8 Amps from a 400w Inverter ?

[ArtM]

Originally Posted by xymotic

So.... why did I measure 8 amps DC with only the airport and WD TV on it?

That's 96 watts (8 aDC x 12VDC). What are the ratings on your TV and AirPort? This seems like an appropriate level.

[ArtM]

Originally Posted by Ex-Calif

The product is rated at 400w @ 110vac = 3.6amps. That's what it pulls from the 12vdc batteries (losses ignored)

Sorry, this isn't correct. 3.6amps on the 12VDC line would be only 43W, so 43W would be the output of the device at 110VAC (ignoring losses)

Speaking of ignoring losses, there are other inaccuracies. a 12V battery does not really provide 12V all the time. At higher voltages, the current draw will be less. At lower voltages, the current draw will be more.

Also, in response to another poster, the line limit on the source (cigarette lighter) does not limit the current available to be drawn, it is the maximum rating for that line. If it is correctly fused, then the fuse will blow. If not, then exceeding the line current will simply overheat the wires and circuits until they fail or catch fire.

Until failure, however, the line will provide the level of current demanded by the device (the inverter, in this case).

[noelex 77]

Originally Posted by ArtM

Also, in response to another poster, the line limit on the source (cigarette lighter) does not limit the current available to be drawn, it is the maximum rating for that line. If it is correctly fused, then the fuse will blow. If not, then exceeding the line current will simply overheat the wires and circuits until they fail or catch fire.

Until failure, however, the line will provide the level of current demanded by the device (the inverter, in this case).

ArtM you are totally ignoring voltage drop and the practicalities of the situation.

Inverters will shut down at about 10.5-11.5v
There are very few cigarette lighter plugs and the associated wiring that can deliver 10A at much more than these voltages. There is just too much resistance and voltage drop in the socket and wiring.

I have installed the very best Sutars cigarette lighter socket in my boat with some heavy duty wiring. This is unusual. I still get so much voltage drop that I will not power anything above 15A

There is no way you can get more than 200w out of cigarette lighter plug (at 12v).
Most boat installations will not deliver much more than 1/2 of this.

Inverters cannot make power they have to draw more power than they can deliver. Attach a 400w inverter to cigarette lighter plug and it will only deliver as much as the plug can, at wattages around 100w (maybe 150w if you have an very good instalation) output the voltage drop will cause the inverter to shut down.

[hellosailor]

"But I'm still confused; @ 400 watts the thing would be pulling 32a DC out of a 15 amp cigarette lighter plug!? "
Which is probably why the instructions that you threw away without reading said "WARNING: MUST BE HARDWIRED DIRECTLY TO BATTERY WITH 0-AWG CABLES NOT TO EXCEED 12" LONG" or something similar.

The lighter plugs are for convenience, and the higher-power inverters don't even come with them because of that problem.

Now, if you really want a headache, forget about watts because you need to figure all the AC loads in volt-amps, which aren't quite watts anyway. (VBG)

noelex-
"Inverters will shut down at about 10.5-11.5v " Don't you wish! That used to be a good thing, because when they shut down you still had some battery left, i.e. to restart an engine. Now I've seen more specs saying the "better" inverters will not shut down until 10 volts, even 9 volts. Well...that makes them seem more capable and in a power outage, perhaps better. But I sure wouldn't want one of those in a "routine service" application!

Sometimes I just wish they'd take a lesson from the software industry: Don't publish any instructions, no one reads them, no one checks them for accuracy, just save money and trees and leave everyone in the dark. (sigh)

[xymotic]

Thanks everyone for clearing it up. I actually consider myself pretty good with Electrical, but you know sometimes it's one of those things where I know enough to get myself into REAL trouble

This is all sort of a temporary installation using some of the boat's equipment in my RV to liveaboard for the next year or two.

[noelex 77]

Originally Posted by hellosailor

Now I've seen more specs saying the "better" inverters will not shut down until 10 volts, even 9 volts. Well...that makes them seem more capable and in a power outage, perhaps better. But I sure wouldn't want one of those in a "routine service" application!
)

I agree. It becomes a very slipery slope. As the voltage drops the current demanded becomes higher which drops the voltage further........

If the inverter is not shutting down untill 9v there is a lot of power being converted to heat somewhere.

It still does not help the inverter deliver more power @9v the cigarette lighter plug would be drawing 16A @ 150w. Most plugs will melt at not much more than 1/2 this.

[Zzzeke]

Originally Posted by Ex-Calif

I have been stomped on this before but I will try again. I have a 400w inverter. It has a cigarette lighter plug.

If - 400w / 12vdc = We presume 33 amps are going through the lighter plug - This is ridiculous and ain't happening. You would smoke every lighter plug I have ever seen.

What is really happening is this:

400w / 110vac = 3.6amps

3.6 amps X 12vdc = 43.7w

Many are presuming the input and output watts are constant. That is not the case. The amps are constant. watts are a calculated product measuring "power". only voltage and amperage are "real"

If we were putting in 12vdc X 33amps = 396 we would get 110 X 33 = 3630 on the output.

Don't beleive it? Plug a hair dryer or kettle into your 400w inverter and see what happens.

The product is rated at 400w @ 110vac = 3.6amps. That's what it pulls from the 12vdc batteries (losses ignored)

400w / 110vac = 3.6amps, yes...

But on the DC side of the inverter that translates into:

400W / 12V = 33.33A

It's the power (Watts) that stays constant on both side of the inverter, not the current (Amps).

[Ex-Calif]

No one is putting 33 amps through a cigarette lighter plug. Sorry. Aint happening...

[Ex-Calif]

Originally Posted by xymotic

So.... why did I measure 8 amps DC with only the airport and WD TV on it?

