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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(tr) = exp(tr/τ)/(T0·f·a). For each problem, f= 25 x 106, a= 106, 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 trto 20ns. For the 74LS74, T0 = 0.4 and τ= 1.5. MTBF(20) = exp(20/1.5)/(0.4·25 x 106·106) = 6.17438 x 10-08s (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 trto 30ns. For the 74ALS74, T0 = 0.0000087 and τ= 1. MTBF(30) = exp(30/1)/( 0.0000087 ·25 x 106·106) = 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 trto 35ns. For the 74F74, T0 = 0.0002 and τ= 0.4. MTBF(35) = exp(35/0.4)/( 0.0002 ·25 x 106·106) = 2.00354 x 1029 s The 74F74 has the longest MTBF so would give you the most job security. 1
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