Question

3.Consider a N-bit carry-skip adder with B bits per block. Assume that one stage of carry ripplery) has the same delay of one skip logic stage (tskip) and both are 1. Determine the block size Bopt that mininizes the worse-case delay time.

***Please be coherent and conscience.

Answer 1

The Delay time in a N-bit carry-skip adder with B bits per block is-

T = t_setup + B t_carry + ((N/B) - 1) t_skip + B t_carry + t_sum

Assuming one stage of carry ripple ( t_carry ) has the same delay of one skip logic stage ( t_skip ) and both are 1.

t_carry = t_skip = 1.

For worst-case delay time,

t_setup = t_sum = 1.

so, delay time,

T = 1 + B + (N/B-1) + B + 1

T = 2B + (N/B) + 1

For optimum block size,

dT/dB = 0

2 - (N/B² ) = 0

B_opt =