I have a Surface Pro 3 and am on the run a lot so I'm thinking about getting a deep cycle AGM battery to put inside my car so I have a constant power.

I need to know and confirm some measurements regarding it so I can buy the proper battery.

  1. power adapter says INPUT: 100-240V @ 1A, OUPUT: 12V @ 2.58A So if I can somehow supply DC power (12V) directly from battery to the surface, it would draw maximum of 2.58 A/h? if yes, if load is lighter then less Ah, right? and so under its heaviest use it would never pull more than 2.58Ah? and INPUT/OUTPUT rating kinda omits hour in Amp so that it actually means Ah?

  2. If I'm just gonna use inverter to plug in via AC power then, with battery source being 12v, it would draw 100V / 12V = 8.33 so 8.33 x 1A = 8.33Ah? And what does 100-240V exactly mean? does it mean it is capable of pulling from 100 to 240V and if in US it can pull 110v? and in Europe it can pull with the help of only prong adapter 240V? if yes, then is it pulling 240V x 1A = 240W of power where in US it would only pull 110W? or does in Europe's case if it's pulling 240V then does A go down to 110/240=0.458Ah?


3 Answers 3


The Amp-hours of a battery gives the number of hours it can deliver 1 amp, or the number of amps it can deliver for one hour.

Amp-hours = amps x hours.

So a 50Ah battery can run for 50 hours at one amp, or 50 amps for one hour. Or 2 amps for 25 hours, or 25 amps for 2 hours.

Slight detour:

I suspect for your use you are correct in wanting a deep-cycle battery. Normal car batteries, for example, can typically only use 25% of their amp-hour capacity. This is because it is not healthy for the battery to be discharged to less than 50%; it will shorten its life (the number of charge-discharge cycles will be reduced). Also, car battery charging systems are not intelligent, and tend not to charge them effectively above 75%. Partly to reduce complexity, partly to increase lifespan of the battery.

As for AC-powered battery chargers, none are going to be 100% efficient. Some may only be 90% efficient, and some far less efficient.

Back to your question, answering part 1:

For your "Surface" device: If its power adapter says INPUT: 100-240v @ 1A, OUTPUT: 12v, 2.58A, That means the Surface would draw 2.58A maximum. 2.58A is continuous; this does not mean 2.58A per hour, it means 2.58A maximum every second, every minute, every hour. Same thing. But yes, per hour, it would use 2.58Ah (amp-hours, not amps). (Maximum.) And yes, with less load, fewer amps would be drawn at that moment.

So if you wanted to use the Surface for 1 hour at full load, then use the formula above to convert amps to amp-hours:

Amp-hours = amps x hours = 2.58 x 1 = 2.58.

So you would need a 2.58Ah capacity battery. Or if you wanted to run it for 10 hours with the full load, you would need a 25.8Ah capacity battery.

Answering part 2:

Regarding "100-240v". Seeing this on a PSU usually means it can work on any voltage in that range, completely automatically, without needing a switch to be flipped on it, and without needing an external voltage converter.

However, exactly how much current draws on the different voltages, at the INPUT, in other words, the amount it takes from the mains supply, is a big question mark. Some PSUs may be 90% efficient under all circumstances: different supply voltages, and different loads on its output. However, some will not. I have seen some where it is enormously inefficient (50%) at the higher voltages. Similarly, some are only 90% efficient under full load. The best bet is to take the worst-case input current that is stated on the device, (amps) and assume it will draw that for all circumstances.

Your calculations of 8.33Ah and 0.458Ah are wrong, I'm afraid. I'll try to work out what you wanted to know:

Assuming you have a 100% efficient inverter, and wanted to run the Surface's PSU on it, here's how you would work out the power of the inverter that you'll need, and the battery you'd need to supply it. But you should note that the inverter will be at most 90% efficient.

If you use a 110v inverter, and the Surface PSU Draws 1A at its INPUT, then the Power of the inverter will have to be at least:

Power = Amps x Volts = 1 x 110 = 110 Watts.

As for the battery capacity required, this could be done in different ways, but perhaps most logical from your point of view would be to start by working out how much the inverter would draw from the battery. Assume 100% efficiency of the inverter. Therefore, power in = power out. Above, we calculated the power. So now we use the above formula to calculate the current (amps) that the inverter will take from the battery.

Power = Amps x Volts 110 watts = amps x 12 Therefore amps (every second, every hour, same thing; it's continuous) = 110/12 = 9.16 amps.

So at any moment, the inverter will need to draw 9.16 amps from the battery. If you need to power the Surface for one hour, it will use 9.16 Amp-hours of the battery's capacity. If you need to run the Surface for 10 hours, it will use 91.6 Amp-hours of the battery's capacity. (If you're using it for 10 hours, it will still only be drawing 9.16 amps. If you're using it for 10 minutes, it will also still be drawing 9.16 amps).

To answer your comments on my answer:

Think of amps as speed, and amp-hours as distance. If a runner runs at 3mph, then in one hour he will run 3 miles. If he runs for 10 hours, he will run 30 miles. But whether for 1 or 10 hours, he will constantly be running at 3 miles per hour.

Let's say he can run a total of 15 miles, like the capacity of a battery. He could spread it out, and run at 1mph for 15 hours, or he could run flat out (his peak), at 5mph for 3 hours.

Or, say he sprints and slowly jogs alternating every half hour: So he would run at his peak for half an hour at 5mph, then a slow jog at 1mph for half an hour, and repeat. Each hour he would run a total of 3 miles, so his average speed would be 3 mph. So he could do this alternating for a total of 5 hours.

Now back to the "inefficiency" you talk about. Imagine our runner is going up a hill. The voltage is equivalent to the steepness of the hill. So if he is going 4mph up a hill with a 1:20 incline, he will be exerting the same energy per X amount of time as if he was going 2mph up a 1:10 incline.

