# Why some phones need a 5V charger although the battery needs 3.7V only?

I've many mobile (smart phone) chargers. The output of them is 5 volts but the battery needs 3.7v only.

Why is it better to make a 5v charger then convert the 5v to 3.7v ?

Why don't we just make a 3.7v charger?

• If you apply 3.7 volts to a Li-ion battery, it will discharge to 3.7 volts, not charge to 4.2. If you apply 4.2 volts, it will take infinite time to charge it fully. – Dmitry Grigoryev Mar 2 '16 at 17:41
• In one word, Overhead. – Passerby Mar 2 '16 at 22:06
• @Passerby The word is headroom, I believe. (Electricity bill is overhead.) – Nick Alexeev Mar 3 '16 at 4:51
• Well, overhead is what it needs to operate. Headroom would be anything extra used/saved for worst case scenarios. Overhead would be like the minimum dropout on a regulator. Headroom would be minimum dropout + 1~2 Volts, to compensate for inrush or battery drain. See en.wikipedia.org/wiki/Headroom_(audio_signal_processing) – Passerby Mar 3 '16 at 5:05
• Because users want to plug the phone into a USB port, and USB ports provide 5V. (At least historically) – immibis Mar 3 '16 at 6:01

If you look at the charge profile of a lithium battery you'll see that at certain points it changes from constant current to constant voltage charging: -

This means that some form of "in series" charge control mechanism needs to be present to act initially as a constant current source then change to a constant voltage source. This charge control circuit comes with an overhead - it needs maybe 0.5 volts across it (minimum) to do its job.

Given that the final part of the charge regime is constant voltage at usually 4.2 volts, it makes sense to use a wall wart with 5 volts output. Even if the wall wart dropped to 4.7 volts the charge circuit would still have enough overhead to deliver 4.2 volts.

• Worth emphasizing the point that in order to control current, you need in essence to place a "variable resistor" in the path of the current; whatever circuit element this is, it will cause a small voltage drop (which you refer to as "overhead"). The ability to measure and control the current depends of there being some minimum voltage available. There is also the question of voltage drops across corroded terminals, power source fluctuations, ... It all comes down to headroom. – Floris Mar 3 '16 at 13:47

Another reason is standarisation - a few years ago the EU pushed for standarisation of the charging connector in phones which ended being micro USB. USB implies 5V.

And as a personal note I worry about those devices which require over an amp to charge properly charging through USB 2.0.

When you have battery charged at 3.7V it's basically means it is charged to 0%.(once load applied voltage drops)

When it is charged to 4.2V it is full charged(100%).

4.2 =100% 4.1 =80% 4.0 =60% 3.9 =40% 3.8 =20% 3.7 =0%

This is how we get how much battery is charged by calculating the voltage of battery.

Why don't we just make a 3.7v charger?

if you charge with 3.7V charger it's basically means your battery always will be at 3.7V (0% charged). you won't able to power on your mobile.

• The voltage of a Li-ion battery over a discharge it not linear as implied here. The discharge curve is mostly flat. – Gorloth Mar 2 '16 at 23:58
• This is wrong - 3.7 is not fully discharged, it's around 50% or nominal - a fully discharge is around 2-3v (protections that prevent over-discharge should cut in before it reaches 2v to prevent damage to the battery, usually around 3v). – user2813274 Mar 3 '16 at 14:42
• @user2813274 3.7 is around 0% because when you put load(cell phone) to 3.7V it voltage drops around 3.6 or less. actually checked my LG phone when it has the voltage of 3.6V i tried to on my phone but i was not able to then i charged manually around 20 sec voltage was above 3.7V then my phone got powered on. Please check with your phone battery i hope you will find something new.(but it won't be around 2V) – editinit Mar 3 '16 at 17:24