## 4 Interleaved Inverter  
### Mathematical Model  
With reference to Figure 1, as already pointed out, the inverter is composed of two identical three-phase half bridges. Each of the two modules is a converter itself, and the superscripts \(p\) and \(n\) are used respectively for the upper and lower ones. For clarity, in Figure 1, only the first phase voltages and currents are indicated. For the generic \(k^{th}\) phase, the following voltage balance can be written as follows:  
\[\left\{ \begin{array}{l}v_{s,k}=L_{R}\frac{\mathrm{d}i_{s,k}^{p}}{\mathrm{d} t}+v_{r,k}^{p}+v_{NO}\\ v_{s,k}=L_{R}\frac{\mathrm{d}i_{s,k}^{p}}{\mathrm{d}t}+v_{r,k}^{n}+v_{NO} \end{array} \right.\quad\text{with}:\;k\in\left\{1,2,3\right\}, \tag{17}\]  
where \(v_{NO}\) is the voltage of the point, \(N\) referred to the grid neutral point \(O\). Referring to the same index \(k\), the semi-sum of the first and second of Equation (17) gives the following:  
\[v_{s,k}=\frac{L_{R}}{2}\frac{\mathrm{d}i_{s,k}}{\mathrm{d}t}+\frac{\left(v_{r,k}^{p}+v_{r,k}^{n}\right)}{2}+v_{NO}\quad\text{with}:\;k\in\left\{1,2,3 \right\}, \tag{18}\]  
The space vectors \(\mathbf{y}\) can now be introduced, together with the homopolar components \(y_{0}\), by means of the following transformation:  
\[y_{1},y_{2},y_{3}\Rightarrow\left\{ \begin{array}{l}\mathbf{y}=\frac{2}{3} \sum\limits_{k=1}^{3}y_{k}\mathrm{e}^{j\frac{2\pi}{3}\left(k-1\right)}\\ y_{0}=\frac{1}{3}\sum\limits_{k=1}^{3}y_{k}\end{array} \right., \tag{19}\]  
The Equation (18) can then be written as follows:  
\[\left\{ \begin{array}{l}\mathbf{v_{s}}=\frac{L_{R}}{2}\frac{\mathrm{d}i_{s}} {\mathrm{d}t}+\frac{\left(\Psi^{p}_{r}+v_{r}^{n}\right)}{2}\\ v_{NO}=-\frac{\left(v_{r,0}^{p}+v_{r,0}^{n}\right)}{2}\end{array} \right., \tag{20}\]  
The first of Equation (18) gives the grid current space vector behavior as a function of the two converters' space vector voltages, while the second establishes that the neutral point displacement voltage \(v_{NO}\) instantaneously coincides with the semi-sum of the two converter homopolar voltages \(v_{r,0}^{p},v_{r,0}^{n}\). Referring to the same index \(k\), the difference of the first and second of Equation (17) gives the following:  
\[0=L_{R}\frac{\mathrm{d}\left(\frac{\mathrm{d}\left(v_{s,k}^{p}-v_{s,k}^{n}\right) }{\mathrm{d}t}+\left(v_{r,k}^{p}-v_{r,k}^{n}\right)\right.\quad\text{with}:\;k \in\left\{1,2,3\right\}, \tag{21}\]  
The two converters' phase currents can be expressed as follows:  
\[\left\{ \begin{array}{l}i_{s,k}^{p}=i_{s,k}^{c}+\frac{i_{s,k}}{2}\\ i_{s,k}^{n}=-i_{s,k}^{c}+\frac{i_{s,k}}{2}\end{array} \right., \tag{22}\]  
with \(i_{s,k}^{c}\) as the common mode current on the \(k^{th}\) phase of the interleaved inverter, Equation (21) can be rewritten as follows:  
\[0=L_{R}\frac{\mathrm{d}i_{s,k}^{c}}{\mathrm{d}t}+\frac{\left(\frac{v_{r,k}^{p} -v_{r,k}^{n}}{2}\right)}{2}\quad\text{with}:\;k\in\left\{1,2,3\right\}, \tag{23}\]  
Applying the transformation of Equation (19):  
\[\left\{ \begin{array}{l}0=L_{R}\frac{\mathrm{d}i_{s,k}^{c}}{\mathrm{d}t}+ \frac{\left(\frac{\mathbf{v}_{r}^{p}-\mathbf{v}_{r}^{n}}{2}\right)}{2}\\ 0=L_{R}\frac{\mathrm{d}i_{s,0}^{c}}{\mathrm{d}t}+\frac{\left(v_{r,0}^{p}-v_{r,0}^{n}\right)}{2}\end{array} \right., \tag{24}\]  
Finally, the interleaved inverter mathematical model is given as follows:  
\[\left\{ \begin{array}{l}\mathbf{v}_{\mathbf{s}}=\frac{L_{R}}{2}\frac{\mathrm{ d}i_{s}}{\mathrm{d}t}+\mathbf{v}_{\mathbf{r}}\\ 0=L_{R}\frac{\mathrm{d}i_{s,0}^{c}}{\mathrm{d}t}+\Delta\mathbf{v}_{r,0}\end{array}  \right.\quad\text{with}:\;\left| \begin{array}{l}\mathbf{v}_{\mathbf{r}}= \frac{\left(\mathbf{v}_{r}^{p}+\mathbf{v}_{r}^{n}\right)}{2}\\ \Delta\mathbf{v}_{r}=\frac{\left(\mathbf{v}_{r}^{p}-\mathbf{v}_{r,0}^{n} \right)}{2}\end{array} \right., \tag{25}\]  
It is now clear that the inductors \(L_{R}\) are necessary to filter the harmonic content, which is introduced by the modulation of \(\mathbf{v}_{\mathbf{r}}\), limiting the oscillation on the grid currents.