Ready to make a UPS, just need some clarification

Marine components are designed to be more corrosion resistant. Since your setup will be inside of a car and not exposed to salt spray, that shouldn't be a problem for you.

Wire size is an issue because of voltage drop. All normal wire has electrical resistance, and this resistance causes part of the electrical power to be converted to heat. The problem with 12V systems is that you are not starting with much voltage in the first place, and for inverters, the current draw can be very high. The higher the current draw and the larger the resistance, the greater the voltage drop on the wire. This is expressed by Ohm's Law: V=I*R. The bigger the wire the lower the resistance.

You could calculate the voltage drop and determine the required wire gauge, but you probably don't need to do that. If you keep the length of the wire leads the same or shorter than those supplied by the manufacturer, you can use the same gauge wire. If you go longer, you need to use a heavier gauge. Bigger is better.

If keep all of this in mind, automotive grade products should be fine.

Tom

 

I used marine connectors so I could remove the device for service. I dont want the dealer denying my warranty because of a ups device. Trust me they can void your warranty and YOU DONT want to be on the wrong end of that battle. Also the marine fuse go to 600 amp. I used 80 amp square fuses. You can order the parts online for a low price.

 

Yes, you can mix wire sizes. The size of the wire gets bigger as the awg number gets smaller, so 4 awg is larger than 6 awg. For your situation, it is always okay to use a larger wire (smaller awg number), but using a smaller wire will increase the voltage drop. How much voltage drop is acceptable depends on the device, the current, and the voltage of the battery. Try to keep the 12V wires short and use as large a wire as you can conveniently use. When you have an option, keep the 12V short and put the extension on the 110VAC side. Voltage drop isn't much of a problem on the AC side.

Tom

 

You might also look at industrial components, which you might find at your local big electrical supply house. I thought of them belatedly for my installation, mostly because I had the stuff lying around and hadn't found anything I liked much better at the auto supply places. One very nice thing is that the screw-clamp style industrial terminal blocks make very reliable connections with none of the hassle of putting crimp terminals on your wire ends. So I wound up crimping ring terminals only on the wire ends attaching to the battery, relay module studs, and inverter.

The positive-battery-terminal assembly includes a 120 A fuse (in my gen 1 anyway) that bolts in. The load-side fuse bolt works great as a place to attach a ring terminal for the + connection, so your wiring is protected by the 120A fuse without you needing to add any new fuse at that point. A little panel on the side of the terminal assembly pops open for access to the bolt.

Please post pics of your set-up!

-Chap

 

Voltage drop due to wire resistance is an issue primarily for long runs. For the distances we are looking at here, overheating of the wires is a greater risk. For normal home wiring, 14 Gauge wire is rated for 15 Amps, 12 ga wire is rated for 20A. I believe there are insulations available which permit operation at higher temperatures, and thus slightly higher current (maybe 20 & 25A respectively).

Looking at Problemchild's pictures in another thread, it looks as though the leads from the battery to the inverter (12V) are probably about 6 feet long. 6 feet of 6 ga wire has a resistance of about 0.002 Ohms (0.004 Ohms for both leads). If the inverter INPUT is 1000 Watts, it will be drawing 83A, causing a voltage drop of ~ 0.33V. With the battery being charged at 13.6V, the inverter input would be 13.27V.

If the inverter input leads were replaced with 12 ga wires, the voltage drop would be ~ 1.6V. The inverter input would be 12V, until the insulation on the input wires melted or caught fire.

When determining wire size, two separate checks must be made. 1) the total resistance should be kept low enough to avoid an excessive voltage drop (remember a 50 foot extension cord has 100 feet of wire), and 2) the maximum current must not exceed the rated current for the wire (to prevent fire or physical damage).

Norm

 

Norm, in this sort of situation voltage drop becomes a deal breaker long before overheating can come into play. You can't have overheating without voltage drop, and a nominal 12V system starts with such a low voltage that voltage losses quickly reach an unacceptable level. Otherwise your comments are correct, and safety would be the main issue.

Tom

 

Normally the inverter is grounded to the car chassis AND to earth ground (through a ground wire to the house or other AC load). The neutral wire on house wiring is already grounded to earth ground. If you don't ground the inverter to earth ground, you run the risk of having the entire car at 110VAC, so that grabbing the door handle could be fatal. It would take a failure for the car to become electrically hot, but that is why we use protective grounds. They protect against this sort of failure.

Tom

 

Tom,

The voltage drop is a function of the resistance of the leads, which for a given wire size is proportional to length. With ~83A input current to the UPS when drawing 1KW, the voltage drop will be proportional to the length of the input leads. 12 ft (6 ft each for positive & negative) will have a total resistance of ~0.019 Ohms, causing a voltage drop of ~1.6V. If the battery is being charged @ 13.6V (HSD system on), the input voltage to the inverter will be 12V (its rated input voltage). That cannot be considered too much voltage drop.

