Install Power Inverter

[GrantG]

Good Evening,
We have a 2003 Trillium Outback and I would like to install a power inverter. Should I connect it directly to the battery posts or can I connect it to the panel where the battery +ve and -ve are connected? I've attached an image of the panel. Thank you in advance for your help on this.
GG

[Raspy]

If it can draw more amps than those small panel wires can handle, it must be hooked up to the battery. Those wires look like they might be #14, which are rated for 15 amps. What size is your inverter? Even a 500 watt inverter will draw about 46 amps from the battery.

[GrantG]

I don't have an inverter yet but I was thinking just a 300 watt would be all I need for charging phones, baby monitor, etc. In case we plug in something the draws a lot of amps, I suppose it would be best to play it safe and connect the inverter directly to the battery. I have a solar charge controller connected to the battery as well. Are there any issues with having the inverter connected to the battery as well as the solar charge controller?

[Raspy]

Should be no problem connecting both, the inverter and the solar charger to the battery.

A very simple solution for charging phones, computers, etc, is to get a cigarette lighter plug in inverter. I have one that draws less than 12 amps and supplies 120 watts of 120 volt power. It has a household style receptacle on it. I have several similar ones that we use constantly. I also use one to run the DVD player.

It's not very efficient to convert 12 to 120 to charge a phone, etc. But the overall use is still low and it's very convenient.

I just got a new one at Walmart in the car battery/charger section. It is an EverStart 120W Power inverter. It might be just the thing for you.

I also have a 400 watt one with alligator clamps for the battery.

[GrantG]

Thank you for your help on this John. Much appreciated.

[John in Santa Cruz]

for phonews, etc, I use automobile plugs, 12V to 5V USB direct, its much more efficient.

I have a 200W inverter I rarely pull more than 100W out of, typically to charge my wife's $JOB laptop that doesn't have an auto adapter... I put PP30 outlets in the RV on 30A fuses, and I put a PP30 pigtail on the DC inputs of the 200W inverter so I can plug it into this powerpole outlet, and that works great. 200W at 12V is around 15-16 amps, plus a couple extra for the overhead of the inverter, so it will never blow the 30A fuse.

[redbarron55]

a 300 watt inverter would probably draw an extra 6% power if high efficiency so 300 watts at 120 volts would be about 324 at 12 volts or P=IE or P/E=I 324/12= 27 amps.
A fairly significant load.
Of course this is a coarse approximation, but it should indicate the order of magnitude of the current draw.

[Lynn Eberhardt]

The easiest conversion considering losses and safety is just divide by 10.
100 watts @ 110V needs 10 amps @ 12V.
200 watts @ 110V needs 20 amps @ 12V.
Etc.
It may be a bit excessive, but it's safe.

[Raspy]

Originally Posted by Lynn Eberhardt

The easiest conversion considering losses and safety is just divide by 10.
100 watts @ 110V needs 10 amps @ 12V.
200 watts @ 110V needs 20 amps @ 12V.
Etc.
It may be a bit excessive, but it's safe.

Or, to be more accurate and factor in inverter inefficiency, multiply by 11.

120 watts @ 120 volts = 1 amp at 120 volts. 1 amp X 11 = 11 amps at 12 volts.
so, 120 watts at 120 volts (from an inverter) = 11 amp draw on the battery.

A 1200 watt microwave (input) = approximately a 110 amp draw on the battery.

But even that is not real accurate because as the load increases on the battery the fewer overall amp/hours it will deliver. And recharging will have a loss of about 25% So, 1 hour at 110 amp draw might cost about about 140 amp hours to re-charge. A group 27 deep cycle battery is not able to supply more than about 1/2 hour of that kind of load and still remain above a 50% charge. So, a half hour load of 110 amps, might cost about 70 amp hours to re-charge. If one is only getting 7 amps average from solar, for instance, it might take 10 hours to make up for 1/2 hour of microwave use. More than a whole day from a small solar system to make up for 1/2 hour of microwave use, in that example.

Or, one could substitute a hair dryer and an electric coffee pot for some of the microwave time to come up with approximately the same overall amp hour use, at approximately the same amp draw. 10 minutes of hair dryer, 10 minutes of microwave and 10 minutes of electric coffee pot, for example. If it takes more than one full day to make that up from the small solar, there is nothing left for lighting, charging the computer, watching a movie, etc. So doubling the size of the solar would fix that, on nice bright days. Two group 27 bats would reduce the load on each battery.

[Baxter]

For the things you’re talking about the 12v circuit should be more than enough

[ARVZ]

Don't forget to put a fuse or DC circuit breaker right at the battery or hot end for any battery connection or individual branch circuit in the trailer. You size the wire to safely handle the load, then size the fuse or breaker to protect the wire and the load. It never hurts to oversize the wire for less voltage drop and wasted heat; (Watts = amps squared x resistance) Oversizing a wire in a trailer doesn't cost much. But you still size the fuse or breaker for the load, which should have a specified fuse size or maximum amps stated in the manual or printed on the part.


I like these Type III DC trolling motor breakers, which allow you to turn off the circuit like a switch and manually reset it if trips. https://www.ebay.com/itm/40A-300A-12...MyNRMWRfb8a05g


Or these Type III manual reset stud mount breakers: https://www.delcity.net/store/Manual...98672.h_201365


You do not want automatic reset breakers for this use; we need to check out the situation before restoring power to the circuit.



This is a familiy-owned business with great prices on Type SXL primary wire and other DC wiring parts, and they sell it in smaller rolls. Type SXL provides much better safety for your low-voltage DC wiring projects: https://spectrowireandcable.com/c/wi...-primary-wire/


Any time you design an electrical system, use parts with AT LEAST double the rated current-carrying capacity of your expected loads. Heat kills electrical stuff, and the equation above shows why a small increase in amps makes a big difference in heat (watts). Barely stressed electrical parts live much longer and are worth some extra money.