For adding two 16-bit binary numbers using ripple-carry logic, how many half-adders and full-adders are required in the adder chain?

Introduction / Context:
Binary adders are built from two primitives: the half-adder (adds two bits) and the full-adder (adds two bits plus a carry-in). Understanding how many of each are needed for an n-bit addition helps in datapath design and estimating delay through a ripple-carry chain.

Given Data / Assumptions:

• We add two 16-bit operands.
• Ripple-carry architecture is used.
• The least significant stage has no incoming carry.

Concept / Approach:
The least significant bit (LSB) can be summed by a half-adder because there is no carry-in. Every other bit position requires a full-adder because it must include the carry generated by the previous stage. Therefore, for n bits: 1 half-adder and n−1 full-adders. For n = 16: 1 half-adder and 15 full-adders.

Step-by-Step Solution:
Determine the LSB stage requirement: no carry-in → half-adder.For the remaining 15 stages: carry-in present → full-adder per stage.Count devices: 1 half-adder, 15 full-adders.Choose the option that matches these counts.

Verification / Alternative check:
General formula: for n-bit addition, components = 1 HA + (n−1) FA. Substitute n=16 to verify: 1 + 15 = 16 stages total, which is correct for a 16-bit adder chain.

Why Other Options Are Wrong:
8/8 and 4/12 do not follow the n−1 full-adder rule; 16 half-adders cannot handle carry propagation; “None” is false since a correct option exists.

Common Pitfalls:
Using a half-adder at a non-LSB stage where a carry-in exists; forgetting that carry propagation dictates full-adders beyond the first stage.

Final Answer:
1 half-adders, 15 full-adders