Hall effect sensors

Why it matters

Hall sensors are everywhere a magnetic field can stand in for something mechanical or electrical: crankshaft and wheel-speed sensors, brushless motor commutation, lid switches in laptops and phones, and isolated current sensors in drives and battery systems. They are cheap, contactless and work down to DC — but only if you understand why the sensing element must be a thin semiconductor and how offset and temperature limit accuracy.

Key ideas

The Hall effect. A thin plate of thickness t carries a current I along its length. A magnetic flux density B is applied perpendicular to the plate. The Lorentz force q(v × B) pushes the moving carriers towards one edge, where they accumulate until the transverse electric field balances the magnetic force. The resulting Hall voltage V_H appears across the width, perpendicular to both I and B.

• Balance of forces gives V_H = I·B/(n·q·t) and the Hall coefficient R_H = 1/(n·q). The sign of R_H (and of V_H) tells whether the carriers are electrons (negative) or holes (positive).
• V_H is inversely proportional to carrier density n and thickness t. Metals have n ≈ 10²⁸–10²⁹ m⁻³, giving nanovolts; semiconductors (Si, GaAs, InSb, InAs) have far lower n and high mobility, giving millivolts. Hence Hall elements are thin semiconductor films.
• Combining a Hall measurement with resistivity gives the carrier mobility: μ = R_H/ρ. This is the standard way to characterise semiconductor material.

Practical Hall sensors integrate the element with a constant-current bias, amplifier and temperature compensation:

• Linear (ratiometric) sensors give an analog output proportional to B, specified in mV/mT (or mV/G), with an output centred at half the supply at zero field.
• Digital switches and latches add a Schmitt trigger with built-in hysteresis between operate and release points, giving a clean on/off output; used for proximity, gear-tooth speed sensing and BLDC commutation.
• Error sources: offset (misalignment voltage) from imperfect contact placement and stress — reduced by "spinning current" techniques; temperature dependence of n and μ; non-linearity and saturation at high fields.
• A Hall element is a passive transducer in the sense that it needs a bias current.

Applications.

• Position and speed: a magnet or ferrous gear tooth moving past the sensor modulates B; counting pulses gives speed.
• Current sensing: the conductor passes through a ferromagnetic core with a narrow gap that holds the Hall element. Open-loop: B in the gap ∝ current. Closed-loop (compensated): a secondary winding drives the gap flux to zero, and the secondary current is the measurement — better linearity and temperature stability. Both measure DC and AC with galvanic isolation.
• Power/multiplier: since V_H ∝ I·B, making I proportional to the load voltage and B proportional to the load current gives an output proportional to power (Hall wattmeter).
• Magnetic field measurement: gaussmeters.

Formulas

V_H = I·B / (n·q·t) = R_H·I·B / t V_H = Hall voltage (V); I = bias current (A); B = flux density normal to the plate (T); n = carrier density (m⁻³); q = 1.602 × 10⁻¹⁹ C; t = thickness along B (m).

R_H = 1 / (n·q) R_H = Hall coefficient (m³/C).

μ = R_H / ρ = R_H·σ μ = carrier mobility (m²/V·s); ρ = resistivity (Ω·m); σ = conductivity (S/m).

B = μ0·N·I / l_g (gapped core, high-permeability core, reluctance dominated by the gap) N = turns of the primary conductor (1 for a straight-through bus); l_g = gap length (m); μ0 = 4π × 10⁻⁷ H/m.

Worked examples

Example 1 — metal vs semiconductor. (a) A copper strip (n = 8.5 × 10²⁸ m⁻³) 0.5 mm thick carries 3 A in B = 0.2 T. (b) A semiconductor film (n = 1 × 10²² m⁻³) 0.5 mm thick carries 10 mA in the same field. Find V_H in each case.

1. (a) V_H = 0.2 × 3 / (8.5 × 10²⁸ × 1.602 × 10⁻¹⁹ × 5 × 10⁻⁴) = 0.6 / 6.81 × 10⁶ = 8.81 × 10⁻⁸ V → 88 nV.
2. (b) V_H = 0.2 × 0.01 / (1 × 10²² × 1.602 × 10⁻¹⁹ × 5 × 10⁻⁴) = 0.002 / 0.801 = 2.50 × 10⁻³ V → 2.50 mV.
3. With 300 times less current, the semiconductor gives about 28 000 times more voltage — this is why Hall elements are semiconductors.

Example 2 — material characterisation (GATE level). A semiconductor sample 1 mm thick carries 10 mA in B = 0.5 T and shows V_H = 2.5 mV. Its resistivity is 0.01 Ω·m. Find R_H, the carrier density and the mobility.

1. R_H = V_H·t / (I·B) = 2.5 × 10⁻³ × 1 × 10⁻³ / (0.01 × 0.5) = 5 × 10⁻⁴ m³/C.
2. n = 1/(q·R_H) = 1 / (1.602 × 10⁻¹⁹ × 5 × 10⁻⁴) = 1.25 × 10²² m⁻³.
3. μ = R_H/ρ = 5 × 10⁻⁴ / 0.01 = 0.05 m²/V·s (500 cm²/V·s).

Answer: R_H = 5 × 10⁻⁴ m³/C, n = 1.25 × 10²² m⁻³, μ = 0.05 m²/V·s.

Example 3 — open-loop current sensor. A bus bar passes once through a ferrite core with a 2 mm gap containing a linear Hall sensor of sensitivity 5 mV/mT. Neglecting core reluctance, find the output per ampere and at 20 A.

1. B/I = μ0/l_g = 4π × 10⁻⁷ / 2 × 10⁻³ = 0.628 mT/A.
2. Output sensitivity = 0.628 × 5 = 3.14 mV/A; at 20 A: 62.8 mV above the zero-field output.

Common mistakes

• Using the plate width instead of the thickness along B in the denominator.
• Losing powers of ten: a copper Hall voltage is nanovolts, not microvolts.
• Forgetting that only the B component perpendicular to the plate counts.
• Treating the zero-field output as zero: linear sensors have an offset (often Vcc/2) and a misalignment voltage.
• Assuming R_H is positive for all materials; its sign follows the majority carrier.
• Expecting a Hall switch to toggle at a single field — it has deliberate hysteresis.

For GATE IN

Typical questions: compute V_H, R_H, carrier density or mobility from given data; identify carrier type from the sign of V_H; field in a gapped core for current sensing; Hall-effect multiplier/wattmeter principle. Watch units carefully — mm to m, mA to A, and the result in µV or mV.

Quick check

1. If carrier density doubles, what happens to V_H at fixed I, B and t?
2. A sample gives a positive Hall coefficient. What are the majority carriers?
3. Find V_H for I = 1 mA, B = 0.5 T, n = 1 × 10²² m⁻³, t = 0.5 mm.
4. Why can a Hall current sensor measure DC while a current transformer cannot? Answers: 1. it halves; 2. holes (p-type); 3. 0.624 mV; 4. the Hall element responds to static flux density, while a CT relies on changing flux to induce a voltage.