Piezoelectric load cell and charge amplifier

(a) Piezoelectric load cell

Certain crystals and ceramics (quartz, Rochelle salt, barium titanate, lead zirconate titanate (PZT)) develop an electric charge on their faces when a force deforms them (the direct piezoelectric effect):

$$q = d\,F$$

where $d$ is the charge sensitivity (C/N). The crystal is a self-generating (active) transducer, mounted between two plates in a stiff housing. It is equivalent to a charge source $q$ in parallel with a capacitance $C$, so the open-circuit voltage is $V = q/C$.

        F
        v
   [=========]  <- top plate
   | crystal |  charge +q / -q on faces
   [=========]
   housing -> output

Why a charge amplifier: the voltage $q/C$ depends on the cable capacitance, so changing the cable changes the reading. A charge amplifier (an op-amp with a feedback capacitor $C_f$) gives $V_o = -q/C_f$, which is independent of cable length, and it has a high input impedance.

Why not static: the charge leaks away through the insulation and the amplifier input resistance $R$. With a time constant $\tau = RC$, the output decays under a constant force, so only dynamic (and quasi-static) forces can be measured. Advantages of the cell: very high stiffness, high natural frequency, wide dynamic range and small size.

(b) Numerical

Charge generated:

$$q = dF = 2.3\times10^{-12} \times 500 = 1.15\times10^{-9}\ \text{C} = 1.15\ \text{nC}$$

Open-circuit voltage ($C = 1.2$ nF):

$$V = \frac{q}{C} = \frac{1.15\times10^{-9}}{1.2\times10^{-9}} = 0.958\ \text{V}$$

Charge amplifier output ($C_f = 100$ pF), in magnitude:

$$|V_o| = \frac{q}{C_f} = \frac{1.15\times10^{-9}}{100\times10^{-12}} = 11.5\ \text{V}$$

(The sign is negative for an inverting charge amplifier.)

Lower cut-off frequency ($R = 1$ G$\Omega$, $C = 1.2$ nF):

$$f_c = \frac{1}{2\pi RC} = \frac{1}{2\pi(10^{9})(1.2\times10^{-9})} = 0.133\ \text{Hz}$$

Answer: $q = 1.15$ nC; $V = 0.958$ V; charge-amplifier output $= 11.5$ V (magnitude); $f_c = 0.133$ Hz.