Question

In a 8-bit ripple carry adder using identical full adders, each full adder takes 34 ns for computing sum. If the time taken for 8-bit addition is 90 ns, find time taken by each full adder to find carry
1. 6 ns
2. 7 ns
3. 10 ns
4. 8 ns

Answer 1

Correct Answer - Option 4 : 8 ns

Formula:

The total time taken by the for ‘n’ bit ripple carry adder is

T_d = (n – 1) t_c + Maximum(t_c, t_s)

t_c = delay for carry through a single flip flop.

t_s= delay for sum

Data:

Each full adder takes 34 ns for computing

From this, Maximum(t_c, t_s) = t_s = 34 ns

T_d = 90 ns.

n = 8

Calculation:

T_d = (n – 1) t_c + Maximum(t_c, t_s)

90 = (8 – 1)t_c + 34

7t_n = 56 ns.

∴ t_n = 8 ns.

Time taken by each full adder to find carry is 8 ns.

Important Point:

If Maximum(t_c, t_s) = t_c then answer is 11.25 which is not in option

∴ Maximum(t_c, t_s) has to be t_s