Question

Draw a diagram of a MEMS capacitive sensor for pressure and explain how it works. Answer following questions:

1. How does the capacitance of a parallel-plate capacitor depend on area and separation?

2. How does the sensitivity of a capacitive sensor depend on its dimensions and temperature?

3. A MEMS parallel-plate capacitor has a capacitance of 1 pF. If the separation between the plates decreases by 10% what will be its capacitance?

Answer 1

To create a capacitive sensor, conducting layers are deposited on the diaphragm and the bottom of a cavity to create a capacitor. The capacitance is typically a few picofarads.

The deformation of the diaphragm changes the spacing between the conductors and hence changes the capacitance. The change can be measured by including the sensor in a tuned circuit, which changes its frequency with changing pressure.

The sensor can be used with electronic components on the chip to create an oscillator, which generates the output signal. Because of the difficulty of fabricating large inductances on silicon, this will usually be based on an RC circuit.

This approach is well suited for wireless readout because it generates a high-frequency signal that can be detected with a suitable external antenna.

Alternatively, the capacitance can be measured more directly by measuring the time taken to charge the capacitor from a current source. This can be compared with a reference capacitor to account for manufacturing tolerance and to reduce thermal effects.

1)

this equation tells that if the area of capacitive plates is increased, the capacitance also increases. And keeping the area constant if the distance between plates is decreased, then also capacitance increases.

where, is a dielectric constant of the medium.

2) As the capacitance changes, the sensitivity of the sensor changes. It depends on the dimensions of a capacitor ( discussed above: area and separation between plates..) and the temperature.

As the temperature increases, the orientation of dipoles is facilitated and this increases the dielectric constant. So, the capacitance also changes( according to the equation given above.)

3)