Hard Wiring an inverter

Has anyone hard wired a (500 watt or so) inverter into their trucks electrical system to run 110 volt accessories? I am thinking of placing it below the rear seat or molding it into the center console, for use in the rear seats.



I am looking at doing it to power my computer or a TV while camping.



Advice on doing so is greatly appreciated



Thank you,

Michael

 

I hard-wired an inverter for my computer. Are you asking how to hard-wire an inverter? Or whether it's possible? I assume if you are contemplating the project then you already have the skills to make proper electrical connections and you're just wondering where to hook into the vehicle electrical system.



In my opinion, which should never be assumed to be based on any fact or logic, the best way to connect any aftermarket electrical device to the vehicle is via an auxiliary electrical system that runs directly off the main batteries. I use a Painless Wiring Cirkit Boss, but have since found you could build a nearly identical copy of the Painless kit for less than 1/2 the price.



If you insist on connecting directly to the vehicle electrical system, you could simply splice into the "power point" wiring for the center console and put the inverter there (as you suggested). Of course, beware the maximum current draw for the inverter.



Either way, there shouldn't be any particular problems doing this. My inverter has been running flawlessly for 3 years.



Ryan

 

If I remember my electronics 101 class correctly you'll need to be fused at 45 amp at 12V I'd go to the battery and put a circuit breaker within 6 inches of the connection then use the proper size wire for that load... . Maybe a cheap set of jumper cables could provide the wire???

 

Stoops, you obviously ran the quick back-of-the-envelope calculation I did using the definition of power (P=VI). But I don't think that's correct in this case, since we're converting DC to AC power. The real definition of power is p(t)=v(t)i(t).



I don't think it's appropriate to simplify the law to P=VI, since V=V(t) and I=I(t) for AC. I think (I could be wrong here) that for a simple AC wave with the voltage and current in the same phase (and at 60 hz), the instantaneous power would be



p(t)=(1/2)VI(1+cos(2*60*t))



such that V and I are maxima. Then, at time t=0, p(0)=Vmax*Imax. But now that we've converted to 120v AC power, Vmax=120 so Imax=500/120=4. 2 amp.



But I'm not an electrical engineer, and all of the above could be complete crapola! Which is why I didn't mention it in my earlier post. I figured it would be easier to just check the input current specification on the inverter. 45 amp just doesn't seem right to me. 4. 2 seems a lot better.



Ryan

 

I installed an inverter on my truck. I put it under the drivers seat and it powers my son's 13" tv and playstation for the long trips. I used either #8 or 10 wire with 30 amp circuit breaker. Thats the biggest I could find at the time and has worked great. It has never tripped or given me a problem. My inverter is a 700/1400 peak watts. I have used it to run my 1/2" drill to build a deck at the soccer fields. It has came in very handy at times.



Randy

 

Power In always has to equal Power Out. What levels the playing field in comparing AC to DC calculations is the AC sine wave, area under the curve, is added up by rms (root mean square) method. Now the AC sine wave area equals the DC square wave area. Call outs of AC values are rms unless otherwise noted. So 120 volts AC at one Amp has same heating value (power) as 120 volts DC at one amp. Tho' instantaneous peak values of AC sine wave is 1. 414 multiplied by rms value or corallary is . 707 multiplied by peak sine wave equals rms value. So 45 amps dc and corresponding large wire size and circuit breaker size I believe is correct. What skews things a bit for the observer is short peak events get buffered and go unoticed. One or two sizes down on wire works ok but wire a bit warmer. Most circuit breakers are heating elements so there's some time delay and if max current lasts for only a bit (acceleration period of drill motor) then no circuit breaker trip. Similar notion with Inverters. Their internal capacitors (electrical accumulators) draw extra current during initial turn on but quickly charge up and can give that back on short time period larger current draw. BUT there's no getting around 100% duty cycle. If the drill motor is pushing a 2 inch hole saw thru thick material instead of shooting a screw, the inverter needs to supply 500 watts to the hole sawing drill motor for much longer duty time period then screw shooting. Proof of all this can easily be seen with clamp on dc ammeter around the 12 volt wire feeding the inverter. Most Fluke meters have accessory clamp on ammeter probes available. Not cheap but not bad spendy either. JimB.

 

I put one under the front seat facing torwards the back. I have a cig lighter adaptor on it that I plug into the console when I need to use it. I think mine is 600w, I use it to run a laptop that I run Delorme on when going on trips that I need directions. I have also run a small TV and VCR off if it. I have velcro on the bottom if it that "sticks" to the carpet to keep it from moving around.



You have to remember to turn them off while starting the truck, and leave it off until the engine warms up. If the grid heaters are running the voltage will drop low enough to cause them to reset.

 

Several points of interest... . 10 amps at 120 vac is = to 1200 watts... . 1200 watts divided by 12 vdc allowing for no loss in the inverter would mean a 100 amp draw... so the difference from 12 to 120 v is about 10 to 1... . but we are actually dealing with a charging system usually running close to 14 v.



Remember that most fluke meters are designed to measure AC amps and do not have a scale for DC amps... . a good fluke meter that can measure DC amps is close to $300. 00. I personally ran across 2 in a pawn shop for $25 each and bought them...



Most inverters have some kind of ability to dump heat... . this heat will show with an amp meter as extra current draw at 12 vdc so I'm guessing that you should plan for at least 5 to 7 % loss...



And my last comment... I have an inverter that I use all the time with my laptop, small printer, and tv / dvd player in my trailer..... I looked at the current demand on the label of the product and added them up... my little 200 watt inverter will run either my laptop / small printer, or my TV / DVD... but not all 4 items at once... .



BTW if in fact you do install the 500 watt inverter..... that is close to 45 amp draw at 12vdc, you'll need 8 gauge wire to support it... and maybe 6 gauge if you run it close to its rated output all the time... . Your local box store should be able to show you a chart... . 14 gauge is 15 amp, 12 gauge is 20 amp, 10 gauge is 30 amp, 8 gauge is 40 or 50 amp. . can't recall off hand...



Hope this helps.

 

Hi Ryan,

I believe your calculation to be correct in that you've determined that 500W at 120V = 4. 2 amp. If you were to fuse the 120V output of the invertor that 4. 2 would be appropriate. However when you invert or jump voltage up from 12V to 120V the amp draw increases dramatically. My suggestion was to equip for a 45amp 12V load from the invertor.



W divided by V equals A

 

Per Tripp Lite site here is the specs on their 600 watt continuous inverter:



Full continuous load - 64A at 12V DC, No load - 0. 95A at 12V DC,



They also include the correct wire size for the units also.



here is the Link

 

Well, there you go, I was wrong! I thought for sure 45 amp was too high - I didn't trust my own calculations.



Thanks everyone for setting me right.



Ryan