To measure the torque transmitted by a circular shaft with strain gauges, the gauges are bonded:

around the circumference, at 90° to the axis

In an LVDT, the direction of core displacement from the null position is indicated by:

the phase of the output relative to the excitation

the DC offset of the primary current

A piezoelectric force sensor is unsuitable for measuring a steady (static) force because:

the generated charge leaks away through the finite resistance of the sensor and circuit

it responds only to temperature

its natural frequency is too low

it needs an external magnetic field

Absolute optical encoders are commonly coded in Gray code rather than natural binary because:

only one bit changes between adjacent positions, so a reading at a transition is off by at most one step

Gray code needs fewer tracks for the same resolution

Gray code can be added directly by a binary adder

it gives direction without a second channel, unlike binary

Which statements are true?

The resistance of an NTC thermistor falls as its temperature rises

A full bridge of four active gauges gives about four times the output of a quarter bridge at the same strain

A dummy gauge in an adjacent arm compensates for temperature changes of the active gauge

The resistance of an NTC thermistor falls as its temperature rises

A full bridge of four active gauges gives about four times the output of a quarter bridge at the same strain

The output voltage of a tachogenerator is proportional to angular acceleration

A dummy gauge in an adjacent arm compensates for temperature changes of the active gauge

For a parallel-plate capacitive displacement sensor, which statements are true?

Varying the plate gap gives a capacitance proportional to 1/d, a non-linear response

Varying the overlapping area gives a linear response

A differential arrangement improves linearity

Varying the plate gap gives a capacitance proportional to 1/d, a non-linear response

Varying the overlapping area gives a linear response

A differential arrangement improves linearity

The capacitance does not depend on the dielectric between the plates

Sensors, Signal Conditioning and Industrial Transducers

Section A.2 names resistive, capacitive, inductive, piezoelectric and Hall-effect sensors with their signal-conditioning circuits, and the transducers of industrial instrumentation for displacement (linear and angular), velocity, acceleration, force, torque and pressure. A robot or machine knows its own state only through these. The chapter takes each sensing principle, the circuit that turns its small change into a usable voltage, and then the transducer built on it for each measured quantity — with the numbers GATE asks: a gauge’s resistance change, a bridge’s output, an encoder’s resolution.

1. Resistive sensors and the Wheatstone bridge

A potentiometer turns a displacement into a resistance ratio and so a voltage ratio. A strain gauge turns strain into a small resistance change, defined by its gauge factor GF = (ΔR/R)/ε; metal-foil gauges have GF close to 2. A 120 Ω gauge with GF = 2 at 500 microstrain changes by ΔR = GF·ε·R = 2 × 500 × 10⁻⁶ × 120 = 0.12 Ω — a change of 0.1%, which is why a bridge is needed to see it. An RTD follows R = R₀(1 + αT); for platinum, α ≈ 0.00385 per °C, so a Pt100 reads about 119.25 Ω at 50 °C. A thermistor is a semiconductor whose resistance changes strongly and non-linearly; an NTC thermistor’s resistance falls as it warms.

The Wheatstone bridge balances four arms so the output is zero at rest and proportional to the imbalance. With excitation V and one active gauge (a quarter bridge), V_o ≈ V·GF·ε/4 for small strain; two active gauges in adjacent arms with opposite strains (half bridge) double it, and four active gauges (full bridge) quadruple it. At V = 10 V, GF = 2 and 1000 microstrain, a quarter bridge gives 10 × 2 × 10⁻³/4 = 5 mV. A dummy gauge, unstrained but at the same temperature, in the adjacent arm cancels the temperature-induced change of the active gauge, because equal changes in adjacent arms do not unbalance the bridge.

2. Capacitive and inductive sensors

A parallel-plate capacitor has C = εA/d, and a sensor varies one of the three. Varying the gap gives C ∝ 1/d — sensitive but non-linear (halving the gap doubles C, from 10 pF to 20 pF); varying the overlap area gives a linear change; varying the dielectric senses level or moisture. A differential arrangement, a moving plate between two fixed ones, makes one capacitance rise as the other falls and linearises the output. Capacitive sensing detects non-metallic objects and underlies most MEMS accelerometers.

