Chapter 5 · 8 hours
Sensors
Practice questions
Practice questions and answers
6 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.
- Practice · 5 marks
Classify transducers. Differentiate between active and passive transducers, and between primary and secondary transducers, with examples.
Answer
A transducer converts one form of energy or physical quantity into another, usually into an electrical signal. A sensor is the sensing element of a transducer.
Classification
- By energy source: active (self-generating) and passive.
- By function: primary (sensing) and secondary.
- By output signal: analog and digital.
- By principle: resistive, capacitive, inductive, piezoelectric, thermoelectric, photoelectric, Hall effect.
- By quantity measured: displacement, force, pressure, temperature, flow and so on.
- By operating mode: transducers (input-to-output converters) and inverse transducers.
Active and passive
| Point | Active transducer | Passive transducer |
|---|---|---|
| Power | Generates output emf or charge from the measurand | Needs external excitation |
| Also called | Self-generating | Externally powered |
| Output | Voltage, charge, current | Change in R, L or C |
| Examples | Thermocouple, piezoelectric crystal, photovoltaic cell | Strain gauge, thermistor, LVDT, potentiometer |
Primary and secondary
- Primary transducer: first element in contact with the measured quantity; it converts it into a mechanical or other signal. Example: a bourdon tube turns pressure into displacement.
- Secondary transducer: converts the output of the primary into an electrical signal. Example: a potentiometer or LVDT attached to the bourdon tube gives voltage.
A load cell is a combination: an elastic member (primary) with strain gauges (secondary).
- Practice · 4+4 marks
(a) Explain the working of a sliding-contact (potentiometer) displacement transducer. State its advantages and limitations.
(b) A 10 k linear potentiometer with a 10 V supply drives a 50 k load. Find the output voltage and loading error at 50% and 75% of the travel, and state how the loading error can be reduced.
Answer
(a) Potentiometer transducer
A resistance potentiometer (pot) has a resistive element (wire-wound, carbon or conductive plastic) with a sliding contact (wiper) mechanically linked to the moving member. A supply is applied across the full length and the output is taken between the wiper and one end. With no load, the output is linear:
Es o----+----------+
| Rp |
| ____ |
+-|_||_|---+
wiper ->|--- o e_o
Displacement x ---|
Advantages: simple, cheap, high output, no amplifier needed, works with dc, linear or rotary.
Limitations:
- Wear and friction of the sliding contact; limited life.
- Finite resolution for wire-wound types (one turn step).
- Loading error with finite load resistance.
- Limited frequency response because of wiper inertia; noise at contact.
(b) Loading error
With a load across the wiper output, and = fractional position:
, V.
| x | Ideal (V) | Actual (V) | Error (V) | Error % of |
|---|---|---|---|---|
| 0.50 | 5.000 | 0.238 | 2.38 | |
| 0.75 | 7.500 | 0.271 | 2.71 |
Answer: V at 50% (error 0.238 V) and V at 75% (error 0.271 V).
The loading error is largest at about 2/3 travel. To reduce it:
- Make (use a high-input-impedance voltmeter or buffer amplifier).
- Use a smaller (limited by power dissipation).
- Use a tapped (non-linear) pot to compensate.
- Practice · 6 marks
State the laws of thermocouples. A chromel-alumel thermocouple gives the following emf against a 0 C reference: 30 C, 1.203 mV; 100 C, 4.096 mV; 200 C, 8.138 mV; 300 C, 12.209 mV. Find (a) the emf when the junctions are at 300 C and 100 C, and (b) the hot-junction temperature if an instrument whose cold junction is at 30 C reads 7.5 mV (use linear interpolation).
Answer
A thermocouple produces an emf when two dissimilar metals joined at two junctions are at different temperatures (Seebeck effect).
Laws of thermocouples
- Law of homogeneous circuit: a thermoelectric current cannot be sustained in a circuit of a single homogeneous metal by heat alone, however the section varies. A temperature gradient in a uniform wire adds no emf.
- Law of intermediate metals: a third metal inserted in the circuit (e.g. the voltmeter leads) adds no net emf if its two junctions are at the same temperature.
- Law of intermediate temperatures: if the emf between junction temperatures and is and between and is , then . This lets tables with a 0 C reference be used for any cold junction.
(a) Junctions at 300 C and 100 C
(b) Hot junction temperature
The instrument reads 7.5 mV with cold junction at 30 C. Referred to 0 C:
This lies between 200 C (8.138 mV) and 300 C (12.209 mV).
Answer: (a) 8.113 mV; (b) C.
- Practice · 4+4 marks
(a) Explain the working of thermistors. Differentiate between NTC and PTC types and state the advantages and limitations of thermistors.
