Transducers for Energy Harvesting

Abstract

The aim of this chapter is to briefly explain fundamental concepts related to the physics of electromechanical transducers used for vibration energy harvesting. We present only a concise discussion on this problem and refer a reader to the literature cited in this chapter for a more detailed study of this matter. Transducers are capital for the energy harvesting process: this device takes power from one domain (for instance, the mechanical domain) and converts it to another domain (for instance, the electrical domain). In this chapter, we discuss the two most suitable transducer for micro- and nanoscale energy harvesting—piezoelectric and electrostatic transducers.

Appendix I

In this section, we present the demonstration of the fact that the area of a charge-voltage cycle performed by a variable capacitance is numerically equal to the electrical energy generated or absorbed by the capacitance, depending on the cycle direction.

The demonstration starts from the formula (4.18) expressing the work achieved by the capacitive transducer in the mechanical domain

$$\begin{aligned} A=\frac{1}{2}\oint _{\varGamma }\left[ V{\mathrm {d}}Q-Q{\mathrm {d}}V\right] . \end{aligned}$$

(4.29)

The Green theorem states that for a positively oriented, piecewise smooth, simple closed curve $\varGamma$ in a right-handed plane (V, Q), for a region D bounded by $\varGamma$ and for functions L(V, Q), M(V, Q) defined on an open region containing D and having continuous partial derivatives, the following equality is true [15]:

$$\begin{aligned} \oint _{\varGamma }(L{\mathrm {d}}V+M{\mathrm {d}}Q)=\iint \limits _D \left[ \frac{\partial M}{\partial V}-\frac{\partial L}{\partial Q}\right] {\mathrm {d}}V {\mathrm {d}}Q \end{aligned}$$

(4.30)

Applying this theorem to Eq. (4.29), we get

$$\begin{aligned} A=\frac{1}{2}\oint _{\varGamma }\left[ -Q\mathrm {d}V-V\mathrm {d}Q\right] =\frac{1}{2}\iint \limits _D [1+1]\mathrm {d}V \mathrm {d}Q=\iint \limits _D \mathrm {dV} \mathrm {dQ}. \end{aligned}$$

(4.31)

The last double integral expresses the area of the domain D enclosed by the curve. For a positively oriented (counterclockwise) path, A is positive: that means that the energy is transferred from the electrical into the mechanical domain. Conversely, for a negatively inverted (clockwise) path, the transducer’s force work is negative, and the elctrical energy is converted from the mechanical energy.