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Linearising capacitive displacement transducer

Short answer

Nonlinearity from C = εA/d is linearised by an op-amp with the transducer in feedback, giving Vo = −Vi·C1·d/(εA), linear in gap. Alternatively, a differential arrangement gives (C1−C2)/(C1+C2) = x/d, linear with doubled sensitivity and reduced errors.

Explain how non-linear characteristics of a capacitive transducer, used for the measurement of displacement using the principle of change in capacitance due to change in distance between plates, can be made linear.

For parallel plates of area A at distance d with permittivity ε: C = εA/d. So C is inversely proportional to d, which makes it non-linear for a plate displacement x: C = εA/(d ± x).

The linearisation methods are as follows.

1. Op-amp circuit with the transducer in the feedback path

A fixed capacitor C1 is placed at the input, and the displacement transducer Cx is placed in the feedback path.

                 Cx = εA/d (transducer)
           +--------||--------+
           |                  |
 Vi o--||--+----(-)\          |
       C1           >---------+----o Vo
       gnd------(+)/

For an ideal op-amp: Vo = −Vi · (Z_f/Z_in) = −Vi · (1/jωCx)/(1/jωC1) = −Vi · C1/Cx.

Substituting Cx = εA/d:

Vo = −Vi · C1 · d /(εA)

Since C1, ε, A and Vi are constant, the output is directly proportional to the gap d, i.e. to the displacement. The output is therefore linear.

2. Differential (push-pull) arrangement

Two fixed plates with a moving centre plate form C1 = εA/(d − x) and C2 = εA/(d + x). Using them in adjacent bridge arms, or taking the ratio:

(C1 − C2)/(C1 + C2) = x/d

This is linear in x. It also doubles the sensitivity, cancels even-order non-linearity, and reduces temperature and stray effects.

3. Other ways

  • Use variation of the overlapping area (sliding plates) instead, where C ∝ x is linear.
  • Restrict the displacement to a small fraction of d (x ≪ d); the error is then small.

Method 1 or 2 is the usual textbook answer.

✓ Answer checked by Claude Opus 5.5 (high effort)

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