(b) An NTC thermistor has a resistance of 10 k at 25 C and a material constant K. Find its resistance at 50 C and 100 C and its temperature coefficient of resistance at 50 C.
Answer
(a) Thermistors
A thermistor (thermal resistor) is a semiconductor (sintered metal oxides of Mn, Ni, Co, Cu) whose resistance changes strongly with temperature. Its sensing element is a bead, disc or rod.
For an NTC thermistor, the resistance falls as temperature rises:
where and are in kelvin. The temperature coefficient is .
| Point | NTC | PTC |
|---|---|---|
| Coefficient | Negative | Positive (sharp rise above a switching temperature) |
| Material | Metal oxides | Doped barium titanate |
| Use | Temperature sensing, compensation | Overcurrent protection, self-regulating heaters |
Advantages: high sensitivity (about 4%/K), small size and fast response, high resistance so lead resistance is negligible.
Limitations: strongly nonlinear, small range (up to about 300 C), interchangeability poor, self-heating error.
(b) Numerical
Given k at K, K.
At 50 C ( K):
At 100 C ( K):
Temperature coefficient at 50 C:
Answer: k, , C.
- Practice · 8 marks
Explain the construction and working principle of a linear variable differential transformer (LVDT). Draw its output against core displacement, and state its advantages, limitations and applications.
Answer
The LVDT is a passive inductive transducer that converts linear displacement into an ac voltage whose amplitude is proportional to the displacement and whose phase gives the direction.
Construction
- One primary winding at the centre and two identical secondary windings , wound symmetrically on either side, on a hollow non-magnetic former.
- A movable ferromagnetic core (soft iron or nickel-iron) slides inside the former and is attached to the object whose displacement is measured.
- The primary is excited by a 1-10 V, 50 Hz-20 kHz ac supply. The two secondaries are connected in series opposition.
+---------+--------+---------+
~E --| S1 | Primary| S2 |-- e_o = e1 - e2
| ((( | ((( | ((( |
+----[=== core ===]----------+
<-- x -->
Working
The primary field induces emf and in the secondaries, and the output is .
- Core at centre (null position): equal flux links both secondaries, , so .
- Core moved toward : gets more flux, so and is in phase with the primary voltage.
- Core moved toward : and is out of phase with the primary.
The magnitude of is proportional to the displacement over the linear range. A phase-sensitive demodulator is used to obtain a dc signal with sign.
e_o | /
| / <- linear range
----+--/---------- x
|/
(phase reversal at null)
Advantages
- Frictionless, infinite resolution, no contact wear, long life.
- High sensitivity (up to about 2 mV/V/mm); good linearity (about 0.25%).
- Rugged, tolerant to shock, vibration, dirt and environment.
- Small null error, and the core is free to move without touching the coils.
Limitations
- Needs ac excitation and a demodulator.
- Stray magnetic fields and temperature affect it; sensitivity drops at high frequencies (the carrier must be at least about 10 times the signal frequency).
- Limited range, and a residual voltage at null.
Applications
Displacement and position, thickness, force and pressure (with an elastic element such as a diaphragm or bourdon tube), acceleration, liquid level and load cells.
- Practice · 5 marks
Explain the principle of variable-inductance (variable-reluctance) transducers. Describe the single-coil, change-of-air-gap type and the differential (push-pull) type, and state why the differential type is preferred.
Answer
A variable-reluctance (inductance) transducer senses the measurand through a change in the inductance of a coil due to a change in the reluctance of its magnetic path.
Principle
The inductance of a coil of turns is
so . Changing the air gap , the area , or the permeability changes .
Single-coil, change of air-gap type
A U- or E-shaped iron core carries the coil. A ferromagnetic armature is separated from it by a gap . When the armature moves with the measurand, changes, so changes.
+---[ coil ]---+
| |
__|__ __|__ <- iron core
|___| |___|
gap l_g (variable)
=================== <- movable armature
^ x
The relation is nonlinear; it is approximately linear only for small gap changes. The iron armature also exerts a magnetic pull on the moving member.
Differential (push-pull) type
Two identical coils are placed on opposite sides of a common armature. When the armature moves up, the gap of one coil decreases ( increases) and the other increases ( decreases). The coils are placed in adjacent arms of an ac bridge, so the output is proportional to .
[ coil 1 ] gap 1
============ <- armature (moves)
[ coil 2 ] gap 2
Why differential is preferred
- Even-order nonlinearity cancels, giving better linearity and about twice the sensitivity.
- The magnetic pull forces on the armature cancel.
- Effects of temperature, supply variation and stray fields on both coils cancel in the bridge.
Typical uses: small displacement, pressure (diaphragm) and proximity sensing.
Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.
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