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I'm having this disagreement with a good friend of mine regarding charging voltage of the cells inside modern smartphones when quick charge, USB PPS and other techniques are used.

Let's take as an example Samsung's S21.

  1. In my opinion, since its using PPS, the device and charger are regulating either the voltage or current (or both) for different charging phases, which in fact allows the 4.2Vmax cell to be charged with 9V/1.67A.

  2. In my friend's opinion, he insists that the cell cannot be charged with actual 9V at any given moment or the battery will explode. He insists that there is not a single regular smartphone cell which could be charged with over 5V, with each smartphone having some sort of module that will always prevent the cell from charging the cell with more than 5V.

Which theory is correct? I've scoured the Internet back and forth and I couldn't find a definitive, clear answer.

Are smartphone cells actually being charged (even for a moment) with more than 5V or are they actually never charged with more than 5V?

Keep in mind that I'm writing this question here not to prove who's right or wrong, but so both of us would have a better understanding of the modern charging techniques.

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    \$\begingroup\$ If you want to be 100% sure, you need to look at the charging IC/chip's graph, on its datasheet to be sure what voltages/currents it puts into the battery. So find out what kind of battery the said phone charges, open it up, see its IC, google its datasheet and see the specs/graphs. If you don't want to tear your smartphone apart, what you need to know, is that depending on the battery type, there is a charging curve, that must be followed for the battery to get charged "correctly". If you google " fast charging li ion curve" and compare it with normal charging li ion curves, might help \$\endgroup\$ 2 days ago
  • \$\begingroup\$ Thank you @ChristianidisVasileios the problem is that I could not find a single datasheet for modern smartphone batteries, let alone what max voltage it can be charged at. There's no concrete information available. That's why I made this post. S21 5G Cell I could not find a single datasheet or graph that would answer if this cell - or different cell is actually charged above 5v at any given point. \$\endgroup\$
    – olokos
    2 days ago
  • \$\begingroup\$ @olokos But the cell is just like any other cell, and charged just like cells need to be charged, it will have no clue what the power supply outputs as long as the charging circuit charges it properly. \$\endgroup\$
    – Justme
    2 days ago

5 Answers 5

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Your friend is right in that you cannot charge a lithium ion cell with just any old voltage.

The battery university goes into some detail about it.

Short form:

  • A single lithium ion cell has a voltage of a little over 4V.
  • Charging with a voltage much above the cell voltage will cause the cell to go "bang" and probably catch on fire as well.
  • The charge cycle has phases of constant current and constant voltage in order to keep the cell operating in a safe area while charging.

A typical charge cycle looks like this:

enter image description here

Image from the battery university lithium ion charging page.

The phones have an integrated charging circuit that takes care of all that. The USB power supply just delivers power - it isn't actually the charger. All the smarts for safely charging the cell are in the phone.

The cell itself will often include safety cut offs so that it won't charge if the voltage is too high or discharge if its own voltage is too low. Either condition can cause a lithium ion cell to go up in smoke.


Most phones use a single cell. Tool battery packs are usually built of multiple cells.

Battery packs for rechargeable tools might charge at the voltage of the pack,or they might charge each cell separately at the normal cell voltage.

