So I've seen batteries showing their capacity in mAh and I've seen others showing in Wh. Phone batteries are always measured in mAh, while for some reason BatteryInfoViewer on my computer only shows in Wh, but doesn't show any amp levels.

So here's two different batteries.

My phone's battery; 3.1 Ah x 3.85 volts = 11.9 Wh My laptop's battery; 42 Wh / 17 volts = 2470 mAh

When you look at it in mAh, the phone's battery seems to have more capacity than the laptop, which made no sense before. But now with Wh, laptop's battery has 3.5 times more capacity than the phone. Why do we have two different measurements for capacity?

But I really don't get it... Why is it less in mAh but more in Wh? I get that it's because of voltage but that's not the question.

The question is... If I were to somehow use my laptop's 3100 mAh (42Wh) battery on my phone, would it last longer or shorter?

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    \$\begingroup\$ Please check your phones battery, I assume you mean 3.1Ah instead of 3.1mAh \$\endgroup\$ Sep 13, 2018 at 6:34
  • \$\begingroup\$ Yes, sorry, my mistake. I meant to write 3.1 Ah. \$\endgroup\$ Sep 13, 2018 at 6:37

4 Answers 4


Why is it less in mAh but more in Wh? I get that it's because of voltage but that's not the question.

It might not be the question, but it is the answer.

If I were to somehow use my laptop's 3100 mAh (42Wh) battery on my phone, would it last longer or shorter?

Wh is a measure of the amount of energy stored in the battery. With an efficient voltage converter the battery with the higher energy would power your phone for longer.


Wh is energy, Ah is charge.

The difference between them is the voltage at which they're delivered, which of course varies during battery discharge.

In the bad old days when we only had linear regulators, Ah was the more appropriate measure. Regulators reduce the voltage to what the circuit needs, and neglecting any small operating current for the regulator, its output and input currents are equal. This means the load current comes from the battery, and Ah at the load is the same as Ah at the battery. Extra battery voltage is just wasted heat energy in the regulator.

Now switched mode power supplies (SMPS) are commonplace, we can use all (almost all, SMPSs have small but finite losses) of the energy in the battery. When we step down to a lower voltage, the output current of the SMPS can be, and often is, higher than the input current. Ah becomes a less useful figure, and Wh is more relevant. Extra battery voltage means more energy to prolong the load run time.

Even though Wh is more relevant, more people understand, or are at least familiar with, Ah, which tends to continue to be the way people compare and buy batteries.

That's the science bit. On a more cynical note, the battery headline specifications, in the adverts, will be whatever figure makes the product look better. In the case of a nominally 3.7v battery powering a 5v powerbank, the battery's Ah rating looks better that that of the powerbank!


Well, the biggest issue here are traditions.

In general, when we talk about the battery capacity, we should internally think about the amount of energy which is stored inside of the battery, since electrical devices consume energy to work.

Amount of energy is measured in joules. As unit of power, watt is one joule per second. Therefore, if we multiply watt value by a unit used to measure time, we get energy. So one watt*hour is 3.6 kJ.

On the other hand, we have the ampere*hour measurement unit, and for portable devices, its derivative milliampere*hour.
As you noted in your question, we need to multiply the ampere*hour value by battery voltage to get the watt*hour value, and we've already established that the watt*hour is a measurement unit for amount of energy.

Therefore, to properly compare run-times using different battery types, you need to use the energy values, that is to say the watt*hour values. Since there are different "standard" laptop battery voltages, you need to compare them using energy that they store, which is what your software seems to be doing.

This still leaves us with a bit of confusion about famous "mAh" numbers we often see on batteries.

The idea of its use is as follows: My device uses the battery's nominal voltage, and it needs current of x mA to function. Therefore, if I power it with a yy mAh battery, my device will work for yy/x hours.

However, for this approximation to work, you need to fulfill the assumptions. Pretty much all phones today use a single cell lithium based battery, with nominal voltage of around 3.8 V. Therefore, it's useful to use mAh to compare such batteries. Cars often use batteries with nominal voltage of 12 V, and Ah is used to compare them. However, when you're trying to compare two dissimilar battery types, you can't use the Ah values, since the battery voltages are different.

To finally answer your question: If you had a way to lower voltage of your laptop battery to the voltage your phone needs to work, and you could do that without wasting any energy, then the laptop battery would last longer, since it has ~3 times more energy than your phone battery.

In real life, we could achieve efficiencies in the 80% to 90% range, perhaps even a bit more.
If you were, however, to just turn the extra voltage of laptop battery into heat, then the phone battery would last longer.

  • \$\begingroup\$ But... Lowering the voltage level would cause a drop in Wh levels, right? For instance, now it has 2.47 amps x 17 volts = 42 Wh. But you're saying that if you lowered the voltage to 3.85, it would last longer than your phone's battery. My phone's battery capacity is: 3.1 amps x 3.85 volts = 12 Wh Lowered voltage of my laptop's battery's capacity would be: 2.47 amps x 3.85 volts = 9.5 Wh So how is it that it's gonna have more Wh in it when we lower the voltage? \$\endgroup\$ Sep 13, 2018 at 6:45
  • \$\begingroup\$ @Burak Okuroğlu No, you need to be careful here. I said that, if you could lower the voltage without wasting any energy, then it would last longer. So I'm not actually lowering the voltage of the battery itself, which would cause Wh decrease. I'm lowering voltage after battery. This would lower the voltage consumer sees, but it will also affect the current. \$\endgroup\$
    – AndrejaKo
    Sep 13, 2018 at 6:52
  • \$\begingroup\$ Hmm... And that's exactly what my phone does when it charges, right? Because my phone charges deliver between 5 - 12.5 volts but my phone is always at 3.8 and 4.2 volts. \$\endgroup\$ Sep 13, 2018 at 6:55
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    \$\begingroup\$ @Burak Okuroğlu Well, that could be one way of looking at things. Let's say your phone needs 100 mA at 3.8 V to work, and you have a 38 V power supply + 100% efficient converter. Since phone needs power to work (3.8 * 0.1 = 0.38 W), it would consume 10 mA from the 38 V power supply (38*0.01=0.38). \$\endgroup\$
    – AndrejaKo
    Sep 13, 2018 at 7:00
  • \$\begingroup\$ So, even though my phone's battery voltage can only go up to 4.2, the adapter can somehow make use of deliverigng 12.5 voltage? How does that work? \$\endgroup\$ Sep 13, 2018 at 21:15

One is capacity and one is Energy.

The European Commistion Regulation No 1103/2010 states in Article 3:

The capacity of portable secondary (rechargeable) batteries and accumulators shall be expressed in ‘milliampere-hour(s)’ or ‘ampere-hour(s)’, using the abbreviations mAh or Ah respectively.


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