Difficulty: Easy
Correct Answer: It requires an external trigger to enter a temporary quasi-stable state before returning to its stable state
Explanation:
Introduction / Context:Monostable, astable, and bistable multivibrators serve different timing and storage roles. The monostable (one-shot) is used to generate a single pulse of defined width when triggered, making it fundamental in timing, debounce, and pulse-stretch applications.
Given Data / Assumptions:
Concept / Approach:A monostable has one stable state. Upon receiving a trigger, it transitions to a quasi-stable state for a duration T set by circuit parameters, then automatically returns to the stable state. It does not free-run. Bistable devices, by contrast, require inputs to change state and remain there until commanded; astables free-run between states with no stable resting state.
Step-by-Step Solution:
Identify: only one stable state → needs a trigger to leave it.Trigger received → output pulse of duration T (e.g., T ≈ 1.1 * R * C in 555).After T, circuit returns to its stable state automatically.Thus, the defining trait is the need for an external trigger.Verification / Alternative check:Scope measurements show a single pulse following each trigger edge; with no triggers, the output remains in its stable level indefinitely.
Why Other Options Are Wrong:
Continuous oscillation: describes an astable, not a monostable.Two stable states: that is a bistable (flip-flop).DC–AC conversion/duty claims: unrelated to the definition.Cannot be implemented with logic: false—one-shots exist as standard logic functions and as 555 timers.Common Pitfalls:Confusing retriggerable vs non-retriggerable modes; ignoring input conditioning and debounce for clean triggering; miscalculating pulse width due to component tolerances.
Final Answer:It requires an external trigger to enter a temporary quasi-stable state before returning to its stable state
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