Potentiometers and strain gauges

Why it matters

Potentiometers and strain gauges are the two classic resistive transducers. Potentiometers still read throttle, valve and robot-joint positions; strain gauges sit inside almost every load cell, pressure transmitter and torque sensor. Both are simple to use but have well-known error mechanisms — loading for the potentiometer, temperature and small signal size for the strain gauge — that you must be able to quantify.

Key ideas

Resistive potentiometer. A resistive track of total resistance Rp with a sliding wiper, excited by a voltage Vs. The wiper position x (fraction of full travel, 0 to 1) sets the output: ideally Vo = x·Vs, a zero-order transducer. Linear (translational) potentiometers measure linear displacement; rotary and helical (multi-turn) ones measure angle.

• Track types: wire-wound (resolution limited by turn spacing — the output moves in steps of one turn), cermet and conductive plastic (essentially infinite resolution, long life).
• Resolution of a wire-wound pot = 1/N of full travel, where N is the number of turns.
• Loading: a meter or ADC of resistance Rm across the wiper draws current and pulls the output low. The error is zero at both ends and maximum near x ≈ 2/3. Keep Rm/Rp ≥ 10 or buffer the wiper.
• Choice of Rp: a high Rp reduces power dissipation and self-heating; a low Rp reduces loading error and noise pick-up. This trade-off is a standard exam question.
• Limitations: friction and wear (finite life), wiper noise, contact bounce under vibration, needs mechanical coupling to the measurand.

Strain gauge. A fine wire or etched foil grid (or a semiconductor element) bonded to a surface. When the surface is strained, the grid's length, cross-section and resistivity all change, changing R = ρL/A.

• Gauge factor GF = (ΔR/R)/ε. Differentiating R = ρL/A gives GF = 1 + 2ν + (Δρ/ρ)/ε. The 1 + 2ν part is purely dimensional (≈1.6 for ν = 0.3); the resistivity (piezoresistive) part makes metal foil gauges about 2 and semiconductor gauges about 50–200 (positive for p-type, negative for n-type).
• Typical values: R = 120 Ω or 350 Ω; GF ≈ 2.0–2.1 for constantan foil; working strains of a few hundred to a few thousand microstrain (1 µε = 10⁻⁶ m/m). The resulting ΔR/R is of order 10⁻³, so a Wheatstone bridge (next topic) is essential.
• Strain and stress: on a uniaxially loaded member in its elastic range, ε = σ/E. Transverse strain is −ν·ε, which is why gauge orientation matters.
• Temperature effects: the gauge's own temperature coefficient and the difference in thermal expansion between gauge and specimen both give an apparent strain. Remedies are self-temperature-compensated foil matched to the specimen material, and a dummy gauge in an adjacent bridge arm.
• Semiconductor gauges trade high GF for poor linearity, strong temperature sensitivity and fragility.
• Bonding and installation: adhesive creep, poor bonding and moisture cause drift; gauges should be mounted along the principal strain direction.

Formulas

Vo = x·Vs (unloaded potentiometer) Vo = output (V); Vs = supply (V); x = wiper fraction of travel (0–1).

Vo = x·Vs / [1 + x(1 − x)·(Rp/Rm)] (potentiometer loaded by Rm) Rp = total pot resistance (Ω); Rm = load resistance (Ω).

Resolution = 1/N (wire-wound, fraction of full travel) N = number of turns.

GF = (ΔR/R) / ε = 1 + 2ν + (Δρ/ρ)/ε ΔR = resistance change (Ω); R = unstrained resistance (Ω); ε = longitudinal strain (m/m); ν = Poisson's ratio; Δρ/ρ = fractional resistivity change.

ε = σ / E (uniaxial elastic stress) σ = stress (Pa); E = Young's modulus (Pa).

Worked examples

Example 1 — potentiometer loading. A 10 kΩ linear potentiometer is supplied with 10 V. Its wiper is at mid-travel (x = 0.5) and is read by a voltmeter of 50 kΩ. Find the reading and the error.

1. Rp/Rm = 10/50 = 0.2.
2. Vo = x·Vs / [1 + x(1 − x)·Rp/Rm] = 0.5 × 10 / (1 + 0.25 × 0.2) = 5/1.05 = 4.762 V.
3. Ideal Vo = 5 V, so error = −0.238 V = −4.76 % of reading = −2.38 % of full scale.
4. Scanning x shows the largest full-scale error for this ratio is 2.84 % FS, at x ≈ 0.67.

Answer: 4.76 V; error −0.24 V (−2.38 % FS).

Example 2 — strain gauge on a steel bar (GATE level). A 120 Ω foil gauge is bonded axially to a steel bar (E = 200 GPa) carrying a uniform tensile stress of 100 MPa. The foil material has ν = 0.3 and its resistivity changes by Δρ/ρ = 0.4ε. Find the gauge factor, the strain and the resistance change.

1. GF = 1 + 2ν + (Δρ/ρ)/ε = 1 + 0.6 + 0.4 = 2.0.
2. ε = σ/E = 100 × 10⁶ / 200 × 10⁹ = 5 × 10⁻⁴ = 500 µε.
3. ΔR = GF·ε·R = 2.0 × 5 × 10⁻⁴ × 120 = 0.12 Ω.

Answer: GF = 2.0, ε = 500 µε, ΔR = 0.12 Ω (only 0.1 % of R — hence the bridge).

Example 3 — resolution. A wire-wound linear pot has 500 turns over a 50 mm stroke. Resolution = 50/500 = 0.1 mm (0.2 % of travel).

Common mistakes

• Using Vo = x·Vs when the readout has finite resistance; the loaded output is always lower, worst near x ≈ 2/3.
• Thinking a higher Rp is always better — it increases loading error.
• Taking GF as purely geometric (1 + 2ν); for metals the resistivity term is significant, for semiconductors it dominates.
• Forgetting the micro in microstrain: 500 µε is 5 × 10⁻⁴, not 5 × 10⁻².
• Ignoring temperature: a few degrees of change can produce an apparent strain comparable to the real one.
• Mounting the gauge across the load direction and expecting the full axial strain (you get −ν·ε).

For GATE IN

Expect NAT problems on potentiometer loading error, ΔR from GF and strain, stress-to-strain-to-ΔR chains, and gauge factor from Poisson's ratio. MCQs ask why semiconductor gauges have high GF, what limits wire-wound resolution, and how temperature compensation works. Practise the loaded-potentiometer formula until it is automatic.

Quick check

1. A 5 kΩ pot at x = 0.4 with 10 V supply is read by a 50 kΩ meter. What is the reading?
2. What is GF for a material with ν = 0.5 and no piezoresistive change?
3. A 350 Ω gauge with GF = 2 sees 1000 µε. Find ΔR.
4. Why is a semiconductor strain gauge less linear than a foil gauge? Answers: 1. 4/(1 + 0.24 × 0.1) = 3.906 V; 2. 2.0; 3. 0.7 Ω; 4. its resistivity change (piezoresistance) is itself non-linear and strongly temperature-dependent.