899 views
1 votes
1 votes

 

3 Answers

0 votes
0 votes
3*14+44= 86
0 votes
0 votes
Time after which output sum bit becomes available from the last full adder   

= Time taken for its carry in to become available + Sum propagation delay of full adder

= { Total number of full adders before last full adder X Carry propagation delay of full adder } + Sum propagation delay of full adder

= 3 X 14 +44

= 86
0 votes
0 votes

let T(n) is the time taken by nth full adder as it has to wait for 14ns(PD$_C$) for previous adder to calculate its carry

so we can form a recurrence

T(n)  =  T(n-1) + PD$_C$  (Propagation delay of carry)

T(1) = (PD$_s$) Propagation delay of sum (as for only 1 full adder time taken = 44ns(=(PD$_s$)

solving by back substitution 

T(n) = T(n-2) + (PD$_C$)+ (PD$_C$)

 T(n) = T(n-2) + 2*(PD$_C$)

  •  
  •  
  •  

 T(n) = T(n-k) + k*(PD$_C$)

if n-k=1 ,k=n-1

 T(n) = T(n-(n-1)) + (n-1)*(PD$_C$)

 T(n) = T(1) + (n-1)*(PD$_C$)

 T(n) = (PD$_C$) + (n-1)*(PD$_C$)  we can derive this formula by ton a ways 

here n=4 (PD$_C$) = 14ns (PD$_s$) =44ns

 T(4) = 44 + (4-1)*(14)

T(4) =86ns

Related questions

1 votes
1 votes
0 answers
1
0 votes
0 votes
2 answers
2
Na462 asked Oct 1, 2018
811 views
0 votes
0 votes
1 answer
3
srestha asked May 21, 2018
1,276 views
One ripple carry adder is adding two n-bit integers. The time complexity to perform addition using this adder is (We know carry look ahead adder takes time log n. Is it s...
3 votes
3 votes
3 answers
4
admin asked Mar 31, 2020
3,871 views
In which of the following adder circuits, the carry look ripple delay is eliminated?Half adderFull adderParallel adderCarry-look ahead adder