Inductive sensors change a magnetic path. The LVDT (linear variable differential transformer) has a primary and two series-opposed secondaries on a movable core: at the central null the secondary voltages cancel; off-centre the output amplitude is proportional to displacement and its phase (0° or 180° relative to the primary) gives the direction, so a phase-sensitive demodulator recovers a signed DC signal. It is frictionless and has effectively infinite resolution. Variable-reluctance pickups sense passing ferrous teeth, and eddy-current and inductive proximity sensors detect metal targets only.

3. Piezoelectric and Hall-effect sensors

A piezoelectric crystal (quartz) or ceramic (PZT) develops a charge proportional to force, q = d·F, d the charge coefficient in coulombs per newton, and the voltage is q/C across its capacitance. The charge leaks away through any finite resistance, so a piezoelectric sensor measures dynamic force, pressure and acceleration well and static force not at all; a charge amplifier converts the charge to a voltage independent of cable capacitance. Its stiffness gives it a high natural frequency, which is why it is the standard vibration accelerometer.

A Hall-effect sensor passes a current I through a thin conductor of thickness t in a field B normal to it; the magnetic force pushes carriers to one edge and a transverse voltage V_H = IB/(n e t) appears, n the carrier density. It is inversely proportional to n, so semiconductors, with few carriers, give usable voltages. With B = 0.5 T, I = 10 mA, n = 10²¹ m⁻³ and t = 0.5 mm, V_H = 0.005/(10²¹ × 1.6 × 10⁻¹⁹ × 5 × 10⁻⁴) = 62.5 mV. Hall sensors give contactless position and speed sensing, current measurement, and the rotor-position signals that commutate brushless DC motors.

4. Transducers for displacement, velocity, acceleration, force, torque and pressure

Industrial transducers by measured quantity
Quantity	Common transducers	Point to remember
Linear displacement	Potentiometer, LVDT, capacitive, linear encoder	LVDT phase gives direction.
Angular displacement	Rotary potentiometer, incremental and absolute encoders, resolver	Absolute encoders use Gray code.
Velocity	Tachogenerator (V = Kω), encoder pulse rate	Output proportional to speed, not acceleration.
Acceleration	Seismic-mass accelerometers: piezoelectric, capacitive MEMS, strain-gauge	Used well below its natural frequency.
Force	Strain-gauge load cell, piezoelectric load cell	Full bridge for sensitivity and compensation.
Torque	Strain gauges at ±45° on a shaft, reaction torque cells	±45° are the principal directions in torsion.
Pressure	Diaphragm with strain gauges, piezoresistive, capacitive, Bourdon tube	Piezoelectric only for changing pressure.

An incremental encoder with N lines per revolution gives N pulses per turn on each of two channels A and B in quadrature (90° apart); the order of the edges gives the direction, and counting every edge of both channels (×4 decoding) resolves 360/(4N) degrees — 0.09° for 1000 lines. It loses position at power-off and needs a reference (index) pulse. An absolute encoder reads a unique code for each position, in Gray code so that only one bit changes between adjacent positions and a reading taken mid-transition is wrong by at most one step.

5. Signal conditioning

Conditioning turns a sensor’s raw output into a clean voltage in the range an ADC expects: bridge circuits turn resistance or capacitance changes into voltage; the instrumentation amplifier amplifies the small bridge difference while rejecting the large common-mode voltage, its gain set by one resistor, G = 1 + 2R/R_G for the three-op-amp form (51 for R = 25 kΩ and R_G = 1 kΩ); filters remove noise and, before sampling, frequencies above half the sampling rate; isolation and linearisation follow where needed. For piezoelectric sensors the first stage is a charge amplifier; for LVDTs, a phase-sensitive demodulator; for thermocouples, cold-junction compensation.

