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Input. Combinational Logic. Output. Excitation Vectors. States. Memory. RS Latch Q+ = S + R’ Q. Two Problems: R=S= 1 Not allowed, Data is transparent. Q. D. CLK. The D Latch. Q + = D. Problem: Level sensitive. JK Latch : Universal, Level sensitive, - PowerPoint PPT Presentation
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Excitation Vectors
InputCombinational Logic
Memory
Output
States
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R S Qt+1 0 0 tq
0 1 1 1 0 0 1 1 —
R
SQ
Q
RS Latch
Q+ = S + R’ Q
Two Problems:
R=S= 1 Not allowed, Data is transparent
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Q
D
C
Q
QC
D
RS Flip Flop
The D Latch
Problem: Level sensitive
D Q+
0 0
1 1
Q+ = D
DLatch
D
CLK
Q
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QKQJQ 1t tt
JK Flip Flop:
S
R
J
K
Q
Q
JK Latch : Universal, Level sensitive,
Timing Constraints due to feed back.
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DLatch
DLatch
Q
QD
C
QD
C
D
C
Q
Master Slave
Master and Slave Flip Flop :
A D Flip Flop with a falling-edge trigger.
Q
D
C
Master Slave Flip Flop Edge sensitive,Set up and Hold time
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Concordia UniversityMaster Slave Flip Flop Edge sensitive,-Falling Edge
Path to setup data
Path to hold data
DLatch
DLatch
Q
QD
C
QD
C
D
C
Q
Master Slave
Set Up and Hold Time constraints
Q
D
C
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S
R
Clk
Q
Q
`
Edge triggered Flip Flop:
Set up and Hold time Constraints
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D Flip-Flop
clkreset
D Q
Edge Triggered, D Flip Flop,
With Reset
clk
D
reset
Q
NAND1
NAND2
NAND3
NAND4
NAND5
NAND6
S
R
Active Low
Reset C D Q Q’
0 X X 0 1
1 0 0 1
1 1 1 0
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D Flip-Flop
clkreset
D Q
Edge Triggered, D Flip Flop When CLK=0
Clk=0
D
reset
Q
NAND1
NAND2
NAND3
NAND4
NAND5
NAND6
S 1
R 1
Active Low
0
0
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clk
D
reset
Q
NAND1
NAND2
NAND3
NAND4
NAND5
NAND6
S
R
When CLK changes from 0 to 1
Case1, D=0: tsetup= t4, thold=t3
Path for set up
Path for hold
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clk
D
reset
Q
NAND1
NAND2
NAND3
NAND4
NAND5
NAND6
S
R
When CLK changes from 0 to 1
Case2, D=1 tsetup=t4 + t1 thold= t2
Path to set up
Path to hold
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clk
D
reset
Q
NAND1
NAND2
NAND3
NAND4
NAND5
NAND6
S
R
When CLK changes from 0 to 1
Case1, D=0: tsetup= t4, thold=t3
Case2, D=1 tsetup=t4 + t1 thold= t2
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JK Flip Flop with a rising-edge :
C
J
K
J
K
Q
Q
JK FFJ
K
Q
Q
C
Q+ = JQ’ + K’Q
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JK FFJ
K
Q
Q
C
Q+ = JQ’ + K’Q
J=K=T
Q+ =TQ’ + T’ Q
T=1 Q+ = Q’
T=0 Q+ = Q
QK
J
CLK
T
T-Flip Flop
T
CLK
Q
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Q+ = JQ’ + K’Q
J=K=T
Q+ =TQ’ + T’ Q
T=1 Q+ = Q’
T=0 Q+ = Q
T-Flip Flop
QD
CLK
T
D
CLK
Q
T
CLK
Q
