working out 12 volt draw through inverter

I use a laptop through an inverter, and the inverter draws 1 amp when working. The charger for the laptop says 100-230 volts 1.5 amps. However when using the laptop, it draws not 1.5 amps, but 6.5.
What's going on?
Is there a way of finding out how much current an item is going to draw when used through an inverter?

You've left out the voltage side of the equation. I would guess that your 1A output is at 120VAC and your 1A computer draw is also at 120VAC while your 6.5A draw is at 14.4vdc.

Ultimately, you need to convert everything to power (watts) to have an idea of what the 12vdc side of the inverter will draw. 1A at 120VAC is 120 watts (Current x voltage = watts). Your wattage on the input side should be higher then your actual output wattage due to efficiency drops in the inversion process. However, at 6.5A at 14.4vdc you are only using 93.6 watts on the input. This makes me think that you are actually drawing less then 1A at 120VAC, otherwise you are actually creating energy inside your inverter and I believe there is a physics law somewhere that states that that is not possible!

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Kevin Pierson

Adding to KP's reply, a general rule of thumb (ROT) for 12V loads is divide the output power by 10 to get the approximate input current.

FYI...
That /10 instead of using the "correct" 12V or 14.4V etc is an easy approximation that factors in typical efficiencies of ~80% - eg, amplifier or inverter output versus their input power.


FYI...
Your laptop uses a max of 150W (100VAC x 1A), hence 150/10 = 15A input. You are under half that because the 1A plugpack is a standard size and over-rated for your laptop.
The 1A "idling" current means your inverter consumes ~12W without anything connected, hence why people don't leave inverters on (for long) unless the battery is charging.   

The 1 amp output is at 230 volts, so that's 230 watts?
But yes, 6.5 amps at 14.4 = 93.6 watts. Is that nearer to what you'd expect?

This can't be right.
1.5 amps at 230 volts, 1.5 x 230 = 345 watts.
Then 345 / 12 volts = 28.75 amps....

Am I missing something?

No, that's right.

Since 240:12V = 20, you'd expect 20x the 12V current compared to its 240V current (close enough to 230V). It's 10x for 120V systems.


But remember, just because a supply can supply certain power does NOT mean it supplies that power. (IE - it only supplies what the load consumes.)

So really, there isn't an accurate method of knowing what anything will draw at 12 volts without measuring it inline?

Well, that's relative.

Knowing the 230V power you can guesstimate the 12V current - ie, assuming 80% efficiency and deciding if it's at 14.4V or 12.6V etc.
But you don't seem to know the laptop's load. It's probably around 90W but consumption varies markedly with laptop specs etc.

It depends on why you want the info. If it's to design the distribution, near enough should be good enough.
You could base that on the inverter's max output - eg, if 300W max out => ~400W in => ~30A hence a 40A circuit.
But you could base it on the expected consumption, but you need to find out what the laptop consumes. EG, if 90W => ~120W input => 10A hence a 15A (or 20A) circuit. (Ensure the fuse is rated for or less then the cable in case anyone adds a load to the inverter. You want the fuse to blow, not the wire to melt.)


If power (efficiency) is a concern, the a dc-dc converter is the way to go.
And dc-dc converters designed for automotive use should handle cranking voltage dips etc - most inverters cut out around 10.5V or higher where as dc-dc convs can tolerate down to 8V and sometimes 6V. A dc-dc solution should mean ~20% less power consumed.

I need to know simply so I can design a PC that runs from 12 volts, but one that won't draw more than the one I have currently, which uses approx 6 amps.

6A at 12V? (72W)

So what can you measure - DC Amps? Measure the inverter input current with the laptop running.

Or find what your laptop consumes - maybe by measuring its input voltage and current (on the DC side).


Dare I ask why it can't draw more than the present device?

What I do is measure the DC current coming from the inverter as the lappie is the only thing I run from it.
But the lappie is 230 volts 1.5 amps, so should pull 1.5 through the inverter plus some inverter current, but it doesn't, it pulls about 6

So it 's hard to tell when thinking of a new computer what will fit and what won't.
I tried an imac recently and that pulled 10 amps from the inverter.

Surely you mean INTO the inverter? Its output is 230VAC, however if you are measuring that current (which can't be more than ~1.5A AC), then the meter may be reading wrong because it is not on a DC range or not true-RMS (I assume the inverter is a stepped wave, ie "pseudo" sinewave).


IMO all you need do is design your new PC to match (or not exceed) the laptop's power consumption - ie, if the laptop is 75W, then the PC should be 75W or less - I don't see why you need to know the 12V current.
The PC consumption can be increased a bit if you use a DC-ATX supply or a dc-dc converter rather than an inverter (depending on why you are specifying that power limit).

The inverter is a 1kw pure sine wave. Far bigger than I need I know but all I've got now.
And I'm sorry, measuring the 6 amps is the flow form the batteries to the inverter, not coming out of the inverter. That's what i always measure to work out what effect a device has on the batteries.

I see your point about sticking to a wattage that I know is safe. The problem is this one is 230 volts at 1.5 amps which is 345 watts. An iMac is rated at about 171 watts, yet when switched on the drain from the batteries to the inverter is 10 amps. Luckily I had one to try before buying.

I have considered using a dc to dc converter, but I recall when I had my old Dell. It drew 3.5 amps using it's mains charger through the inverter, and almost 5 amps using the 12v Air charger!

