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Old 07-01-2018, 01:17 AM   #1
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Default Mysterious voltage drop under load at 12v outlet?

My '96 Roadtrek has a 12v outlet. If I plug an 400 watt inverter into it, it will read the proper battery voltage... 12.5 volts, for example.

But, when I USE the inverter to charge my laptop, it drops down to 10.8 and the invert essentially stops. If I connect the wires of the inverter to the wires plugged into the socket, same thing.

But, if I attach the inverter directly to the battery, there is no problem!

Assuming there is some resistance in the line, I searched for small wires... couldn't find any. I pulled the fuse, cleaned it and put it back in, no change.

Anyone ever have this happen? Oh, if I run the van (charging the battery). there is still a voltage drop, but the overall voltage is high enough to run the inverter.


-Pete
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Old 07-01-2018, 02:40 AM   #2
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Quote:
Originally Posted by VTPete View Post
My '96 Roadtrek has a 12v outlet. If I plug an 400 watt inverter into it, it will read the proper battery voltage... 12.5 volts, for example.

But, when I USE the inverter to charge my laptop, it drops down to 10.8 and the invert essentially stops. If I connect the wires of the inverter to the wires plugged into the socket, same thing.

But, if I attach the inverter directly to the battery, there is no problem!

Assuming there is some resistance in the line, I searched for small wires... couldn't find any. I pulled the fuse, cleaned it and put it back in, no change.

Anyone ever have this happen? Oh, if I run the van (charging the battery). there is still a voltage drop, but the overall voltage is high enough to run the inverter.


-Pete

The cigarette outlet (and the wires) is good for 150w.
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Old 07-01-2018, 02:47 AM   #3
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Did you check the voltage at the fuse with the inverter charging your laptop? If it's low, the voltage drop is between the battery and the fuse. If it's OK, the drop is between the fuse and the outlet. Determining that would be a start.
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Old 07-01-2018, 10:21 AM   #4
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Did you check the voltage at the fuse with the inverter charging your laptop? If it's low, the voltage drop is between the battery and the fuse. If it's OK, the drop is between the fuse and the outlet. Determining that would be a start.
Good idea. I'll do that. The other post mentioned the cigarette lighter plug and wires was only good for 150 watts. I'm presuming it's the gauge of the wire that might be the limiting factor here.

If the answer is just that the wire gauge is too small, I can live with that. Sadly, I had pulled and replaced all the walls in my rig, so if I had been smart, would have tackled the wiring as well!

-Pete
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Old 07-01-2018, 01:00 PM   #5
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Some (most?) Plugs are rated significantly less than 150. I wired a 300 watt inverter directly to the wiring in the converter box and mounted it to the door on it with 3M dual lock. Works great.

The inverter I have has 2 sets of leads on it and automatically de-rates if the cigarette lighter set is used. Have you measured the current draw?
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Old 07-01-2018, 01:12 PM   #6
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While the draw fluctuates, it averages 85 watts.
I checked the voltage at the fuse box and it's fine. It looks like I just need to do what you did, Steve!
Thanks, guys.
Now if I can just stop the yellow jackets from flying into the RV when I have th doors open!
-Pete


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Originally Posted by SteveJ View Post
Some (most?) Plugs are rated significantly less than 150. I wired a 300 watt inverter directly to the wiring in the converter box and mounted it to the door on it with 3M dual lock. Works great.

The inverter I have has 2 sets of leads on it and automatically de-rates if the cigarette lighter set is used. Have you measured the current draw?
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Old 07-05-2018, 06:48 PM   #7
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I think some who read this thread may be getting confused.


My purpose is to explain and clarify. If I confuse you more, I'm sorry.



Let's start with basics and talk in generalizations.


Power = Current X Voltage


On the AC side (output of inverter) if you have 400 watts available and running a 120 volt appliance, the maximum amount of current available to that appliance would be 3.33 amps. Divide 400 by 120 = 3.33



Assuming that power can neither be created or destroyed, in order to produce 400 watts on the AC side, you have to input 400 watts on the DC side. This means at maximum power of 400 watts on the AC side, the DC (cigarette outlet) side needs this from the battery:



Power 400 = ?current X 12 volts or 400/12 = 33 amps. No factory cigarette lighter plug is going to produce that. Even if we assume the best scenario with the alternator running and producing 14.5 volts 400/14.5 = 27.5 amps needed at outlet.


Old cigarette lighter plugs were rated at 15 amps in order to heat the lighter element. This is where the rule of thumb of 150 watts maximum comes in. P = 15 amps X 12 volts or 180 watts absolute maximum.


