# # Wires and Fuses

## # Wires

For any current carrying conductor, resistance ($R$) of the conductor is given by

$R = \frac{\rho \cdot L}{A}$

Hence the efficiency of current carrying conductor depends mainly on length ($L$), cross-sectional area ($A$) and specific resistance ($\rho$) of the material. In addition, parameters like ambient temperature and installation conditions play an important role in maximum current carrying capacity (MCCC) of the wire. MCCC ensures that neither the metal (wire) nor the insulation around it reaches its melting point.

### # Cross section area

For higher cross-sectional areas of the conductor, more area will be there for the electrons to flow and hence lesser resistance offered to the current flow. So its current carrying capacity and efficiency will be more. Also, less heat will be dissipated in case of higher cross-sectional area. On the downside, more material means higher costs.

### # Length

The higher the length of the conductor, the greater is the effective resistance offered by the path and hence more power will be dissipated. This limits the MCCC of the conductor and reduces its efficiency as well.

### # Specific resistance

Specific resistance is the property of the material used for the wire. Different materials like copper, aluminium, silver and others exhibit different value of specific resistance. Materials like silver with low specific resistance are more conductive in nature, but are more expensive as well. Hence wire of suitable material should be used according to the requirement. The following table gives the value of specific resistance for a few materials.

Material Specific Resistance ( Ω mm²/m)
Silver 0.0159
Copper 0.0168
Copper, annealed 0.0172
Iron 0.0971

### # Ambient temperature

Higher ambient temperatures require less heat (generated by current flowing through wire) for the insulation to melt. Hence ambient temperature has to be kept low to achieve higher current carrying capacity of the wire. The deviation of current carrying capacity (from 30°C) depending on ambient temperature is given in the below table, which is according to the VDE 0298 norm.

Ambient Temperature (°C) Deviating Factor
10 1.22
20 1.12
30 1.00
35 0.94
40 0.87
45 0.79
55 0.61
65 0.35

### # Installation conditions

Enclosing the wires in duct, trays or conduit will decrease the current carrying capacity of the wire, since heat builds up in such enclosed system. This issue can be relaxed by providing proper ventilation or cooling system. In contrast, installing wires in open air allows for higher current carrying capacity of the wire. When installing in open air, UV and moisture resistance has to be considered for the insulation material.

The interactive graph given below, helps us to understand how parameters like specific resistance ($\rho$), length ($L$) and cross-sectional area of the conductor ($A$) affects efficiency and voltage drop across it.

The voltage drop across the wire has to be considered when the charge controller is configured to maintain the proper voltage on the battery to charge/discharge.

### # Example

The distance between the panels and the charge-controller shall be $D=5m$, which means a total length (two directions) of $l_{wire} = 10m$. For a copper wire with a diameter of $d=1.5mm$, the total resistance of the wire (for an ambient temperature of 30°C is

$R=\frac{0.0172 \cdot 10}{1.5} \Omega = 0.114 \Omega$

If we assume a maximum current of $I_{max} = 10A$ like charge controller MPPT 1210 HUS (opens new window) can handle, the maximum power dissipation in the wire is

$P_{wire} = I_{max}^2 * R = 100A^2 * 0.114 \Omega = 11.4W$

In most cases, this approach is rather used to determine the minimal cross section for the wires. If a power loss of 2% of the peak power is tolerated and the maximum expected temperature in the wire is assumed to be 50°C, the cross section can be determined using the above formulas. Let the peak power be $P_{peak}=120W$. The specific resistance at 50°C is $\rho_{50} = 0.0185\frac{\Omega mm^2}{m}$.

$P_{wire} = 2\%\cdot P_{peak} = 2.4W$
$R_{wire} = \frac{P_{wire}}{I_{max}^2} = \frac{2.4W}{100A} = 0.024\Omega$
$A_{wire} = \frac{\rho \cdot l_{wire}}{R_{wire}} = 7.7mm^2$

So for this example, a wire with a cross section of $A=7.7mm^2$ should suffice.

For short connections below 5 meters, you can use:

Rule of Thumb:

current / 3 = cable cross section area

Table 2 lists common cross section for some situations. Note that the length is given as the total of both positive and negative lines.

length=5m length=10m
Cross section in [mm²] max. current [A] max. current [A]
0.75 2.3 1.1
1.5 4.5 2.3
10 30 15
35 105 53
50 150 75
120 360 180

## # Fuses

A Fuse is an important component in electrical systems, used to prevent excessive flow of current which can lead to damage of electrical components or even accidents like fire. Proper care must be taken in deciding the placement of the fuse to prevent any damage. Generally it is always advisable to place the fuse as close as possible to the energy source. If the fuse is placed away from the energy source as shown in Fig.2a, the short-circuit which occurs at point X would go undetected, resulting in excessive current. Whereas in the case of Fig.2b, this problem can be avoided.

### # Branch

In case of branches where multiple lines are drawn from the main wire as shown in the Fig.3, a fuse has to be placed on each individual branch lines. Failure in doing so might result in excessive current in individual lines of the branch beyond their limit even when total current in the main line is within the limit. i.e, Fuse $F$ might have higher current rating than fuse $F_1$. In this case, even though when fuse $F$ doesn't trip and current is within the safe limit of the main line, fuse $F_1$ might trip due to excessive current in branch 1 beyond the limit of branch 1 line. Absence of $F_1$ in this case leads to damage of the systems.