# How can I interpret the readings of a USB voltage tester considering the input voltage is 5V and I am charging a 3.7V battery?

I have this USB voltage tester that I place between my usb wall charger and a USB charging cable. This device gives me the total current that went through over a period of time. Let's say after 2.5h of charging I get this:

IN: 5.22V TIME: 150min 0.835A

And that means over the duration of two hours and a half 835mA passed through at roughly 5V (that somewhat varies a bit)

The thing is, the device I am charging has a 3.7V battery with a total capacity of 800 mA. And this is where it all gets confusing to me:

The total power that went through should be ~5 x 0.835 = 4.174W

However, the 3.7V battery can take 3.7 x 0.8 = 2.96W = ~3W

That's over 1 Watt difference that I don't know where it went. The charging takes place by plugging a miniUSB cable directly in the device that holds the battery. The device has a display that briefly gets lit the first 15 minutes but then I managed to shut it down and remains black throughout the charging. There is also one red LED that stays on throughout the charging duration and then turns off.

Should I believe over a quarter (over 1W) of the power passed through is waste? Or is my math wrong?

Edit: full charging video available here https://www.youtube.com/watch?v=5kYaG2KwxDw

Learn this: Energy = Power x Time.

This device gives me the total current that went through over a period of time. Let's say after 2.5h of charging I get this:

IN: 5.22V TIME: 150min 0.835A

No, that's the average current, not the total. To try and "total" a current would be like trying to total your speed on a 150 minute journey. It makes no sense.

And that means over the duration of two hours and a half 835 mAh passed through at roughly 5V (that somewhat varies a bit).

No. The mAh is calculated, as suggested by the units, by mA x time (in hours). 835mA for 2.5 hours gives 2088mAh.

The device I am charging has a 3.7V battery with a total capacity of 800 mAh. And this is where it all gets confusing to me:

The total power that went through should be ~5 x 0.835 = 4.174W

No, you want to calculate the total energy that went in. That's power by time, so $$\ E = VIt = 5 \times 0.835 \times 2.5 = 10.4 \text {Wh} \$$

However, the 3.7V battery can take 3.7 x 0.8 = 2.96W = ~3W.

Again, W is power, Wh is energy. The battery can store 3 Wh during charge.

That's over 1 Watt difference that I don't know where it went.

No, it's 1 Wh difference.

Should I believe over a quarter (over 1W) of the power passed through is waste? Or is my math wrong?

That seems reasonable. Your battery will get warm during charging and the missing energy will have been give off as heat.

• Hey there! Thanks for your answer. I believe I may have not described correctly the device that I used to measure with. I updated my question with the video showing exactly how my numbers were produced, hopefully that will clear things a bit. Aug 18 '18 at 14:12
• I didn't have any problems with your readings - only your units of measure. Have I helped at all? Aug 18 '18 at 15:14
• I do not understand why would you multiply 835 mA with 2.5 when I specifically mentioned that the 835 mA were gradually registered by the device during those 2.5h and the video I added shows the same thing. Or, let me say this. The device registered an average of 334mAh for 2.5h. Aug 18 '18 at 15:46
• Because you had confused your units in the question (and I had answered before you posted the video). You have called it both 0.835 A and 835 mAh. One is current and the other is current x time. Aug 18 '18 at 15:56
• Thank you for the correction. Considering there was no considerable heating of the battery (I never noticed it to heat during several charges), do you still believe over 1 Watt out of 4 Watts got wasted through Joules effect? That is the thing that bugs me. Aug 18 '18 at 16:03