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How do I add 3 bits using full adder and basic gates, when the initial carry bit is 0??

I have tried using two full adders where two terminals of the first adder gets the first two inputs then the output of this adder goes to of the inputs of the second adder and the third input is given to the second input terminal of the second adder. The carry out from the first adder goes to the carry input of the second adder.

I've added a picture of the circuit which was created by using Logisim(software for designing digital logic circuit).

But it doesn't give the right answer.

normally in binary system, for addition of three bit sum is 1 and carry output bit is 1.

1+1+1 = 1 & carry bit = 1

But as you can see in the picture of the circuit I designed below, output of my circuit for adding three binary digit, sum is 0 and carry output bit is 1. which is not the right answer

1+1+1, is sum = 0 & carry = 1

The reason of this is clear to me. But I can't design the exact circuit. How do I design the circuit that can add 3 bit when initial carry is 0?

enter image description here

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  • $\begingroup$ If the initial carry is 0, why don't you use it as the 3rd bit? $\endgroup$ – Baku Sep 28 at 4:23
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    $\begingroup$ Trying to use the first adder just makes it a lot harder, you'd have to connect the carry outs of both adders with an or. Totally pointless. $\endgroup$ – gnasher729 Sep 28 at 15:48
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A full adder adds three bits and outputs a sum and a carry. If you want to add three bits, you just connect them to the inputs x, y, and carry of the full adder.

If for some reason there is a requirement that the carry input of the first full adder must be zero, then you leave that first full adder totally unconnected and connect to the three inputs of the second full adder.

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