Winter 2015 – Q.2

2. (a) Explain piezoelectric effect. Explain how piezoelectric oscillator is used to produce ultrasonic waves, with the help of a neat circuit diagram. [6M]

Piezoelectric Effect: When an AC voltage is applied across a piezoelectric crystal it vibrates with the frequency of the applied voltage. This effect is called as piezoelectric crystal. The amplitude of the vibrations becomes maximum at the natural resonant frequency of the oscillations of the crystal. The natural frequency of the crystal is determined by it’s physical dimension and the way in which it is cut. The frequency of the vibration is given by,

$latex f=\frac n{2L}\sqrt{\frac Y\rho}$

Where, L is thee length of the crystal plate, Y is the Young’s modulus in a particular direction and ρ is the density of the crystal.

The piezoelectric crystal may oscillate in fundamental mode or in overtones. The fundamental frequency is the lowest frequency of the crystal. And there is upper limit to this fundamental frequency of oscillation, since the crystal cannot be cut too thin without fracturing. The upper limit in these crystal is about 20 MHz, for most of the crystals. Higher frequencies up to 500 MHz can be obtained only in overtones which are generally odd multiple of the fundamental.

Piezoelectric oscillator can be constructed using triode valve or transistor circuit. The fundamental triode valve oscillator first designed by Langevin is shown in the figure.

The plate coil L₂ is inductively coupled to the coil L₁. The capacitor C and the inductor coil L₁ act as a tank circuit which oscillates with the frequency given by

$latex f=\frac1{2\mathrm\pi\sqrt{\mathrm{LC}}}$

The frequency of oscillations can be controlled by varying the capacitor C. By mutual induction emf is induced in the coil L₃ which is parallel to the crystal. The frequency of the capacitor is varied frequency of the oscillation of the oscillator matches with the natural frequency of the oscillation of the crystal. At resonance the crystal produces the ultrasonic waves of desired frequency in the surrounding medium.

The resultant amplitude is given as, $latex E_\theta=E_m\frac{\sin\alpha}\alpha,\;where\;\alpha=\frac\phi2$

Therefore,

$latex E_\theta=E_m\frac{\sin{\displaystyle\frac\phi2}}{\displaystyle\frac\phi2}$

$latex I_\theta=I_0\left(\frac{\sin{\displaystyle\frac\phi2}}{\displaystyle\frac\phi2}\right)^2$

Since, $latex I_\theta\;\propto\;E_\theta^2$

For minima to occur, $latex \phi=2\pi$

$latex \begin{array}{l}i.e.\;\frac{2\mathrm\pi}\lambda d\;\sin\theta=2\pi\\\\i.e.\;d\;\sin\;\theta=\lambda\end{array}$

which is the condition of first minima.

Given: Angle of reflection $latex =35^\circ\;so\;i=35^\circ,\;\mu=1.33$

$latex \begin{array}{l}\sin\;r=\frac{\sin\;i}\mu\\\\\;\;\;\;\;\;\;\;=\frac{\sin\;35^\circ}{1.33}\\\\\;\;\;\;\;\;\;\;=\frac{0.57}{1.33}\\\\\;\;\;\;\;\;\;\;=0.43\end{array}$

$latex \begin{array}{l}\cos\;r=\sqrt{\left(1-\sin^2\;r\right)}\\\\\;\;\;\;\;\;\;\;\;=\sqrt{1-0.186}\\\\\;\;\;\;\;\;\;\;\;=0.902\end{array}$

Condition for destructive interference

$latex \begin{array}{l}2\mu t\;\cos\;r=m\lambda\\\\2\times1.33\times5\times10^{-7}\times0.902=m\lambda\\\\12\times10^{-7}m=m\lambda\\\\12000\;\overset\circ A=m\lambda\end{array}$

For $latex m=1,\;\lambda=12000\;\overset\circ A,\;$ which is Not in the visible range.

For $latex m=2,\;\lambda=6000\;\overset\circ A,\;$ Visible range.

For $latex m=3,\;\lambda=4000\;\overset\circ A,\;$ Visible range.

Wavelength $latex 6000\;\overset\circ A\;and\;4000\;\overset\circ A$ will be absent from the visible spectrum.