Positive edge-triggered D flip-flop capture A positive edge-triggered D flip-flop will store a logic 1 under which timing condition?
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Athe D input is HIGH and the clock transitions from HIGH to LOW
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Bthe D input is HIGH and the clock transitions from LOW to HIGH
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Cthe D input is HIGH and the clock is LOW
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Dthe D input is HIGH and the clock is HIGH
Answer
Correct Answer: the D input is HIGH and the clock transitions from LOW to HIGH
Explanation
Introduction / Context:Edge-triggered flip-flops sample inputs only at a specific clock transition. For a positive edge-triggered device, the LOW-to-HIGH transition is the critical instant. Understanding this sampling instant is essential for meeting setup and hold time constraints and avoiding metastability.
Given Data / Assumptions:
- Device: positive edge-triggered D flip-flop.
- Input: D must be stable around the sampling edge.
- Clock: the active edge is rising (LOW → HIGH).
Concept / Approach:At the rising edge of the clock, the flip-flop captures the logic level present on D (if setup/hold requirements are satisfied). If D = 1 at that instant, Q becomes 1 after the device propagation delay and remains 1 until the next active edge (unless asynchronously overridden by preset/clear signals, if present).
Step-by-Step Solution:
Ensure D = 1 during the setup interval before the rising edge.Apply a LOW-to-HIGH clock transition.Maintain D stable through the hold interval immediately after the edge.Observe Q transitioning to 1 after t_pd (propagation delay) and remaining latched until a later edge.Verification / Alternative check:Timing diagrams show the sampling window centered on the rising edge, with setup/hold slacks specified. Simulations using HDL primitives confirm that Q follows D only on that edge, not during steady clock levels.
Why Other Options Are Wrong:
- HIGH-to-LOW: Describes negative-edge devices.
- Clock LOW or HIGH (steady levels): Edge-triggered devices do not capture simply due to level; they sample on transitions.
Common Pitfalls:Violating setup/hold times, which may cause metastability or incorrect capture, and assuming level-sensitive behavior instead of edge-triggered.
Final Answer:the D input is HIGH and the clock transitions from LOW to HIGH