comp541 flip-flop timing
DESCRIPTION
COMP541 Flip-Flop Timing. Montek Singh Oct 6, 2014. Topics. Timing analysis flip - flops sequential systems clock skew. Lab 7 : VGA Display. Anyone having trouble with Lab 7? Be careful about the “Sync Polarity” A “1” means a downward going pulse - PowerPoint PPT PresentationTRANSCRIPT
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COMP541
Flip-Flop Timing
Montek Singh
Oct 6, 2014
Topics Timing analysis
flip-flopssequential systemsclock skew
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Lab 7: VGA Display Anyone having trouble with Lab 7?
Be careful about the “Sync Polarity”A “1” means a downward going pulse
sync signal is normally high, but goes low during the pulseA “0” means an upward going pulse
Use my self-checking text bench!simulates my VGA driver …… and compares your outputs with mineflags any mismatches
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Timing of sequential circuits
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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)
CLK
tsetup
D
thold
ta
Output Timing Constraints Propagation delay: tpcq = max time after clock
edge by which output Q is guaranteed to have stabilized (i.e., not changing anymore)
Contamination delay: tccq = min time after clock edge during which Q will not have started changing yetCLK
tccq
tpcq
Q
Dynamic Discipline The input to a synchronous sequential circuit
must be stable during the aperture (setup and hold) time around the clock edge
Specifically, the input must be stableat least tsetup before the clock edgeat least until thold after the clock edge
Implications on Design Constrains operation
Given a clock period, constrains circuit delays
Given a circuit, constraints clock periodThe delay between
registers (which impacts clock period) has a minimum and maximum delay, dependent on the delays of the circuit elements
Delays of both comb. logic and flip-flops must be taken into account
CL
CLKCLK
R1 R2
Q1 D2
(a)
CLK
Q1
D2
(b)
Tc
Setup Time Constraint Setup time
input to R2 must be stable at least tsetup before the clock edge
constrains max delay from R1 through combinational logic
What’s min clock period?
CLK
Q1
D2
Tc
tpcq tpd tsetup
CL
CLKCLK
Q1 D2
R1 R2
What’s Tc?
Tc ≥ tpcq + tpd + tsetup
tpd ≤ Tc – (tpcq + tsetup)
So, clock period constrained by:• Delay in CL• Delay in previous reg (R1)• Setup requirement in next reg (R2)
Hold Time Constraint Hold time
input to R2 must be stable for at least thold after clock edge
constrains the minimum delay from register R1 through the combinational logic
often try to design circuits with 0 hold time requirement
CLK
Q1
D2
tccq tcd
thold
CL
CLKCLK
Q1 D2
R1 R2
thold < tccq + tcd
tcd > thold - tccq
Timing Analysis
CLK CLK
A
B
C
D
X'
Y'
X
Y
per
gate
Timing Characteristicstccq = 30 ps (FF contamination)
tpcq = 50 ps (FF propagation)
tsetup = 60 ps
thold = 70 ps
tpd = 35 ps
tcd = 25 pstpd =
tcd =
Setup time constraint:
Tc ≥
fc =
Hold time constraint:
tccq + tcd > thold ?
tpd = 3 x 35 ps = 105 ps
tcd = 25 ps
Setup time constraint:
Tc ≥ (50 + 105 + 60) ps = 215 ps
fc = 1/Tc = 4.65 GHz
(30 + 25) ps > 70 ps ? No!
Fixing Hold Time Violation
per
gate
CLK CLK
A
B
C
D
X'
Y'
X
Y
Timing Characteristicstccq = 30 ps
tpcq = 50 ps
tsetup = 60 ps
thold = 70 ps
tpd = 35 ps
tcd = 25 pstpd = 3 x 35 ps = 105 ps
tcd = 2 x 25 ps = 50 ps
Setup time constraint:
Tc ≥ (50 + 105 + 60) ps = 215 ps
fc = 1/Tc = 4.65 GHz
Hold time constraint:
tccq + tpd > thold ?
(30 + 50) ps > 70 ps ? Yes!
Add buffers to the short paths:
Hold Time Often flip-flops are designed for a hold time of
zeroTo avoid these tricky problems
Clock Skew Clock doesn’t arrive at all registers at the
same timeSkew is the difference between the arrival times of the
clock edge at two different (typically neighboring) flip-flops
Examine the worst case:guarantee that discipline is not violated for any
register pairmany registers in a system!
t skew
CLK1
CLK2
CL
CLK2CLK1
R1 R2
Q1 D2
CLKdelay
CLK
Setup Time Constraint with Clock Skew
Worst case: CLK2 is earlier than CLK1
CLK1
Q1
D2
Tc
tpcq tpd tsetuptskew
CL
CLK2CLK1
R1 R2
Q1 D2
CLK2
Tc ≥ tpcq + tpd + tsetup + tskew
tpd ≤ Tc – (tpcq + tsetup + tskew)
Similar Issue w/ Hold Time We won’t go over example
Have a look in book
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Next Time Read Section 3.5.1-3.5.3
Then we’ll move on to memoriesSection 5.5
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