# Why do the Wh and mAh ratings of a USB battery pack appear to be inconsistent?

I received a battery pack that I had ordered today, one of those ones that provide 5 volts using a USB connector, and plug into a USB socket to charge.

I noticed that on the box, it said it was rated for 6000 mAh. The battery itself said 22.2 Wh, and that the output voltage was 5V DC. Converting that back to mAh was: (22.2 / 5) x 1000 = 4,400 mAh. About 75% of the 6,000 mAh printed on the box!

Feeling shortchanged, I picked up another (Li-Polimer) battery pack from a different manufacturer I had lying around and saw it's capacity was: 15 Wh, 4,000 mAh. The voltage again being 5 volts. Converting the Wh reading to mAh was: (15.0 / 5) * 1000 = 3,000 mAh (about 75% of the mAh reading).

My guess is that as the batteries deplete, like the lead-acid batteries of old, the voltage decreases. So if the voltage dropped to 3 volts, it would still deliver the milliamps for the mAh reading, but the Wh for the watt hour reading would be reduced.

I suppose this means that the useful capacity of the battery (the period of time where it actually delivers 5 volts) is even less... perhaps only just half of the mAh rating?

• you are mixing up the actual battery capacity for the output charge capacity. The cells inside the package would be passing through a buck boost converter where it is likely to lose 10-20% capacity. In addition the nominal cell voltage is likely 4.2 V when fully charged ... try the maths again using those figures Oct 8, 2015 at 3:16
• Sloppy general practice /marketing / convention. Battery Ah x 3.6 of 3.7V = Battery Wh. Output Wh = Battery Wh x converter efficincy BUT about mobody states output Wh. Generally there are enough unknowns that knowing ~~ Ah of internal 1W LiIon/. LiPo cell is enough. Converter MAY be 90%+ efficint but is likely 75%-85% and may be ;other'. Rhen the phone etc MAY downconvert from 5V to internal LiIon with an internal buch but many don't. Internal LiIon may be at 3V when fully discharged and should be 4.2V when charged so IF converter feeds 5.0V efficieny of resistive dropping inside phone ... Oct 8, 2015 at 8:06
• ... ranges from 3/5 = 60% at flat internal battery to 4.2/5 = 85% with fully charged internal battery. This los is in addition to boost converter loss in external power pack. Oct 8, 2015 at 8:08

Firstly the battery is not $5\mathrm{V}$, it is nominally $3.7\mathrm{V}$ as pointed out already.

However the confusion lies in misinterpreting the $\mathrm{mAh}$ rating of the battery, and it has nothing to do with converter losses, and everything to do with

$$P=I\times V$$

If the battery is rated at $6\mathrm{Ah}$, it means you can draw $1\mathrm{A}$ for $6\mathrm{\space hours}$. So lets say you are drawing $1\mathrm{A}$ from the battery. This means it is delivering:

$$P=I\times V = 1 \times 3.7 = 3.7\mathrm{W}$$

Now lets say you feed this through an ideal boost converter. In this case $P_{in}=P_{out}$. So assuming we boost it up to $5\mathrm{V}$, this means that the current that must be drawn from the output to discharge the battery at this rate is:

$$I=P/V=3.7/5=0.74\mathrm{A}$$

So what this means is at the higher voltage, you can draw a current $0.74\mathrm{A}$ for $6\mathrm{\space h}$ for that battery capacity - so the output capacity is $4.44\mathrm{Ah}$.

Again, this is not converter losses - in fact with losses the number would be lower. Instead it has to do with the fact that you are sacrificing current capacity for a voltage gain which is how a boost converter works - if this wasn't the case then you would have invented free energy.

Essentially at a higher voltage, the energy delivered by each unit of charge $\left(\mathrm{V}=\mathrm{J}/\mathrm{C}\right)$ is higher, hence at a higher voltage but lower current you are still delivering the same energy.