🧠 Where the ±45° comes from

A shaft in pure torsion is in pure shear at its surface, and Mohr’s circle for pure shear puts the principal stresses, +τ and −τ, at ±45° to the axis. Gauges along those directions see equal and opposite strains, so a four-gauge bridge doubles the signal of each pair and cancels bending and axial strain.

Key takeaways

GF = (ΔR/R)/ε ≈ 2 for metal foil; a quarter bridge gives V·GF·ε/4, a full bridge four times that; a dummy gauge compensates temperature.
C = εA/d: gap variation is non-linear, area variation linear; an LVDT gives amplitude for distance and phase for direction.
Piezoelectric sensors measure dynamic quantities only; V_H = IB/(net), so Hall sensors are semiconductors.
Torque gauges sit at ±45°; an incremental encoder with ×4 decoding resolves 360/(4N) degrees; absolute encoders use Gray code.
An instrumentation amplifier has G = 1 + 2R/R_G and high common-mode rejection; filter before sampling.

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A 120 Ω strain gauge with gauge factor 2 is bonded to a member strained to 500 microstrain. Its change in resistance is ____ Ω.
Numerical answer — type the value.
Show answer
Answer: 0.12
ΔR = GF × ε × R = 2 × 500 × 10⁻⁶ × 120 = 0.12 Ω. Forgetting that a microstrain is 10⁻⁶ gives 120 000 Ω, an absurd value that is itself the check.
A quarter-bridge circuit with one active strain gauge (gauge factor 2) is excited by 10 V. When the gauge is strained to 1000 microstrain, the bridge output, using the small-strain approximation, is ____ mV.
Numerical answer — type the value.
Show answer
Answer: 5
V_o ≈ V × GF × ε/4 = 10 × 2 × 10⁻³/4 = 5 × 10⁻³ V = 5 mV. Omitting the 1/4 gives 20 mV, the full-bridge figure with four active gauges.
To measure the torque transmitted by a circular shaft with strain gauges, the gauges are bonded:
1. at ±45° to the shaft axis
2. along the shaft axis
3. around the circumference, at 90° to the axis
4. on the end faces of the shaft
Show answer
Answer: A — at ±45° to the shaft axis
Torsion puts the surface in pure shear, whose principal directions are at ±45° to the axis, where the strains are equal and opposite. Along the axis and around the circumference the normal strain from pure shear is zero, so gauges there see bending or axial load instead.
In an LVDT, the direction of core displacement from the null position is indicated by:
1. the phase of the output relative to the excitation
2. the amplitude of the output
3. the frequency of the output
4. the DC offset of the primary current
Show answer
Answer: A — the phase of the output relative to the excitation
The two secondaries are connected in opposition, so the output amplitude grows with displacement either way; which secondary dominates flips the phase by 180°. The output frequency is the excitation frequency whatever the position.
A parallel-plate capacitive displacement sensor reads 10 pF. If the gap between the plates is halved, with everything else unchanged, it reads ____ pF.
Numerical answer — type the value.
Show answer
Answer: 20
C = εA/d, so halving d doubles C to 20 pF. The inverse dependence is what makes gap-variation sensing non-linear: equal steps in d do not give equal steps in C.
A piezoelectric force sensor is unsuitable for measuring a steady (static) force because:
1. the generated charge leaks away through the finite resistance of the sensor and circuit
2. it responds only to temperature
3. its natural frequency is too low
4. it needs an external magnetic field
Show answer
Answer: A — the generated charge leaks away through the finite resistance of the sensor and circuit
The crystal produces a charge only when the force changes; held constant, that charge decays through the insulation and amplifier resistance with a finite time constant, and the output returns to zero. Piezoelectric sensors are in fact very stiff with high natural frequencies — the reason they suit dynamic measurement.
