DIY Manual
38 | P a g e 10-543-1 REV A
DIY Manual
Factors which determine the appropriate wire gauge:
1. The anticipated current draw of a load or circuit;
2. The temperature to which the wire is subjected, influenced by
a. Ambient temperature,
b. If conductor conduit is exposed to direct sunlight,
c. Type of wire insulator, and
d. Equipment termination temperature rating;
3. Wire rating in terms of voltage and temperature;
4. Derating of wire based on conduit fill (Number of wires in the conduit); and
5. Ambient temperature correction factors.
NOTE: Refer to the NEC for more information on determining the correct wire gauge for your
specific application. Special rules apply such as 4 or more conductors in the same conduit or
conduit mounted on a rooftop as opposed to conduit mounted on the side of a building, away
from direct sunlight. Refer to NEC Table 310.15(B)(16).
Examples
• Disregarding (for simplicity) the possible correction factors mentioned above, determine
the correct wire gauge needed to connect a 4000W inverter to a 24V battery bank,
located 8 feet apart.
1. 4000W / 24V = 166A,
2. Reference Figure 6, and
3. #2/0 AWG wire is required (180A at 8’).
• Determine the correct wire gauge for a 4000W inverter with input power range of 22-24V
with an efficiency rating of 85%, 8 feet apart.
1. 4000W / 22V / 0.85 = 214A,
2. Reference Figure 6, and
3. #2/0 AWG wire is still correct (200A at 8’).
NOTE: In the two examples above, if we apply correction factors for conduit fill and
temperature, the values of 166A from Example #1 and 214A from Example #2 will be
reduced by the correction factors. This means either less current can safely pass through
the selected wire OR you as the installer need to select a larger size (smaller gauge #) wire.
• Determine the correct wire gauge to connect a PV array to the charge controller, based
on PV short-circuit current (Isc) of 9.2A and a distance from array to charge controller of
45 feet.
1. Maximum PV circuit current: 9.2A x 125%** = 11.5A,
2. Conductor ampacity: 11.5A x 125% = 14.4A,