The capacity of a battery is measured in mAh, for example a 1000mAh battery can provide 1A for 1 hour. Why is voltage not considered in this equation? 1A at 5V draws twice as much power (P = V*I) as 1A at 2.5V, which would surely drain the battery twice as quickly? The only reason I could think is the mAh is calculated on the output voltage of the battery in which case how can we compare batteries using mAh ratings?
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asked Apr 28, 2016 at 9:06
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1\$\begingroup\$ Because as you have found, the capacity in mAh does not tell the complete story. It does not tell you the amount of energy in the battery. For that we use Wh (Watt Hour, 1 Wh means 1 Watt during 1 hour). Consider a hypothetical 1 mV battery with a 1000 Ah capacity. 1000 Ah sounds like a lot but at 1 mV you're only getting 1 Wh. \$\endgroup\$– BimpelrekkieCommented Apr 28, 2016 at 9:13
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\$\begingroup\$ Of course, the capacity is measured in A·h simply because it's defined like that. I read the question as "Why are batteries characterized by capacity rather than energy?" \$\endgroup\$– Dmitry GrigoryevCommented Apr 28, 2016 at 9:42
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\$\begingroup\$ Knowing both the voltage AND the capacity of a battery allows you to do all the necessary calculations when supplying a load. If the voltage was mixed into the value (VmAh) you would still need to know the voltage separately. \$\endgroup\$– JIm DeardenCommented Apr 28, 2016 at 9:43
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1\$\begingroup\$ Because it's a measure of charge ( 1 mAh = 3.6Coulombs) \$\endgroup\$– user16324Commented Apr 28, 2016 at 11:10
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\$\begingroup\$ And it doesn't make any sense to measure a voltage per time. Current and power per time do make sense since they are both a flow in electrons and energy per second. \$\endgroup\$– lucas92Commented Apr 28, 2016 at 12:47
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4 Answers
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There are two reasons why batteries are rated for capacity (A·h) instead of energy (W·h):
Cells using the same chemistry will have equal (or very close) voltage ratings, so voltage can be factored out when comparing their capacity.
In many chemistries, the voltage changes significantly during charge-discharge cycle, but also with temperature and load. Factoring that variable voltage in the battery energy will make many calculations complex:
- a cell discharged to 50% of its energy would require e.g. 80% of its nominal energy to recharge back to 100%, while a cell discharged at 0% may require 200%. With capacity, charge and discharge figures are much closer, and the losses are nearly the same regardless of the charge level.
- the same cell at 50% energy would gain some energy by simply being heated up. Its remaining capacity, on the other hand, would not change significantly.
- the same battery would deliver 95% of its energy to load X, but only 70% to load Y. This is also true for the capacity in some chemistries, but to a lesser extent.
As you can see, if you say "I have a batter of 1 W·h", you'll need to specify in which conditions this energy has to be consumed, while "a battery of 1 A·h" characterizes the battery itself, not the environment where it will be used.
answered Apr 28, 2016 at 9:28
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\$\begingroup\$ I think the statement "the same battery would deliver 95% of its energy to load X, but only 70% to load Y" needs a bit more information to explain why that might happen. Do you mean that one load requires a higher voltage to operate, or something else? \$\endgroup\$– gbulmerCommented Apr 28, 2016 at 12:23
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1\$\begingroup\$ He means that if load Y discharges the battery at a current, say, ten times faster than load X then the battery Ah is de-rated. The Ah rating is specified for a particular discharge current. Decrease the current and battery capacity goes up. Increase it and it goes down. (More heat losses in the battery.) \$\endgroup\$ Commented Apr 28, 2016 at 13:03
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The marketed mAh value is obtained as follows: The manufacturer puts the battery in a constant-current sink circuit (e.g. 100mA), and checks the voltage of the battery, which then decreases slowly. Once the voltage reaches a minimum threshold (let's say 1.2V for an alkaline battery), The manufacturer multiplies the current by the time it took to reach this value and says it's the capacity of the battery.
So the mAh value actually depends on at least those two parameters: current drawn and voltage threshold (it also depends on temperature, etc...). But if the current drawn is smaller, they show better capacity, for example.
Now for the voltage: The manufacturers usually provide graphs that show how the voltage evolve during this test. Here is one:
So, the mAh is given for the voltage of the battery all througout this test. Now, if you want to have the total energy that the battery can give (expressed in Wh : Watt-hour), which is the value you seemed to be interested in, you integrate this graph (not easy), but it's not really necessary anyway, since the voltage stays within reasonable bounds. Just multiply the mAh by the average battery voltage.
To sum up and go back to you question
If you want to compare batteries: if they are specified with the same voltage, you can directly compare the given mAh values. If they have different nominal voltages, multiply the mAh values by the voltage of the batteries, you'll have mWh values that you can compare to each other.
Now, whether the battery energy is drawn "quicker" at 5V than at 2.5V is not really relevant. It depends solely on the load type.
Finally, why the manufacturers indicate the mAh rather than the mWh ? I don't really know. It's like that. Maybe because the mAh value is easier to get from the above graph.
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\$\begingroup\$ Interesting - though when, say, using a USB battery - isn't it supposed to be able to provide 5V at all times?! Many divides require at least that (or have some other minimum requirement). \$\endgroup\$– niicoCommented Dec 12, 2017 at 17:09
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1\$\begingroup\$ @niico I don't know what you mean by "USB battery". But I guess it is those battery-powered portable devices that are able to charge a phone through USB, right? In this case, it is not just a battery. It is a battery followed by a boost converter that is able to output 5V from whatever voltage the battery provides. The battery itself within this device isn't providing 5V at all times. \$\endgroup\$– dimCommented Dec 12, 2017 at 19:34
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\$\begingroup\$ USB battery packs - they are very common now, Anker for example make them. People use them to charge their phones. Your point on batteries v boost converters is interesting. However, these devices also state their power in mAh - eventhough they do have a boost converter. \$\endgroup\$– niicoCommented Dec 13, 2017 at 16:04
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As I understand it, most lithium ion batteries are rated at 3.6 volts. Since the voltage is standard, you can compare the mah of two lithium ion batteries and determine how much relative power they will each provide. As others have pointed out, the actual usable power is less than the full capacity and how they are used also effects the power they will provide. For instance, the slower you use the power, the more total power the battery will provide. The quicker you draw it down, the less power it will provide. So that mah rating is only nominal, the actual power provided will be different.
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1\$\begingroup\$ Your answer is, in general, correct. A few details: 'mAh', not 'mah'. "... the actual usable power ..." You mean the actual usable energy. (Energy = power x time.) Similarly, "... also effects the power they will provide." should read "... also affects the energy they will provide. Welcome to EE.SE. \$\endgroup\$ Commented Jan 11, 2019 at 14:45
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Since the Voltage of a battery is fixed, a 2.5 V with 1000 mAh battery will drain at the same speed as a 5 V with 1000 mAh, when both provide a current of 1A. But the power of both batteries are surely different. How much a Power a battery can provide depends on each single battery. Hope this could help.
