Skip to main content

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

  1. By energy source: active (self-generating) and passive.
  2. By function: primary (sensing) and secondary.
  3. By output signal: analog and digital.
  4. By principle: resistive, capacitive, inductive, piezoelectric, thermoelectric, photoelectric, Hall effect.
  5. By quantity measured: displacement, force, pressure, temperature, flow and so on.
  6. By operating mode: transducers (input-to-output converters) and inverse transducers.

Active and passive

PointActive transducerPassive transducer
PowerGenerates output emf or charge from the measurandNeeds external excitation
Also calledSelf-generatingExternally powered
OutputVoltage, charge, currentChange in R, L or C
ExamplesThermocouple, piezoelectric crystal, photovoltaic cellStrain 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 kΩ\Omega linear potentiometer with a 10 V supply drives a 50 kΩ\Omega 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 EsE_s is applied across the full length LL and the output is taken between the wiper and one end. With no load, the output is linear:

eo=EsxLe_o = E_s\frac{x}{L}
   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 RmR_m across the wiper output, and xx = fractional position:

eoEs=x1+x(1−x)RpRm\frac{e_o}{E_s} = \frac{x}{1 + x(1-x)\dfrac{R_p}{R_m}}

Rp/Rm=10/50=0.2R_p/R_m = 10/50 = 0.2, Es=10E_s = 10 V.

xIdeal (V)Actual eoe_o (V)Error (V)Error % of EsE_s
0.505.00051+0.25(0.2)=4.762\dfrac{5}{1 + 0.25(0.2)} = 4.7620.2382.38
0.757.5007.51+0.1875(0.2)=7.229\dfrac{7.5}{1 + 0.1875(0.2)} = 7.2290.2712.71

Answer: eo=4.76e_o = 4.76 V at 50% (error 0.238 V) and 7.237.23 V at 75% (error 0.271 V).

The loading error is largest at about 2/3 travel. To reduce it:

  • Make Rm≫RpR_m \gg R_p (use a high-input-impedance voltmeter or buffer amplifier).
  • Use a smaller RpR_p (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 ∘^\circC reference: 30 ∘^\circC, 1.203 mV; 100 ∘^\circC, 4.096 mV; 200 ∘^\circC, 8.138 mV; 300 ∘^\circC, 12.209 mV. Find (a) the emf when the junctions are at 300 ∘^\circC and 100 ∘^\circC, and (b) the hot-junction temperature if an instrument whose cold junction is at 30 ∘^\circC 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

  1. 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.
  2. 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.
  3. Law of intermediate temperatures: if the emf between junction temperatures T1T_1 and T2T_2 is E1,2E_{1,2} and between T2T_2 and T3T_3 is E2,3E_{2,3}, then E1,3=E1,2+E2,3E_{1,3} = E_{1,2} + E_{2,3}. This lets tables with a 0 ∘^\circC reference be used for any cold junction.

(a) Junctions at 300 ∘^\circC and 100 ∘^\circC

E100,300=E0,300−E0,100=12.209−4.096=8.113 mVE_{100,300} = E_{0,300} - E_{0,100} = 12.209 - 4.096 = 8.113\ \text{mV}

(b) Hot junction temperature

The instrument reads 7.5 mV with cold junction at 30 ∘^\circC. Referred to 0 ∘^\circC:

E0,T=E30,T+E0,30=7.5+1.203=8.703 mVE_{0,T} = E_{30,T} + E_{0,30} = 7.5 + 1.203 = 8.703\ \text{mV}

This lies between 200 ∘^\circC (8.138 mV) and 300 ∘^\circC (12.209 mV).

T=200+8.703−8.13812.209−8.138×100=200+0.5654.071×100=213.9 ∘CT = 200 + \frac{8.703 - 8.138}{12.209 - 8.138}\times 100 = 200 + \frac{0.565}{4.071}\times 100 = 213.9\ ^\circ\text{C}

Answer: (a) 8.113 mV; (b) T≈213.9 ∘T \approx 213.9\ ^\circC.

