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

Production and Use of Ultrasound (Cambridge IGCSE/A‑Level Physics 9702 – Section 24.1)

Learning objective

Explain how a piezoelectric transducer generates and detects ultrasound, and relate the principal physical quantities (frequency, acoustic impedance, attenuation, reflection) to medical and industrial applications.

1. What is ultrasound?

• Sound waves with frequencies above the upper limit of human hearing (≈ 20 kHz).
• In the A‑Level syllabus the useful range is typically 1 MHz – 10 MHz.

2. The piezoelectric effect (two‑way conversion)

Certain crystals (e.g. quartz, PZT – lead zirconate titanate) show a reversible conversion between mechanical stress and electrical charge.

• Converse effect: When a potential difference (p.d.) is applied the crystal changes shape (expands or contracts).
• Direct effect: When the crystal’s shape is changed by an external mechanical stress it generates an e.m.f. (produces a voltage).

This wording mirrors the Cambridge syllabus wording: “the crystal changes shape when a p.d. is applied … and generates an e.m.f. when its shape changes.”

3. Generation of ultrasound – voltage → mechanical → acoustic

1. An alternating voltage is applied across the crystal $$V(t)=V_0\sin(2\pi ft)$$
2. Because of the converse piezoelectric effect the crystal repeatedly expands and contracts, acting as a source of an electromotive force (e.m.f.) that drives a longitudinal pressure wave into the surrounding medium.
3. The acoustic wave propagates through the coupling medium with speed c (≈ 1540 m s⁻¹ in soft tissue).

The efficiency of the conversion is characterised by the electromechanical coupling factor k (0 < k < 1).

4. Detection of ultrasound – mechanical → electrical

An incoming ultrasound pulse compresses (or stretches) the crystal. By the direct piezoelectric effect this mechanical deformation produces a charge Q(t), which appears as a voltage that can be amplified and recorded.

For a linear transducer the received voltage is proportional to the acoustic pressure:

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

where S is the sensitivity (V Pa⁻¹).

5. Acoustic impedance and reflection

• Acoustic impedance of a medium: Z = \rho c (kg m⁻² s⁻¹, Rayl).
• At a boundary between two media (impedances Z₁ and Z₂) part of the wave is reflected. The intensity reflection coefficient is $$\frac{I_R}{I_0}= \left(\frac{Z_1-Z_2}{Z_1+Z_2}\right)^2$$ and the transmission coefficient is 1 – I_R/I_0.
• Large impedance mismatches (e.g. bone vs. soft tissue) give strong echoes; good matching reduces loss.

6. Attenuation of ultrasound

The intensity of a travelling wave decreases exponentially:

$$I = I_0 e^{-\mu x}$$

• I₀ – initial intensity.
• μ – attenuation coefficient (Np m⁻¹ or dB cm⁻¹).
• x – distance travelled in the medium.

Attenuation rises with frequency, which explains why higher‑frequency probes give better resolution but poorer depth of penetration.

7. Key parameters of a piezoelectric transducer

Parameter	Symbol	Typical medical value	Physical significance
Resonant frequency	fr	2 – 10 MHz	Frequency at which the crystal vibrates most efficiently (maximum k).
Bandwidth	Δf	≈ 30 % of fr	Range of frequencies over which the transducer works effectively; larger Δf → shorter pulse → better axial resolution.
Electromechanical coupling factor	k	0.4 – 0.7	Fraction of electrical energy converted to mechanical energy (and vice‑versa).
Acoustic impedance	Z	≈ 1.5 MRayl (soft tissue)	Product ρc; determines how well the crystal is matched to the load.
Attenuation coefficient	μ	0.5 – 1 dB cm⁻¹ MHz⁻¹ (soft tissue)	Rate at which intensity falls with distance; larger for higher frequency.

8. Matching layer and backing material

To maximise transmission into the target medium a thin matching layer is placed between the crystal (impedance Z_c) and the tissue (impedance Z_t).

Optimal impedance of the matching layer:

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

A backing material with high acoustic attenuation absorbs the wave that travels back into the crystal, reducing “ringing” and improving axial resolution.

9. Applications of ultrasound

• Medical imaging (sonography): Real‑time pictures of organs, foetus, heart.
• Doppler ultrasound: Measures blood‑flow velocity using the frequency shift $$\Delta f = \frac{2 v f_0 \cos\theta}{c}$$
• Non‑destructive testing (NDT): Detects cracks, voids and thickness variations in metals and composites.
• Therapeutic ultrasound: Focused high‑intensity beams produce local heating for physiotherapy or tumour ablation.

10. Example calculations

10.1 Axial (depth) resolution

For a transducer with centre frequency f = 5 MHz and bandwidth Δf = 2 MHz:

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

Axial resolution ≈ ½ SPL ≈ 0.4 mm.

10.2 Doppler shift

Probe frequency f₀ = 3 MHz, blood speed v = 0.5 m s⁻¹ directly towards the probe (θ = 0°).

$$\Delta f = \frac{2 v f_0}{c} = \frac{2(0.5)(3\times10^{6})}{1540} \approx 1.95\times10^{3}\ \text{Hz}$$

The received echo is shifted by ≈ 2 kHz.

10.3 Reflection at a bone–soft‑tissue interface

Typical impedances: soft tissue Z₁ ≈ 1.5 MRayl, bone Z₂ ≈ 7.8 MRayl.

$$\frac{I_R}{I_0}= \left(\frac{7.8-1.5}{7.8+1.5}\right)^2 \approx \left(\frac{6.3}{9.3}\right)^2 \approx 0.46$$

≈ 46 % of the incident intensity is reflected – a strong echo.

10.4 Intensity after attenuation

Attenuation coefficient = 0.7 dB cm⁻¹ MHz⁻¹, frequency = 5 MHz, path length = 3 cm.

Total attenuation (dB) = 0.7 × 5 × 3 = 10.5 dB.

Intensity remaining: $$I = I_0 \times 10^{-10.5/10} \approx 0.089\,I_0$$ ≈ 9 % of the original intensity.

11. Summary

• Piezoelectric crystals convert electrical energy ↔ mechanical vibration (converse and direct effects). The crystal changes shape when a p.d. is applied and generates an e.m.f. when its shape changes.
• Key design features – resonant frequency, bandwidth, coupling factor, matching layer, backing – control conversion efficiency, resolution and penetration.
• Acoustic impedance, reflection, and attenuation dictate how much of the wave reaches the target and returns as a usable echo.
• Ultrasound is essential for medical imaging, Doppler flow measurement, industrial inspection, and therapeutic heating.

12. Suggested diagram

13. Practice questions

1. Explain why a matching layer improves the transmission of ultrasound into soft tissue.
2. A Doppler probe operates at 3 MHz. Blood flows at 0.5 m s⁻¹ directly towards the probe. Calculate the Doppler frequency shift (take c = 1540 m s⁻¹).
3. Describe how the bandwidth of a transducer influences axial resolution.
4. Calculate the intensity that remains after a 3 cm path in soft tissue for a 5 MHz wave, given an attenuation coefficient of 0.7 dB cm⁻¹ MHz⁻¹.
5. Why does a high‑frequency probe give better image detail but poorer depth of penetration?