understand how ultrasound waves are generated and detected by a piezoelectric transducer

Production and Use of Ultrasound

Learning Objective

Understand how ultrasound waves are generated and detected by a piezoelectric transducer, and appreciate their practical applications in medicine and industry.

1. What is Ultrasound?

Ultrasound refers to sound waves with frequencies above the upper limit of human hearing (≈ 20 kHz). In most A‑Level applications the frequencies lie between 1 MHz and 10 MHz.

2. The Piezoelectric Effect

The piezoelectric effect is the reversible conversion between mechanical stress and electrical charge in certain crystals (e.g., quartz, PZT – lead zirconate titanate).

• Direct effect: Mechanical stress → electric potential.
• Converse effect: Applied voltage → mechanical deformation.

When a voltage is applied across a piezoelectric crystal, it expands or contracts at the frequency of the voltage, producing a longitudinal acoustic wave.

3. Generation of Ultrasound with a Piezoelectric Transducer

1. A high‑frequency alternating voltage $V(t)=V_0\sin(2\pi f t)$ is applied.
2. The crystal undergoes rapid cyclic strain, creating a pressure wave in the surrounding medium.
3. The acoustic wave propagates with speed $c$ (≈ 1540 m s⁻¹ in soft tissue).

The efficiency of conversion is characterised by the electromechanical coupling factor $k$, where $0

4. Detection of Ultrasound

When an incoming ultrasound wave reaches the piezoelectric crystal, it exerts a pressure $p(t)$ that compresses the crystal, generating a charge $Q(t)$ via the direct piezoelectric effect. This charge is amplified and recorded as an electrical signal.

Mathematically, the received voltage $V_{\text{rec}}$ is proportional to the pressure:

$$V_{\text{rec}} = S \, p(t)$$

where $S$ is the sensitivity of the transducer (V Pa⁻¹).

5. Key Parameters of a Piezoelectric Transducer

Parameter	Symbol	Typical \cdot alue (Medical)	Physical Significance
Resonant frequency	$f_r$	2–10 MHz	Frequency at which the crystal vibrates most efficiently.
Bandwidth	$\Delta f$	≈ 30 % of $f_r$	Range of frequencies over which the transducer operates effectively.
Electromechanical coupling factor	$k$	0.4–0.7	Measure of conversion efficiency between electrical and mechanical energy.
Acoustic impedance	$Z$	≈ 1.5 MRayl (soft tissue)	Product of medium density and sound speed; determines matching to the load.

6. Matching Layers and Backing Materials

To maximise transmission into the target medium, a thin matching layer with acoustic impedance $Z_m$ is placed between the crystal ($Z_c$) and the medium ($Z_t$). The optimal $Z_m$ satisfies:

$$Z_m = \sqrt{Z_c Z_t}$$

A backing material with high attenuation absorbs the backward‑propagating wave, reducing ringing and improving axial resolution.

7. Applications of Ultrasound

• Medical imaging (sonography): Real‑time visualization of internal organs.
• Doppler ultrasound: Measurement of blood flow velocity using the frequency shift $\Delta f = \frac{2 v f_0 \cos\theta}{c}$.
• Non‑destructive testing (NDT): Detection of cracks and voids in metals.
• Therapeutic ultrasound: Targeted heating for physiotherapy.

8. Example Calculation – Depth Resolution

For a transducer with centre frequency $f = 5\,$MHz and bandwidth $\Delta f = 2\,$MHz, the spatial pulse length (SPL) is approximately:

$$\text{SPL} \approx \frac{c}{\Delta f} = \frac{1540\ \text{m s}^{-1}}{2\times10^{6}\ \text{Hz}} \approx 0.77\ \text{mm}$$

The axial resolution is roughly half the SPL, i.e. $\approx 0.4\,$mm.

9. Summary

• Piezoelectric crystals convert electrical energy to mechanical vibrations (generation) and vice‑versa (detection).
• Key design features – resonant frequency, coupling factor, matching layer, and backing – determine performance.
• Ultrasound finds widespread use in medical diagnostics, industrial inspection, and therapy.

10. Suggested Diagram

11. Practice Questions

1. Explain why a matching layer improves the transmission of ultrasound into soft tissue.
2. A Doppler ultrasound probe operates at 3 MHz. If blood flows at 0.5 m s⁻¹ directly towards the probe, calculate the Doppler shift. (Take $c = 1540\,$m s⁻¹.)
3. Describe how the bandwidth of a transducer influences axial resolution.