Difficulty: Medium
Correct Answer: Connect A’s P < Q, P = Q, P > Q outputs to B’s <, =, > inputs, and also tie A’s = input high (VCC) with < and > inputs low (ground).
Explanation:
Introduction / Context:
The 7485 is a 4-bit magnitude comparator with cascade inputs (<, =, >) and outputs (P < Q, P = Q, P > Q). Cascading two devices allows comparison of wider words such as 8-bit values.
Given Data / Assumptions:
Concept / Approach:
The lower nibble determines equality and less/greater status only if the higher nibbles are equal. Therefore, the outputs from the lower comparator must feed the cascade inputs of the higher comparator. The lowest-level comparator’s cascade inputs must be initialized to “equal” true and “less/greater” false to start the chain.
Step-by-Step Solution:
Assign COMP A to bits [3..0] and COMP B to bits [7..4].Initialize COMP A cascade inputs: = input = 1 (VCC), < input = 0 (GND), > input = 0 (GND).Connect COMP A outputs (P < Q, P = Q, P > Q) to COMP B inputs (<, =, >), respectively.Read the final comparison results from COMP B outputs.
Verification / Alternative check:
This is the standard cascading method in the 7485 data sheet truth table and application diagrams.
Why Other Options Are Wrong:
Connecting outputs to outputs (option a) is invalid and may damage parts.Option b lacks the necessary initialization of the lowest stage inputs.Option c initializes but fails to cascade the stages.
Common Pitfalls:
Accidentally swapping less/greater lines or forgetting to tie initial cascade inputs, resulting in indeterminate comparisons.
Final Answer:
Connect A’s P < Q, P = Q, P > Q outputs to B’s <, =, > inputs, and also tie A’s = input high (VCC) with < and > inputs low (ground).
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