Inductive transducers and LVDT

Why it matters

Inductive transducers measure displacement without a sliding contact, so they survive millions of cycles, dirt, oil and vibration. The LVDT in particular is the standard sensor for valve positions, servo-actuator feedback, gauging of machined parts and the secondary element in many pressure and force transducers.

Key ideas

An inductive transducer converts displacement into a change of self-inductance, mutual inductance or reluctance, and needs AC excitation (passive transducer).

Self-inductance and reluctance. For a coil of N turns on a magnetic circuit of reluctance ℜ, L = N²/ℜ. Displacement can change N (tapped coil), the permeability (moving core) or, most often, the air-gap length. In a magnetic circuit dominated by the air gap, ℜ ≈ (total gap length)/(μ0·A), so L is inversely proportional to the gap — sensitive but non-linear.

• Variable-reluctance (variable-gap) sensor: an armature moves relative to a C- or E-core. Fractional sensitivity dL/L = −dg/g.
• Differential arrangement: two coils with the armature between them; one gap shrinks while the other grows. Used in a bridge, the even-order non-linear terms cancel, sensitivity doubles and temperature effects cancel.
• Eddy-current (proximity) sensor: an RF coil near a conducting target induces eddy currents that reduce the coil's effective inductance and raise its losses. Non-contact measurement of gap, shaft vibration and runout; the target must be conductive and its material affects calibration.

LVDT (linear variable differential transformer): one primary and two identical secondaries on a hollow former, with a free-moving ferromagnetic core. The primary is excited with a sine wave (typically 1–10 kHz, a few volts). The secondaries are connected in series opposition, so the output is Vo = Vs1 − Vs2.

• At the null (core central) Vs1 = Vs2 and Vo is ideally zero. Moving the core one way raises Vs1 and lowers Vs2; the other way does the reverse.
• The magnitude of Vo is proportional to displacement over the linear range (typically ±0.5 mm to ±250 mm, depending on the model); the phase flips by 180° as the core passes through null. A plain AC voltmeter reads only magnitude, so direction is lost — a phase-sensitive (synchronous) demodulator recovers a signed DC output.
• Residual (null) voltage: in practice a small output remains at null due to harmonics in the excitation and stray capacitance between windings; it limits measurement near zero.
• Advantages: no friction or contact, practically infinite resolution (limited by electronics), high output, good linearity, rugged, electrical isolation between core and coils, tolerates temperature and vibration.
• Limitations: needs AC excitation and demodulation; core mass limits dynamic response; frequency response is limited to about one-tenth of the excitation frequency; sensitive to stray magnetic fields (shielding needed).
• RVDT: a rotary version for angles, linear over about ±40°.

LVDT sensitivity is usually given ratiometrically in mV/V/mm (output per volt of excitation per millimetre).

Formulas

L = N² / ℜ, ℜ = l_g / (μ0·A) (air-gap dominated) L = inductance (H); N = turns; ℜ = reluctance (A·turn/Wb, i.e. H⁻¹); l_g = total air-gap length (m); A = gap area (m²); μ0 = 4π × 10⁻⁷ H/m.

dL/L = −dg/g (variable gap)

ΔL_diff = L1 − L2 ≈ 2·L0·d/g (differential pair, small d) L0 = inductance at centre (H); d = armature displacement (m); g = nominal gap (m).

Vo = Vs1 − Vs2 (LVDT, series opposition)

Vo = S·Vex·x (LVDT linear range) S = sensitivity (V/V/mm); Vex = excitation (V rms); x = core displacement (mm).

Worked examples

Example 1 — LVDT output and resolution. An LVDT has sensitivity 40 mV/V/mm and is excited at 5 V rms. (a) Find the output for 2.5 mm displacement. (b) The readout resolves 1 mV. What is the displacement resolution?

1. Sensitivity at 5 V: 40 mV/V/mm × 5 V = 200 mV/mm = 0.2 V/mm.
2. Vo = 0.2 × 2.5 = 0.5 V rms.
3. Resolution = 1 mV / 200 mV/mm = 0.005 mm = 5 µm.

Answer: (a) 0.5 V rms; (b) 5 µm.

Example 2 — variable-gap sensor (GATE level). A coil of 500 turns sits on a high-permeability core with two air gaps in series, each g = 1 mm long and of area 4 cm². Neglect core reluctance and fringing. (a) Find L. (b) Find L when each gap closes to 0.9 mm. (c) Two such coils are used differentially (one gap 0.9 mm, the other 1.1 mm). Find ΔL and compare with the linear estimate.

1. Total gap = 2g = 2 mm; L = N²·μ0·A/(2g) = 500² × 4π × 10⁻⁷ × 4 × 10⁻⁴ / (2 × 10⁻³).
2. L = 250000 × 1.2566 × 10⁻⁶ × 4 × 10⁻⁴ / 2 × 10⁻³ = 62.83 mH.
3. At g = 0.9 mm: L = 62.83 × 1/0.9 = 69.81 mH (+11.1 % for a 10 % gap change, already non-linear).
4. At g = 1.1 mm: L = 62.83/1.1 = 57.12 mH.
5. Differential: ΔL = 69.81 − 57.12 = 12.69 mH; linear estimate 2L0·d/g = 2 × 62.83 × 0.1 = 12.57 mH (within 1 %).

Answer: (a) 62.8 mH; (b) 69.8 mH; (c) 12.69 mH vs 12.57 mH linear.

Common mistakes

• Adding the secondary voltages: they are in series opposition, so the output is their difference.
• Expecting an AC voltmeter to tell the direction of displacement — it cannot; use a phase-sensitive demodulator.
• Assuming the output at null is exactly zero; there is a residual null voltage.
• Forgetting that a single variable-gap sensor is non-linear (L ∝ 1/g) and treating it as linear over a large travel.
• Exciting an LVDT at a frequency too close to the measurement bandwidth — the carrier must be about ten times faster than the motion.
• Calling an LVDT a self-inductance transducer; it works on mutual inductance.

For GATE IN

Expect numerical questions on LVDT output from sensitivity and excitation, displacement resolution, reluctance and inductance of gap-type sensors, and differential linearisation. Conceptual questions ask about series opposition, phase reversal at null, residual voltage, need for demodulation and excitation frequency. Practise the reluctance formula with unit conversions (cm² and mm).

Quick check

1. Vs1 = 2.4 V and Vs2 = 1.6 V in an LVDT. What is the output?
2. What happens to the phase of an LVDT output when the core crosses null?
3. A gap-type sensor has L = 10 mH at g = 2 mm. Estimate L at g = 1.9 mm.
4. Why is the excitation frequency of an LVDT kept much higher than the frequency of the displacement? Answers: 1. 0.8 V; 2. it shifts by 180°; 3. 10 × 2/1.9 = 10.53 mH; 4. the output is an amplitude-modulated carrier, and demodulation needs several carrier cycles per cycle of motion (about 10:1).