Real battery capacity is around 2300mAh.

Roph

From empty to 100% charge, using a USB power monitor. Final result, just over 2300 mAh. Certainly not 3,000 mAh as advertised.

MarcinP

Burs replied at 2017-10-13 19:50
Edited by Burs at 2017-10-13 22:28 \n\nA bit off-topic, but have to do with the battery: does the p ...

In this price range not too many phones supporting fast charging. Umi Crystal not supporting fast charging that why standard charger was supplied with phone.
If you connect more powerful charge but still 5V then nothing change to charging as phone is not able receive more than 1A. Of course higher voltage charger will damage device.

MarcinP

This information is calculated by USB adapter which is only calculating charging power by time. Please don't forget that system is judging charging voltage, current and time which could give completely different reading. Very good example is every phone with fast charging which is giving completely different reading.

penamas

Try Accubattery app, but not sure if the device allows the app to calibrate and check the battery capacity. Tried the Accubattery app on UMI Super but not successful

Roph

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MarcinP

Roph replied at 2017-10-7 08:36
Edited by Roph at 2017-10-7 08:42 \n\n Edited by Roph at 2017-10-7 08:39 \n\n
... You just describe ...

Unfortunately to you I know very well what I talking about electric and electronic as I am electronic engineering.
As you mentioned that I don't know what I talking about this is very simple example that you are completely wrong with your statement that 12V and 1A is the same as 5V and 1A. Power which is delivered to battery for charging is 12W in first example and 5W in seconds example which is showing that the battery 3,7 V battery will be charged way faster.
Unfortunately USB adapter can't calculate this correctly as charging power which is delivered to battery is completely different than this which is consumed by device.
That why you have different reading.
To calculate more accurately in this situation you need divide charging voltage 12V to more accurate charging voltage which is 5V so 12 : 5 = 2,4 multiply your reading from USB adapter.
For fully correct reading you need open battery and put charging reading adapter between charging converter (which is built in battery) and battery it self.

MarcinP

Roph replied at 2017-10-7 08:36
Edited by Roph at 2017-10-7 08:42 \n\n Edited by Roph at 2017-10-7 08:39 \n\n
... You just describe ...

Also just to let you know I do not trying disagree your thread just only explain why you have such reading on USB adapter etc.

Localhorst86

Roph replied at 2017-10-7 08:36
Edited by Roph at 2017-10-7 08:42 \n\n Edited by Roph at 2017-10-7 08:39 \n\n
... You just describe ...

I am usualy not the one to defend a company with a track record like UMI/UMIDIGI, but people seem to forget the technicalities and physics behind everything. And if you understand physics and electronics you'd know that giving the capacity of a battery is useless without also stating the voltage. What's important for a battery is its electric power, given in watts.

Your charger was pumping 2300mA at 5V. That's an an electric power of 11500mW or 11.5W (P = I * U)

Cellphone batteries are usually rated at 3.7V, at least I found most of my cellphone batteries to be rated at that. Most likely cellphones run at that voltage internaly. So if we take the power you pumped into the battery (P = 11.5W) and divide it by 3.7V (U) we receive the current that was used to charge the battery. So taking the physical formula P = I * U and solving for I, we can calculate that your battery now holds 3108mA at 3.7V

If you're going to acuse someone of lying or hiding the truth, at least get your facts straight. UMI certainly isn't perfect and are sometimes using lies or misleading statements to advertise their product, but you have proven with your measurements that their battery capacity is probably not something they lied about (they might be misleading you about the manufacturer, though).

Please also note that I was basing my calculations on the common 3.7V voltage of cellphone batteries, I do not own a Crystal (and certainly never will) so I can't tell you if this assumption is correct.

AlastairWalker

Localhorst86 replied at 2017-10-9 19:17
I am usualy not the one to defend a company with a track record like UMI/UMIDIGI, but people seem t ...

You're not far wrong though 3.7V is the NOMINAL voltage of a typical Li-ion cell, the voltage will vary from around 2.5V** with zero charge up to around 4.2V*** at full charge.

** depending on specific cell chemistry and device safety cut-out voltage this value will vary significantly.
*** a few manufacturers now have cell technologies allowing higher than the usual 4.2V

Localhorst86

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AlastairWalker replied at 2017-10-9 20:08
You're not far wrong though 3.7V is the NOMINAL voltage of a typical Li-ion cell, the voltage will ...

I am not following the battery technology sector too closely. I was expecting some fluctuation of the voltage as with any battery technology, I was not aware of the fact that the range here was so huge. But the electrical power should still be calculated at the nominal voltage, shouldn't it? So the maths would still check out and in practice it should be realy close to the calculated results, or should we use some other value?

Always willing to learn something new

AlastairWalker

Localhorst86 replied at 2017-10-9 20:43
Edited by Localhorst86 at 2017-10-9 20:53 \n\n
I am not following the battery technology sector too ...

Calculating at the nominal voltage will give a reasonable, but not exact, calculation.  It is the best number to use for such calculations as the alternative is to measure the current and voltage changes over time and break the calculation down into many small pieces with a separate calculation for each voltage then add them all together for a total value.