Full-adder single-bit case — Cin = 0 For a full-adder with Cin = 0 and inputs A = 1, B = 1, what are the outputs for sum (∑) and carry-out (Cout)?

Difficulty: Easy

Correct Answer: ∑ = 0, Cout = 1

Explanation:


Introduction / Context:
A full-adder produces a 1-bit sum and a carry-out from three inputs: A, B, and Cin. Recognizing the mapping for simple combinations builds intuition for designing adders and debugging arithmetic logic units (ALUs).


Given Data / Assumptions:

  • Inputs: A = 1, B = 1, Cin = 0.
  • Sum function: A XOR B XOR Cin.
  • Carry-out function: majority of inputs or AB + Cin(A XOR B).


Concept / Approach:
Two ones with Cin = 0 already create a carry. The XOR-based sum toggles each time an odd number of inputs are 1. With two ones, the sum is 0; the carry-out is 1 because at least two inputs are 1.


Step-by-Step Solution:

1) Compute sum: A XOR B XOR Cin = 1 XOR 1 XOR 0 = 0 XOR 0 = 0.2) Compute carry: AB + Cin(A XOR B) = 11 + 0(0) = 1.3) Therefore, ∑ = 0 and Cout = 1.4) Check consistency with truth table of a full-adder.


Verification / Alternative check:
Interpret as decimal: 1 + 1 + 0 = 2 → binary 10, so sum bit 0, carry-out 1.


Why Other Options Are Wrong:

  • ∑ = 1, Cout = 0: corresponds to adding 1 with 0 only, not two ones.
  • ∑ = 1, Cout = 1 or ∑ = 0, Cout = 0: do not match the full-adder equations for A = B = 1, Cin = 0.


Common Pitfalls:
Confusing half-adder and full-adder equations or forgetting that XOR of two ones equals zero.


Final Answer:
∑ = 0, Cout = 1

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