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Possible Duplicate:
Why are deep cycle batteries rated in amp hours instead of watts hours?

I don't understand why it is so common for specifications of battery capacity to be given in mA·h, when you have no clue how much power that is and how long a given device can be powered through such a battery, or to compare batteries with different voltages (and try doing that when the voltage is conveniently omitted from the spec altogether).

Why don't everyone give out capacity specs in W·h?


Similarly, although seemingly not nearly as common as with batteries, some devices likewise only list their power consumption in mA again, where you again have no clue how much power they'd be consuming, unless you happen to know what voltage they operate on.

Why isn't consumption always given in watts instead of milliamperes?


Isn't it error-prone to be giving capacity and power consumption specs in units which cannot be directly compared without, (a), knowing the voltage of both the battery and the to-be-powered device, and, (b), having an expensive calculator or a piece of paper.

Why aren't Watts universally used here, and what kind of sense does it make to be giving mA or aA·h ratings without ever mentioning the voltage (happens all the time when you look at various specs online, apparently).

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  • \$\begingroup\$ It would behoove you to get some in-depth understanding of batteries. I can recommend Battery University web site. Doing some experimentation would behoove you as well. That could help stave off analysis paralysis. Finally, I don't see why you are complaining about the cost of paper and a calculator. After all, you seem to own a computer. \$\endgroup\$ – Nick Alexeev Jan 23 '13 at 21:09
  • \$\begingroup\$ -1 for attitude. You don't know why it's being done the way it is, but are sure it's "useless"!? Rants are off topic here and need to be closed. Asking why is fine, but passing judgement before understanding why is immature at best. Grow up. \$\endgroup\$ – Olin Lathrop Jan 23 '13 at 21:29
  • \$\begingroup\$ @OlinLathrop, I've edited the post to change the wording to make it more open-minded. Thanks for your comment. \$\endgroup\$ – cnst Jan 23 '13 at 21:53
  • \$\begingroup\$ No, the title is still quite judgemental. Do you want to rant, or truly ask for information? Pick one. Hint: rants are off topic and will be closed. Only two more close votes to go. Time is quickly running out as you are trying to play games instead of asking properly. \$\endgroup\$ – Olin Lathrop Jan 23 '13 at 22:50
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If you assume that a battery or power supply is a perfect voltage source, than the two are equivalent. Multiply capacity in milliamp-hours by the battery voltage to get capacity in milliwatt-hours, or multiply current by input voltage to get power in watts.

For many devices, the current (mA) will be the same regardless of input voltage. This is true of anything with a linear voltage regulator; current remains constant, but as input voltage goes up, power (W) increases because it is the product of current and voltage. The linear regulator simply converts excess voltage into heat.

For batteries I don't have a good excuse. We could also give capacity in watt-seconds or joules (same thing, a unit of energy), but that would introduce other problems. I suppose a battery is a complex device, and the available energy depends on the load (high current? low current?), temperature, and many other factors. In the absence of a simple and accurate model that covers all cases, the convention has favored units that make common calculations convenient.

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  • \$\begingroup\$ So if the voltage is different, you can still do the calculations with amperes alone, without any regard to the difference in voltage? \$\endgroup\$ – cnst Jan 23 '13 at 20:20
  • \$\begingroup\$ @cnst which calculations? \$\endgroup\$ – Phil Frost Jan 23 '13 at 20:21
  • \$\begingroup\$ the calculations to get the unit of hours \$\endgroup\$ – cnst Jan 23 '13 at 20:22
  • \$\begingroup\$ @cnst I'm not sure what you mean. A specific example would help. \$\endgroup\$ – Phil Frost Jan 23 '13 at 20:24
  • \$\begingroup\$ Some device with the model INS-TK103B has the following specs: <<Working Voltage: DC 10V-36V Standby Work Current: < 30mA Battery: Chargeable changeable 3.7V / 1.2Ah Li-ion battery>>. If I connect it to the always-on circuitry on an average car, how many hours can I leave the car undriven for? (Apparently, this question now became electronics.stackexchange.com/questions/55925/…) \$\endgroup\$ – cnst Jan 23 '13 at 20:35
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The Ampere-hour units actually express charge. An ampere is a Coulomb of electricity flowing per second, so when that is multiplied by time, you get Coulombs. Ampere-hours express how many electrons are displaced in the battery.

