Digital Design_ Principles and Practices, Chegg Solution Manual_parte_103

Prévia do material em texto

Step 3.57DP The LOW- state and state fan out can be calculated by using the following formulae: (i) The fanout where is the the maximum output current the TTL which is used to drive another gate we can say that given as input to another gate and is equal the maxi mum input current required LOW (ii) The HIGH-state fanout where the maximum output current the TTL which is used drive another gate can say that given as input to another gate and the maximum input current required pull HIGH. (iii) The Overall fan out is minimum of the LOW-state and HIGH-state Step (a) mine the LOW- state fanout for 74HCT driving 74LS The LOW- state fanout The of 74HCT given by mA and of given 0.4 mA which are required to operate Substitute for and for The LOW-state fanout 4 0.4 =10 Thus, the state fanout 10. Step Determine the HIGH- state fanout for 74HCT driving 74LS HIGH-state fanout The OHmax of 74HCT given by 4000 and of 74LS given by 20 UA which are to operate Substitute 400 for and 20 for HIGH-state fanout 4000 20 200 the HIGH state fanout is The maximum fanout the minimum of OW state fan-out and HIGH state fan-outs, that 10. Thus, the maximum fanout of 74HCT driving 74LS is The excess current is in HIGH state and 3800 Step (b) Dete the state fanout for 74VHCT driving 74S The LOW. state fanout The of 74VHCT is given by mA and of 74S given by mA which are required to operate Substitute for and for The LOW-state fanout the state fanout Step Determine the HIGH state fanout for 74VHCT driving HIGH-state fanout The of given by UA and of 74S is given by which are required to operate Substitute 8000 and 50 for HIGH-state fanout 8000 50 160 Thus, the HIGH state fanout The maximum fanout the minimum of LOW state fan-out and HIGH state fan-outs, that 4. the maximum fanout of 74VHCT driving 74S The excess current in HIGH state and value 7800 UA Step of 9 Determine the LOW-state fanout for 74VHCT driving The LOW-state fanout The of 74VHCT given mA and of 74ALS is 0.2 mA which are required to operate it. Substitute for and for The LOW-state fanout 0.2 40 Thus, the state fanout Step of9 Determine the HIGH-state fanout 74VHCT driving 74ALS HIGH fanout The of 74VHCT is given by 8000 UA and of 74ALS is 20 UA which are required to operate it. Substitute 8000 for and 20 for HIGH state fanout 8000 20 400 the HIGH 400 The maximum fanout the minimum of state fan-out and HIGH state outs, that 40. the maximum fanout of 74VHCT driving 74ALS is The excess current is in HIGH state and value 3840 Step of 9 (d) Dete mine the state fanout for 74HCT driving The LOW- state fanout The of 74HCT given mA and of 74AS given by 0.5 mA which are required to operate Substitute for and 0.5 for The LOW state fanout 0.5 =8 the LOW-state fanout Step of 9 Determine the HIGH state fanout 74HCT driving HIGH-state fanout The of 74VHCT given by and of 74AS given by which are required to operate Substitute 4000 and 20 for HIGH-state fanout 4000 20 200 Thus, the HIGH -state fanout 200 The maximum fanout the minimum of state fan-out and HIGH state fan outs, that 8. the maximum fanout of 74HCT driving 74AS is The excess current in HIGH state and value is 3840