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Full Adder Logic — The carry-out (Cout) of a 1-bit full adder in terms of inputs A, B, and Cin is which Boolean expression?

Difficulty: Easy

Correct Answer: AB + ACin + BCin

Explanation:


Introduction:
A full adder sums three one-bit inputs (A, B, Cin) and produces a sum bit (S) and a carry-out (Cout). Identifying the correct Boolean form of Cout is essential for ripple-carry adders and arithmetic logic design.

Given Data / Assumptions:

  • Inputs: A, B, Cin ∈ {0,1}.
  • Outputs: S and Cout.
  • Standard full-adder truth table semantics.


Concept / Approach:
Cout is 1 if at least two of the three inputs are 1 (majority function). The canonical sum-of-products for a majority of three is AB + ACin + BCin. By contrast, the sum bit is S = A XOR B XOR Cin and is not the carry-out.

Step-by-Step Solution:

Step 1: Enumerate the cases where two or more inputs are 1: (A=1,B=1), (A=1,Cin=1), (B=1,Cin=1).Step 2: Write SOP terms: AB, ACin, BCin.Step 3: OR the terms: Cout = AB + ACin + BCin.


Verification / Alternative check:

Truth-table check confirms Cout = 1 for 011, 101, 110, 111; the expression AB + ACin + BCin evaluates to 1 in exactly those rows.


Why Other Options Are Wrong:

A XOR B XOR Cin: That is the SUM output, not carry-out.A + B + Cin: Would assert carry whenever any single input is 1, which is incorrect.ABCin: True only for the 111 row; misses many carry cases.


Common Pitfalls:

Confusing the sum XOR relation with the carry majority logic.Forgetting one of the pairwise terms in the majority function.


Final Answer:

AB + ACin + BCin
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