Why these units are used may be traditional, and also because certain formulas related to batteries, such as Peukert's Law are based in them. Peukert's Law is tied to current because that is convenient, since the variable which varies is discharge rate measured in Amperes.

Ampere-hours are more convenient for electronics calculations. We often know how many milliamps a circuit draws and so if we have an ideal battery of such and such mAh capacity, we can almost instantly tell how many hours of life we can expect, without any conversion back and forth to energy units.

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In the olden days people used a galvanometer to measure current and voltage. To measure power one would have to read a needle on a dial for two measurements and multiply them together on paper or using a slide rule. To measure total energy delivered from a battery one would have to do that repeatedly and integrate over time using Newton's method by hand calculation.

Being able to take just one reading was a big shortcut, especially for people not trained in engineering, or for whom a rough estimate was good enough. Digital hand calculators, let alone multimeters with built-in microcontroller to do multiplication, did not exist for most of the history of electrical engineering. Considering that lead-acid or nicad battery voltage only varies some 20-30% from full to empty, such may have been an acceptable measurement error in order to allow technicians to take usable measurements without having to do math.

Even battery manufacturers faced the same effort to deliver a W-hr measurement. A constant-resistance load for runtime measurement requires one component: connect a resistor across a battery and measure how long it takes for the battery to drop below the "dead" voltage. For extra credit, plot the voltage over time on a graph. In the 1970's a consumer would have approximated this test by putting new batteries in a flashlight or AM radio and leaving it on to see how long until it stopped working.

A constant-current load requires a few active parts, maybe one opamp and one power transistor, plus a few resistors and capacitors. Measure the time to dead and you have, exactly to the extent conditions are duplicable, a mA-hr rating. Again, measure and plot the voltage over time (ideally at constant temperature) for a more complete picture of battery performance. Such data is sufficient to estimate battery runtime to within maybe 20% under what used to be ordinary conditions: runtime of 5 to 100 hours at room temperature. Just measure current from the battery and divide mA-hr by mA = hours. Good enough for choosing between D cells and AA, quality vs cheap junk alkalines, are your NiCd worn out, etc.

A constant-power load requires multiplication in the load itself (voltage x current in the feedback path), complicating the design and driving up the cost of battery test equipment. Measuring energy delivered requires logging voltage and current, multiplying at each data point, and integrating over time. Possible in a spreadsheet and perhaps trivial with one of today's microcontrollers in the equipment, but a tremendous amount of work 30+ years ago.

In practice, the amount of energy a battery can deliver to a load is highly dependent on the discharge rate anyway, so there would never be one single W-hr rating. If you do enough testing you'll get a chart with power on the x axis and energy on the y axis, where you'd see a moderate to severe downward slope. Additional factors affecting energy storage are temperature, age and usage history of the battery, etc. Further complicating the picture is that modern devices may present an infrequent but heavy pulsed load, which may not "average out" or integrate to the same response from a chemical battery as it would from an ideal voltage source.

Which is not to say that more data from battery manufacturers wouldn't be useful, certainly it would be to those who can make sense of it. Rather that more accurate information would look like a 3D graph, energy vs power and temperature, and that's just the for first charge cycle. The graph could look very different at 100, 300, 1000 charge cycles, under lab conditions where variables are held constant. As most novice engineers don't understand that batteries are not ideal voltage sources, such a tide of data might serve to confuse rather than enlighten most users. In real life, load varies depending on what the multifunction device is doing, temperature varies by tens of degrees, usage varies from minutes to hours per day, on and on. Building a reasonably accurate model of battery behavior (energy storage and state of charge) under such varied circumstances is complex and time-consuming and thus expensive, the domain of university researchers and companies with a business designing battery fuel gauge chips.

In summary, a device does not generally present a constant-power load to the battery irrespective of voltage, and a battery does not deliver a constant amount of energy irrespective of load. Measuring power is (or was) harder than measuring current, and measuring total energy stored/delivered is much harder than measuring rate of energy flow. To a rough estimate current can tell most people what they need to know.

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