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# Full Adder in Digital Logic

• Difficulty Level : Easy
• Last Updated : 09 Jun, 2022

Logical Expression for SUM: = A’ B’ C-IN + A’ B C-IN’ + A B’ C-IN’ + A B C-IN = C-IN (A’ B’ + A B) + C-IN’ (A’ B + A B’) = C-IN XOR (A XOR B) = (1,2,4,7)

Logical Expression for C-OUT: = A’ B C-IN + A B’ C-IN + A B C-IN’ + A B C-IN = A B + B C-IN + A C-IN = (3,5,6,7)

Another form in which C-OUT can be implemented: = A B + A C-IN + B C-IN (A + A’) = A B C-IN + A B + A C-IN + A’ B C-IN = A B (1 +C-IN) + A C-IN + A’ B C-IN = A B + A C-IN + A’ B C-IN = A B + A C-IN (B + B’) + A’ B C-IN = A B C-IN + A B + A B’ C-IN + A’ B C-IN = A B (C-IN + 1) + A B’ C-IN + A’ B C-IN = A B + A B’ C-IN + A’ B C-IN = AB + C-IN (A’ B + A B’)

Therefore COUT = AB + C-IN (A EX – OR B)

2 Half Adders and an OR gate is required to implement a Full Adder.

With this logic circuit, two bits can be added together, taking a carry from the next lower order of magnitude, and sending a carry to the next higher order of magnitude.

### Implementation of Full Adder using NAND gates: Implementation of Full Adder using NOR gates:

Total 9 NOR gates are required to implement a Full Adder. In the logic expression above, one would recognize the logic expressions of a 1-bit half-adder. A 1-bit full adder can be accomplished by cascading two 1-bit half adders.

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