Full adder

Question

 

Comments

yaa 86

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Well please explain the concept not the formula I know that too how the formula came ? A little hint will also suffice.

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86 is correct how'd you get that /@magma

Answer 1

3*14+44= 86

Comments

time when 4 th bit will get carry is 42 nsec because 14*3 so it takes 42 nsec to generate third carry after that fourth bit needs 44 sec more to generate sum so 42+44 =86 nsec answer.Draw like Gantt chart as we are doing in operating system you will easily get answer no need of mugging formula.

Answer 2

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

Answer 3

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$_C$)

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 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