# DC Cable Sizing in Solar Power Projects

The DC cable selection is an important factor in designing of solar power projects.

Typically, a solar power project has about 5-10 km of DC cable and contributes to about 2-3% of overall plant losses. The DC cable is used at multiple levels in case of large ground mounted solar projects. Typically, 4 mm^{2} DC cable is used in string formation and if multiple strings are taken together up to string combiner box from array, a 6 mm^{2} is used as a standard design practice.

It is important to understand the cable sizing calculations for evaluating various losses in the DC cables. The DC cable sizing is done as per **BS EN 50618 code.**

In this article we highlight the DC cable sizing calculations from string to array junction box (AJB).

Before starting the design process following are the major steps to be considered for the DC cable sizing calculations.

__Step – I__

Estimate the current to be carried in the DC cable. The total current carried by a DC cable in string is dependent on overall string size.

Typically, in a 1500 V DC system about 30 modules are connected in series and considering a 540 W_{p} Mono Perc module the typical string current is about 13.51 Amps as per module datasheet.

Here we have considered 540 W_{p} module.

Parameters | Values |

Maximum Power (P_{MAX}) | 540 W_{p} |

Voltage at maximum power (V_{MP}) | 42.81 V |

Voltage at maximum power (I_{MP}) | 12.63 A |

Short circuit current (I_{SC}) | 13.51 A |

Open circuit voltage (V_{OC}) | 49.30 V |

- Calculate maximum circuit current I
_{MAX }with formula:

I_{MAX} = I_{SC} × 1.25 (safety factor)

I_{MAX} = 13.51 × 1.25 (With respect to our consideration)

I_{MAX} = 16.89 Amps

- Calculate minimum required conductor ampacitywith formula:

= I_{MAX} × 1.25 (safety factor)

= 16.89 × 1.25 (With respect to our consideration)

= 21.11 Amps

Note: – 1.25 is the safety factor.

**Apply Correction Factor (De-rating Factor) for degradation as well as ampacity for 4 mm ^{2}.**

- Calculate total Deration Factor (DF)with the below formula:

Total DF = For 50^{o }Cambient temperature with allowable ampacity × DF for more than 3 current carrying conductors

Total DF = .82 × .7 (With respect to our consideration)

Total DF = .57

- For 4 mm
^{2}cable ampacity @ 60^{o}C is 44 Amps:

= Ampacity × Total Derating Factor

= 44 Amps × .57 (With respect to our consideration)

= 25.08 Amps

**Apply Correction Factor (De-rating Factor) for degradation as well as ampacity for 6 mm ^{2}.**

- Calculate total Deration Factor (DF) with the below formula:

Total DF = For 50^{o }Cambient temperature with allowable ampacity × DF for more than 3 current carrying conductors

Total DF = .82 × .7 (With respect to our consideration)

Total DF = .57

- For 6 mm
^{2}cable ampacity @ 60^{o}C is 57 Amps:

= Ampacity × Total Derating Factor

= 57 Amps × .57 (With respect to our consideration)

= 32.49 Amps

__Step – II__

Consider the voltage drop allowed in cable sizing while deciding the cable selection the designer should know the acceptable voltage drop in the DC cable string length.

Typically, a voltage drop of 2-3% is considered while selecting the DC cable. The voltage drop in DC cable selection is provided by the client as per their expected power loss parameter.

**Here in our design calculations a voltage drop of 2% has been considered.**

__Step – III__

Estimate the maximum cable length in your plant design.

Typically, a 4 mm^{2} DC cable which is used for string formation has to run along the length of the string and is about 30 modules connected in series and some additional cable bending/looping, a reasonable estimate for **DC cable length can be considered as 50 meters**.

Generally, two strings are connected together to a Y connecter and the cable selection beyond Y connecter carries a current of two strings up to array junction box (AJB).

In this case we will do cable sizing of DC 4 mm^{2} cable up to Y connecter and Y connector to array junction box (AJB) with 6 mm^{2}. A reasonable estimate for the cable length is done through cable scheduling.

