Capacitive transducers

Variable-gap, variable-area and variable-dielectric capacitive transducers, differential bridges, coaxial level probes and the stray-capacitance problem, with worked numericals.

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Why it matters

Capacitive sensing is behind industrial pressure transmitters, liquid-level probes, humidity sensors, MEMS accelerometers and touch screens. It can resolve displacements of nanometres with almost no force on the measurand, but the capacitances are only picofarads, so understanding geometry, linearity and stray capacitance is essential.

Key ideas

A capacitor's value depends on geometry and the material between the electrodes. For parallel plates, C = ε0·εr·A/d. A capacitive transducer makes the measurand change one of the three variables:

  • Variable gap (d): very sensitive to small movements, but C ∝ 1/d is non-linear. The fractional sensitivity is dC/C = −dd/d, so the absolute sensitivity rises sharply as the gap closes. Used for diaphragm pressure sensors and proximity probes.
  • Variable area (A): one plate slides or rotates over the other; C is proportional to overlap, so the output is linear in displacement or angle (a half-disc rotary capacitor gives C ∝ θ). Lower sensitivity, larger travel.
  • Variable dielectric (εr): a material of different permittivity enters the gap — the classic liquid-level probe, where a non-conducting liquid rises between coaxial electrodes, and thin-film humidity sensors whose polymer absorbs water.

Differential (push–pull) arrangement. A centre plate moves between two fixed plates: one gap becomes d − x and the other d + x. In an AC bridge with two equal fixed arms, the output becomes Vo = Vs·x/(2d) — exactly linear in x for a gap-type sensor, with doubled sensitivity and cancellation of common effects such as temperature and dielectric changes. Most capacitive pressure transmitters use this "delta cell" principle.

Measurement circuits. Because C is small, its impedance 1/(ωC) is very high (about 90 kΩ for 177 pF at 10 kHz), so:

  • excitation is AC, usually kHz to MHz;
  • the readout must have very high input impedance, or use a charge amplifier / capacitance-to-digital converter;
  • cable and stray capacitances add directly to C and can swamp the change — keep leads short, use driven guards (guard rings also suppress fringing at the plate edges);
  • putting the gap-type sensor in the feedback path of an inverting amplifier gives an output proportional to d, i.e. linear in displacement.

Advantages: very small force needed, high resolution, good frequency response, low power, no contact, works for static and dynamic inputs. Limitations: sensitive to stray capacitance, humidity and dirt in the gap, temperature effects on dimensions and εr, need for shielding and fairly complex electronics, non-linearity of single gap-type sensors.

Cylindrical (coaxial) electrodes have C = 2π·ε0·εr·L / ln(b/a), which is the basis of level measurement.

Formulas

C = ε0·εr·A / d (parallel plate, fringing neglected) C = capacitance (F); ε0 = 8.854 × 10⁻¹² F/m; εr = relative permittivity; A = overlap area (m²); d = gap (m).

dC/C = −dd/d (variable gap, small change)

C = 2π·ε0·εr·L / ln(b/a) (coaxial) L = length (m); a = outer radius of inner electrode (m); b = inner radius of outer electrode (m).

C(h) = 2π·ε0·[εr·h + (H − h)] / ln(b/a) (coaxial level probe, non-conducting liquid) H = probe length (m); h = liquid height (m).

Vo = Vs·x / (2d) (differential gap sensor in a bridge with two equal fixed arms) Vs = bridge supply (V); x = displacement of centre plate (m).

X_C = 1 / (2π·f·C) X_C = capacitive reactance (Ω); f = excitation frequency (Hz).

Worked examples

Example 1 — variable-gap sensor. Two parallel plates of area 0.02 m² are 1 mm apart in air (εr = 1). (a) Find C. (b) The gap closes by 0.1 mm. Find the new C and the change.

  1. C = 8.854 × 10⁻¹² × 0.02 / 1 × 10⁻³ = 1.771 × 10⁻¹⁰ F = 177.1 pF.
  2. New gap 0.9 mm: C = 177.1 × 1.0/0.9 = 196.8 pF.
  3. ΔC = 196.8 − 177.1 = 19.7 pF (+11.1 % for a 10 % gap change — non-linear).

Answer: (a) 177.1 pF; (b) 196.8 pF, ΔC ≈ 19.7 pF.

Example 2 — capacitive level probe (GATE level). A coaxial probe 2 m long has inner electrode radius 5 mm and outer tube inner radius 25 mm. It is immersed in oil with εr = 3. Find C when the tank is empty, when the oil is 1.2 m deep, and the sensitivity in pF/m.

  1. ln(b/a) = ln(25/5) = ln 5 = 1.6094.
  2. Empty: C0 = 2π × 8.854 × 10⁻¹² × 2 / 1.6094 = 69.1 pF.
  3. At h = 1.2 m: C = 2π·ε0·[3 × 1.2 + (2 − 1.2)] / 1.6094 = 2π × 8.854 × 10⁻¹² × 4.4 / 1.6094 = 152.1 pF.
  4. Sensitivity: dC/dh = 2π·ε0·(εr − 1)/ln(b/a) = 2π × 8.854 × 10⁻¹² × 2 / 1.6094 = 69.1 pF/m.

Answer: 69.1 pF empty, 152.1 pF at 1.2 m, sensitivity 69.1 pF/m (linear in level).

Example 3 — differential sensor. A push–pull sensor with nominal gaps d = 1 mm is in a bridge supplied at 10 V. The centre plate moves 0.05 mm. Vo = Vs·x/(2d) = 10 × 0.05/(2 × 1) = 0.25 V.

