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I've noticed an odd trend in the lithium-ion batteries used in smartphones and tablets: rather than the 3.6V or 3.7V per cell typical of most Li-ion batteries in other types of consumer devices, they use 3.8V batteries that are charged to a maximum voltage of 4.35V (this is the case with both my Nexus 5X and Nexus 9). In at least one case (the LG G5 battery), the battery has a nominal voltage of 3.85V and is charged to 4.4V.

What's with these high-voltage Li-ion cells? I can understand that the higher voltage translates to more overall energy, but why pursue higher voltage instead of just higher capacity (as is done with 18650 cells)? Is there a drawback to using this type of battery?


A chat discussion starting here suggests that this higher voltage is specific to Li-poly batteries and does not apply to cylindrical cells like 18650 or prismatic cells like the sort used in compact camera batteries. Is this indeed the case?

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  • \$\begingroup\$ Possibly improvements in construction and/or chemistry that permit this? Higher charged-state voltage translates to higher internal mechanical forces in Li-po batteries, I think. \$\endgroup\$ – user2943160 Jul 6 '16 at 22:52
  • \$\begingroup\$ I am not sure we can provide an exact answer to your question. These details are at the manufacturing level and as process and materials improve the higher voltage is a natural side effect. The fine details are proprietary and copyright protected, so even Wikipedia is not likely to offer help. \$\endgroup\$ – Sparky256 Jul 6 '16 at 23:23
  • \$\begingroup\$ @Sparky256: I'm not necessarily looking for information on the underlying chemistry. I'm looking for information about the practical reasons for preferring this kind of chemistry in mobile devices and the tradeoffs involved. \$\endgroup\$ – bwDraco Jul 6 '16 at 23:34
  • \$\begingroup\$ @bwDraco. It is a refined chemistry, not a new chemistry, or the change in voltage would be more dramatic than 10th's of a volt. The battery manufactures protect these process details. They are not made public, as 15 minutes of searching has yielded no results. It is like asking how capacitors are smaller but with higher or the same capacitance. It is about a better manufacturing process and more pure materials. \$\endgroup\$ – Sparky256 Jul 6 '16 at 23:50
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    \$\begingroup\$ I'm going to guess that if they leave the voltage the same, they can only increase capacity by making the cells larger/thicker versus some chemistry voodoo they found to increase voltage and increase overall capacity without increasing physical size. \$\endgroup\$ – Vince Patron Jul 7 '16 at 0:42
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So I did some research and found out that there's a recent advance in battery technology that allows LiPo cells, used in both mobile devices and hobbyist/RC applications, to operate at higher voltages. Specifically, a silicon-graphene additive is used in the anode to protect against corrosion at higher voltages, allowing them to be charged to 4.35V or even 4.4V. This results in slightly higher energy density, but charging the battery to higher voltages can reduce its service life.

The high power consumption of mobile devices means that high energy density is more important than any other characteristic. This means that reduced service life is an acceptable trade-off; since the typical consumer replaces their smartphone every two years, service life is not a major requirement.

In essence, the higher voltage is just another avenue of increasing overall energy density.

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    \$\begingroup\$ +1 for a nice summary of all the above comments, and good intuition. \$\endgroup\$ – Sparky256 Jul 7 '16 at 1:07
  • \$\begingroup\$ Practically, they cheat by overcharging them. That's why they die faster. I can go 4.4V with any Li-Ion cell but it won't last as long as usual. \$\endgroup\$ – Overmind Feb 12 '18 at 6:20
  • \$\begingroup\$ @Overmind, no, they don't "cheat". This answer specifically states that advances in technology allows this to happen. Modern 4.35-V cells have the same or higher number of charge-discharge cycles, which is guaranteed by research, characterization, and production tests. \$\endgroup\$ – Ale..chenski Apr 17 '18 at 21:58
  • \$\begingroup\$ Yes, they cheat. Charge them to 4.2 only and you'll see highly significant improvements in battery life. I can OC any 18650 cell to 4.35 and it works fine, but it will end up with a significant shorter life-span. It's proven and tested for LG, Samsung, Sony and Sanyo/Panasonic cells. \$\endgroup\$ – Overmind Apr 18 '18 at 12:56
  • \$\begingroup\$ @Overmind: No. There's an actual change in cell chemistry which allow this. For normal cells, 4.35V is unsafe to charge to on a regular basis and is basically the safety margin. For the new cells, 4.35V is a safe voltage, even if charging to that voltage accelerates cell degradation. It's a matter of cell venting/fire risk more than anything else; the improved chemistry mitigates this risk and allows higher voltages. \$\endgroup\$ – bwDraco Apr 18 '18 at 15:52
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this number 3.6-3.7-3.8V is the nominal voltage of the cell during its discharge. example: a battery goes full 4.2V to empty 3.0V at a linear rate, will have a nominal voltage of 3.6V. A second battery goes from 4.3V full tot 3.3V empty will have a nominal voltage of 3.8V

if your device is using 3Watts of power, then the battery will have to deliver 714mA at 4,2V but when it is almost empty at 3.0V the battery has to deliver 1000mA. the battery capacity of (example=>) 1500mAh will be empty more fast. The second battery will deliver 697mA at 4,3V up to 909mA at 3,3V when almost empty.

a 3.8V 1500mAh battery will work longer than a 3.6V 1500mAh battery. a more steady discharge voltage is better then a bigger capacity on a battery. the most important for your device is the Wh rate.

3.8V x 1800mAh= 6.8Wh

3.6V x 1900mAh= 6.8Wh

A device using 1W will work 6.8hours with both battery's

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