ISRO2015-7

Question

If half adders and full adders are implements using gates, then for the addition of two $17$ bit numbers (using minimum gates) the number of half adders and full adders required will be

• A. $0,17$
• B. $16,1$
• C. $1,16$
• D. $8,8$

Comments

explain it in detail

 

 

plz

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We have two 17 bit numbers, one adder is used for adding one bit so for the least significant bit we don't need any full adder as we always  have 0 as initial carry for the LSB so we can add LSB using half adder but for rest of the 16 bits from 2nd bit from right to MSB we need full adder because carry can be generated by them hence there is need of 16 full adder.

Hence option C) is correct

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same question - https://gateoverflow.in/1494/gate1999-2-16

Answer 1

Option c ) 1 half adder 16 Full adder

Comments

yes

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Why cant we achieve it using 0,17 ?

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yes you can.. but ur solution must be cost effective.. right?

with 1 HA instead of 1 FA, you can reduce number of gates.

Answer 2

Answer : Option C 

 

Comments

great!

Answer 3

Ans (c)

1 H.A    and   (n-1) F.A

Comments

is it a formula?

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how??? is this any formula?!! pls explain

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i think u should make diagram for clear understanding:)

Answer 4

Let the two 17 bit numbers be,

(a0 a1 a2 a3....a16)

(b0 b1 b2.........b16)

1 half adder will take input a16, b16 and return a sum and a carry. This carry will be forwarded to the next adder (full adder). The first full adder will take three inputs (the carry from previous stage, a15, b15) and generate the next sum and next carry for the second full adder. In this way, 16 full adders are used.

So, 1 H.A and 16 F.A (ans)

Answer 5

for implementation n-bit of parallel adder n bit full adder required.

n bit full adder = (n-1) full adder and 1 half adder required.

so accoring to the question 16 full adder and 1 half adder required.

so option c is correct.