"Michael A. Powers" <mapowers@IEEE.ORG> wrote:
> I was thinking of the opposite of that.  Suppose I put dielectric foam
> between two conductive parallel plates.  When force is applied to the
> plates, the foam will deform and cause an increase in capacitance.  The only
> trick is finding foam with a coefficient of deformation and thickness that
> allows for capacitance changes that are significant to the measurement
> instrument. The permittivity of the foam isn't important.

I think you'll be fairly disappointed with the repeatability and hysteresis
associated with using foam (conductive or dielectric) as the spring element
of your force sensor. Mechanically speaking, foams are extremely lossy.

Metal springs (leaf, coil or thin diaphrams) give the best performance.
Use a rigid dielectric to separate two metal elements and measure the
capacitance change as they deform.

Sensing small changes in resistance or capacitance is generally not a
problem if you build the appropriate (DC or AC) bridge circuit. For
example, the following circuit measures small changes in a differential
capacitor used in a tilt sensor. It was originally written up as a quiz
question for Circuit Cellar Magazine.

   Question

   The following circuit is used to read a tilt sensor that is implemented
   as a differential capacitor -- when the sensor is tilted, C1 increases
   while C2 decreases (or vice-versa). Both capacitors vary between 4 and
   30 pF.
                         1 nF
                      +---||---+---------------+----------o A
                      |        |               |
                      |     D1 v               - D3
                      |        -               ^
                      |        |               |
        +-o 10VAC o---+        +---||--+--||---+
        |   1 MHz     |        |   C1  |  C2   |
        V             |     D2 v       V       - D4
       Gnd            |        -      Gnd      ^
                      |        |               |
                      +---||---+---------------+----------o B
                         1 nF

   The voltage differential between points A and B varies with the position
   of the sensor. How would you analyze the operation of this circuit?

   Answer

   The 10V, 1MHz source causes the diodes to conduct in pairs -- D1 and
   D4 conduct on the positive peaks, and D2 and D3 conduct on the negative
   peaks. Since D1 and D2 never conduct at the same time, the net current
   through them is directly porportional to the value of C1. Similarly, the
   current through D3 and D4 is proportional to the value of C2.

   If C1 and C2 have the same value, the D1/D2 current equals the D3/D4
   current, and the average voltage difference between points A and B is
   zero (although both are swinging up and down at 1 MHz). On the other hand,
   if the sensor is unbalanced, say C1 increases and C2 decreases, more
   current will flow in D1/D2 than in D3/D4, causing the average voltage at B
   to rise relative to A.

   Note that the difference between A and B can't exceed two forward diode
   drops (about 1.5 V) in either direction, and in fact, the voltage between
   them will be related to the net current flow by the diode equation. For
   values less than 1 V, the voltage varies nearly linearly with the
   capacitance difference.

   Additional information on this topic can be found in INK #61 ConnectTime.

-- Dave Tweed

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