7 votes 7 votes If a 3 bit multiplicand is multiplied to a 2 bit multiplier, minimum number of two input AND, XOR, and OR gates, needed in the design are respectively?A) 16,14,1B) 7, 3, 1C) 10, 4, 1D) 8, 4, 1 Digital Logic digital-logic + – admin asked Sep 16, 2015 admin 5.5k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 5 votes 5 votes let the three bit multiplicand be $111$ and two bit multiplier be $11$ $1 1 1$ $ * $ $1 1$ ${\color{Yellow} 1}$ ${\color{Green} 1}$ $1$ $1$ ${\color{Yellow} 1}$ ${\color{Green} 1}$ here we need 6 AND gates TO MULTIPLY ${\color{Red} 1}$ $0$ $1$ $0$ $1 $ TO ADD GREEN BITS WE NEED 1 HALF ADDER AND FOR YELLOW BIT WE NEED 1 FULL ADDER AND ONE MORE HALF ADDER FOR LAST BIT AS WE KNOW HALF ADDER NEEDS 1 XOR(FOR SUM) AND 1 AND GATE(FOR CARRY BIT) FULL ADDER NEED 2 XOR (FOR SUM)AND 2 AND GATE AND 1 OR GATE(FOR CARRY BIT) THEREFORE 2 HALF ADDER =2-XOR GATE +2 AND GATE 1 FULL ADDER =2-XOR GATE+2-AND GATE+1-OR GATE THEREFORE TOTAL 10 AND GATES 4 XOR GATES AND 1 OR GATE THEN IT WOULD BE 10 AND GATE 4 XOR GATE AND 1 OR GATE.(OPTION C) Umang Raman answered Sep 26, 2015 edited May 18, 2020 by srestha Umang Raman comment Share Follow See all 13 Comments See all 13 13 Comments reply Show 10 previous comments sushmita commented Jan 28, 2017 reply Follow Share CAN U FURTHER TELL THE TOTAL DELAY? 0 votes 0 votes sushmita commented Jan 28, 2017 reply Follow Share IT SHOULD BE 10 AND GATES RIGHT?? CAN SOMEONE PLZ TELL THE TOTAL DELAY TOO?? ASSUMING EVERY GATE HAS 1 UNIT OF DELAY.?? 0 votes 0 votes Kaluti commented Oct 8, 2017 reply Follow Share can someone elaborate more why do we need here 2 half adder and 1 full adder instead of 2 full adder and one half adder i am not getting it 1 votes 1 votes Please log in or register to add a comment.