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In this adder we have 2 inputs of N-bit numbers and one output of N-bit number with a carry. We can achieve this either using

This is the one which would accept bit by bit input of the n-bit numbers and there is a bit by bit output of the n-bit Sum. In this adder we would be required one full adder and a memory element.

Hence we see we require lesser hardware. The circuit for serial addition is as follow: Parallel adder is the one where we input the all the bits of two given numbers and we don’t need any memory element.

Let’s now calculate the time required for the carry to propagate from adder 1 to last adder and when we get the final result.

If at time t=0 we input the variables, we’ll the carry of 1st adder at t=2Δ which would be propagated to 2nd adder and at t=3Δ we get the sum variable S1. When at t=2Δ carry C1 is propagated to 2nd adder, we get the carry output of 2nd adder at t=4Δ and at t=5Δ we get the S2. At t=4Δ we have carry available at

3rd adder so its carry output comes at t=6Δ and sum output comes at t=7Δ. Similarly we get the final carry of 4 bit parallel adder at t=8Δ and sum S4 & hence complete output at t=9Δ.

And for n-bit adder we have the total time taken as 2 * (n-1) Δ + 3Δ = (2n+1) Δ

For 16-bit adder we have the

Time delay= (2*16+1) Δ = 33Δ

which is quiet large