Training Reactive (C & L) Sensors Inductive Sensors — LVDT and Eddy-Current
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Inductive Sensors — LVDT and Eddy-Current

30 min Reactive (C & L) Sensors

Inductive Sensors — LVDT and Eddy-Current Probes

Inductive sensors use magnetic coupling to translate motion or material properties into an electrical signal. The Linear Variable Differential Transformer (LVDT) is the most celebrated example: two secondary coils feed a moving core, and their difference voltage is proportional to displacement over a remarkably linear range. Eddy-current probes detect the presence of a conductive target by the reflected impedance in a sensing coil.

Webster and Pallàs-Areny treat these sensors with explicit equations for mutual inductance and reflected impedance. The beauty of the LVDT is its infinite resolution, frictionless operation, and the mechanical robustness that has kept it in service for more than eighty years in hydraulic actuators, aircraft control surfaces, and civil-engineering instrumentation.

We will derive the LVDT output equation, discuss phase-sensitive demodulation, and compute the characteristic frequency of an eddy-current target.

LVDT Transfer Function

For a core displacement $x$ from the centre:

$$V_{\text{out}} = k\,x\,V_s \cos(\omega t + \phi)$$

$k$ is the sensitivity (mV per mm per volt excitation), $V_s$ is the primary excitation (typically 1–10 V RMS at 1–10 kHz). Phase $\phi$ reverses sign across the null point, enabling direction sensing.

Eddy-Current Characteristic Frequency

$$f_c = \frac{1}{\pi \mu_0 \sigma r^2}$$

$\sigma$ = target conductivity, $r$ = coil radius. Operation near $f_c$ maximizes phase shift between the coil voltage and current, giving best sensitivity.

Example 1 — LVDT output

An LVDT has sensitivity $k = 50\,\text{mV/mm/V}$ and runs from $V_s = 3\,\text{V RMS}$. Find the RMS output when $x = +2\,\text{mm}$.

$V_{\text{out}} = k\cdot x\cdot V_s = 50\,\text{mV/mm/V}\cdot 2\,\text{mm}\cdot 3\,\text{V} = 300\,\text{mV RMS}$.

Example 2 — Phase reversal

The core moves to $x = -2\,\text{mm}$. Compare the output with Example 1.

Magnitude is the same (300 mV RMS), but the carrier phase is 180° shifted relative to the primary. Phase-sensitive demodulation converts this into a negative DC output — unambiguous direction.

Example 3 — Eddy-current probe frequency

A 5-mm-radius coil over an aluminium target ($\sigma = 3.5\times10^7\,\text{S/m}$). Find $f_c$.

$f_c = 1/(\pi\cdot 4\pi\times10^{-7}\cdot 3.5\times10^7\cdot (5\times10^{-3})^2)$.

$= 1/(\pi\cdot 4\pi\times10^{-7}\cdot 3.5\times10^7\cdot 2.5\times10^{-5})$

$= 1/(3.45\times10^{-3}) \approx 290\,\text{Hz}$. Most commercial eddy-current probes therefore operate at a few kilohertz to low megahertz to balance sensitivity and lift-off immunity.

Interactive Demo: LVDT Demonstrator
|V_out| =300.0mV RMS
Phase =0°
Scale factor =150mV/mm
Full-scale output =1500mV RMS

Practice Problems

1. LVDT: $k = 80\,\text{mV/mm/V}$, $V_s = 5\,\text{V}$, $x = 3\,\text{mm}$. Find $V_{\text{out}}$.
2. Same LVDT, $x = -1\,\text{mm}$. Magnitude and phase?
3. Why is an LVDT frictionless?
4. Compute $f_c$ for a 2-mm-radius coil over copper ($\sigma = 5.8\times10^7\,\text{S/m}$).
5. Why do LVDT signal conditioners use phase-sensitive demodulation rather than a simple rectifier?
6. Name one application where an LVDT is preferred over a potentiometer.
Show Answer Key

1. $V_{\text{out}} = 80\cdot 3\cdot 5 = 1200\,\text{mV} = 1.2\,\text{V RMS}$.

2. 400 mV RMS, 180° out of phase relative to +x output.

3. The core moves without mechanical contact — no wiper or bearing surface wears out.

4. $f_c = 1/(\pi\cdot 4\pi\times10^{-7}\cdot 5.8\times10^7\cdot 4\times10^{-6}) \approx 1.09\,\text{kHz}$.

5. To recover the sign of the core displacement — a rectifier gives only magnitude.

6. Aircraft control-surface position sensing (harsh environment, long life, no wearout).

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