Past exam of the materials science course of the University of Cambridge / 2017 / Paper 1 / 2

Take the longitudinal scalar piezoelectric coefficient appropriate to the load and electrode direction. For a uniformly stressed slab of face area $A$ and thickness $t$, the source charge is $\Delta Q=d\Delta F=dA\Delta \sigma$. Its capacitance is $C={ϵ}_0\kappa A/t$, so the ideal initially uncharged, electrically isolated sensor givesThe sign depends on the poling direction and electrode convention. Equivalently, the incremental constitutive law is $\Delta D=d\Delta \sigma +{ϵ}_0\kappa \Delta E$; an ideal open circuit keeps free electrode charge, hence $D$, fixed and gives $\Delta E=-d\Delta \sigma /({ϵ}_0\kappa )$. area cancels because both generated charge and capacitance scale with area. Thickness does not cancel: the material comparison is through $d/\kappa$ only when geometry is fixed. Here $\kappa$ must be the relative permittivity under the relevant mechanical and frequency conditions.

For identical thicknesses, compare $d/\kappa$ using the supplied coefficients:for barium titanate, lead zirconate titanate and quartz respectively. Thus barium titanate, then lead zirconate titanate, then quartz is the increasing voltage sensitivity of a piezoelectric sensor order. A large charge coefficient alone does not guarantee a large voltage: the large permittivity of a ferroelectric also increases its capacitance.

For quartz, $A=10^{-4}{\mathrm{m}}^2$ and $t=5.0\times 10^{-4}\mathrm{m}$. With $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the load and stress are $F=4.905\mathrm{N}$ and $\Delta \sigma =4.905\times 10^4\mathrm{Pa}$. ThereforeAs a check, $Q=dF=11.28\mathrm{pC}$ and $C=6.73\mathrm{pF}$ give the same result. This is the ideal maximum open-circuit change on applying the load. Input capacitance reduces it to $Q/(C+C_{\mathrm{in}})$, while finite leakage makes it decay: it is not a guaranteed permanent DC output.

The sensor must survive the stress range and repeated loading without fracture, fatigue or excessive mechanical compliance. The tensor character of the piezoelectric coefficient makes crystal orientation important; a ceramic's poling quality and domain stability also matter. Compare linearity, hysteresis, reproducibility, aging and the usable frequency bandwidth, including resonance and dynamic response.

Electrical design is equally important: leakage resistance and readout input impedance determine low-frequency loss, and cable/input capacitance loads a voltage output. A charge amplifier may favour a different material ranking from a passive open-circuit voltage measurement. temperature changes can alter the coefficients, produce a pyroelectric signal in a polar material and depole a ferroelectric near its transition. Humidity, electrode bonding, electrical breakdown, manufacturability, cost and environmental constraints such as lead content also affect a practical choice. Calibration should separate applied stress from temperature drift and readout loading.

ferroelectric poling is essential. An initially unpoled ceramic has differently oriented crystal grains and domains whose charge responses largely cancel. An applied electric field favours suitably oriented domains and leaves a remanent macroscopic polarization after removal; it does not rotate all crystal grains into a single crystal.

In the linear coefficient model the added source polarization isThis is the stress-generated contribution at fixed electric field, equivalent to the short-circuit charge per face area. Under open-circuit conditions the induced field supplies an opposing dielectric contribution, so $\Delta D=0$; the source quantity must not be confused with a change of total free electrode charge. For scale, room-temperature single-domain barium titanate has spontaneous polarization of order $0.26\mathrm{C}{\mathrm{m}}^{-2}$: the calculated increment is about $3.0\times 10^{-4}$ of that value, a small response rather than wholesale reversal of the spontaneous polarization.^[ref]

For the requested displacement estimate, assign the whole additional dipole moment of one cell to movement of its single $\mathrm{T}{\mathrm{i}}^{4+}$ ion relative to a fixed background. Then $\Delta P=(4e)\Delta x/(a^{2}c)$, givingThis formal-charge estimate is only an equivalent displacement: real polarization changes involve oxygen and barium displacements, electronic polarization and Born effective charges, as well as possible domain-wall motion. The coefficient supplied for the ceramic cannot uniquely determine an actual microscopic titanium trajectory.