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+22 votes

The number of full and half-adders required to add $16$-bit numbers is

  1. $8$ half-adders, $8$ full-adders

  2. $1$ half-adder, $15$ full-adders

  3. $16$ half-adders, $0$ full-adders

  4. $4$ half-adders, $12$ full-adders

asked in Digital Logic by Veteran (59.8k points)
edited by | 5k views

2 Answers

+35 votes
Best answer
Answer is B.

Ffor LSB addition we do not need a full adder.

For addition of subsequent bits we need full adders since carry from previous addition has to be fed into the addition operation.
answered by Loyal (8.7k points)
edited by
But why to take C0 as 0.
Maybe because,if we don not take C0 as 0 then we will need 16 full adders which is not in option.
ppl here are not explaining the answers.. just writing the answrs.. very bad
The explanation is present in the selected answer. The Carry In of the LSB Adder will always be 0. So, the LSB adder need not be full. It can be a half adder. Therefore 15 full adders and 1 half adder.
C0(carry at LSB) will always be 0 for addition , it will be 1 in case we want to do subtraction using an adder, so we need 15 FA for Most significant 15 bits & 1 half adder for two bits at least significant position.

we know that HA adds 2 bits & FA adds 3 bits, at LS position, there will be two bits only(we can ignore C0).

take 16 bit 2 numbers for first 2 bit of each number we need one half adder and for rest(15 bit of each number) we will have one carry of previous sum and two bits of each number so we need one full adder for each rest of bits. 



ma'am is vishalshrm539 above comment reasoning is right ?

We need 16 Full adders + 0 Half adders when it is Adder cum subtractor

Bur for Pure binary adder its 15 FA + 1 HA

what is meaning of adder -cum subtractor?



@jatin khachane 1 while subtraction borrow is generated only in one case when 1 is being subtracted from 0. 

subtraction = A ex or B (it will give 1 as A =0 and B=1)

Borrow = ~A.B (It will give one for the same input.)

So in LSB of result we get 1 which is desired(10-1) and for all the next bits we have to deal with 3 bits in worst case so full adder.

Hence 15 full adder + 1 half adder.


In subtraction we add $\bar{B}$  to A by giving 1 to M that make $B{}'$ (2s complement of B)

Hence $C_{0}$ also 1..At LSB we have to add 3 Bits and from LSB onward we may have to add 3 bits

check 5 - 1

Circuit is built in such a way that it takes sum with 1s complement first and then add 1 to it. So possibility of 3 1s at LSB is not possible. 

+6 votes

Answer : Option B


answered by Active (5k points)
Thanks for this detailed explanation.
Thank you buddy :)
One small doubt is it acting as ripple carry adder?

@Shamim Ahmed

yes it is ripple carry. You can check carry look ahead adder for better version of this.


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