LVDT with amplifier: An LVDT outputs 2 mV for a 0.5 mm displacement and feeds a 0–5 V voltmeter through a gain-250 amplifier. Determine the instrument sensitivity in V/mm.

Difficulty: Easy

Correct Answer: 1 V/mm

Explanation:


Introduction / Context:
A Linear Variable Differential Transformer (LVDT) is a highly sensitive displacement transducer. In practice, its small output is amplified before display or digitization. Sensitivity of the overall instrument chain (display volts per millimetre) is a key calibration figure for users.


Given Data / Assumptions:

  • LVDT raw output: 2 mV at 0.5 mm displacement.
  • Amplifier gain: 250 (unitless voltage gain).
  • Voltmeter range: 0–5 V (for context; not needed directly for sensitivity computation).


Concept / Approach:

Total sensitivity equals (amplifier gain) * (sensor output per mm). First compute the LVDT's millivolts per millimetre, then multiply by the gain to obtain the final volts per millimetre at the meter terminals.


Step-by-Step Solution:

LVDT raw sensitivity: 2 mV / 0.5 mm = 4 mV/mm.After amplification: 4 mV/mm * 250 = 1000 mV/mm = 1 V/mm.Therefore, instrument sensitivity = 1 V/mm.


Verification / Alternative check:

At 0.5 mm, the chain output is 1 V/mm * 0.5 mm = 0.5 V. Also, raw 2 mV * 250 = 0.5 V—consistent. This leaves ample headroom on a 0–5 V meter for larger displacements within linear range, improving resolution.


Why Other Options Are Wrong:

  • 0.1 V/mm and 0.5 V/mm: understate the amplified sensitivity by factors of 10 and 2.
  • 5 V/mm: overstates the gain result by 5×.
  • 0.02 V/mm: confuses raw LVDT output with post-gain value.


Common Pitfalls:

  • Forgetting to convert millivolts to volts after applying gain.
  • Confusing raw sensor sensitivity with system sensitivity.


Final Answer:

1 V/mm

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