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Inverter - fusing for continuous or peak power?

Hiya. Quick question, when sizing cable and fuse for an inverter, should I be sizing for continuous power or peak power?

I have the Victron 12v 250w inverter with continuous power of 200w and peak power of 400w, connecting to a 12v 405Ah battery array with a cable run of less than two meters both ways.

Here's my (possibly misguided?) calculations... assuming I'm looking a peak power then 400w divided by 12 volts equals 33.4 amps. Then divided by efficiency factor of 0.87 equals 38.3 amps? Added on a 25% safety margin to reach 47.9 amps. The maximum cable size for this inverter is 10mm so I'm figuring a 50 amp fuse should be sufficient to protect this size cable?

Was planning on using 35mm cable for the battery array to future proof any subsequent upgrades.

If anyone can confirm I'm on the right track or if not kindly put me on the right track it would be much appreciated. Also are there any definite guides for cable and fuse sizing out there please?

inverter current draw
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3 Answers
rickp avatar image
rickp answered ·

More expert minds will add to this, I’m sure, but I think of fusing as a protection of the cabling more than the devices connected in a properly sized system. You are on the right track about figuring the load, and should size cable to that. I use the nearest size fuse above that highest expected load, which puts me safely under the cable rating, while accounting for the anticipated start/run loads of the circuit.

Best wishes! Rick

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myhomebuzz avatar image
myhomebuzz answered ·

Hi

For the section of the cables, this also depends on the outward and return length between the two elements to be connected. You can use the following formula:


Ω = copper 0.0179 (but be careful with aluminum: 0.028!)

S = ρ x A / 0.03 x V or

S = (Ω x (L x 2)) x A / 0.03 x V or

S = (Ω x (L x 2)) x (W / V) / 0.03 x V


Example with 2 meters of cable, outward and return, in 12V:

S = 0.0179 x (2 x 2) x (400/12) / 0.03 x 12 -> S = 6.62 mm2 -> 10 mm2 is the section to be installed because the previous standardized section (6 mm2) is exceeded.


For the battery, with 1 meter of cable, outward and return, in 12V:

S = 0.0179 x (1 x 2) x 405 / 0.03 x 12 -> S = 40.275 mm2 -> 50 mm2


For fuses, the formula is often used:

A x 1.25 or

(W / V) x 1.25


Example with 400W peak and 12V:

(400/12) x 1.25 = 41.667 -> 45A. But it also depends on the connected elements. This is for the W peak. If elements do not have a strong load at start-up, the 250W gives 30A rounding. Everything must be taken into account. But in all cases, the wiring in mm2 is sectioned according to the W peak.

Hope this helps you.

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Eduard avatar image
Eduard answered ·

Over the years I have collected some info on sizing.

One is: Fuse size is 1.52 x the amps you use(33.4) gives 50.77A.

Round off to 50Amp.

Althoug 35mm calbe looks oversized, this will be also my choice.

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