ENGRD2300: Introduction to Digital Logic
Fall 2008
Homework 7 Solutions
Problem 1.
Calculate the MTBF of a synchronizer built according to Wakerly Figure 876 using (a) 74LS74,
(b)74ALS74 and (c) 74F74 flipflops, assuming a clock frequency of 25MHz and an asynchronous
transition rate of 1MHz.
Assume that the setup time of the 74LS74 is 20ns, the 74ALS74 is 10ns and the
74F74 is 5ns.
Assume that the hold time for each type of flipflop is 0ns.
Which flipflop gives you the
most job security?
Why?
This problem is solved using the formula for MTBF(t
r
) = exp(t
r
/
τ
)/(
T
0
·
f
·
a)
.
For each problem,
f
= 25 x 10
6
,
a
= 10
6
, and the other parameters are specific to the flipflops.
(a) At 25MHz the clock period is 40ns and the setup time for the 74LS74 is 20ns, so this leaves 20ns for
the first flipflop to stabilize.
Thus we will set t
r
to 20ns.
For the 74LS74,
T
0
= 0.4 and
τ
= 1.5.
MTBF(20) = exp(20/1.5)/(0.4·25 x 10
6
·10
6
) = 6.17438 x 10
08
s
(b) At 25MHz the clock period is 40ns and the setup time for the 74ALS74 is 10ns, so this leaves 30ns for
the first flipflop to stabilize.
Thus we will set t
r
to 30ns.
For the 74ALS74,
T
0
= 0.0000087 and
τ
= 1.
MTBF(30) = exp(30/1)/( 0.0000087 ·25 x 10
6
·10
6
) = 491332.1647s
(c) At 25MHz the clock period is 40ns and the setup time for the 74F74 is 5ns, so this leaves 35ns for the
first flipflop to stabilize.
Thus we will set t
r
to 35ns.
For the 74F74,
T
0
= 0.0002 and
τ
= 0.4.
MTBF(35) = exp(35/0.4)/( 0.0002 ·25 x 10
6
·10
6
) = 2.00354 x 10
29
s
The 74F74 has the longest MTBF so would give you the most job security.
1
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ENGRD2300: Introduction to Digital Logic
Fall 2008
Problem 2.
In the style of Wakerly Figure 99, describe a possible layout of a 64K x 4 ROM.
Remember we like
“square” chips!
The storage array is512 bits by 512 bits, which satisfies our desire for a “square” chip.
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 Fall '08
 LONG
 Frequency, Flipflop, A7 A6 A5, A6 A5 A4, A5 A4 A3, A9 A8 A7, A8 A7 A6

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