Oscillators for Energy Harvesting

Abstract

The aim of this chapter is to briefly explain fundamental concepts such as linear and nonlinear oscillators and transducers since they are required for understanding the energy harvesting principle. We discuss the model of free and harmonically driven linear and nonlinear oscillators. In particular, we show that the solution of a forced linear oscillator is a harmonic oscillation at the frequency of the external signal. We discuss the principles of nonlinear oscillators, resonance in nonlinear oscillator and multistability. In order to use an oscillator for the energy harvesting process, one requires a transducer, a device that 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 transducers for micro- and nanoscale harvesting—piezoelectric and electrostatic transducers.

Appendices

Appendix I

In this appendix, we calculate the fundamental harmonic of the force generated by a nonlinear spring submitted to sinusoidal deformation. Let the spring be characterised by the relationship between the force $F_{spring}$ and the deformation x

$$\begin{aligned} F_{spring}=F(x), \end{aligned}$$

(3.115)

and x is given by

$$\begin{aligned} x=X\sin (\omega t). \end{aligned}$$

(3.116)

The complex amplitude of the fundamental harmonic of the force $\dot{F}_{spring}$ is given by

$$\begin{aligned} \dot{F}_{spring}=\frac{\omega }{\pi }\int _{0}^{T}f[X\sin (\omega t)]e^{-j\omega t}\mathrm dt \end{aligned}$$

(3.117)

(Do not confuse $\dot{F}_{spring}$ used in this section with the notation for the derivative.)

We first calculate the real part of this integral

$$\begin{aligned} Re(\dot{F}_{spring})=\frac{\omega }{\pi }\int _{0}^{T}f[X\sin (\omega t)]\cos (\omega t)\mathrm dt=\nonumber \\ \frac{1}{\pi }\int _{0}^{T}f[X\sin (\omega t)]\mathrm d\sin (\omega t)= \frac{1}{\pi }\oint _{y}f[Xy]\mathrm dy \end{aligned}$$

(3.118)

Since the force of a spring is potential, this integral on a closed path is zero, so that

$$\begin{aligned} Re(\dot{F}_{spring})=0. \end{aligned}$$

(3.119)

Now we calculate the imaginary part

$$\begin{aligned} Im(\dot{F}_{spring})=-\frac{\omega }{\pi }\int _{0}^{T}f[X\sin (\omega t)]\sin (\omega t)\mathrm dt=\nonumber \\ \frac{1}{\pi }\int _{0}^{T}f[X\sin (\omega t)]\mathrm d\cos (\omega t)=\nonumber \\ \frac{1}{\pi }\int _{1}^{-1}f[X\sqrt{1-y^2}]\mathrm dy+\frac{1}{\pi }\int _{-1}^{1}f[-X\sqrt{1-y^2}]\mathrm dy \end{aligned}$$

(3.120)

For the complex amplitude of the spring force, we get

$$\begin{aligned} \dot{F}_{spring}=j\left[ \frac{1}{\pi }\int _{1}^{-1}f[X\sqrt{1-y^2}]\mathrm dy+\frac{1}{\pi }\int _{-1}^{1}f[-X\sqrt{1-y^2}]\mathrm dy\right] \end{aligned}$$

(3.121)

and as a consequence, the force in the time domain is given by

$$\begin{aligned} \dot{F}_{spring}=\left[ \frac{1}{\pi }\int _{1}^{-1}f[X\sqrt{1-y^2}]\mathrm dy+\frac{1}{\pi }\int _{-1}^{1}f[-X\sqrt{1-y^2}]\mathrm dy\right] \sin (\omega t). \end{aligned}$$

(3.122)

Appendix II

In this appendix, we calculate the fundamental harmonic of the force generated by a nonlinear damper submitted to sinusoidal deformation. Let the damper characterised by the relationship between the force $F_{damper}$ and the deformation x

$$\begin{aligned} F_{damper}=F(\dot{x}), \end{aligned}$$

(3.123)

and x is given by

$$\begin{aligned} x=X\sin (\omega t). \end{aligned}$$

(3.124)

The complex amplitude of the fundamental harmonic of the force $F_{damper}$ is given by

$$\begin{aligned} \dot{F}_{damper}=\frac{\omega }{\pi }\int _{0}^{T}f[X\omega \cos (\omega t)]e^{-j\omega t}\mathrm dt \end{aligned}$$

(3.125)

We first calculate the real part of this integral

$$\begin{aligned} Re(\dot{F}_{damper})=\frac{\omega }{\pi }\int _{0}^{T}f[X\omega \cos (\omega t)]\cos (\omega t)\mathrm dt=\nonumber \\ \frac{1}{\pi }\int _{0}^{T}f[X\omega \cos (\omega t)]\mathrm d\sin (\omega t)=\nonumber \\ \frac{1}{\pi }\int _{T/4}^{5T/4}f[X\omega \cos (\omega t)]\mathrm d\sin (\omega t)=\nonumber \\ \frac{1}{\pi }\int _{1}^{-1}f[X\omega \sqrt{1-y^2}]\mathrm dy+\frac{1}{\pi }\int _{-1}^{1}f[-X\omega \sqrt{1-y^2}]\mathrm dy \end{aligned}$$

(3.126)

Now we calculate the imaginary part

$$\begin{aligned} Im(\dot{F}_{damper})=-\frac{\omega }{\pi }\int _{0}^{T}f[X\omega \cos (\omega t)]\sin (\omega t)\mathrm dt=\nonumber \\ \frac{1}{\pi }\int _{0}^{T}f[X\omega \cos (\omega t)]\mathrm d\cos (\omega t)=\nonumber \\ \frac{1}{\pi }\oint _{y}f[X\omega y]\mathrm dy \end{aligned}$$

(3.127)

According to the Green-Riemann theorem [29], this integral is zero.

The expression of the reaction force of a damper is

$$\begin{aligned} \dot{F}_{damper}=\left[ \frac{1}{\pi }\int _{1}^{-1}f[X\omega \sqrt{1-y^2}]\mathrm dy+\frac{1}{\pi }\int _{-1}^{1}f[-X\omega \sqrt{1-y^2}]\mathrm dy\right] \cos (\omega t). \end{aligned}$$

(3.128)