The problem is that 83 amps flowing through 0.019 Ohms of resistance will dissipate about 130W, or nearly 10W per foot. This is over 16 times as much as the wire is rated for.

Note: if you were able to mount the inverter right next to the battery, with each #12 lead only about 1-1/2 ft long, the total resistance, and therefore voltage drop, would be the same as with the two 6 ft leads of #6 wire supplied with the inverter. The total power lost in the wire resistance would be the same, ~34W. With #6 wire, the power dissipated would be less than 3W/ft, with #12 wire, it would be nearly 11W/ft. #6 wire can safely dissipate 3W/ft, #12 cannot handle 11W/ft.

Norm

 

Hey Norm,

Where can I find the W/ft wire ratings you're using? I've always wanted to know that. I expect they'd depend on the insulation material, so I assume you're talking about the stuff that's sold as "primary wire" for automotive use? Do the tables also involve adjustments for ambient temperature or number of wires bundled together?
I know the N.E.C. does, for the sorts of wiring materials used in structures, but I've never really found an official source of the same info for auto wiring.

Thanks,
-Chap

 

Whoa there fella, back up the bus. I never suggested using smaller wire. This whole conversation is about extending the supplied wiring. Obviously you are going to burn the wires if you use under-rated wiring. Perhaps I assume too much (being an electrical engineer), but only an idiot is going to replace the supplied cable with something smaller.

As for your example, #6 wire is rated for 101A for chassis wiring, which is well over the 83A in your example. If you are going to bury it inside a wooden wall, then you may have problems.

It's perfectly acceptable to shorten the existing leads. It is also acceptable to lengthen the leads with the same or larger wire, but you need to watch the voltage drop. That was the point I was trying to make.

On my sailboat, I have a 1KW inverter installed with about 10 ft of cable (there and back). I used #00. I don't need that for safety, but it keeps the I^2*R losses down, which improves performance. The heavier wire also allows the system to function even after some degradation of the cable and connectors, which often happens in a marine environment.

As for the question about wire ratings, search for AWG wire. The AWG tables are available all over the Internet.

Tom

 

Hi Tom,

Have you tried that search? In Google the first three hits have AWG tables and give several different ampacity figures. One of them even explains why there's such a variety of ampacity tables:

The ability of a wire to carry a given amount of current is affected by a number of additional factors, which are not accounted for in the AWG table above. The ambient temperature of the surrounding air, wire insulation, and number of other wires bundled together [provided below].

Ampacity relates to the ability of the conductor to carry current [amps] before the cable over heats. I understand there are hundreds of Ampacity tables for many different conditions. ​

I was just wondering where Norm got the watts-per-foot dissipation figures he quoted, and what assumptions about insulation type, ambient, and environment they were based on. I haven't seen anything that quite so neatly rates wire for watts per foot before.

-Chap

 

The Watts per foot figure depends on the assumptions that you state, as well as the type of cable and how the cables are bundled. Capacity rating will fall into two broad groups: those from the manufacturer, and those set by regulation or industry standards. Ultimate failure will occur when the wire in the cable melts, or when the insulation melts or catches fire, whichever comes first. Safety standards and good practices are set well below the ultimate figures.

The best place to get dissipation figures is from the cable manufacturer.

Tom

 

Thanks, Tom, that makes sense. Probably only Norm will be in a position to reply with the sources and assumptions for the specific numbers he used.

-Chap

 

Tom,

I wasn't trying to imply that you were suggesting the use of smaller wire, in fact you stated the same or larger wire. If someone only half understood your post, and used a smaller, but shorter, wire while keeping the total resistance the same, there could be an overheating problem.

I was trying to point out that there are two limiting factors for the wire size to be used, 1) Voltage drop (IR losses) and 2) ampacity of the wire.

For 12 AWG wire, I assumed 20A capacity (the value allowed for household wiring). I know the capacity depends on many factors, and this value falls between the "chassis wiring" and "Power transmission" values given on the 'powerstream.com' website.

Chap,

Tom is right about where I got my figures. Power dissipated by a resistor (including a piece of wire) can be calculated by the formula I^2 * R. The resistance per foot is calculated from the table (which provides Ohms per 1000 ft).

My Watts per foot values were calculated from the current and the resistance-per-foot value from the table

Norm

 

Norm, thanks for the clarification. I sometimes simplify things to make them easier to understand, but that can leave room for misinterpretation. The engineer in me wants to spell it all out in detail, but then my posts start to read like a contract: "The heretofore said mentioned party of the first part..."

Tom

 

Hi Norm,

Thanks for responding - as for how you calculated the watts per foot that would be dissipated at a certain current, yes that's straightforward, I followed that.

My question was more about:

That is, the source of your W/ft limits. Where did you find that #6 can safely dissipate 3 W/ft, #12 can't handle 11 W/ft, and the wire originally in question (was that #12 also?) can't handle 10/16 W/foot?

It sort of makes sense that there would be references giving wire dissipation limits in that form, but I don't know where to find them.

Thanks,
-Chap