Piezoelectric and Photodiode Sensors
Piezoelectric and Photodiode Sensors
Piezoelectric crystals convert mechanical force or acceleration directly into electrical charge — the same effect that made quartz watches possible and that underlies every accelerometer in your phone. Photodiodes do the analogous job for photons: each absorbed photon releases a charge carrier, producing a photocurrent proportional to the incident light. Both are self-generating sensors with high output impedance, demanding charge amplifiers or transimpedance front-ends for clean readout.
Following Pallàs-Areny and Webster, we treat the piezoelectric element as a charge source in parallel with a capacitance, and the photodiode as a current source with a shunt capacitance. Both pictures lead naturally to the signal-conditioning circuits you will see in Module 4.
This lesson derives the charge, voltage, and current equations for both devices and works through three quantitative examples covering accelerometer output, dynamic pressure sensing, and photodiode responsivity.
$$q = d\,F$$
where $d$ is the piezoelectric charge constant (C/N) of the material (~2.3 pC/N for quartz, ~400 pC/N for PZT). For an accelerometer with seismic mass $m$:
$$q = d\,m\,a$$
$$I_\text{ph} = R(\lambda)\,P_{\text{opt}}$$
$R(\lambda)$ is the responsivity in A/W. Silicon peaks near 0.5 A/W around 900 nm.
A PZT accelerometer has $d = 350\,\text{pC/N}$ and a seismic mass $m = 2\,\text{g}$. Compute the charge for $a = 10\,g \approx 98.1\,\text{m/s}^2$.
$F = ma = 2\times10^{-3}\cdot 98.1 = 0.196\,\text{N}$.
$q = 350\times10^{-12}\cdot 0.196 = 6.87\times10^{-11}\,\text{C} = 68.7\,\text{pC}$.
The accelerometer has an internal capacitance $C_s = 1\,\text{nF}$. Find the open-circuit output voltage.
$V = q/C_s = 68.7\times10^{-12}/10^{-9} = 68.7\,\text{mV}$.
This is why piezoelectric accelerometers need charge amplifiers — cable capacitance in parallel would otherwise dilute the voltage dramatically.
A silicon photodiode with $R = 0.45\,\text{A/W}$ sees optical power $P_{\text{opt}} = 100\,\mu\text{W}$ of red laser light. Find the photocurrent.
$I_\text{ph} = 0.45\cdot 100\times10^{-6} = 45\,\mu\text{A}$.
Through a $10\,\text{k}\Omega$ transimpedance resistor this becomes $V = 0.45\,\text{V}$ — a clean, easily-processed signal.
Practice Problems
Show Answer Key
1. $F = 5\times10^{-3}\cdot 490.5 = 2.45\,\text{N}$; $q = 2.3\times10^{-12}\cdot 2.45 = 5.64\,\text{pC}$.
2. $V = 5.64\times10^{-12}/500\times10^{-12} = 11.3\,\text{mV}$.
3. $I_\text{ph} = 6\,\mu\text{A}$.
4. To present near-zero input impedance and keep the diode at zero bias — eliminates photodiode shunt capacitance from the frequency response.
5. Long-term stability of piezoelectric coefficient; quartz does not depole.
6. A charge amplifier is insensitive to cable capacitance (virtual ground), while a voltage amp sees $V = q/(C_s + C_{\text{cable}})$.