Positive edge-triggered D flip-flop — operational description Which statement correctly describes how a positive edge-triggered D flip-flop captures and presents data at its output?

Difficulty: Easy

Correct Answer: The output will follow the input on the leading edge of the clock.

Explanation:


Introduction / Context:
The positive edge-triggered D (data) flip-flop is the workhorse memory element of synchronous logic. Its behavior is simple: sample D at the rising edge of the clock and hold that value until the next rising edge. Understanding this sampling action is essential for designing pipelines, registers, and safe clock-domain crossings (with additional synchronizers).


Given Data / Assumptions:

  • Device: positive edge-triggered D flip-flop.
  • Inputs: D and CLK (plus optional asynchronous set/clear not discussed here).
  • Output: Q (and often Q').


Concept / Approach:
At each LOW-to-HIGH clock transition, the flip-flop captures the current logic level present at D, provided setup and hold times are satisfied. Immediately after the edge (allowing for propagation delay), Q updates to that sampled value and remains stable for the rest of the clock period, regardless of subsequent changes on D.


Step-by-Step Solution:

1) Identify the sampling instant: the leading (rising) edge.2) Apply the sampling rule: Q(next) = D(at the edge).3) Between edges: Q holds its previous state even if D toggles.4) Repeat at the next rising edge to update Q again.


Verification / Alternative check:
Review a timing diagram: align D transitions to meet setup/hold around the rising edge; Q transitions occur shortly after each rising edge only.


Why Other Options Are Wrong:

  • Toggle behavior with both inputs HIGH describes a J–K flip-flop, not a D FF.
  • Invalid state with both LOW applies to S–R latches, not D FFs.
  • Leading-edge capture and trailing-edge output transfer describes master–slave staging, not a single edge-triggered D FF.


Common Pitfalls:
Ignoring setup/hold requirements leading to metastability, or assuming Q follows D continuously during the HIGH level of the clock.


Final Answer:
The output will follow the input on the leading edge of the clock.

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