IC monostable multivibrator timing behavior At the falling edge of the trigger, the monostable output goes HIGH for a time interval determined primarily by what parameter(s)?

Difficulty: Easy

Correct Answer: value of the RC timing components

Explanation:


Introduction / Context:
A monostable multivibrator produces a single output pulse of fixed width for each valid trigger. In common IC monostables (including 555-based configurations), the pulse width is governed by an RC timing network rather than the trigger amplitude or repetition rate, provided the trigger meets logic-level requirements.


Given Data / Assumptions:

  • Edge-triggered monostable, output HIGH after a falling trigger edge.
  • Timing set by resistor R and capacitor C.
  • Trigger pulses are valid logic levels and narrow compared to the output pulse.


Concept / Approach:
The monostable timing interval T is approximately proportional to R * C for a wide range of supply voltages and within component tolerance limits. The trigger only initiates the cycle; it does not set its duration once initiated.


Step-by-Step Solution:

1) Recognize monostable action: one fixed-width pulse per trigger.2) Identify the timing network: R and C determine the capacitor charge/discharge path.3) Conclude that T ≈ k * R * C (k depends on topology, often around 1.1 for 555).4) Therefore, pulse width depends on the value of the RC timing components.


Verification / Alternative check:
Datasheets provide formulas for T versus R and C, showing near-linear scaling across practical ranges.


Why Other Options Are Wrong:

  • Amplitude of the input trigger: once above logic threshold, amplitude does not set T.
  • Frequency of the input trigger: frequency affects repetition, not individual pulse width.
  • Magnitude of the dc supply: within limits it has minor effect compared to RC.


Common Pitfalls:
Relying on trigger width or level to control pulse width instead of properly sizing R and C.


Final Answer:
value of the RC timing components

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