/0
0/0
1/0
1/0
1/1
0/00/1
4343
Mealy and Moore ModelsMealy and Moore Models
The Mealy model: the outputs are functions of both the present state and inputs (Fig. 5-15).
● The outputs may change if the inputs change during the clock pulse period.
♦ The outputs may have momentary false values unless the inputs are synchronized with the clocks.
The Moore model: the outputs are functions of the present state only (Fig. 5-20).
● The outputs are synchronous with the clocks.
4444
Mealy and Moore ModelsMealy and Moore Models
Fig. 5.21 Block diagram of Mealy and Moore state machine
4545
Mealy and Moore ModelsMealy and Moore Models
Present State
I/PNext State
O/P
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
O/P
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
4646
Moore State DiagramMoore State Diagram
State / Output
0 0 / 0 0 1 / 0
1 1 / 1 1 0 / 0
0
1
1
1
00
01
4747
State Reduction and AssignmentState Reduction and Assignment
State Reduction Reductions on the number of flip-flops and the number of gates.
● A reduction in the number of states may result in a reduction in the number of flip-flops.
● An example state diagram showing in Fig. 5.25.
Fig. 5.25 State diagram
4848
State ReductionState Reduction
● Only the input-output sequences are important.
● Two circuits are equivalent
♦ Have identical outputs for all input sequences;
♦ The number of states is not important.
Fig. 5.25 State diagram
State:a a b c d e f f g f g a
Input:0 1 0 1 0 1 1 0 1 0 0
Output:
0 0 0 0 0 1 1 0 1 0 0
4949
Equivalent states
● Two states are said to be equivalent
♦ For each member of the set of inputs, they give exactly the same output and send the circuit to the same state or to an equivalent state.
♦ One of them can be removed.
5050
Reducing the state table
● e = g (remove g);
● d = f (remove f);
5151
● The reduced finite state machine
State:a a b c d e d d e d e a
Input:0 1 0 1 0 1 1 0 1 0 0
Output:
0 0 0 0 0 1 1 0 1 0 0
5252
● The unused states are treated as don't-care condition fewer combinational gates.
Fig. 5.26 Reduced State diagram
5353
State AssignmentState Assignment
State Assignment
To minimize the cost of the combinational circuits.
● Three possible binary state assignments. (m states need n-bits, where 2n > m)
5454
● Any binary number assignment is satisfactory as long as each state is assigned a unique number.
● Use binary assignment 1.
5555
Design Procedure Design Procedure
Design Procedure for sequential circuit
● The word description of the circuit behavior to get a state diagram;
● State reduction if necessary;
● Assign binary values to the states;
● Obtain the binary-coded state table;
● Choose the type of flip-flops;
● Derive the simplified flip-flop input equations and output equations;
● Draw the logic diagram;
5656
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
0
1
1
0 0
1
0
1
State A BS0 0 0
S1 0 1S2 1 0
S3 1 1
5757
Design of Clocked Sequential CircuitsDesign of Clocked Sequential Circuits
Example:
Detect 3 or more consecutive 1’s
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
0
1
1
0 0
1
0
1
5858
Design of Clocked Sequential CircuitsDesign of Clocked Sequential Circuits
Example:
Detect 3 or more consecutive 1’s
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
5959
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with DD F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
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)
= A B
Synthesis using DD Flip-Flops
B
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
6060
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with DD F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
DA = A x + B x
DB = A x + B’ x
y = A B
Synthesis using DD Flip-Flops
D Q
Q
A
CLK
x
BD Q
Q
y
6161
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)
0 x
1 x
x 1
x 0
0
1
0
1
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
0
1
1
0
6262
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with JKJK F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
Present State
InputNext State
Flip-FlopInputs
A B x A B JA KA JB KB
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
6363
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with JKJK F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
JA = B x KA = x’
JB = x KB = A’ + x’
Synthesis using JKJK Flip-Flops
B
0 0 1 0
A x x x xx
B
x x x x
A 1 0 0 1x
B
0 1 x x
A 0 1 x xx
B
x x 1 1
A x x 0 1x
CLK
J Q
QK
x
A
B
J Q
QK y
6464
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with TT F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
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)
6565
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with TT F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
TA = A x’ + A’ B x
TB = A’ B + B x
Synthesis using TT Flip-Flops
B
0 0 1 0
A 1 0 0 1x
B
0 1 1 1
A 0 1 0 1x
A
B
y
T Q
Q
x
CLK
T Q
Q
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