# DC Cabling of Large-Scale Photovoltaic Power Plants
## 3 Array Cabling on the DC Side
### DC Cabling
#### 3.1.2 Configuration with One Type of Cable  
With one type of cable, strings are connected in the same way as with two cable types. The second type of cable is replaced by connecting the strings to the input of the block for a parallel connection of strings, SCB (String Combiner Box). This block also contains elements of overcurrent protection (fuses), blocking diodes, measuring and monitoring systems, etc. [30]. If the inverter is placed in an array (minimum length of DC cables), then the SCB is placed directly next to the inverter. The cable that connects the SCB and the inverter is then short (with required cross-section), and its losses can be neglected. The conductors of individual strings are connected through one blocking diode ((+) pole) and three connectors (the connection between the cable and the anode of the blocking diode, the connection between the cathode of the blocking diode, and the common point for (+) input to the inverter, the connection between the (-) pole of the string and the common point (-) input in the inverters) to the bus of the inverter input. Therefore, losses on connectors and blocking diodes can be calculated in the same way as with two types of cables. The total length of cables that replace the type 2 cable depends on the position of the inverter. It is shown herein that the minimum length of this part of the cabling is achieved when the inverter is placed in the middle of the cable tray, which connects the strings of one array.  
Let it be assumed that the number of strings is \(N_{\rm S}\) (\(S_{\rm tr1},S_{\rm tr2},S_{\rm tr3},\ldots,S_{\rm trN_{S}-1},S_{\rm trN_{S}}\)), and that the inverter is placed in the middle of the \(k\)-th string (Figure 6).  
Then, the sum of the lengths of the strings' distances from the inverter is given by:  
\[d_{\rm v} = 2D\sum_{i=1}^{k-1}i+2D\sum_{i=1}^{N_{\rm S}-k}i+2D\sum_{i=1}^{k- 1}i+Dk+2D\sum_{i=1}^{N_{\rm S}-k}i-D(N_{\rm S}-k) \tag{26}\] \[= 2\cdot 2D\frac{k(k-1)}{2}+2\cdot 2D\frac{(N_{\rm S}-k)(N_{\rm S }-k+1)}{2}+2 Dk+DN_{\rm S}\] \[= D\Big{(}4k^{2}-4 kN_{\rm S}-2k+2N_{\rm S}^{2}+N_{\rm S}\Big{)}.\]  
The first two terms are the sum of the lengths for the (+) pole of the cables, the third and fourth terms are the sum of the lengths for the (-) pole on the left side of the inverter, and the fifth and sixth terms are the sum of the lengths for the (-) pole on the right side  
Figure 6: Determining the cable length of parallel connections for one type of cable.  
of the inverter. The result is a quadratic function of \(k\). The minimum is obtained for \(4k-2N_{\text{S}}-1=0\), and then \(k\) can be expressed as follows:  
\[k=\frac{N_{\text{S}}}{2}+\frac{1}{4}. \tag{27}\]  
It is shown that the optimal location of the inverter is in the middle of the cable tray when a minimum cable length on the DC side is required. In Equation (27), \(1/4\) is a consequence of string widths and asymmetry of (\(+\)) and (\(-\)) cable lengths.  
Let it be assumed that \(k=N_{\text{S}}/2\), that is, the inverter is in the middle of the cable tray. By replacing \(k\) in Equation (26), the total length of the cable in the cable tray is:  
\[d_{\text{v}}=N_{\text{S}}^{2}\cdot D,\quad N_{\text{S}}>3. \tag{28}\]  
The cable length in one array is:  
\[d_{\text{P}}=d_{\text{S}}+d_{\text{v}}, \tag{29}\]  
where \(d_{\text{S}}\) is determined by Equation (13).