# SVPWM Method for Multilevel Indirect Matrix Converter with Eliminate Common Mode Voltage
## 2 Multilevel Indirect Matrix Converter and the Proposed SVPWM Method
### The SVPWM Method for Elimination the CMV
#### 2.2.2 Inverter Stage Control
The reference voltage vector \(V_{ref}\) can be obtained as:  
\[V_{ref}=d_{01-1}V_{01-1}+d_{10-1}V_{10-1}+d_{1-10}V_{1-10}+d_{0-11}V_{0-11} \tag{9}\]  
where \(d_{01-1}\), \(d_{01-1}\), \(d_{1-10}\) and \(d_{0-11}\) are duty cycles of four active vectors \(V_{01-1}\), \(V_{10-1}\), \(V_{1-10}\) and \(V_{0-11}\), respectively. Figure 6: Space vector diagram and the selected vectors for zero common voltage. \begin{table}
\begin{tabular}{c c} \hline \hline
**Inverter Stage’s Switching State** & **Common Mode Voltage** \\ \hline (0,0,0) & \(0\) \\ (\(-1\),\(-1\),\(-1\)) & \(-V_{\text{DC}}\)/2 \\ (1,1,1) & \(V_{\text{DC}}\)/2 \\ (1,0,0); (0,1,0); (0,0,1); (\(-1\),\(1\),\(1\)); (\(-1\),\(-1\)); (\(1\),\(-1\)) & \(V_{\text{DC}}\)/6 \\ (0,0,\(-1\)); (\(-1\),0,0); (\(-1\),\(0\)); (\(-1\),\(-1\),\(-1\)); (\(-1\),\(-1\)); (\(-1\),\(-1\),\(1\)) & \(-V_{\text{DC}}\)/6 \\ (1,0,\(-1\)); (0,1,\(-1\)); (\(-1\),\(1\)); (\(-1\),0,1); (\(0\),\(-1\),\(1\)); (\(-1\),\(0\)) & \(0\) \\ (1,1,0); (1,0,1); (0,1,1) & \(V_{\text{DC}}\)/3 \\ (0,\(-1\),\(-1\)); (\(-1\),\(-1\),\(0\),\(-1\)); (\(-1\),\(-1\),0) & \(-V_{\text{DC}}\)/3 \\ \hline \hline \end{tabular}
\end{table} 
Table 5: The CMV of the 3-level T-Type NPC inverter. When one phase leg is clamped to the neutral-point, its output current is injected to this point. Therefore, the average neutral-point current within sampling period is determined as:  
\[I_{O}=d_{01-1}i_{A}+d_{10-1}i_{B}+d_{1-10}i_{C}+d_{0-11}i_{A} \tag{10}\]  
The neutral-point current can be rewritten as:  
\[I_{O}=d_{01-1}i_{A}+d^{\prime}_{10-1}i_{B}+d^{\prime}_{1-10}i_{C}+d^{\prime \prime}_{10-1}i_{B}+d^{\prime\prime}_{1-10}i_{C}+d_{01-1}i_{A} \tag{11}\]  
where  
\[d_{10-1}=d^{\prime}_{10-1}+d^{\prime\prime}_{10-1} \tag{12}\]  
\[d_{1-10}=d^{\prime}_{1-10}+d^{\prime\prime}_{1-10} \tag{13}\]  
To find the duty cycles of active vectors for obtaining the zero-average value of neutral-point current, the following two equations are used:  
\[d_{01-1}=d^{\prime}_{10-1}=d^{\prime}_{1-10} \tag{14}\]  
\[d^{\prime\prime}_{10-1}=d^{\prime}_{1-10}=d_{01-1} \tag{15}\]  
Substituting the (12), (13) to (8), we have:  
\[V_{ref}=d_{01-1}V_{01-1}+d_{10-1}V_{10-1}+d_{1-10}V_{1-10}+d_{01-1}V_{01-1} \tag{16}\]  
From (10)-(14), the duty cycles of the active and zero vectors is calculated as following:  
\[d_{1-10}=d^{\prime}_{1-10}+d^{\prime\prime}_{1-10} \tag{17}\]  
\[d_{10-1}=d^{\prime}_{10-1}+d^{\prime\prime}_{10-1} \tag{18}\]  
\[I_{O}=d_{01-1}i_{A}+d_{10-1}i_{B}+d_{1-10}i_{C}+d_{0-11}i_{A} \tag{10}\]0  
\[I_{O}=d_{01-1}i_{A}+d_{10-1}i_{B}+d_{1-10}i_{C}+d_{0-11}i_{A} \tag{10}\]1  
\[I_{O}=d_{01-1}i_{A}+d_{10-1}i_{B}+d_{1-10}i_{C}+d_{0-11}i_{A} \tag{10}\]2  
To complete one sampling period, the zero vector \(V_{000}\) is inserted and its duty cycle is calculated as:  
\[I_{O}=d_{01-1}i_{A}+d_{10-1}i_{B}+d_{1-10}i_{C}+d_{0-11}i_{A} \tag{10}\]3  
where \(\beta_{v}\) is a phase angle of reference output voltage vector (\(-\pi/6\leq\beta_{v}\leq\pi/6\)) and \(d_{000}\) are duty cycle of zero vector \(V_{000}\).