• OK that makes sense, but it doesn't make sense that they provide readings on the outside of the battery pack for 6Ah when someone connecting 1A circuit to the battery pack would never be able to see 6 Ah in practice. Surely those ratings belong to the batteries inside, and the rating on the outside should be adjusted to take into account of the boost converter?
– xirt
Oct 8, 2015 at 6:49
• @xirt If you draw $1\mathrm{A}$ from the output of the converter, you are drawing $1.35\mathrm{A}$ from the battery, so with $6\mathrm{Ah}$ capacity you get $6/1.35=4.43\mathrm{\space h}$ run time. I can't comment on the labelling of the pack, its either marketing or safety (knowing how much lithium in the battery?), don't know. Oct 8, 2015 at 6:55
• @TomCarpenter the answer is 100% true but I want to add, besides boost converter efficiency that is 80% (most likelly), in case of a portable battery you boost at 5V and then you regulate at 3.7V inside your phone with efficiency 90% so with a battery with 6Ah at output you have (empirically) 3.5 Ah and you charge your battery with 3.15Ah. The battery efficiency is 60% (LiPo) so in your battery you will have 1.89Ah... This is the truth about energy... In case of a portable battery you can use only 31% of your battery capacity and you consume 11Ah to charge that battery. Oct 8, 2015 at 8:01
• @FlorinPutrea I'd like to see a source of those claimed 60% battery efficiency. The numbers I usually see are like 96%+ charge efficiency, or is your battery efficiency not charge efficiency? Oct 8, 2015 at 14:32
• @Arsenal The battery efficiency is the difference between the amount of energy you give to battery (the output of the source) and after disconection what amount of energy that you will have in the battery. The difference is because of the chemical processes that happens inside the battery during charging, actually this efficiency is caused by the chemical processes. What is different from your question is 96% - 90% charge efficiency... 96%+ is for phone chargers from 220V/110V alternating current that gives you 5V DC After this you have a voltage regulator on chip that have a randament of 90% Oct 9, 2015 at 11:12

Marketing trick, like hard drive capacity being base 2 in an OS while manufacturers state the size in base 10 (gibibyte vs gigabyte). The capacity in mAh is raw capacity of the battery, before any regulator loss. The raw battery in the USB battery pack is typically a 3.7V Lipo. As such, 22.2 Wh / 3.7 V = 6 Ah (6000 mAh). Figure 80% boost efficiency, you'd see 22.2 Wh * .8 = 17.76 Wh at the 5V output, giving 17.76 Wh / 5V = 3.55Ah, not adjusting for the Lipo's voltage range or any (if your lucky) under voltage protection.

You got played by standard consumer marketing tactics.

• See my answer for why this is wrong. It has nothing to do with converter losses. Oct 8, 2015 at 5:51

I noticed that on the box, it said it was rated for 6000 mAh. The battery itself said 22.2 Wh, and that the output voltage was 5V DC. Converting that back to mAh was: (22.2 / 5) x 1000 = 4,400 mAh. About 75% of the 6,000 mAh printed on the box!

There are two parts to the "battery pack". The first is the internal cell, usually some variant of a lithium ion chemistry. The second is a boost converter to change that cell's voltage to the 5V output you need.

The specifications you are quoting refer only to the cell, not the output of the regulator.

So the cell itself is a 6,000mAH cell, and can hold 22.2Wh of energy when fully charged and new.

Given that P = I * V we can find the cell's voltage: V = P / I --> V = 22.2 / 6.000 --> V = 3.7.

3.7V is a common voltage for Lithium chemistry cells. So the marketing packaging is proclaiming the battery's actual capacity. This is a good thing for consumers because regardless of the voltage regulator's output, the capacity is comparable between devices.

So keep in mind that the 5V output of the regulator never enters into the picture for these specifications - only the cell's voltage (which may vary depending on exact chemistry). You can still compare two packs using the watt-hour rating for capacity, and to a lesser degree the mAH rating, and find which one holds more energy.