So you have measured 1.5A on the inverter output to the laptop's plugpack?

Hmmm. No I haven't. I only ever measure what goes in. I don't think the inverter is bust though as I used to have a 600w pseudo sine and the current draw was almost the same on that one too.
Is it worth measuring what's coming off the inverter? Will it help in any way?

So where is that 1.5A from?
As per before, plugpack ratings are not the same as their load ratings...

Why worry about the AC current? All you need to know is the laptop's DC input, else the inverter's DC input.
But surely the laptop has a power consumption figure whether on some label, or manual, or spec sheet?

This is getting confusing so perhaps best to go back.

The laptop charger unit has stamped on it input is AC 100-240v, 1.5 a. So that's 240 x 1.5 = 360 watts.

The output is DC 19v 4.7a. 19 x 4.7 = 89.3 watts.

Measuring the load to the inverter from the batteries when the laptop is in operation, I measure a load of 6 amps coming from the DC 12v batteries. 6 x 12 = 72 watts.

So, I've got 3 completely different figures, but I thought watts would always be equal?

Now, lets say I want a new computer and I know that it runs at AC 240v 2 amps. 240 x 2 = 480 watts. Because all of the other figures are inconsistent, how can I tell how many amps will be drawn when using the new computer?

The 240V x 1.5V is totally irrelevant so forget that figure. (Forget entirely what the plugpack says. That is only useful as an upper limit to what a load could draw.)

Since KP has a point about energy not being created out of nothing, I suspect your output 19V 4.7A is also from the plugpack data and not from actual measurement.

That leaves you with 6A probably at ~12.5V, hence 75W. I would therefore assume the laptop itself takes ~60W (assuming 80% inverter efficiency).


And again, what is the PC's 240V @ 2A based on - it's 480W PSU rating? Irrelevant! What is the PC actually taking?
But a PC with a 480W ATX supply (or any PC for that matter) is unlikely to draw as little as a 60W laptop. You need to design your PC with equivalent low-power laptop chips & components.

And if the PC is drawing 480W, then you have your answer - the PC is drawing about 8x what the laptop draws.


PS - get a dc-dc converter for the laptop and bring the inverter & plugpack as a backup. DC laptop converters are available for ~$30 upwards. (I bought a 120W 12V DC converter WITH AC input (that combination is rare else more expensive) for $29. Granted, it runs hot, but it works well.)

PPS - KP - have you considered writing a paper on that energy theory?

I think some old guy beat me to it!

If you KNOW the current on the DC side and you KNOW the current on the AC side you can calculate the actual efficiency of the inverter. You can then determine roughly what an actual AC load will do to the DC side. As oldspark said the numbers listed on the power supply for current is absolute MAX current under worst operating condition. I would expect the current to drop in half when you use 230VAC as opposed to 100VAC.

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Kevin Pierson

KPierson wrote:

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I think some old guy beat me to it!

If you KNOW the current on the DC side and you KNOW the current on the AC side you can calculate the actual efficiency of the inverter. You can then determine roughly what an actual AC load will do to the DC side. As oldspark said the numbers listed on the power supply for current is absolute MAX current under worst operating condition. I would expect the current to drop in half when you use 230VAC as opposed to 100VAC.

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Is there a formula for calculating the efficiency then?

Yes, power-out divided by power-in. (Times 100 if expressed as a percentage.)
[It's the same as any efficiency formula - ie, output/input.]

oldspark wrote:

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Yes, power-out divided by power-in. (Times 100 if expressed as a percentage.)
[It's the same as any efficiency formula - ie, output/input.]

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:)

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Kevin Pierson

KPierson wrote:

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oldspark wrote:

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Yes, power-out divided by power-in. (Times 100 if expressed as a percentage.)
[It's the same as any efficiency formula - ie, output/input.]

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:)

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I don't understand a lot of what you are saying, and I feel that my question isn't being heard let alone answered.
It must be that I simply don't understand enough to be able to grasp what you are saying.
Thanks for trying to help anyway.

KP - your inbox is full. (Minor technical issue/trivia.)

Sorry boristhemoggy, it is your latter. Read through the thread again. Many things have been repeated - eg, forget nameplate ratings as they have NOTHING to do with what your PCs & laptops are actually consuming, & power P = VI [hence Pout/Pin = (Vout x Iout)/(Vin x Iin) etc].


But if you want my short conclusion - forget it - you will not be able to design a PC that consumes less or equal to that of a laptop unless you use laptop components or wait for the next generation of chips if you want a PC that is more powerful than your laptop.

But hence why I asked you WHY you had your limited power (or current?) specification.
I assume it isn't a battery reserve issue because then you probably would not be using a inverter. Nor would you use an inverter if you needed cranking ride-thru etc (not needed for a laptop, but essential for a car PC).

You should be able to google your laptop consumption... Then add ~30% to get an inverter's power input (and I = P/V from P-VI) or ~10% if using a dc-dc converter.


But read thru again and then feel free to ask for clarification or more detail.
And please use only relevant measurements and specs - not plugpack etc ratings.


PS - "efficiency" is a basic definition. See f.ex wikipedia Electrical efficiency. So too formulas like P=VI.

PPS - wiki has IMO made many pages more complex - too mathematical etc. EG, only the first part of the above link is relevant. Maybe their Electric power page is better for you (again, first section only).