## 3 DC-Side Capacitance Analysis
### Effect of DC Voltage Fluctuations on Load Voltage  
Due to the presence of the midline, when the three-phase load is unbalanced, the unbalanced current charges and discharges the DC-side capacitor, which, in turn, causes fluctuations in the DC-side capacitance voltage [18; 19], which can be expressed as follows:  
\[\left\{\begin{array}{l}u_{c1}=\frac{U_{c1}}{2}+u_{c}\\ u_{c2}=\frac{U_{dc}}{2}-u_{c}\end{array}\right. \tag{8}\]  
In Equation (8), \(u_{c}\) is the AC component of the capacitive voltage on the DC-side caused by the midline current, which lags the midline current by \(90^{\circ}\) in the phase, and in order not to lose generality the following is set:  
\[u_{c}=U_{\rm cm}\cos(\omega t+\varphi) \tag{9}\]  
Figure 10: Phase function curve between the DC-side capacitance and CA phase.  
Figure 9: Phase function curve between the DC-side capacitance and BC phase.  
\(U_{\rm cm}\) is the amplitude of the capacitive voltage fluctuation on the DC-side, using SPWM control and assuming a modulation system of \(m\). The duty cycle can be expressed as follows:  
\[D(\omega t)=m\cdot\sin(\omega t) \tag{10}\]  
Due to the fluctuation in the capacitor voltage on the DC-side, combined with Equation (8), the output load phase voltage can be expressed as follows:  
\[u_{\rm o}(t)=(\frac{U_{\rm dc}}{2}+u_{\rm c})\cdot D(\omega t) \tag{11}\]  
The three-phase load phase voltage can be expressed as follows:  
\[\left\{\begin{array}{l}u_{\rm an}=[\frac{U_{\rm dc}}{2}+U_{\rm cm}\cos( \omega t+\varphi)]\cdot m\cdot\sin(\omega t)\\ u_{\rm bn}=[\frac{U_{\rm dc}}{2}+U_{\rm cm}\cos(\omega t+\varphi)]\cdot m\cdot \sin(\omega t-120^{\rm o})\\ u_{\rm cn}=[\frac{U_{\rm dc}}{2}+U_{\rm cm}\cos(\omega t+\varphi)]\cdot m\cdot \sin(\omega t+120^{\rm o})\end{array}\right. \tag{12}\]  
Transforming the above equation further provides the following:  
\[\left\{\begin{array}{l}u_{\rm an}=\frac{U_{\rm dc}}{2}m\sin(\omega t)+\frac {m}{2}U_{\rm cm}\sin(2\omega t+\varphi)-\frac{m}{2}U_{\rm cm}\sin\varphi\\ u_{\rm bn}=\frac{U_{\rm dc}}{2}m\sin(\omega t-120^{\rm o})+\frac{m}{2}U_{\rm cm }\sin(2\omega t+\varphi-120^{\rm o})-\frac{m}{2}U_{\rm cm}\sin(\varphi+120^{ \rm o})\\ u_{\rm cn}=\frac{U_{\rm dc}}{2}m\sin(\omega t+120^{\rm o})+\frac{m}{2}U_{\rm cm }\sin(2\omega t+\varphi+120^{\rm o})-\frac{m}{2}U_{\rm cm}\sin(\varphi-120^{ \rm o})\end{array}\right. \tag{13}\]  
In Equation (13), the first term is the positive sequence's fundamental component, the second term is the second harmonic, and the third term is the DC component. In a three-phase, four-wire inverter, unbalanced currents caused by unbalanced loads flow through the center line, resulting in fluctuations in the capacitance voltage on the DC-side. The DC-side capacitor experiences voltage fluctuations, which, in turn, cause second harmonics and DC components in the load voltage, so the DC-side capacitor design directly affects the system's performance and phase imbalance. Usually, the larger the capacitance, the smaller the voltage fluctuation. In order to effectively suppress voltage fluctuations, the capacitance value can be sized according to the requirements of the voltage fluctuations of the capacitor on the DC-side. And the magnitude of the DC-side's capacitance voltage fluctuation is related to the unbalanced current.