555 timer in monostable mode — compute R1 for a target pulse width A 555 timer is configured as a monostable multivibrator with timing capacitor C1 = 0.01 µF. Determine the resistor R1 required to obtain a pulse width of 2 ms (use T = 1.1 * R1 * C1).

Difficulty: Easy

Correct Answer: 182 kΩ

Explanation:


Introduction / Context:
The 555 timer in monostable mode generates a single pulse whose duration depends on an external resistor and capacitor. The standard timing relation is T = 1.1 * R * C for the high output interval. This question tests your ability to rearrange the formula and select a practical resistor value for a specified pulse width and known capacitance, a frequent task in embedded timing and pulse-generation circuits.


Given Data / Assumptions:

  • Monostable mode (one-shot) using a 555 timer IC.
  • Target pulse width T = 2 ms.
  • Timing capacitor C1 = 0.01 µF = 10 nF = 10 * 10^-9 F.
  • Timing relation: T = 1.1 * R1 * C1.


Concept / Approach:
Rearrange the timing equation to solve for R1. Ensure unit consistency (seconds, farads, ohms). After computing R1, compare the result to the offered options, rounding to a standard resistor value where appropriate.


Step-by-Step Solution:

1) Start with T = 1.1 * R1 * C1.2) Solve for R1: R1 = T / (1.1 * C1).3) Substitute values: R1 = 0.002 s / (1.1 * 10e-9 F).4) Compute denominator: 1.1 * 10e-9 = 11e-9.5) R1 = 0.002 / 11e-9 = (2e-3) / (11e-9) ≈ 181,818 Ω ≈ 182 kΩ.


Verification / Alternative check:
Back-calculate the pulse: T_est = 1.1 * 182,000 * 10e-9 ≈ 1.1 * 0.00182 ≈ 0.002002 s ≈ 2.00 ms, matching the requirement.


Why Other Options Are Wrong:

  • 200 kΩ: gives T ≈ 2.2 ms (too long).
  • 91 kΩ: gives T ≈ 1.0 ms (half the target).
  • 182 Ω: off by three orders of magnitude; pulse would be microseconds.


Common Pitfalls:
Confusing microfarads with nanofarads, omitting the 1.1 constant, or rounding to a non-standard value that misses timing by a wide margin.


Final Answer:
182 kΩ

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