Training Reactive (C & L) Sensors Capacitive Sensors — Gap, Area, Dielectric
1 / 2

Capacitive Sensors — Gap, Area, Dielectric

30 min Reactive (C & L) Sensors

Capacitive Sensors — Proximity, Level, Humidity

A capacitive sensor converts a displacement, fluid level, or material property into a change in capacitance, and then into a measurable voltage or frequency. Because capacitance depends on geometry ($A$ and $d$) and permittivity ($\varepsilon_r$), the same basic structure can serve as a proximity switch, a liquid-level gauge, a humidity sensor, or a touch button.

Pallàs-Areny and Webster emphasize that capacitive sensors typically operate with AC excitation — at DC, a capacitor is an open circuit. The signal chain therefore involves an oscillator, a charge amplifier, and often a synchronous demodulator to recover the envelope.

In this lesson you will work out the capacitance of parallel-plate and coaxial geometries, compute the change produced by a moving target, and estimate the sensitivity in pF per millimetre or per percent relative humidity.

Parallel-Plate Capacitance

$$C = \varepsilon_0 \varepsilon_r \frac{A}{d}$$

$\varepsilon_0 = 8.854\times10^{-12}\,\text{F/m}$, $A$ = plate area, $d$ = gap, $\varepsilon_r$ = relative permittivity.

Variable-Gap Sensitivity

For small gap changes $\Delta d \ll d$:

$$\frac{\Delta C}{C} = -\frac{\Delta d}{d}$$

The response is highly nonlinear over large ranges — a capacitive sensor is most useful as a small-signal differential device, often driven in a balanced (two-plate) configuration to linearize output.

Example 1 — Parallel plates

Two 20 × 20 mm plates sit 1.0 mm apart in air. Find $C$.

$A = 0.020\cdot 0.020 = 4.0\times10^{-4}\,\text{m}^2$. $C = 8.854\times10^{-12}\cdot 1\cdot 4\times10^{-4}/10^{-3} = 3.54\times10^{-12}\,\text{F} = 3.54\,\text{pF}$.

Example 2 — Proximity response

The gap of the above sensor closes to $0.9\,\text{mm}$. Find $\Delta C$.

$C_{\text{new}} = 8.854\times10^{-12}\cdot 4\times10^{-4}/0.9\times10^{-3} = 3.935\,\text{pF}$.

$\Delta C = 0.395\,\text{pF}$, $\Delta C/C = 11.2\%$ — easily resolvable with a charge amplifier.

Example 3 — Humidity sensor

A polymer capacitive humidity sensor has $C = 150\,\text{pF}$ at 0 % RH. The dielectric constant changes with humidity as $\varepsilon_r(H) = \varepsilon_0(1 + 0.004 H)$ where $H$ is in %RH. Find $C$ at 50 % RH.

Ratio: $\varepsilon_r(50)/\varepsilon_r(0) = 1 + 0.004\cdot 50 = 1.20$.

$C(50) = 150\cdot 1.20 = 180\,\text{pF}$, a 30 pF change — well within reach of a modern capacitance-to-digital converter.

Interactive Demo: Parallel-Plate Capacitive Gap Sensor
C =3.54pF
|dC/dd| =3.54pF/mm
Oscillator f (with 100 µH) =26.8MHz

Practice Problems

1. Two 10×10 mm plates, 0.5 mm apart, air. Find $C$.
2. Same sensor, gap reduces to 0.25 mm. New $C$?
3. For the gap sensor, derive $dC/dd$.
4. Humidity sensor: $C_0 = 100\,\text{pF}$, linear 0.003/%RH. Find $C$ at 80 %RH.
5. Why do capacitive sensors use AC excitation?
6. What geometric parameter change produces the most linear response: $A$ or $d$?
Show Answer Key

1. $A = 10^{-4}\,\text{m}^2$, $C = 8.854\times10^{-12}\cdot 10^{-4}/5\times10^{-4} = 1.77\,\text{pF}$.

2. $C = 1.77\cdot 2 = 3.54\,\text{pF}$ (C scales as $1/d$).

3. $dC/dd = -\varepsilon_0\varepsilon_r A/d^2$.

4. $C = 100(1 + 0.003\cdot 80) = 100\cdot 1.24 = 124\,\text{pF}$.

5. A capacitor is open at DC; AC excitation allows displacement current to flow and be measured.

6. Variable-area sensors (sliding plates) give linear $C(x)$; variable-gap is inherently nonlinear.

Inductive Sensors — LVDT and Eddy-Current