Difficulty: Easy
Correct Answer: On the next active clock edge
Explanation:
Introduction: The 74195 is a universal shift register supporting both serial shifting and parallel loading via a SHIFT/LOAD control input. Understanding whether the parallel load is synchronous or asynchronous is essential for correct timing design.Given Data / Assumptions:
Concept / Approach: In parallel load mode, each stage takes its input from the parallel data bus. However, loading occurs on the defined active clock edge (typically the rising edge) so that all bits are captured simultaneously and deterministically. Therefore, outputs do not change “immediately”; they update synchronously on the next clock edge while SHIFT/LOAD remains LOW.Step-by-Step Solution:
Set SHIFT/LOAD = LOW to select parallel inputs.Apply the desired parallel data on the input pins.Toggle the clock to its active edge; the register outputs update together.Verification / Alternative check:
Consult timing diagrams: output transitions are edge-aligned; there is no level-sensitive, asynchronous load in normal operation.Why Other Options Are Wrong:
Immediately: Would imply asynchronous load, which is not the case.Only if CLOCK is LOW: Level of clock is irrelevant; the edge triggers the load.Depends on J and K: The 74195 does not use J/K controls; that wording confuses with JK flip-flops.Common Pitfalls:
Assuming “parallel load” equals “asynchronous load.”Forgetting to hold SHIFT/LOAD LOW through the active edge.Final Answer:
On the next active clock edge
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