How to Section AC Cable for Solar PV systems

Background

In PV systems, we need to consider three types of cables: PV cables, AC cables, and grounding cables.

PV cables are usually laid outdoors and need to be protected from moisture, direct sunlight, cold temperatures, and ultraviolet. It is essential to choose PV-certified cabling, which cannot be replaced by conventional cabling(PV-Certified Cable’s UV protection, insulation protection, and DC voltage resistance (usually 600VDC) are superior to conventional cables). The more commonly used cabling is PV1-F*4mm2.

Grounding cables are mainly used for system lightning strike protection grounding. We just need to make sure that the grounding cable we used to make the system’s ground resistor requirements.

AC cables are used to connect the AC output of the inverter to the grid. They are usually installed outdoors, so they also need the same protective characteristics as the DC cables. Due to the different output currents of the inverter, the selection of AC cabling becomes more complicated. At present, the main basis for the selection of AC cabling is the relationship between cable diameter and ampacity, but the influence of ambient temperature, voltage loss, and laying method on the current-carrying capacity of the cable is generally ignored. In this Solis Seminar, we will discuss how to properly choose the right AC cabling in the PV system.

AC cable selection

The cable selection for a solar PV system needs to consider the following:

1. Voltage Loss

The voltage loss in a solar PV system can be expressed as:

Voltage loss = passing current * cable length * voltage factor

Voltage loss is proportional to the length of the cable.

When designing and installing the system, we should follow the principle that the distance of the PV module array to the inverter and the inverter to the grid connection point should be as close as possible.

We need to ensure that the DC voltage loss between the PV array and the inverter is less than 3% of the output voltage of the array, and the AC voltage loss between the inverter and the grid connection point does not exceed 2% of the output voltage of the inverter.

The calculation formula：△U=（I*L*2）/(r*S)

Note: △U ：Cable voltage drop -V

I ：The maximum current that the cable needs to withstand -A

L ：Cable laying length -m

S ：Cross-sectional area of the cable -mm²

r ：Conductivity of conductor -m/（Ω*mm²），r of copper=57，r of aluminum=34

2. Carrying Capacity Calculation

When we calculate the current carrying capacity of the cable, in addition to referring to the parameters in the current carrying table, we also need to consider the wire type, installation method, ambient temperature, and obtain the actual current value through these correction factors.

Table-1 Cable size and normal current rating

The current carrying capacity of the cable will change depending on the ambient temperature. The datasheet of each manufacturers cable will have a corresponding temperature correction factor table so that the correct selection can be made.

Table-2 correction factors for ambient temperature

3. Parallel Laying Problem of Multiple Multi-Core Cables

In an actual installation scenario, the AC cables of the PV system may be laid in parallel with multiple multi-core cables. For example, in a small-capacity three-phase system, the AC cable adopts “1 four-core cable” or “1 five-core cable” cable. A single-phase system will use “1 two-core cable” or “1 three-core cable” cable; In a large-capacity three-phase system, multiple cables in parallel are used for AC wiring instead of single-core large-diameter cables. In this case, the current carrying capacity of the actual cable will be attenuated. We need to consider this attenuation at the beginning of the project design, as shown in Table 2.

Table-3 the current load correction factor of multiple parallel or multi-core cables

	The current load correction factor of multiple parallel or multi-core cables
1 core	2 core	3 core	4 core	5 core	6 core	7 core	8 core	9 core	12 core	16 core	20 core
1.00	0.85	0.79	0.75	0.73	0.72	0.72	0.71	0.70	0.70	0.70	0.70

Example System

We use an example of a residential project installed with S5-GR1P6K single phase inverter to calculate the AC cable. The AC cable on site is 30 meters away from the grid connection point. We use AC cables with PVC protective shells.

For full inverter data, please refer to the S5-GR1P6K datasheet. This shows::

•Rated output current = 26.0A

•Maximum output current = 27.3A

Cable type: 1 two-core AC cable with PVC protection;

• Cable section: The maximum AC output current of S5-GR1P6K is 27.3A, and the 4mm² cable normal current rating is 39A (in the air) obtained from table-1.
• In the case of an ambient temperature of 45°C, the temperature correction factor is 0.79;
• The single-phase inverter uses 1 two-core AC cable, and the correction factor is 0.85;

Actual current-carrying capacity calculation (coefficient correction)：

39A*0.79*0.85≈26.2A ＜ 27.3A

Voltage loss: △U=（I*L*2）/(r*S)=（27.3*30*2）/（57*4）≈7.18V；

The grid voltage is 230V, so the voltage loss is greater than 230V*2%=4.6V.

Example conclusion:

Since the maximum current carrying capacity for fault-free operation is lower than the maximum output current of the inverter used, the selected AC cable cannot be used in this example.

Example solution：

Use 6mm² cable

The 6mm² cable normal current rating is 50A (in the air) obtained from table-1.

Actual current-carrying capacity calculation (coefficient correction)：

50A*0.79*0.85= 33.575A ＞ 27.3A

Voltage loss:△U=（I*L*2）/(r*S)=（27.3*30*2）/（57*6）≈4.78V；The grid voltage is 230V, So the voltage loss is close to 230*2%=4.6V。

Therefore, 6mm² cable is the best choice.