# A Dual Stage Low Power Converter Driving for Piezoelectric Actuator Applied in Micro Mobile Robot
## 3 Design and Control of the Drive Circuit
### DC/AC High-Voltage Drive Stage  
The DC/AC drive stage can convert DC voltage into high-voltage unipolar drive signals. Moreover, the energy stored in a piezoelectric actuator should be recovered to increase conversion efficiency. Many prior works have been conducted on the high-voltage driving of piezoelectric actuators. [PERSON] et al. proposed a drive stage without energy recovery [15]. [PERSON] et al. proposed a method for driving a piezoelectric actuator using a low-frequency square wave signal at the cost of increasing the complexity of a circuit [20]. Unipolar signals of any shape can be generated by utilizing a half-bridge inverter that is composed of a high-quality factor inductor \(L\) and two inherent capacitors of the electrostatic actuators. Different operating modes of the high-voltage half-bridge drive stage are illustrated in Figure 6.  
Figure 5: Switching waveforms of the flyback converter during (**a**) one switching period; (**b**) steady-state operation. Figure 6: Six switching modes of the high-voltage half-bridge drive stage. In mode (a), the inductor current \(i_{L}\) begins to rise once the switch Q\({}_{\rm H}\) is turned on. The high-side equivalent capacitor \(C_{1}\) of the piezoelectric actuator starts to transfer the most unused energy to the low side, whereas the remaining energy is recovered in capacitor \(C\). When the equivalent series resistance in a circuit is considered, the governing differential equations can be obtained as follows according to [PERSON]'s voltage law:  
\[\left\{ \begin{array}{l}L\cdot\frac{\partial i_{L}}{\partial t}+R_{L}\cdot i_ {L}+v_{c2}=V_{\rm C}\\ C_{2}\cdot\frac{\partial v_{c2}}{\partial t}=i_{c2}=i_{L}-i_{c1}=i_{L}-C_{1} \cdot\frac{\partial v_{c1}}{\partial t}\end{array} \right.. \tag{5}\]  
When the two equations are combined, a second-order differential equation in this mode can be obtained as  
\[\frac{\partial^{2}v_{c2}}{\partial t^{2}}=2\beta\cdot\frac{\partial v_{c2}}{ \partial t}+\omega_{0}^{2}\cdot v_{c2}=\omega_{0}^{2}\cdot V_{C}, \tag{6}\]  
where \(\beta=R_{L}/2L\) is the neper frequency, and \(\omega_{0}=1/(2 LC_{2})^{0.5}\) is the angular resonance frequency. Under an underdamped oscillation (\(\beta^{2}-\omega_{0}{}^{2}<0\)),  
\[v_{c2}(t)=V_{\rm C}-V_{\rm C}\cdot e^{-\beta t}\left(\cos\omega_{d}t+\frac{ \beta}{\omega_{d}}\cdot\sin\omega_{d}t\right), \tag{7}\]  
where the drive signal voltage \(v_{c2}\) is an underdamped voltage with a decaying oscillation at frequency \(\omega_{d}=(\omega_{0}{}^{2}-\beta^{2})^{0.5}\). Mode (a) ends as soon as \(v_{c2}\) increases and becomes equal to \(V_{C}\) in the first oscillation cycle. At the end of mode (a), Q\({}_{\rm H}\) is turned off and the inductor current \(i_{L}\) starts to decrease via the freewheeling diode D\({}_{\rm L}\). A new oscillation occurs between inductor \(L\) and the capacitive loads. The circuit starts operating alternatively in modes (b) and (e). The initial voltage in mode (b) is \(V_{C}\). The governing equations can be obtained as  
\[\left\{ \begin{array}{l}L\cdot\frac{\partial i_{L}}{\partial t}+R_{L}\cdot i_ {L}+v_{c2}=0\\ C_{2}\cdot\frac{\partial v_{c2}}{\partial t}=i_{c2}=i_{L}-i_{c1}=i_{L}-C_{1} \cdot\frac{\partial v_{c1}}{\partial t}\end{array} \right.. \tag{8}\]  
Under an underdamped oscillation (\(\beta^{2}-\omega_{0}{}^{2}<0\)), the drive signal voltage becomes  
\[v_{c2}(t)=v_{c2}(t_{a})\cdot e^{-\beta(t-t_{a})}\cdot\left[\cos\omega_{d}(t- t_{a})+\frac{\beta}{\omega_{d}}\cdot\sin\omega_{d}(t-t_{a})\right], \tag{9}\]  
where (\(v_{c2}\) (\(t_{a}\)) = \(V_{C}\)) represents the initial voltage condition at the end of mode (a) at \(t_{a}\). The drive signal voltage \(v_{c2}\) continues oscillating until it becomes stable and equal to \(V_{C}\) during the rest of the period. In mode (c), the lower side electrostatic actuator is completely charged, whereas the upper side electrostatic actuator is empty. The drive signal voltage \(v_{c2}\) is the same as \(V_{C}\) during this period.