CONTINUOUSWAVE

This is not for a boat, but is for 12-Volt DC wiring for an inverter on my tow vehicle, but the same principles apply as for boat wiring. I have a big inverter (1500-Watt steady state, 3000-Watt surge) that I want to wire in my Suburban. I'm already set-up with a second battery under the hood with AWG-1 wire running from that battery into the interior.

I would like to use a very short run of AWG-4 wire to connect from the inverter itself to a junction under the dash. The system is fused at 200-Amperes but would not be expected to pass that much current for any length of time. I don't know a specific current draw for sure, but it is rated at 88% efficient. If I figure 3,000-Watt at 12-Volts and 88% efficient, I get 284-Ampere maximum, while at the 1,500-Watat steady state that translates into 142-Ampere continuous. I think that the surge load could be reasonably analogized to the starting load on an outboard motor, where you don't size the system for continuous operation at that load, as evidenced by the only 200-Ampere fuse recommendation from the manufacturer. I have similar size fuses on my Optimax 150 engines.

The manufacturer recommends 10-feet or less of 2-AWG cable. I have about 8-feet of 1-AWG cable at the moment and would like to add a one-foot connector of 4-AWG cable due to a very tight bend area that I am working in. If I use this voltage calculator the manufacturer specification gives a voltage drop of 5-percent at 200 A.

When I look at the same web page, it recommends no less than 2-AWG gauge wire for a 200A load. However, I suspect that is because rarely would someone ever use as short a piece as one fooet. Marinco has a similar recommendation and for 200A recommends 2-AWG with 4-AWG gauge topping out at 160-Amperes.

If I am doing my math correctly, I think I could reasonably use a short 4-AWG connector, since my maximum continuous load would be 142-Ampere. Voltage drop per one foot of 4=AWG at 200-Amperes is just under one-percent.

--Am I doing my math correctly on figuring out the load in Amperes?

--Does it seem like a very short piece of 4-AWG would be reasonable in this system?

Thanks for your input and quality assurance. I know we have a lot of wiring whizzes on the page here.

If 12-Volt DC is converted to 120-VAC at 88-percent efficiency, and 1,500-Watts of AC current it provided, then the DC input must be 1500/0.88 = 1700-Watts. If the input voltage is 12.0-Volts, then the current needed for 1,700-Watts would be 1700/12 = 142-Ampere.

Copper wire of 4-AWG has a resistance of 0.2485-Ohms/1000-feet, and a 2-foot length (i.e., the total wire in a 1-foot 2-conductor circuit) would have a resistance of 0.000497-Ohms. If a current of 142-Ampere flows, the voltage drop across that 2-feet of 4-AWG wire would be 0.070574-Volts. Expressed as a percentage of 12-Volts the voltage drop across the conductor would be 0.070574/12 = 0.6-percent.

The normal rating of 4-AWG for maximum current in wiring in a chassis is 135-Amperes. A maximum current rating is based on heating effects and can vary with the type of insulation on the wire.

In making a splice connection between two conductors, there will be some added resistance in the splice. I would expect the spice to have about as much resistance as the one-foot length of 4-AWG wire.

For powering an inverter, I would consider that the voltage drop should be held to about 3-percent. Using my method of "Ampere-feet", a unit I invented for allowing fast calculation, for a distance of 9-feet and a current of 142-Amperes, the conductor must be selected for 9x142=1270-Ampere-feet. From my tables in my article

Conductor Size for Power Distribution
http://continuouswave.com/whaler/refere ... rSize.html

we enter the table with 1270-Ampere-feet and look for the appropriate wire size. From the table for 13.2-Volts and 3-percent drop, we select 2-AWG as the conductor size for this application.

If we consider the system voltage to be 12.0-Volts and allow for 10-percent drop, we can use the 12.0-Volt table, but the values have to be scaled for 10-percent instead of 3-percent. We divide the required Ampere-feet of 1270 by the ratio of the voltage drops, 10/3, to find the new Ampere-feet value:

1270-Ampere-feet / (10/3) = 381-Ampere-feet

Entering the 12.0-Volt and 3-percent table with 381-Ampere-feet, we find that a conductor of 6-AWG would suffice, that is, for a distance of 9-feet, a current of 142-Amperes, and a system voltage of 12.0-Volts, the voltage drop would be less than 1.2-Volts.

Just to verify that, we compute the resistance in 18-feet of 6-AWG wire. From the earlier mentioned linked resource, 6-AWG has a resistance 0.3951-Ohms per 1000-feet. In 18-feet (i.e., the total wire in a 9-foot two conductor circuit) the resistance will be

0.3951/1000-feet x 18-feet = 0.0071118-Ohm

When 142-Amperes flows in this circuit the voltage drop will be

0.0071118 x 142 = 1.0098756-Volt

In terms of percentage of 12-Volts, that is a drop of

1.0098756/12 = 0.0841563 x100 = 8.4-percent.

My conclusion: you could wire the inverter to the battery with 9-feet of 6-AWG conductor and have only 8.4-percent voltage drop at a system voltage of 12.0-Volts. Using any wire larger than 6-AWG reduces the voltage drop.

Thanks for your input jimh.