# AC Fault Ride-Through Strategy of a Transformerless Soft Normally Open Point
## 4 Characteristic Analysis of Single-Phase Grounding Fault  
In order to further analyze the asymmetric fault characteristics of the system quantitatively, the phase C metallic grounding fault which happens on the feeder connected to MMC1 in Figure 1, is taken as an example. Assuming that the three-phase voltage is symmetrical before faults, the amplitude of the rated phase-to-ground voltage is \(U_{\text{m}}\), and the initial phase of phase A is \(\varphi_{1}\). After the grounding fault occurs, the voltage of the fault phase is zero, and the voltage of the non-fault phase rises to line-to-line voltage. \[\left\{ \begin{array}{l}u_{a1}=\sqrt{3}U_{\text{m}}\sin(\omega_{0}t+\varphi_{ 1}-\frac{\pi}{5})\\ u_{b1}=\sqrt{3}U_{\text{m}}\sin(\omega_{0}t+\varphi_{1}-\frac{\pi}{2})\\ u_{c1}=0\end{array} \right. \tag{10}\]  
Adopting the symmetrical component method, the positive-, negative-, and zero-sequence voltages are given as follows. \[\left\{ \begin{array}{l}u_{\bar{a1}}^{+}=U_{\text{m}}\sin(\omega_{0}t+\varphi _{1})\\ u_{\bar{a1}}^{0}=U_{\text{m}}\sin(\omega_{0}t+\varphi_{1}-\frac{\pi}{3})\\ \end{array} \right. \tag{11}\]  
It can be seen from Equation (11) that when phase C is in a grounding fault, the amplitude of the positive- and zero-sequence component is the same, which is the rated phase-to-ground voltage. Combining Equation (9), the following equation can be obtained. \[u_{0}\approx u^{0}=U_{\text{m}}\sin(\omega_{0}t+\varphi_{1}-\frac{\pi}{3}) \tag{12}\]  
Making the voltage modulation ratio \(k\) (\(k<1\)) as follows, the positive DC voltage \(U_{\text{dp}}\) and the negative DC voltage \(U_{\text{dn}}\) is  
\[k=\frac{U_{\text{m}}}{U_{\text{d}}/2} \tag{13}\]  
\[\left\{ \begin{array}{l}U_{\text{dp}}=\frac{U_{\text{d}}}{2}(1+k\sin(\omega_{ 0}t+\varphi_{1}-\frac{\pi}{3}))\\ U_{\text{dn}}=\frac{U_{\text{d}}}{2}(-1+k\sin(\omega_{0}t+\varphi_{1}-\frac{\pi} {3}))\end{array} \right. \tag{14}\]  
It can be seen from Equation (14) that the positive and negative DC voltages fluctuate in power frequency, and the amplitude of the fluctuation is equal to the AC phase voltage \(U_{\text{m}}\). Although the positive and negative poles have voltage fluctuation, the fluctuation is still in the range of the rated DC voltage, and there is no polarity reversal. The maximum and minimum value of the positive DC voltage is \(0.5(1+k)U_{\text{d}}\) and \(0.5(1-k)U_{\text{d}}\). The maximum and minimum value of the negative DC voltage is \(-0.5(1+k)U_{\text{d}}\) and \(-0.5(1-k)U_{\text{d}}\). It is obvious that the maximum voltage of the positive and negative pole is close to the rated pole-to-pole voltage, which has an impact on the safe operation of the system. For MMC2, due to the voltage fluctuation of \(u_{o}\) in Equation (12), it is known from Equations (8) and (9) that the zero-sequence voltage shown in Equation (12) will be introduced into the AC system of MMC2. Before the grounding fault in MMC1 happens, the  
Figure 4: Zero-sequence current path. AC system of MMC2 only contains positive-sequence components. (In order to simplify the analysis, it is assumed that the rated phase voltage amplitude of MMC2 is the same as that of MMC1.) \[\left\{ \begin{array}{l}u_{a2}=u_{a2}^{+}=U_{\text{m}}\sin(\omega_{0}t+\varphi_{ 2})\\ u_{b2}=u_{b2}^{+}=U_{\text{m}}\sin(\omega_{0}t+\varphi_{2}-\frac{2\pi}{3})\\ u_{c2}=u_{c2}^{+}=U_{\text{m}}\sin(\omega_{0}t+\varphi_{2}+\frac{2\pi}{3}) \end{array} \right. \tag{15}\]  
where \(\varphi_{2}\) is the initial phase. Using the inverse transform of the symmetric component method, the AC voltage of MMC2 can be calculated as follows, when the AC system of MMC1 is in fault. \[\left[ \begin{array}{c}u_{a2}\\ u_{b2}\\ u_{c2}\end{array} \right]=\left[ \begin{array}{ccc}1&1&1\\ 2^{a}&a&1\\ a&a^{2}&1\end{array} \right]\left[ \begin{array}{c}U_{m}\sin(\omega_{0}t+ \varphi_{2})\\ 0\\ U_{m}\sin(\omega_{0}t+\varphi_{1}-\frac{\pi}{3})\end{array} \right] \tag{16}\]  
where \(a=\text{e}^{\text{i}120}\).