Common mistakes

  • Forgetting ε0: using εr alone as the permittivity, or using ε0 when a dielectric is present.
  • Unit slips: 10⁻¹⁰ F is 100 pF, not 1 pF; mm² to m² is 10⁻⁶.
  • Saying ΔC is proportional to displacement for a gap-type sensor — only the area type, or the differential arrangement, is linear.
  • Measuring with a low-impedance meter or a long unguarded cable; stray capacitance can exceed the sensor's own change.
  • Forgetting that a conductive liquid shorts a bare probe — use an insulated probe and a different model.

For GATE IN

NAT problems ask for parallel-plate and coaxial capacitance, change of C with gap or level, differential-bridge output and sensitivity, and the effect of a dielectric slab partly filling a gap (treat as series capacitors). MCQs test which configuration is linear, why guards are used, and why capacitive sensors need high-impedance readout. Practise unit conversion between m², mm² and pF.

Quick check

  1. A gap-type sensor reads 50 pF at 2 mm. What does it read at 1 mm?
  2. Which of the three variable types gives a naturally linear output?
  3. What is the bridge output of a differential sensor with d = 0.5 mm, x = 0.01 mm, Vs = 5 V?
  4. Why does a capacitive sensor need a high input-impedance readout? Answers: 1. 100 pF; 2. variable area; 3. 5 × 0.01/1 = 0.05 V; 4. its reactance 1/(ωC) is very high (tens of kΩ to MΩ), so a low-impedance readout loads it heavily.

Interview questions

All Sensors and Transducers interview questions

Try answering each one aloud before you open it.

  1. 1.What is a capacitive transducer?Concept

    A capacitive transducer is a device that converts a physical quantity, such as displacement, pressure, or humidity, into an electrical signal by changing the capacitance of a capacitor. It typically consists of two parallel plates separated by a dielectric material, where the capacitance changes with the variation in the physical quantity.

  2. 2.Explain the working principle of a capacitive transducer.Concept

    The working principle of a capacitive transducer is based on the change in capacitance due to the variation in the distance between the plates, the area of overlap of the plates, or the dielectric constant of the material between the plates. The capacitance C is given by the formula C = ε·A/d, where ε is the permittivity of the dielectric material, A is the area of overlap of the plates, and d is the distance between the plates. Any change in these parameters due to a physical quantity will result in a change in capacitance, which can be measured as an electrical signal.

  3. 3.What are the advantages of using capacitive transducers?Concept

    Capacitive transducers need very little force to operate, so they barely load the measurand, and they can resolve extremely small displacements. They consume little power, have good frequency response, work for both static and dynamic inputs and are non-contact. Variable-area types and differential (push–pull) gap types give linear output; single variable-gap types are sensitive but non-linear.

  4. 4.Why are capacitive transducers used in pressure measurement?Application

    A thin diaphragm forms one electrode; pressure deflects it and changes the gap to a fixed electrode, so capacitance changes. Because tiny deflections give measurable ΔC, the diaphragm can be stiff, which gives good linearity, low hysteresis and fast response. Industrial transmitters use a differential cell with a sensing diaphragm between two fixed plates, so one capacitance rises while the other falls; the bridge output is then linear in deflection and common effects such as temperature cancel.

  5. 5.What happens if the dielectric material in a capacitive transducer is changed?Application

    If the dielectric material in a capacitive transducer is changed, the permittivity (ε) of the material will change, which directly affects the capacitance of the transducer. Since capacitance is given by C = ε·A/d, a change in permittivity will result in a change in capacitance for the same physical displacement or pressure. This can alter the sensitivity and calibration of the transducer, requiring recalibration for accurate measurements.

  6. 6.How does temperature affect the performance of capacitive transducers?Application

    Temperature can affect the performance of capacitive transducers by causing changes in the dielectric constant of the material between the plates and by causing thermal expansion or contraction of the plates themselves. These changes can lead to variations in capacitance that are not due to the measured physical quantity, potentially introducing errors. To mitigate this, temperature compensation techniques are often employed in capacitive transducer designs.

  7. 7.Explain how a capacitive transducer can be used to measure displacement.Application

    The moving object changes either the gap or the overlap area of a capacitor. With a variable gap, C = ε0·εr·A/d is inversely proportional to the gap, which is very sensitive but non-linear; with a variable area, C is proportional to overlap and the output is linear. A differential arrangement with the moving plate between two fixed plates, read in an AC bridge, gives an output Vs·x/(2d) that is linear in displacement. Short guarded leads and a high-impedance readout are needed because the capacitances are only picofarads.

  8. 8.Calculate the capacitance of a parallel plate capacitor with a plate area of 0.02 m², a plate separation of 0.001 m, and a dielectric of relative permittivity 5.Numerical

    C = ε0·εr·A/d = 8.854 × 10⁻¹² × 5 × 0.02 / 0.001 = 8.854 × 10⁻¹² × 100 = 8.854 × 10⁻¹⁰ F, i.e. about 885 pF. A common slip is to drop a factor of ten when converting 10⁻¹⁰ F to picofarads (1 pF = 10⁻¹² F).

  9. 9.A capacitive transducer has a capacitance of 100 pF when the plate separation is 0.5 mm. What will be the capacitance if the plate separation is reduced to 0.25 mm?Numerical

    The capacitance C is inversely proportional to the distance d between the plates, given by C = ε·A/d. If the initial capacitance is 100 pF at 0.5 mm, reducing the distance to 0.25 mm will double the capacitance, as the distance is halved. Therefore, the new capacitance will be 200 pF.

  10. 10.What are some limitations of capacitive transducers?Concept

    Capacitive transducers have limitations such as sensitivity to environmental changes like temperature and humidity, which can affect the dielectric constant and lead to measurement errors. They also require careful shielding to prevent interference from external electric fields. Additionally, they may not be suitable for measuring large displacements due to non-linearity issues and can be affected by mechanical vibrations.

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