hw2s - ENGRD2300: Introduction to Digital Logic Fall 2008...

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ENGRD2300: Introduction to Digital Logic Fall 2008 Homework 2 Solutions Problem 1. You need to drive a large number of 74HC00-like inputs with 74HC00 NAND gates. What is the maximum number of 74HC00-like inputs that you can connect to one single 74HC00 output without exceeding its worst-case loading specifications (fanout)? Use Wakerly Table 3-3. Hint: Calculate BOTH LOW-state and HIGH-state DC fanouts. The maximum amount of current that the output of the 74HC00 can sink or source, I OLmax and I OHmax , cannot be lower than the sum of the maximum input currents of the 74HC00, I ILmax and I IHmax . There are two possible solutions to this problem, depending on which value of I OLmax and I OHmax you choose. If you choose the typical value for connections to “CMOS” inputs, then from Table 3-3 we obtain I OLmax = 20 μ A, I OHmax = -20 μ A, I ILmax = -1 μ A and I IHmax = 1 μ A. The LOW-state DC fanout is: Fanout LOW = | I OLmax / I ILmax | = 20 μ A / 1 μ A = 20 The HIGH-state DC fanout is: Fanout HIGH = | I OHmax / I IHmax | = 20 μ A / 1 μ A = 20 The device fanout is min( Fanout LOW , Fanout HIGH ) = 20, so one should theoretically connect no more than 20 74HC00-like inputs to one output of a 74HC00. If you choose the larger allowable current values for connection to “non-CMOS” inputs, then from Table 3-3 we obtain I OLmax = 4mA, I OHmax = -4mA, I ILmax = -1 μ
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This note was uploaded on 05/09/2009 for the course ENGRD 2300 taught by Professor Albonesi/long during the Fall '07 term at Cornell University (Engineering School).

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hw2s - ENGRD2300: Introduction to Digital Logic Fall 2008...

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