Using 74184 BCD-to-binary converters — two-digit BCD to binary How many SN74184 BCD-to-binary converter ICs are required to convert two complete BCD digits (00–99) into a pure binary value for general-purpose decoding?

Introduction / Context:The classic SN74184 implements BCD-to-binary conversion using internal ROM-style tables. Practical converters for multi-digit BCD often require multiple devices to cover the full decimal range without invalid gaps. This question targets architectural understanding rather than memorizing a specific pinout.

Given Data / Assumptions:

• We must handle two BCD digits (00–99).
• The 74184 device provides a 6-bit BCD-to-binary table designed around typical 0–63 ranges; wider ranges require staged or conditional conversion.
• The goal is a continuous binary output covering 0–99 (requires 7 bits).

Concept / Approach:A common discrete-logic approach uses one converter network for the lower ranges and another (gated by the tens digit) to extend coverage to the higher ranges, then combines results. Rule-of-thumb designs use two 74184 devices per decimal digit for robust full-range conversion; thus, two digits generally require four devices. This scales with known reference designs where three digits use six devices (two per digit), maintaining uniform methodology and timing.

Step-by-Step Solution:

Verification / Alternative check:Reference multi-digit implementations: three BCD digits (000–999) often employ six 74184 ICs (two per digit), which aligns with the two-digit case requiring four.

Why Other Options Are Wrong:

• 1 or 2: insufficient to realize seamless 00–99 binary coverage with clean timing and decoding.
• 8: excessive for two digits and not minimal.

Common Pitfalls:Assuming a single small converter IC always maps any number of BCD digits; multi-digit conversion needs either arithmetic (add/scale) or multiple table devices.

Final Answer:4