A Hall element of thickness 0.5 mm carries 10 mA in a magnetic field of 0.5 T normal to it. Its carrier density is 10²¹ m⁻³ and the electronic charge is 1.6 × 10⁻¹⁹ C. The Hall voltage is ____ mV.
Numerical answer — type the value.
Show answer
Answer: 62.5
V_H = IB/(n e t) = (0.01 × 0.5)/(10²¹ × 1.6 × 10⁻¹⁹ × 5 × 10⁻⁴) = 0.005/0.08 = 0.0625 V = 62.5 mV. Using t = 0.5 m instead of 0.5 mm makes it a thousand times smaller.
An incremental encoder has 1000 lines per revolution and its two quadrature channels are decoded ×4. The angular resolution is ____ degrees.
Numerical answer — type the value.
Show answer
Answer: 0.09
×4 decoding counts both edges of both channels, 4 × 1000 = 4000 counts per revolution, so the resolution is 360/4000 = 0.09°. Without the ×4 it would be 0.36°.
Absolute optical encoders are commonly coded in Gray code rather than natural binary because:
1. only one bit changes between adjacent positions, so a reading at a transition is off by at most one step
2. Gray code needs fewer tracks for the same resolution
3. Gray code can be added directly by a binary adder
4. it gives direction without a second channel, unlike binary
Show answer
Answer: A — only one bit changes between adjacent positions, so a reading at a transition is off by at most one step
In binary, going from 0111 to 1000 changes four bits, and slightly misaligned read heads could momentarily report any of sixteen values. Gray code needs the same number of tracks, cannot be added directly, and an absolute reading already implies direction.
A Pt100 resistance thermometer (100 Ω at 0 °C) has a temperature coefficient of 0.00385 per °C. Assuming linearity, its resistance at 50 °C is ____ Ω.
Numerical answer — type the value.
Show answer
Answer: 119.25
R = R₀(1 + αT) = 100(1 + 0.00385 × 50) = 100 × 1.1925 = 119.25 Ω. Adding only αT = 0.1925 Ω forgets to multiply by R₀.
Which statements are true?
1. The resistance of an NTC thermistor falls as its temperature rises
2. A full bridge of four active gauges gives about four times the output of a quarter bridge at the same strain
3. The output voltage of a tachogenerator is proportional to angular acceleration
4. A dummy gauge in an adjacent arm compensates for temperature changes of the active gauge
Show answer
Answer: A — The resistance of an NTC thermistor falls as its temperature rises; B — A full bridge of four active gauges gives about four times the output of a quarter bridge at the same strain; D — A dummy gauge in an adjacent arm compensates for temperature changes of the active gauge
(A) Negative temperature coefficient. (B) Each active arm adds its contribution, with signs arranged to add. (C) False: V = Kω, proportional to speed. (D) Equal resistance changes in adjacent arms leave the bridge balanced, so only the mechanical strain unbalances it.
A three-op-amp instrumentation amplifier has a differential-stage gain of 1 and input-stage resistors R = 25 kΩ with a gain resistor R_G = 1 kΩ, so that G = 1 + 2R/R_G. Its gain is ____.
Numerical answer — type the value.
Show answer
Answer: 51
G = 1 + 2 × 25/1 = 51. Writing R/R_G gives 26 and dropping the 1 gives 50; the 2 is there because the gain resistor is shared by the two input op-amps.
For a parallel-plate capacitive displacement sensor, which statements are true?
1. Varying the plate gap gives a capacitance proportional to 1/d, a non-linear response
2. Varying the overlapping area gives a linear response
3. A differential arrangement improves linearity
4. The capacitance does not depend on the dielectric between the plates
Show answer
Answer: A — Varying the plate gap gives a capacitance proportional to 1/d, a non-linear response; B — Varying the overlapping area gives a linear response; C — A differential arrangement improves linearity
(A)–(B) Follow from C = εA/d. (C) One capacitance rises as the other falls, and the difference is linear to first order with the even-order errors cancelling. (D) False: C is proportional to ε, which is how level and moisture sensors work.