Still doesbt increase back the mah..

Overcharging Lithium-ion

Lithium-ion operates safely within the designated operating voltages; however, the battery becomes unstable if inadvertently charged to a higher than specified voltage. Prolonged charging above 4.30V forms plating of metallic lithium on the anode, while the cathode material becomes an oxidizing agent, loses stability and produces carbon dioxide (CO2). The cell pressure rises, and if charging is allowed to continue the current interrupt device (CID) responsible for cell safety disconnects the current at 1,380kPa (200psi).

must be crazy to pump 5V directly into a 3.7V battery..

still the excess v doesnt increase back the mah..

er.. so you saying 5V with 2000mAH = 3.7V with 2000mAH ? (after 3-5% conversion loss)

2800mah x 3.7V / 5V = 2072mah..
should be able to charge ur phone at least one times with phone off..
if on wifi, play video etc while charging, wont fully charge of course.

powerbank mah times nominal powebank battery output, 3.7V, divide by its output voltage which is 5V..

simply said, the 2800mah is at 3.7V,
if change voltage to 5V, mah will drop..

2800mah x 3.7V / 5V = 2072mah..

that;'s why I said from that 5V going into the phone circuit will be reduced back to 4.2-3.7V charging.. the capacity wont be just 2072mah..

think in terms of WH, 2800mah x 3.7 = 10.36WH  at 5V it'll still be 10.36WH

when it goes back into the phone battery of 3.7V it'll still be 10.36WH (less conversion inefficiencies)

that;'s why I said from that 5V going into the phone circuit will be reduced back to 4.2-3.7V charging.. the capacity wont be just 2072mah..

think in terms of WH, 2800mah x 3.7 = 10.36WH  at 5V it'll still be 10.36WH

when it goes back into the phone battery of 3.7V it'll still be 10.36WH (less conversion inefficiencies)

no, cos the excess voltage is not used, dissipate as heat or whatever..
its like splash/pour water from a full bucket of water directly into a small bottle opening..

whatever amount of water can go in thru the small opening, go in the bottle.. Excess will spill..

Abstract: Arguably the most useful part of USB's power capabilities is the ability to charge batteries in portable devices, but there is more to battery charging than picking a power source, USB or otherwise. This is particularly true for Li+ batteries, where improper charging can not only shorten battery life, but also can be a safety hazard. A well-designed charger optimizes safety and the user experience. It also lowers cost by reducing customer returns and warranty repairs. Charging batteries from USB requires balancing battery "care and feeding" with the power limitations of USB as well as the size and cost barriers ever present in portable consumer device designs. This article discusses how to achieve this balance.

so i guess the proof will be in the empirical testing? You can test your bank charging and see whether it's x0.9 (as you say 3.7>5.0>3.7) or x0.75 (3.7>5.0 x efficiency).

there is a reason why powerbank say 10000mAH charge 3000mAH battery 3times..  cos you can ignore that USB 5V thing..  it steps up to 5V then IC controller in the phone steps it down to whatever the battery needs. (cos it assumes storage is same type of battery like the phone battery 3.7V type.)

the more accurate way is state the WH rating..  cos if cross comparing different chemical batteries then using WH is more proper.

I have an 11200mah battery and it can only charge a little less than 2.5 times.

There's a seller on ebay for Pineng powerbanks, which has got some literature BY PINENG themselves explaining that it is 3.7/5 x efficiency and that works out to about 0.7 or so x mah rating.

so maybe 11200 battery to your battery is only 0.7 efficiency when charging battery to battery lor..
what has that got to do with step up or step down voltage via USB interface losing 30% of the power?
those up n down function using ICs are in the range of 95% efficiencies

Well as I said, you can test your own powerbank and verify the numbers. TS did his own testing and has data to show for it.

Theoretically it's 90-95% efficiency from the datasheet. Practically?

definitely there is gonna be losses.. but where are those losses coming from, dont just say oh it's because 3.7 change up to 5V USB interface where 30% losses went there.. so just calculate as 3.7/5 ?

This is from Anker, a reputable powerbank manufacturer in the US.

http://www.ianker.com/download.php?file=do...26%20-%20V2.jpg

Read Notice no.4.

This is in line with test results and what I said about Pineng's literature as well.

yes notice 4 says total losses "including" heating and voltage conversion.. it doesnt say 30% loss formula is derived from 3.7/5 figure did it?

So tell me what's a feasible empirical formula to arrive at 30% ? Did you read the first few posts?

what i tested here, i am measuring using power and not mAh, which is accurate.
As you know, VXI= Power
Under 5ohm load, 4.93VX0.94A=4.6342Wh
Because of the measuring time is quite long, i take measurement on every half hour.
Total discharging time is 7 hours 20 minutes.
From calculation, i can get a total power dissipation of 34.13Wh,
the rated power is 41.44Wh ( 3.7V x 11200mAh)
34.13/41.44x100%= 82.36% is the real capacity