Mechanical stress on a crystal produces electric charge on its electrodes.
\[
Q = d\,F
\]
where d is the piezo‑electric coefficient (C N⁻¹).
If the stress varies sinusoidally (as in an incoming ultrasound wave), the generated voltage is proportional to the acoustic pressure p(t):
\[
V_{\text{out}}(t)=g\,p(t)
\]
g is the electromechanical coupling factor (V Pa⁻¹). This is the basis of a receiver transducer.
An electric field across the crystal causes it to deform.
\[
E = \frac{V}{t},\qquad S = dE = d\frac{V}{t}
\]
\(S\) is the strain (relative change in length). For a crystal of original length \(L\):
\[
\Delta L = S L = d\frac{V}{t}\,L
\]
The deformation is longitudinal – the crystal surface moves back and forth, creating alternating compressions and rarefactions in the surrounding medium. These pressure variations are the ultrasound wave that propagates away from the crystal.
\[
Z = \rho\,c
\]
Large differences in \(Z\) between two media cause strong reflections, which are recorded as echoes in diagnostic scanners.
When a wave of intensity \(I_0\) travelling in medium 1 meets a boundary with medium 2, the reflected fraction is
\[
\frac{IR}{I0}= \left(\frac{Z1-Z2}{Z1+Z2}\right)^{\!2}
\]
The square arises because intensity is proportional to the square of the pressure (or amplitude) of the wave.
The intensity of an ultrasound beam decays exponentially with distance:
\[
I(x)=I_0\,e^{-2\alpha x}
\]
where \(\alpha\) is the attenuation coefficient (Np m⁻¹) and the factor 2 accounts for the loss of both pressure amplitude and intensity. Typical values are:
| Medium | \(\alpha\) (dB cm⁻¹ MHz⁻¹) |
|---|---|
| Soft tissue | 0.5–0.7 |
| Fat | 0.6–0.9 |
| Bone | 15–20 |
| Water | ≈0.0 |
Attenuation limits the depth that can be imaged and favours the use of higher‑frequency transducers for superficial structures (better resolution, more attenuation) and lower‑frequency transducers for deeper targets (less resolution, less attenuation).
Maximum conversion efficiency occurs when the crystal thickness \(t\) is an integer multiple of half a wavelength \(\lambda\) of the sound inside the crystal:
\[
t = n\frac{\lambda}{2}\qquad (n=1,2,3,\dots)
\]
Since \(\lambda = v/f\) (where \(v\) is the speed of sound in the crystal), the resonant frequencies are
\[
f_n = \frac{n v}{2t}
\]
Bandwidth and axial resolution – Higher‑order modes (larger \(n\)) give narrower bandwidths, which degrade axial resolution because the pulse length increases. Designers therefore aim for a single, well‑defined fundamental resonance for high‑resolution imaging.
Incoming ultrasound compresses and expands the crystal surface, invoking the direct effect. The alternating pressure \(p(t)\) produces an alternating voltage:
\[
V_{\text{out}}(t)=g\,p(t)
\]
The amplitude of the received signal depends on the intensity‑reflection coefficient at the reflecting interface and on the attenuation the wave suffers on its round‑trip.
| Material | Piezo‑electric coefficient \(d\) (pC N⁻¹) | Usable frequency range (MHz) | Comments (trade‑offs) |
|---|---|---|---|
| Quartz (SiO₂) | 2.3 | 0.1 – 5 | Very stable, low temperature drift; low \(d\) → modest output but high linearity – ideal for precision sensors. |
| PZT (lead zirconate titanate) | 300 – 600 | 1 – 30 | High \(d\) → large displacement and strong signal; however, higher dielectric losses and greater temperature sensitivity – common in medical/industrial transducers. |
| LiNbO₃ (lithium niobate) | 7 – 10 | 0.5 – 10 | Good for high‑frequency (>10 MHz) applications; moderate \(d\) and relatively low loss. |
Given: PZT crystal, thickness \(t = 1\;\text{mm}\), driving voltage amplitude \(V_0 = 100\;\text{V}\), speed of sound in PZT \(v = 4.0\times10^{3}\;\text{m s}^{-1}\), piezo‑electric coefficient \(d = 500\;\text{pC N}^{-1}=5.0\times10^{-10}\;\text{C N}^{-1}\), crystal length \(L = 10\;\text{mm}\).
\[
f_1 = \frac{v}{2t}= \frac{4.0\times10^{3}}{2\times1\times10^{-3}} = 2.0\times10^{6}\;\text{Hz}=2\;\text{MHz}
\]
\[
E = \frac{V_0}{t}= \frac{100}{1\times10^{-3}} = 1.0\times10^{5}\;\text{V m}^{-1}
\]
\[
S = dE = 5.0\times10^{-10}\times1.0\times10^{5}=5.0\times10^{-5}
\]
\[
\Delta L_{\max}=SL = 5.0\times10^{-5}\times10\times10^{-3}=5.0\times10^{-7}\;\text{m}=0.5\;\mu\text{m}
\]
Using the impedance values from the table below:
\[
Z_{\text{tissue}} = 1.62\;\text{MRayl},\qquad
Z_{\text{bone}} = 5.92\;\text{MRayl}
\]
\[
\frac{IR}{I0}= \left(\frac{5.92-1.62}{5.92+1.62}\right)^{2}
= \left(\frac{4.30}{7.54}\right)^{2}
\approx 0.33
\]
Approximately 33 % of the incident intensity is reflected, producing a bright echo that appears as a high‑intensity (white) region on the image.
Typical attenuation coefficient for soft tissue: \(\alpha = 0.6\;\text{dB cm}^{-1}\,\text{MHz}^{-1}\).
For a 5 MHz beam travelling 6 cm into tissue and back (total path = 12 cm):
\[
\text{Total loss}= 0.6 \times 5 \times 12 = 36\;\text{dB}
\]
Thus the returning signal is reduced to about \(10^{-3.6}\approx 2.5\times10^{-4}\) of its original intensity – illustrating why deeper structures require lower frequencies.
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