  • 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 kΩ\Omega at 25 ∘^\circC and a material constant β=3950\beta = 3950 K. Find its resistance at 50 ∘^\circC and 100 ∘^\circC and its temperature coefficient of resistance at 50 ∘^\circC.

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:

R=R0exp⁡[β(1T−1T0)]R = R_0\exp\left[\beta\left(\frac{1}{T} - \frac{1}{T_0}\right)\right]

where TT and T0T_0 are in kelvin. The temperature coefficient is α=1RdRdT=−βT2\alpha = \dfrac{1}{R}\dfrac{dR}{dT} = -\dfrac{\beta}{T^2}.

PointNTCPTC
CoefficientNegativePositive (sharp rise above a switching temperature)
MaterialMetal oxidesDoped barium titanate
UseTemperature sensing, compensationOvercurrent 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 ∘^\circC), interchangeability poor, self-heating error.

(b) Numerical

Given R0=10R_0 = 10 kΩ\Omega at T0=298.15T_0 = 298.15 K, β=3950\beta = 3950 K.

At 50 ∘^\circC (T=323.15T = 323.15 K):

R50=10,000exp⁡[3950(1323.15−1298.15)]=10,000 e−1.0249=3588 ΩR_{50} = 10{,}000\exp\left[3950\left(\frac{1}{323.15} - \frac{1}{298.15}\right)\right] = 10{,}000\,e^{-1.0249} = 3588\ \Omega

At 100 ∘^\circC (T=373.15T = 373.15 K):

R100=10,000exp⁡[3950(1373.15−1298.15)]=10,000 e−2.6628=697.5 ΩR_{100} = 10{,}000\exp\left[3950\left(\frac{1}{373.15} - \frac{1}{298.15}\right)\right] = 10{,}000\,e^{-2.6628} = 697.5\ \Omega

Temperature coefficient at 50 ∘^\circC:

α=−3950323.152=−0.0378 K−1=−3.78%/K\alpha = -\frac{3950}{323.15^2} = -0.0378\ \text{K}^{-1} = -3.78\%/\text{K}

Answer: R50=3.59R_{50} = 3.59 kΩ\Omega, R100=697.5 ΩR_{100} = 697.5\ \Omega, α50=−3.78%/∘\alpha_{50} = -3.78\%/^\circC.

  • 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 S1S_1, S2S_2 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 e1e_1 and e2e_2 in the secondaries, and the output is eo=e1−e2e_o = e_1 - e_2.

  • Core at centre (null position): equal flux links both secondaries, e1=e2e_1 = e_2, so eo=0e_o = 0.
  • Core moved toward S1S_1: S1S_1 gets more flux, so e1>e2e_1 > e_2 and eoe_o is in phase with the primary voltage.
  • Core moved toward S2S_2: e2>e1e_2 > e_1 and eoe_o is 180∘180^\circ out of phase with the primary.

The magnitude of eoe_o 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 NN turns is

L=N2R,R=lgμ0A (air gap dominates)L = \frac{N^2}{\mathcal{R}}, \qquad \mathcal{R} = \frac{l_g}{\mu_0 A}\ \text{(air gap dominates)}

so L=N2μ0AlgL = \dfrac{N^2\mu_0 A}{l_g}. Changing the air gap lgl_g, the area AA, or the permeability μ\mu changes LL.

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 lgl_g. When the armature moves with the measurand, lgl_g changes, so LL changes.

     +---[ coil ]---+
     |              |
   __|__          __|__   <- iron core
   |___|          |___|
      gap l_g (variable)
   ===================    <- movable armature
         ^ x

The relation L∝1/lgL \propto 1/l_g 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 (L1L_1 increases) and the other increases (L2L_2 decreases). The coils are placed in adjacent arms of an ac bridge, so the output is proportional to L1−L2L_1 - L_2.

   [ 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.

Chapter titles and hours from the IOE syllabus ↗