Short circuit and overload protection

When designing a photovoltaic system, it is necessary to provide disconnection and protection devices for all installed products, even if some already have protections inside, such as the inverter and charge controller.

The International Electrotechnical Commission (IEC) acknowledges that the protection of photovoltaic systems is different from standard electrical installations. This is highlighted in the IEC 60269-6 document, which defines the specific characteristics which must have available in the fuse to protect photovoltaic systems.

Figure 1

In general, to select fuses, for example, for string protection, even if all parameters should be considered and studied, the following criteria can be used:

• Multiply the current by 1.56
• Multiply the voltage by 1.2

The values obtained cover most of the variations in system current and voltage.

Usually, for all photovoltaic systems that have three or more strings in parallel, it is recommended for each string to protect cables and modules from faults due to overcurrent. Suppose the fuse is reduced (less ampere) according to the maximum current which can generate the module. In that case, the current itself cannot damage the modules in case of failure or short circuits (including the overload). We suggest putting them equally on both wires (positive and negative). For safety, in addition to the fuse, a disconnector is also useful that allows you to work peacefully downstream of the panels in the event of a fault.

Fuse Specifications 

If the strings in parallel are less than or equal to three, it could be enough for the cable size that can withstand a current at least equal to: 1.56 x Isc (Short circuit current), but as previously said, for our projects we apply the fuse for each string, which must have the following requirements:

• At short-circuit current Isc * 1.56
• The voltage Voc * 1.20 * Ns (Ns = a number of modules in series per string).

The same requirements apply if the strings are greater than three.

In our project the Isc = 6.15 A, Voc = 49.40 Volts. Therefore the fuse must have the following characteristics:

• Nominal current: 1.56 * 6.15 = 10 Amps (approximately)
• Nominal voltage: 1.20 * 49.40 * 2 = 120 Volts (approximately)

Fuses can be plugged into MP4 connectors, as the Figure 2 below

Figure 2

or in the fuse holder as the Figure 3 below

Figure 3

The disconnector divides the electrical system's upstream and downstream parts without the risk of unintentional re-energization and allows maintenance without risk to the installer. Consider also further protection of the equipment from overvoltages or short circuits. In our case, it is 50 A.

Figure 4

Photovoltaic generator protections

As previously said, if the strings in parallel are less than or equal to three, it could be enough that the cable is sized appropriately. For safety, in our project, we would put a fuse in each string on both wires (positive and negative), which in this case they are 10 A each (see Figure 5 below)

 Figure 5

• The specific fuse 1 for DC will be inserted between the charge controller and the battery package. The amperage (A) must not exceed the range of the regulator.

In our case, the current produced by photovoltaics is calculated using Ohm (Ω) law: I = P / V = 1000 / 24 = 41A, which we will get to 50A (see Figure 6 below)

Figure 6

• Fuse 2 should be installed between the charge controller output (light bulb symbol) and 12, 24, or 48 Volts users. The amperage (A) must be calculated based on the maximum absorption in Watts of the connected users, which must not exceed the maximum capacity of the regulator using the formula: I = P / V = A (see Figure 7)

Figure 7

• Protection 3 can be a fuse or a magneto-thermic switch (specific for direct current) installed between the battery and inverter input. The flow rate (A) must be calculated based on the power Watts of the inverter and the supply voltage (Volt) of the battery package, according to Ohm law: P = V x I, which is obtained from the circulating current I = P / V = Ampere (example: if we use an Inverter with a power of 1000 Watt powered at 24 Volt, we will have that the circulating current will be: I = P / V = 1000/24 = 41 Ampere) (see Figure 8 below)

Figure 8

• A circuit breaker suitable for AC must be installed between the inverter output and the sockets of the system where the 230 Volt powered equipment (Computer, TV, Router, Washing Machine, etc.) The capacity in Ampere will be calculated based on the Power Watt of the Inverter and Output Voltage by using the formula: I = W / V (for example: if the Power of the Inverter is 1000 Watts and 230 Volts is the output voltage we will have a current of I = W / V = 1000/230 = 4.3 Ampere, which will be the capacity of the circuit breaker (see Figure 9 below)

Figure 9

Surge arresters for medium-sized photovoltaic generators

The solar panels of photovoltaic systems occupy a space proportional to the power that can be obtained. When the occupied area becomes significant, the systems are more subject to the effects of lightning strikes, especially indirect ones. Installing surge arresters for each polarity closest to the strings would be a good idea to avoid breakups. The choice of the voltage of the SPD (Surge Protection Device) on the DC side systems, which are isolated from the ground, can be calculated using the following formula:

VC (SPD) = VOC STC * K

In our case, the open-circuit voltage (VOC) is 49.40 Volts which multiplied by K (1.20) results in the VC (SPD) being equal to 60 Volts, and we could use the OVR PV arresters (for more details see Figure 10 below)

Figure 10

Note: in many photovoltaic systems, the SPD (Surge Protection Device) is not considered, or it is not calculated correctly to protect the system fully. In my opinion, the SPD is very important to protect the system fully, our projects are included more than one (for each string).

For any questions, you can contact our expert here.