Sizing cascaded adders for required operand width How many 4-bit parallel adders must be cascaded to add two unsigned binary numbers when each number can represent decimal values up to 300 (base 10)?

Introduction / Context:
Digital designers often build wider adders by cascading standard 4-bit parallel adder ICs (such as 74xx283). The key is to determine the minimum operand bit width required by the application and then count how many 4-bit slices are needed to cover that width, including carry propagation between slices. This question asks for the number of 4-bit adders required when each operand may be as large as decimal 300.

Given Data / Assumptions:

• Each operand is an unsigned integer that can be as large as 300 (base 10).
• Adders available: 4-bit parallel adder slices that can be cascaded via carry-in/carry-out.
• Goal: compute the necessary operand bit width, then determine how many 4-bit slices cover that width.

Concept / Approach:

Find the smallest n such that 2^n > 300 to ensure every value up to 300 is encodable. The binary width for one operand determines the number of 4-bit slices needed: slices = ceil(n / 4). Note that the number of slices is driven by operand width, not by the sum's width; a standard ripple connection carries through slices.

Step-by-Step Solution:

Verification / Alternative check:

Binary of 300 is 100101100 (9 bits). Two slices (8 bits) would be insufficient; the ninth bit would be lost. Using three slices gives 12 available bits, more than enough for each 9-bit operand, with carries rippling across slice boundaries.

Why Other Options Are Wrong:

1: would only provide 4 bits; far too small for values up to 300.

2: gives 8 bits; maximum unsigned value 255, still insufficient.

4: provides 16 bits; works but is unnecessary overkill compared to the minimal 3-slice solution.

Common Pitfalls:

Confusing operand width with the maximum sum width; while the sum of two 9-bit numbers can be up to 10 bits, the number of 4-bit slices is still determined by the operand width because each slice adds corresponding operand bits and propagates carry. Another pitfall is assuming 300 requires only 8 bits (it does not; 8 bits top out at 255).

Final Answer:

3