CS 140L Lecture 4 FlipFlops, Shifters and Counters - PowerPoint PPT Presentation

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CS 140L Lecture 4 FlipFlops, Shifters and Counters

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... The maximum delay from register R1 through the combinational logic. ... The input to register R2 must be stable at least tsetup before the clock edge. ... – PowerPoint PPT presentation

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Title: CS 140L Lecture 4 FlipFlops, Shifters and Counters


1
CS 140L Lecture 4Flip-Flops, Shifters and
Counters
  • Professor CK Cheng
  • CSE Dept.
  • UC San Diego

2
  • F-F
  • Shift register
  • Counter (Asynchronous)
  • Counter (Synchronous)

3
Flip-Flops
DFF
D
Asynchronous Clear
Q
Inputs Output
CE
C
CLR CE D C Q 1 X X X 0
0 0 X X No change 0 1 1
1 0 1 0 0
CLR
Clock Enable
CLK 1
CLK 0
4
DFF Timing
CLK
tsetup
thold
t
D
t
Q
t
tcq
5
Input Timing Constraints
  • Setup time tsetup time before the clock edge
    that data must be stable (i.e. not changing)
  • Hold time thold time after the clock edge that
    data must be stable
  • Aperture time ta time around clock edge that
    data must be stable (ta tsetup thold)

6
Output Timing Constraints
  • Propagation delay tpcq time after clock edge
    that the output Q is guaranteed to be stable
    (i.e., to stop changing)
  • Contamination delay tccq time after clock edge
    that Q might be unstable (i.e., start changing)

7
Setup Time Constraint
  • The setup time constraint The maximum delay from
    register R1 through the combinational logic.
  • The input to register R2 must be stable at least
    tsetup before the clock edge.

Tc tpcq tpd tsetup tpd Tc (tpcq
tsetup)
8
Hold Time Constraint
  • The hold time constraint depends on the minimum
    delay from register R1 through the combinational
    logic.
  • The input to register R2 must be stable for at
    least thold after the clock edge.

thold lt tccq tcd tcd gt thold - tccq
9
2) A 3 Bit Shift Register
B
D
C
A
Q
Q
Q
D
D
D
CLK
Time Steps A B C D
  • 0 0 X X X
  • 1 0 X X
  • 2 0 1 0 X
  • 1 0 1 0
  • 1 1 0 1
  • 0 1 1 0
  • 0 0 1 1
  • 1 0 0 1

10
3) A 3 Bit Counter (Asynchronous)
B
A
C
Q
Q
Q
1
1
T
T
T
1
CLK
Assume A(0) B(0) C(0) 0
Time C B A
  • 0 0 0 0 0
  • 1 1 1 7
  • 2 1 1 0 6
  • 1 0 1 5
  • 1 0 0 4
  • 0 1 1 3
  • 0 1 0 2
  • 0 0 1 1

CLK
t
A
1
0
1
0
t
B
1
1
0 0
t
C
1 1 1 1
t
7 6 5 4
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