If you look at 400w inverter ratings they all have a peak output of 800-1000w

[Ex-Calif]

Originally Posted by hellosailor

"But I'm still confused; @ 400 watts the thing would be pulling 32a DC out of a 15 amp cigarette lighter plug!? "
Which is probably why the instructions that you threw away without reading said "WARNING: MUST BE HARDWIRED DIRECTLY TO BATTERY WITH 0-AWG CABLES NOT TO EXCEED 12" LONG" or something similar.

The lighter plugs are for convenience, and the higher-power inverters don't even come with them because of that problem.

Now, if you really want a headache, forget about watts because you need to figure all the AC loads in volt-amps, which aren't quite watts anyway. (VBG)

noelex-
"Inverters will shut down at about 10.5-11.5v " Don't you wish! That used to be a good thing, because when they shut down you still had some battery left, i.e. to restart an engine. Now I've seen more specs saying the "better" inverters will not shut down until 10 volts, even 9 volts. Well...that makes them seem more capable and in a power outage, perhaps better. But I sure wouldn't want one of those in a "routine service" application!

Sometimes I just wish they'd take a lesson from the software industry: Don't publish any instructions, no one reads them, no one checks them for accuracy, just save money and trees and leave everyone in the dark. (sigh)

So it comes with lighter plug attached that I am supposed to cut off? That makes sense...

[ArtM]

Originally Posted by noelex 77

ArtM you are totally ignoring voltage drop and the practicalities of the situation.

I'm not ignoring the voltage drop, I'm just saying that the if a line is rated at 10a, it does NOT mean that it will somehow limit the current flowing through it to 10a. Unless it is fused, it will attempt to deliver as much current as the source (battery) will delivery up to the demand being requested.

Voltage drop is a function of the source supply (the battery or alternator), not of the line. If the battery does not drop the voltage, then the line will not drop the voltage either, at least not until it begins to fail.

To put it simply, if you draw 11 amps from a 10amp line, you will probably get away with it if there is no fuse. The line will deliver 11amps, not 10amps. Overheating is likely to be little or slight, since the "10a" line was probably engineered with a little overhead.

If you try to push 12a or 13a through that 10a line, it will deliver that 12 or 13 amps until it reaches failure point. There is nothing in that 10a line that says "Nope, I'm sending you 10a and no more." (except the fuse)

This does not mean that a 10a line will deliver 30a. It won't. The voltage will drop, the current will spike, and the wire or circuits will burn out.

[ArtM]

Originally Posted by Ex-Calif

No one is putting 33 amps through a cigarette lighter plug. Sorry. Aint happening...

I haven't seen an actual "cigarette lighter" in years. It seems to me they used to be fused at 30a, but my memory might be bad on that.

What most people call a "cigarette lighter" today is a light duty 12V power port, and is typically limited to 10 or 15a.

[Sailmonkey]

Originally Posted by Ex-Calif

No one is putting 33 amps through a cigarette lighter plug. Sorry. Aint happening...

That's part of the point of this whole thread. Marketing meets engineering

[cal40john]

Yes you are ignoring voltage drop. As I found on my boat when my 300 W inverter wouldn't work in the cigarrete plug running a power tool, 20 ft round trip, 16 gage wire, 0.095 ohms, at 10 amps is almost a volt drop. I found another 0.5 volt drop on my ancient circuit breaker, and I didn't measure the drop on the plug connection. Inverter beeps indicating low input voltage and turns off output. Works fine at lower wattages. Previous owner probably only planned on a light or some other light load. Lots of things would work OK with a volt or more drop. This is close to one of the tables they produce for boat wiring where a 10% voltage drop is acceptable and planned for.

John

Originally Posted by ArtM

I'm not ignoring the voltage drop, I'm just saying that the if a line is rated at 10a, it does NOT mean that it will somehow limit the current flowing through it to 10a. Unless it is fused, it will attempt to deliver as much current as the source (battery) will delivery up to the demand being requested.

Voltage drop is a function of the source supply (the battery or alternator), not of the line. If the battery does not drop the voltage, then the line will not drop the voltage either, at least not until it begins to fail.

To put it simply, if you draw 11 amps from a 10amp line, you will probably get away with it if there is no fuse. The line will deliver 11amps, not 10amps. Overheating is likely to be little or slight, since the "10a" line was probably engineered with a little overhead.

If you try to push 12a or 13a through that 10a line, it will deliver that 12 or 13 amps until it reaches failure point. There is nothing in that 10a line that says "Nope, I'm sending you 10a and no more." (except the fuse)

This does not mean that a 10a line will deliver 30a. It won't. The voltage will drop, the current will spike, and the wire or circuits will burn out.

[skipmac]

Originally Posted by Ex-Calif

I have been stomped on this before but I will try again. I have a 400w inverter. It has a cigarette lighter plug.

If - 400w / 12vdc = We presume 33 amps are going through the lighter plug - This is ridiculous and ain't happening. You would smoke every lighter plug I have ever seen.

What is really happening is this:

400w / 110vac = 3.6amps

3.6 amps X 12vdc = 43.7w

Many are presuming the input and output watts are constant. That is not the case. The amps are constant. watts are a calculated product measuring "power". only voltage and amperage are "real"

If we were putting in 12vdc X 33amps = 396 we would get 110 X 33 = 3630 on the output.

Don't beleive it? Plug a hair dryer or kettle into your 400w inverter and see what happens.

The product is rated at 400w @ 110vac = 3.6amps. That's what it pulls from the 12vdc batteries (losses ignored)

Respectfully but you do have it wrong. What is constant is the power in Watts, not the current. You cannot turn 3.6A at 12 V DC into 3.6A at 125V AC.

I have forgotten a lot of what I learned to get my degree in electrical engineering but I do remember that much.