# hw7s - ENGRD2300 Introduction to Digital Logic Fall 2008...

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ENGRD2300: Introduction to Digital Logic Fall 2008 Homework 7 Solutions Problem 1. Calculate the MTBF of a synchronizer built according to Wakerly Figure 8-76 using (a) 74LS74, (b)74ALS74 and (c) 74F74 flip-flops, assuming a clock frequency of 25MHz and an asynchronous transition rate of 1MHz. Assume that the set-up time of the 74LS74 is 20ns, the 74ALS74 is 10ns and the 74F74 is 5ns. Assume that the hold time for each type of flip-flop is 0ns. Which flip-flop 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 flip-flops. (a) At 25MHz the clock period is 40ns and the setup time for the 74LS74 is 20ns, so this leaves 20ns for the first flip-flop 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 flip-flop 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 flip-flop 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 9-9, 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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