# DC Cabling of Large-Scale Photovoltaic Power Plants
## 3 Array Cabling on the DC Side
### DC Cabling
#### 3.1.1 Configuration with Two Types of Cables  
In this configuration, all solar modules are connected in one string with one type of cable (S\({}_{1}\)), and strings are connected in parallel with a different kind of cable (S\({}_{2}\)). The length of the cable S\({}_{1}\) in one string is determined as follows:  
\[d_{\text{S1}}=(N_{\text{M}}-2)\cdot d_{\text{M-M}}+d_{\text{S-S}}+K_{\text{T}} \cdot(L+d_{\text{P-P}}), \tag{12}\]  
where \(d_{\text{P-P}}\) is the distance between two arrays intended for cable laying. The length of the type one cable (S\({}_{1}\)) in one array is determined as follows:  
\[d_{\text{S}}=N_{\text{S}}\cdot d_{\text{S1}}+ \begin{cases}D,&N_{\text{S}}=1\\ 4\cdot D,&N_{\text{S}}=2\cdot\\ 8\cdot D,&N_{\text{S}}=3\end{cases} . \tag{13}\]  
For \(N_{\text{S}}=1,2\), and \(3\) there is no need for a second type of cable in the system. For \(N_{\text{S}}>3\), a type one cable covers a length of \(8D\) (one string on each side with width \(2D\) times \(2\) for both polarities). The length of the second cable type in one array is determined as follows:  
\[d_{\text{S2}}= \begin{cases}4\cdot D\cdot(N_{\text{S}}-3),&N_{\text{S}}>3\\ 0,&N_{\text{S}}=1,2,3\end{cases} . \tag{14}\]  
The length of the (+) pole of the cable along the cable tray is \(2 DN_{\text{S}}-D\). From this length must be subtracted one width of the string on both sides \((2\cdot 2D)\). The factor \(2\) is used to include the (\(-\)) pole of the cable. For a number of strings smaller than \(4\) (\(1,2\), and \(3\)), only one type of cable (S\({}_{1}\)) is used. Power is calculated as follows:  
\[P_{\text{C}}=R_{\text{C}}I^{2}, \tag{15}\]  
where \(R_{\text{C}}\) is the conductor resistance. The conductor resistance for one string (serial and parallel connection) is as follows:  
\[R_{\text{C1}} = d_{\text{S1}}\cdot R_{1}^{\prime}, \tag{16}\] \[R_{\text{C2}} = d_{\text{S2}}\cdot R_{2}^{\prime}, \tag{17}\]  
where \(R_{1}^{\prime}\) and \(R_{2}^{\prime}\) are the resistances of type \(1\) and type \(2\) conductors, respectively. The power losses in the conductors of one string are:  
\[P_{1}=R_{\text{C1}}\cdot I_{1}^{2}, \tag{18}\]  
where \(I_{1}\) is the string current. The power losses on the array are:  
\[P_{1\text{u}}=N_{\text{S}}\cdot P_{1}. \tag{19}\]  
To determine the power losses in a parallel connection, the scheme in Figure 5 will be used. Figure 4: DC cabling of a string by dividing it into two subsets. \(R^{\prime}\) is the resistance of the type 2 conductor between the two strings (\(\mathbb{S}_{2}\)). Since the distances between the strings is equal and since type two cable is used along the whole cable tray (for parallel connection of arrays), then the resistance is calculated as:  
\[R^{\prime}=\frac{R_{\text{C2}}}{N_{\text{S}}-2}. \tag{20}\]  
The inverter is placed in the middle of the cable tray (later, it is shown that this is the optimal position for power loss on DC cables). It is sufficient to calculate losses on one half of the array. Due to the symmetry of the problem, the total losses will be twice those calculated on half the array. According to the scheme from Figure 5, the following applies:  
\[P_{1/2} = R^{\prime}\left(I_{1}^{2}+2^{2}I_{1}^{2}+3^{2}I_{1}^{2}+\ldots+ \left(\frac{N_{\text{S}}}{2}-1\right)^{2}I_{1}^{2}\right) \tag{21}\] \[= R^{\prime}I_{1}^{2}\sum_{n=1}^{\frac{N_{\text{S}}}{2}-1}n^{2}=R ^{\prime}I_{1}^{2}\frac{N_{\text{S}}(N_{\text{S}}-1)(N_{\text{S}}-2)}{24}.\]  
Here, the losses are calculated on half of the array. Therefore the result must be multiplied by two, and then by two again, to include the (\(+\)) and (\(-\)) polarities. Further, \(R^{\prime}\) from Equation (20) is replaced, and the expression \(8D\cdot R_{1}^{\prime}\) is added (because the last strings are connected with the first type of cable).