Equivalent Circuit of a Piezoelectric component
1. Piezoelectric component
When a piezoelectric component is subject to force, equal amounts of opposite charges are generated on its two electrodes, thus the piezoelectric component acts as a charge generator. The piezoelectric medium between the two electrodes is insulated, making the piezoelectric component act as a capacitor, with a capacitance of,
$$C_a = \frac{\varepsilon_r \varepsilon_0 A }{d}$$
where A - the area of Piezoelectric disc (Unit: $ \rm m^2$); d - The thickness of the piezoelectric disc (Unit: m); $\varepsilon$ is the relative permittivity of the piezoelectric material, and $\varepsilon_0$ is the vacuum permittivity ($8.85×10^{−12}F/m$).
(1) The piezoelectric sensor can be represented by an equivalent circuit consisting of an ideal voltage source in serious with a capacitor.
(2) Similarly, a piezoelectric sensor can be equivalently represented by a circuit consisting of a charge source in parallel with a capacitor.
2. Measurement circuit
According to the equivalent circuit, the voltage or charge generated by a piezoelectric sensor in response to applied force can only be sustained for a long period if there is no leakage of the signal charge within the sensor and if the external circuit load is infinite. In reality, it is not possible for a piezoelectric sensor to have zero leakage internally, and the external circuit load cannot be infinite. Only when an external force acts continuously at a relatively high frequency can the charge within the sensor be replenished. Therefore, piezoelectric sensors are not suitable for static measurements. Under the influence of alternating forces, the charge in a pressure sensor can be continuously replenished, allowing a certain current to be generated in the measurement circuit. Hence, piezoelectric sensors are only suitable for dynamic measurements.
In actual, piezoelectric sensor need to be connected to measurement instruments or circuits. Therefore, it is necessary to consider the equivalent capacitance of the connecting cables, the input impedance of the amplifier, input capacitance, and the leakage resistance of the piezoelectric sensor.
In practical, piezoelectric sensors need to be connected to measuring instruments or measuring circuits, hence considerations should be given to:
Piezoelectric sensors typically have high internal impedance and produce low output energy. Therefore, these sensors need to be connected to appropriate measurement circuit to form a complete measurement system. Usually, a high input impedance preamplifier is connected to the output of the piezoelectric sensor.
B) To transform impedance, converting high output impedance of sensors into low output impedance.
The output of the piezoelectric sensor can be either voltage signal or charge signal, hence there are two forms of preamplifiers: Voltage amplifier and charge amplifier.
2.1 Voltage Amplifier
$$R=\frac{R_a R_i}{R_a + R_i}$$
$$C=C_a+C_i$$
Let $F$ be the alternating force acting on the piezoelectric element along the electric axis, with a maximum value of $F_m$ and an angular frequency of $\omega$, i.e.,
$$\dot F = F_m sin(\omega t)$$
If the piezoelectric constant of the piezoelectric element is $d_{11}$, then the charge generated under the force is,
$$q = d_{11} \cdot \dot F = d_{11} F_m sin(\omega t)$$
Thus, the voltage generated on the piezoelectric element is,
$$\dot U_a = \frac{\dot a}{C_a} = \frac{d_{11}F_m sin(\omega t)}{C_a}$$
The voltage at the input terminal of the amplifier is,
$$\dot U_{sr} = \dot U_a \frac{R||Z_c}{Z_{C_a}+R||Z_c} = \frac{d_{11} \dot F}{C_a} \frac{R||Z_c}{Z_{C_a}+R||Z_c} = d_{11} \dot F \frac{j \omega R}{1+j\omega R(C_a+C)}$$
The input voltage amplitude of the preamplifier is,
$$U_{srm} = \frac{d_{11}F_m \omega R}{\sqrt{1+(\omega R (C_a+C_c+C_i))^2}} $$
The phase difference between the input voltage and applied force is,
$$ \varphi = \frac{\pi}{2} - arctan(\omega R (C_a+C_c+C_i)) $$
The voltage sensitivity of the sensor is defined as,
$$S_v = |\frac{\dot U_{sr}}{\dot F}| = \frac{d_{11} \omega R}{\sqrt{1+(\omega R (C_a+C_c+C_i))^2}}$$
(1) when $\omega = 0$, $U_a = 0$, the input voltage of the pre-amplifier $U_{srm} = 0$ and $S_v = 0$, indicating that the piezoelectric sensor cannot measure static signals.
(2) when $(\omega R(C_a+C_c+C_i))^2 >> 1$, we get, $$U_{srm} = \frac{d_{11} F_m}{C_a+C_c+C_i}$$
$$S_v = \frac{d_{11}}{C_a+C_c+C_i}$$. This indicates that the input voltage and the voltage sensitivity of the preamplifier are independent of the force applied on the piezoelectric component, indicating good high-frequency characteristics of the voltage amplifier.
(3) $U_{sr}$, $S_v$ are related to $C_c$, indicating that the connecting cable between the piezoelectric sensor and the preamplifier cannot be used arbitrarily.
2.2 Charge Amplifier
电荷放大器由一个反馈电容和高增益运算放大器构成，当放大器开环增益和输入电阻、反馈电阻相当大时，电荷放大器视为开路。
The ouput of charge amplifier is proportional to q,
$$U_o = -A\frac{q}{C} = -A\frac{q}{C_a + C_c + C_i + (1+A)C_f}$$
通常运放的开环增益为$10^4~10^8$，即$A>>1$，所以，
$$(1+A)C_f >> C_a + C_c + C_i$$
$$U_0 \approx - \frac{q}{C_f}$$
Since the output signal of a voltage amplifier is influenced by the cable capacitance, while the output voltage of a charge amplifier is unaffected by the cable capacitance. The majority of current applications choose charge amplifier.