Similarly, if a jogger runs 4 miles up a 1:20 incline, he will have exerted the same total amount of energy as if he'd run 2 miles up a 1:10 incline.

Regarding your comment, there is actually no difference between AC and DC in all the calculations above - it can basically be ignored for your purposes here when calculating amps, voltages, amp-hours and power. (Just don't try running a DC device off AC power or vice versa).

If you were to power a 12v Surface with a 12v battery, then if the Surface uses 2.58 amps, that is exactly how much is drawn from the battery. (But I would not recommend this, because the battery's voltage could vary from anywhere between 10v to 14.5v depending on its state of charge, among other things. The Surface could permanently damaged).

If, say you were to power the 12v Surface taking 2.58 amps, via a converter, powered by a 6v battery, then with a 100% efficient conversion, 2.58 x 2 = 5.16A would be drawn from the 6v battery.

I hope that clears things up.

  • \$\begingroup\$ Thank you Jodes for your answer. Lots of explanation, I love it! It's helping me to get the concept behind the numbers. But I don't get the every sec, every hr, same thing part. If it's 2.58A "MAXIMUM" at the moment then how do u assume for the duration of 1 hr it will still be 2.58A? If 2.58a has flown thru at the moment of 1 min and another 2.58 the next min then isn't it 5.16a in 2 min? Maybe I'm not understanding the characteristics of electricity but it doesn't make sense. \$\endgroup\$
    – Mike
    Commented May 2, 2015 at 19:48
  • \$\begingroup\$ And for the same device to draw 9a vs 2.5a when it's ac vs dc respectively, that's a lot of waste. Then what is the best way to connect from Agm battery terminals to surface directly? \$\endgroup\$
    – Mike
    Commented May 2, 2015 at 19:49
  • \$\begingroup\$ Please see my edits, I have tried to address your questions \$\endgroup\$
    – Jodes
    Commented May 3, 2015 at 15:04
  • \$\begingroup\$ There is a 12v car adapter for Surface Pro that I can use to hook up Surface to the 12v battery, I guess, but if you say there'd be voltage difference that could lead to its death then what does car cigarette plug have that makes it safe for me to use? Is there like a regulator of some sort between battery and the plug? Can't I use something like that to even the voltage and how would voltage destroy the device (device itself has a battery inside too)? or would it just drain out the surface battery? Sorry for asking too many questions... \$\endgroup\$
    – Mike
    Commented May 3, 2015 at 23:43
  • \$\begingroup\$ Some products that come with a 12v cigarette lighter plug will either have a regulator in the plug, a regulator in the product itself, or just doesn't need a regulator. And not all regulators are equal. If a regulator gives 12v output, only by finding its specifications can you determine what range of input voltages it can accept. Regulators and other electronic components have different properties, and can fail for different reasons. Usually a higher voltage causes excess power, so something overheats and dies/melts/explodes/pops/fizzles/silently passes away, usually quite quickly. \$\endgroup\$
    – Jodes
    Commented May 4, 2015 at 13:33

I have a Surface Pro 3 (SP3) and I connected it to a high precision digital regulated power supply (with built-in meter) and here are my findings:

  1. The SP3 power input is controlled - I measured 9.6V to 17V cut-off with more or less linear current draw.

  2. When writing this message with Wifi ON and Brightness to lowest the power draw is fluctuating between 0.51-0.58, averaging 0.54, or 6.5W

  3. 50% Brightness = 7.2W

  4. 100% Brightness = 9.6W

  5. 50% Brightness and 1080p video on VLC = 12W

Hence, with a 12V, 100Ah car battery, you have 1200Wh total energy for disposal meaning you can;

Write for 1200/6.5 = 184h Play video = 1200/12 = 100h


  1. The Surface Pro 3 power consumption is very moderate, 6W in light use and 12W in high.

  2. For power in your car, you can easily connect this to your existing car battery - no extra battery is needed. Even one days heavy usage -(8h) it will only deplete your car battery 8% assuming you are at standstill with motor off.

  • \$\begingroup\$ You are welcome. I want to add 2 more things; 1) To connect the Surface Pro 3 to a 12V battery, you need another charging cable, for example this one: ebay.com/itm/… 2) When charging and using the SP3 at the same time (basically as soon as you have been using on battery at any given time and connect it again) the power needed is 24W. I have never seen the draw exceed 2A, even when charging AND using it at the same time. \$\endgroup\$ Commented Feb 11, 2016 at 4:52

Power is power

$$P = U \times I$$

and energy is energy.

$$E = U \times I \times \Delta t$$

Apples and oranges.

For simplicities sake, I assume a constant current

  1. The supply can supply \$12V \times 2.58A = 31W\$ of power. that's what the label tells you. It doesn't omit any Ah because time is of no concern to power. Time is not part of the formula for power. And as you can see in the formulas, it's electric energy that depends on time. If your device runs for 1 hour, it will consume 31Wh of energy. It if runs for 2 hours, it will consume 62Wh of energy.
  2. voltages are not easily compared if one is AC and the other DC. The real question is Why don't you use the battery of your car? It provides 12V already and is so much easier than setting up another battery, converting it to AC just to plug in the power supply of the device that goes back to 12V. And all these conversions eat power, too, because they are not perfect.

I mean seriosuly, just ebay for "surface pro 3 12V car adapter". They are like 10$. Keep it simple.

  • \$\begingroup\$ because I don't want to ruin my car battery... I may potentially have other devices or gadgets. \$\endgroup\$
    – Mike
    Commented May 1, 2015 at 18:49
  • \$\begingroup\$ @Mike what exactly do you mean by "ruin"? \$\endgroup\$ Commented May 1, 2015 at 19:51

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