06/08/2026
How to Select the Right DC Cable Size in PV Systems
One of the most common mistakes in PV design is selecting the cable based only on its current rating. In reality, the cable cross-sectional area depends on several design factors. An incorrect cable size can increase power losses, raise cable temperature, reduce system efficiency, and affect the long-term reliability of the entire system.
Factors Affecting DC Cable Selection
When designing DC cables in PV systems, engineers should consider:
- Current Carrying Capacity.
-Allowable Voltage Drop.
-Cable Length.
-Ambient Temperature.
-Installation Method (Conduit, Cable Tray, or Free Air).
-Grouping Factor.
-Conductor Material (Copper or Aluminum).
-Applicable Standards and Codes (IEC).
String Cable vs. Array Cable
* String Cable
The cable connecting each PV string to the Combiner Box, or directly to the inverter if no Combiner Box is used. It carries the current of one string only.
* Array Cable
The cable connecting the Combiner Box to the inverter. It carries the combined current of all parallel strings, which is why it usually requires a larger cross-sectional area.
Calculating the Allowable Voltage Drop
Let's assume the following PV system:
• Number of Strings = 4
• Modules per String = 15
• Operating Voltage = 615 VDC
• String Current = 15 A
• Maximum Allowable Voltage Drop = 1%
Therefore:
ΔV = 615 × 1% = 6.15 V
Calculating the String Cable Cross-Sectional Area
The following equation is commonly used for DC cable sizing:
A = (2 × L × I × ρ) ÷ ΔV
Where:
A = Cable Cross-Sectional Area (mm²)
L = One-way Cable Length (m)
I = Current (A)
ρ = Copper Resistivity = 0.0175 Ω·mm²/m
ΔV = Allowable Voltage Drop (V)
Assume the String Cable length is 45 m.
Substituting the values:
A = (2 × 45 × 15 × 0.0175) ÷ 6.15
A ≈ 3.84 mm²
The next standard cable size is:
4 mm² Copper
Calculating the Array Cable Cross-Sectional Area
Given:
• Number of Strings = 4
• Current per String = 15 A
Therefore:
Array Current = 4 × 15 = 60 A
Assume the Array Cable length is 30 m.
Using the same equation:
A = (2 × 30 × 60 × 0.0175) ÷ 6.15
A ≈ 10.24 mm²
After checking the current carrying capacity and applying the required correction factors, the selected cable size is:
16 mm² Copper
Important Note
The calculations above provide the minimum required cable cross-sectional area. Before finalizing the cable selection, it is essential to verify the current carrying capacity, apply the required correction factors for ambient temperature, installation method, and cable grouping, and ensure compliance with the applicable design code.
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