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handbook of modern sensors

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7.3 Capacitive Sensors 259<br />

(A)<br />

(B)<br />

Fig. 7.5. Operating principle <strong>of</strong> a flat plate capacitive sensorA-balanced position; B-disbalanced<br />

position<br />

<strong>of</strong> a flat capacitor is inversely proportional to the distance between the plates. The<br />

operating principle <strong>of</strong> a capacitive gauge, proximity, and position <strong>sensors</strong> is based<br />

on either changing the geometry (i.e., a distance between the capacitor plates) or<br />

capacitance variations in the presence <strong>of</strong> conductive or dielectric materials. When<br />

the capacitance changes, it can be converted into a variable electrical signal. As with<br />

many <strong>sensors</strong>, a capacitive sensor can be either monopolar (using just one capacitor)<br />

or differential (using two capacitors), or a capacitive bridge can be employed (using<br />

four capacitors). When two or four capacitors are used, one or two capacitors may be<br />

either fixed or variable with the opposite phase.<br />

As an introductory example consider three equally spaced plates, each <strong>of</strong> area A<br />

(Fig. 7.5A). The plates form two capacitors C 1 and C 2 . The upper and lower plates are<br />

fed with the out-<strong>of</strong>-phase sine-wave signals; that is, the signal phases are shifted by<br />

180 ◦ . Both capacitors nearly equal one another and thus the central plate has almost<br />

no voltage because the currents through C 1 and C 2 cancel each other. Now, let us<br />

assume that the central plate moves downward by a distance x (Fig. 7.5B). This results<br />

in changes in the respective capacitance values:<br />

C 1 =<br />

εA<br />

x 0 + x<br />

and C 2 = εA<br />

x 0 − x , (7.3)<br />

and the central plate signal increases in proportion to the displacement and the phase <strong>of</strong><br />

that signal is an indication <strong>of</strong> the central plate direction—up or down. The amplitude<br />

<strong>of</strong> the output signals is<br />

(<br />

V out = V 0 −<br />

x<br />

x 0 + x + C )<br />

. (7.4)<br />

C

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