Now, quite a few vehicles do not have a "cigarette plug" anymore. It is listed as an "aux outlet". Somewhere in your owners manual, you should find the absolute current limit listed for this outlet. Don't assume that it is 15 amps. Many that I saw were only rated for 5 or 10 amps, especially if several outlets were ganged together.



There are inverters that use less DC input: however, I'm going on the assumption that you do not have one of these. Check the DC specifications on your inverter to be sure.



What happens if you continually overload the cig outlet? Simple answer - very bad things. Do NOT assume that your fuse will blow. An overload can super heat your wiring and melt things and set them on fire. This can happen over a period of time and never blow the fuse. Seen it many times.



Since 120 volts on the AC side is 10 times the 12 volt DC input, another rule of thumb is that your DC current is 10x the AC current needed for the AC appliance. If your appliance needs 2 amps of AC current, you will need to produce 20 amps on the DC (battery) side.


This is where you need to pay close attention to maximum values and requirements and make sure your DC wiring will handle the job.



In reality, most inverters (over 150 watts) need to be hard wired directly to the battery.



OK, I'm ready; let the bashing begin. Take it easy on me, I'm almost 70 and may be out of touch with the latest and greatest. Check your inverter specs to be sure.
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Old 07-05-2018, 07:21 PM   #8
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Jim,
Hey, I'm not going to bash you... while I *think* I knew all this, your numbers and explanations are appreciated!

Look, it boils down to this: The wires, being thin, have too much resistance to carry that load. In the end, I did what you all knew I would do: I ran dedicated and fused wiring to the battery and all was well. I also attached a homemade low voltage alarm.

Look at the 12v cables that come with your average 600 watt inverter; they're as thick as my pinky.

In the end, I installed two inverters. A little 125 watt for my cigarette lighter socket and then the hardwired 400 watt one connected directly to the battery. Problem solved.

Thanks!
-Pete

Quote:
Originally Posted by Hatteras Jim View Post
I think some who read this thread may be getting confused.


My purpose is to explain and clarify. If I confuse you more, I'm sorry.



Let's start with basics and talk in generalizations.


Power = Current X Voltage


On the AC side (output of inverter) if you have 400 watts available and running a 120 volt appliance, the maximum amount of current available to that appliance would be 3.33 amps. Divide 400 by 120 = 3.33



Assuming that power can neither be created or destroyed, in order to produce 400 watts on the AC side, you have to input 400 watts on the DC side. This means at maximum power of 400 watts on the AC side, the DC (cigarette outlet) side needs this from the battery:



Power 400 = ?current X 12 volts or 400/12 = 33 amps. No factory cigarette lighter plug is going to produce that. Even if we assume the best scenario with the alternator running and producing 14.5 volts 400/14.5 = 27.5 amps needed at outlet.


Old cigarette lighter plugs were rated at 15 amps in order to heat the lighter element. This is where the rule of thumb of 150 watts maximum comes in. P = 15 amps X 12 volts or 180 watts absolute maximum.


Now, quite a few vehicles do not have a "cigarette plug" anymore. It is listed as an "aux outlet". Somewhere in your owners manual, you should find the absolute current limit listed for this outlet. Don't assume that it is 15 amps. Many that I saw were only rated for 5 or 10 amps, especially if several outlets were ganged together.



There are inverters that use less DC input: however, I'm going on the assumption that you do not have one of these. Check the DC specifications on your inverter to be sure.



What happens if you continually overload the cig outlet? Simple answer - very bad things. Do NOT assume that your fuse will blow. An overload can super heat your wiring and melt things and set them on fire. This can happen over a period of time and never blow the fuse. Seen it many times.



Since 120 volts on the AC side is 10 times the 12 volt DC input, another rule of thumb is that your DC current is 10x the AC current needed for the AC appliance. If your appliance needs 2 amps of AC current, you will need to produce 20 amps on the DC (battery) side.


This is where you need to pay close attention to maximum values and requirements and make sure your DC wiring will handle the job.



In reality, most inverters (over 150 watts) need to be hard wired directly to the battery.



OK, I'm ready; let the bashing begin. Take it easy on me, I'm almost 70 and may be out of touch with the latest and greatest. Check your inverter specs to be sure.
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Old 07-06-2018, 12:41 PM   #9
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Hi Pete:


I do the same thing with a small coffee cup inverter in the cab which I use regularly to charge my computer while driving and have a 500 watt hard wired inverter on board, however, I rarely use that one.


My detailed explanation was for other readers of the thread that may have been confused by some of the comments; or not aware of the massive DC power requirements of large inverters or too afraid to ask "why".



I figured that you were pretty savvy about them since you actually measured and understood voltage drop, loaded and unloaded operation etc.



It's a great topic for discussion, thanks for sharing.
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