Product Manual

Table Of Contents
For a cable of 6 mm² and installation method B2, the maximum current is 30 A x 0.8 =
24 A, which is insufficient. For a cable of 10 mm², the maximum allowed current is 50 A
x 0.8 = 40 A. So 2 parallel cables of 3 x 10 mm² +10 mm² are sufficient.
In case of installation method C, the maximum current of a 6 mm² cable at 40 °C is 36
A x 0.8 = 28.8 A. 2 parallel cables of 3 x 6 + 6 mm² will be sufficient.
Install 32 A fuses on each cable.
Cable sizing according UL/cUL
Calculation method according UL 508A, table 28.1 column 5: allowable ampacities of insulated
copper conductors (75 °C (167 °F)).
Maximum allowed current in function of the wire size (NFPA70 table 310.16)
AWG or kcmil Maximum current
14 20 A
12 25 A
10 35A
8 50 A
6 65 A
4 85 A
3 100 A
2 115 A
1 130 A
Correction factors
Ambient temperature Correction factor
21-25 °C (70-77 °F) 1.05
26-30 °C (78-86 °F) 1.00
31-35 °C (87-95 °F) 0.94
36-40 °C (96-104 °F) 0.88
Calculation method for UL:
Single supply cables (3 phases + 1 PE - configuration (1)):
Add 25 % to the total current (I
tot
from the tables) (see UL 508A 28.3.2: "Ampacity shall
have 125 % of the full load current").
Install the prescribed maximum fuse on each cable
Parallel supply cable (2 x 3 phases + 2 PE - configuration (2)):
Add 25 % to the total current (I
tot
from the tables) and divide by 2
Multiply the ampacity of the cables with 0.8 (see UL 508A table 28.1 continued)
Install fuses of half the size of the recommended maximum fuse size on each cable.
When using 2 x 3 phase + 2 PE as in (3):
Add 25 % to the total current (I
tot
from the tables) and divide by √3
Multiply the ampacity of the cables with 0.8 (see UL 508A table 28.1 continued)
Fuse size: the recommended maximum fuse size divided by √3 on each cable.
Size PE cable:
For supply cables up to AWG8: same size as the supply cables
Instruction book
2920 7140 52 101