However, in this case to understand the calculations** we take 100 meters from Y connector to array junction box (AJB).**

__Step – IV__

Calculate the required cable size based on the string currents and voltage levels. The DC cable resistance is estimated through following formula

Here, we calculate the resistance for 4 mm^{2} as well as 6 mm^{2} cables for the estimated lengths as specified in the previous sections.

**Resistance calculations for 4 mm ^{2 }(String to “Y Connector”):**

Step – I

Resistance/km for 4 mm^{2} at 20^{o} C is:

= 5.09 Ω

Step – II

Resistance for 50 meters at 20^{o} C:

R_{@20}^{o}_{ C} = 5.09 Ω × 0.05 km

R_{@20}^{o}_{ C} = 0.25 Ω

Step – III

Now let’s calculate resistance at 60^{o} Celsius for 50 meters:

R_{@60}^{o}_{ C} = R_{@20}^{o}_{ C }× [1 + α (T_{cable} – 20)] (α = 0.003)

R_{@60}^{o}_{ C} = 0.25× [1 + 0.003 (60 – 20)]

R_{@60}^{o}_{ C} = 0.25× [1 + 0.12]

R_{@60}^{o}_{ C} = 0.25× [1.12]

R_{@60}^{o}_{ C} = 0.28 Ω

Note: –

α = Temperature co-efficient for copper.

**Resistance calculations for 6 mm ^{2 }(“Y Connector” to AJB):**

Step – I

Resistance/km for 6 mm^{2} at 20^{o} C is:

= 3.39 Ω

Step – II

Resistance for 100 meters at 20^{o} C:

R_{@20}^{o}_{ C} = 3.39 Ω × 0.1 km

R_{@20}^{o}_{ C} = 0.34 Ω

Step – III

Now let’s calculate resistance at 60^{o} Celsius for 100 meters:

R_{@60}^{o}_{ C} = R_{@20}^{o}_{ C }× [1 + α (T_{cable} – 20)] (α = 0.003)

R_{@60}^{o}_{ C} = 0.34× [1 + 0.003 (60 – 20)]

R_{@60}^{o}_{ C} = 0.34× [1 + 0.12]

R_{@60}^{o}_{ C} = 0.34× [1.12]

R_{@60}^{o}_{ C} = 0.38 Ω

Note: –

α = Temperature co-efficient for copper.

__Step – V__

**The voltage drop calculations for 4 mm ^{2} (String to “Y Connector”) are as follows: –**

- Calculate I
_{MP}of string

= I_{MP} of SPV module is 12.63 Amps

= I_{MP} of one string is 12.63 Amps (string is connected in series)

- Calculate voltage drop by formula:

Voltage drop = R_{@60}^{o}_{ C}_{ }× I_{MP}

Voltage drop = 0.28 Ω_{ }× 12.63 Amps

**Voltage drop = ****3.54 V **

- Finally, calculating % voltage drop by the below formula:

% Voltage drop = Voltage drop × 100/V_{MP} of module × no. of modules in a string

% voltage drop = 3.54 × 100/42.81 × 30

% voltage drop = 354/1284.3

**% Voltage drop = 0.27 %**

**The voltage drop calculations for 6 mm ^{2} (“Y Connector” to AJB) are as follows: –**

- Calculate I
_{MP}of string

= I_{MP} of SPV module is 12.63 Amps

= I_{MP} of two strings is 25.26 Amps (string is connected in parallel to a “Y connecter”)

- Calculate voltage drop by the below formula:

Voltage drop = R_{@60}^{o}_{ C}_{ }× I_{MP}

Voltage drop = 0.38 Ω_{ }× 25.26 Amps

**Voltage drop = ****9.59 V **

- Finally, calculating % voltage drop by the below formula:

% Voltage drop = Voltage drop × 100/V_{MP} of module × no. of modules in a string

% voltage drop = 9.59 × 100/42.81 × 30

% voltage drop = 959/1284.3

**% Voltage drop = 0.74 %**

**Graphical Representation**