wire size for 1575 inverter 15' from battery

bluemoonshine

I bought a cobra 1575 watt inverter for my boat, I'd like to put the inverter in the cabin about 15' from the battery. The manual reccomends a minimum size of #4 awg copper cable no longer than 4 feet between inverter and battery. that sounds very undersized? comparing it to this post I found,

(((If we assume a maximum power of 1500 Watts out of the inverter, at about 80% efficiency and a low point operating voltage of 10V, then (1500/.8/10 would give about 187.5 Amps. I would probably size it for about 200 Amps of current which would yeild a cable size of about 0 guage minimum)))

Using his math with my 1575/.8/10 = 196.875
so my first question is why would the manufacture reccomend #4awg 90 amp wire when it can draw upto 197 amps?

and what size wire do I ACTUALLY need to run the inverter about 15' from the battery? would 0 gauge work? probly not because its 15' right? so 1/0 awg?
Im having a hard time learning about this DC voltage drop

Julius793

Why are you derating the conductor?

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bluemoonshine

i dont know, I got that from reading this thread
http://www.fieldlines.com/board/inde...ic,131634.html

plummen

Call me crazy but when i divide 1575w by 120v I get around 13a
Id say the #4 wire should be if you use it the way its designed,i dont know about anybody else but a boat is no place i want a fire

jbfan

1575 divided by 12 gives you 131.25.
Inverters are sized by the input voltage, in this case 12 volts.

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Julius793

Read the post again maybe???

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Julius793

As already posted you need wire that can hold about 132a so depending on if its copper or aluminum in a conduit or not look at the NEC charts to find your answer

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plummen

ah,thats why i said call me crazy!

ddawg16

I know exactly what your talking about.....I was just up in Seattle helping a friend with a new 'boat' that has several inverters....EASA unit to convert shore power to DC to charge the batteries....two sets of batteries....2 generators....in additon to the one on the main motor.

For you non-boat guys....boats will typically be 12V, 120Vac and sometimes 24Vdc for some of the nav gear

So...to your original problem.

First off...I would try to get the inverter as close to the batteries as possible. DC drop in the wire is more significant than AC.

#4 wires has 0.2485 ohms of resistance per 1000'.

The one question you don't answer....one battery or several in parallel...I'm going to assume one battery....marine deep cycle....

If we assume that it holds 12v at 1500w....that is 125A. Typically with a lead acid battery that is fully charged....your not going to drop to 10V unless your getting close to the CCA. At 125A, I would expect the voltage to be at least 11V or better at the battery. Even at 11Vdc, your at 136A for a full 1500w load.

So...for the rest of the exercise, lets use 136A

#4 wire (0.2485 ohm/1000 ft) = 0.0002485 ohm/ft...at 136A, your looking at 0.033796 volts (give or take a mv) drop/foot. In other words, about .33 volts drop/10ft.

According to my wire chart....#4 is good for 135A in chassis wiring and 60A in power transmission. Chassis wireing would be a case where the wireing is within a chassis...say maybe 10' total....power transmission is where your going outside of a cabinet (boat) to some field device.

So....back to your original issue....if you stay within 4'...then #4 would do the job....it's right at the edge...but will work. Going 15'....you might be better off with #2 wire.

But to be honest....I think a bigger issue is not so much the voltage drop....but the line impeadance between the inverter and battery. Devices like inverters need a good low impeadance power source for proper operation. As the lead length increase, it creates problems with regulation.

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