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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

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edited
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But why to take C0 as 0.
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Maybe because,if we don not take C0 as 0 then we will need 16 full adders which is not in option.
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ppl here are not explaining the answers.. just writing the answrs.. very bad
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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.
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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).
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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.

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ma'am is vishalshrm539 above comment reasoning is right ?

Bur for Pure binary adder its 15 FA + 1 HA

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what is meaning of adder -cum subtractor?
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+1
Thanks for this detailed explanation.
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Thank you buddy :)
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One small doubt is it acting as ripple carry adder?
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yes it is ripple carry. You can check carry look ahead adder for better version of this.

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