6 Electrical Connection SMA America, LLC
42 SI4548_6048-US-TB_en-13 Technical description
Example for Cable Dimensioning
With a 48 V battery voltage and an outgoing AC power of 4,500 W, a current of up to 100 A flows
through the SI 4548-US-10 battery cable. At the same battery voltage and an outgoing AC power of
6,000 W, a current of up to 130 A flows through the battery cable of the SI 6048-US-10.
The current flowing through the battery line causes a power loss and a voltage drop with every meter
of plain battery cable. You can use the following table to find the power loss and voltage drop
associated with different cable cross-sections.
Example:
For a 33 ft. (10 m) distance between the Sunny Island and the battery, at least 66 ft. (20 m) of line
is needed (distance there and back). Using a cross-section of AWG 2/0 (70 mm²), 100 A (nominal
current of the battery) causes a power loss of 120 W in total and an effective voltage drop of 0.9 V.
Calculation of the averaged nominal current of the battery
You can calculate the averaged nominal current of the connected battery using the following formula:
I
Bat
= Nominal current of the battery
P
AC
= AC power of the inverter
U
Bat
= Nominal voltage of the battery
η
INV
= Inverter efficiency at a given AC power
Cable cross-section Power loss Voltage drop
AWG 2/0 (70 mm²) 1.8 W/ft. (6 W/m) 14 mV/ft. (45 mV/m)
AWG 3/0 (95 mm²) 1.4 W/ft. (4.7 W/m) 11 mV/ft (35 mV/m)
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