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    \$\begingroup\$ @olokos With a higher voltage, you can get more power at lower current. USB PD allows up to 20 V at 5 A for 100 W of power, but if you were going to do that at 5 V you'd need 20 A, which means you need bigger, more expensive cables to make sure the cable itself doesn't heat up too much (which would waste energy as well as potentially being a fire hazard). It's the same reason mains power isn't at 120/230 V until right before it enters the building--using higher voltages for transmission reduces power loss. \$\endgroup\$
    – Hearth
    2 days ago
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    \$\begingroup\$ @olokos The "extra" voltage isn't lost, as it's stepped down using a switching converter, not a linear regulator. Unlike linear regulators, step-down switching converters output more current than is input--an ideal one has 100% efficiency and practical ones can be well over 95%. Also, even at 5 V you need to step the voltage down; applying 5 V to a lithium-ion cell will destroy it just as well as applying 9 V would. \$\endgroup\$
    – Hearth
    2 days ago
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    \$\begingroup\$ @olokos Every smartphone will charge the cell at 4.2 V maximum. 5 V is easily enough to destroy the cell. But yes, that's the main purpose. Not only cheaper, though, as the larger wires will be stiffer and more inconvenient to the user. \$\endgroup\$
    – Hearth
    2 days ago
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    \$\begingroup\$ @olokos You have to understand two things, one thing is the power supply, and other thing is the device which has a built-in battery charger, which takes in power from the power supply, and uses it to charge the battery safely. QC is a protocol between power supply and the phone how much voltage there is between power supply output and charging input. QC has nothing to do with what the battery charger does to charge the lithium cell. You can't charge lithium batteries by connecting them to a power supply. \$\endgroup\$
    – Justme
    2 days ago
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    \$\begingroup\$ @olokos No they provide higher voltage to allow higher power charging with less losses over the same wires/cables you already have and with same current the wires/cables are already rated for. \$\endgroup\$
    – Justme
    2 days ago
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Obviously the 9V will not be directly put on the battery. That's because lithium batteries cannot be charged safely by connecting them to a constant voltage anyway.

Your mains adapter which for example outputs 9V and 1.67A is just a power supply. Just like it can output constant 5V, the devices can negotiate to make the output 9V, to allow more power for charging with same current between the power supply and the phone.

There will be a battery charging circuit in the phone.

It takes in whatever voltage it has negotiated and likely uses a switch mode converter to convert the incoming voltage to charge a lithium battery in a way that lithium batteries need to be charged safely.

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  • \$\begingroup\$ Exactly! So when the handshake between phone and charger does allow for 9v, what happens then? Is the voltage always brought down to 5v and those 9v are converted into higher amperage output instead? Seems like a massive energy loss, instead of just getting 5v and higher current. \$\endgroup\$
    – olokos
    2 days ago
  • \$\begingroup\$ @olokos Minimal energy loss if the voltage is brought down with an SMPS, which as you say, increases the amps to maintain the power/energy more or less the same (less unavoidable losses of 5%-15%). A SMPS is ideal for this, as it minimises the power loss as the battery voltage varies between 3 and 4.2 V during charging, and allows for the charging PSU to be any reasonable voltage, 5, 9, 12, or higher if designed for it. \$\endgroup\$
    – Neil_UK
    2 days ago
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    \$\begingroup\$ @olokos No, the 9V will not be brought down to 5V as 5V is not used for battery charging, and no, there will be no massive energy losses. I already mentioned the input voltage is converted down in a "switch mode converter" and that is specifically done to avoid losses. The voltage is brought down to a level of voltage at battery terminals, to push constant charging current into battery, until battery voltage has risen to maximum level of about 4.2V and then current is allowed to flow into battery at the constant 4.2V charging voltage, until the current drops to a level it is fully charged. \$\endgroup\$
    – Justme
    2 days ago
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It's a programmable source to a Buck CC Charger Regulator.

You are both correct, yet don't know how it is used. The programmable 9V charges an inductor to the required current and then discharges into the battery in repetitive CCM with a hysteretic comparator.

Then it changes current levels when safe and then at Vmax changes regulation to CV then cutoff. Something like this;

enter image description here https://www.pengohome.com/UploadFile/images/PPS-EN.jpg

The profile depends on optimal efficiency.

For giggles, I did a quick design of charging a supercap so that it charges up quicker since the simulation is in slow motion.

Notice how increasing the input voltage increases the frequency, changing the hysteresis ratio on the comparator changes the ripple current and the offset reference controls the current. No cutoff was included.

For brevity, not all the design specs were included, but the schematic shows component values and scope plots. FET values are available with the mouse Lt. click properties. ESR and DCR values were added. Charge current was set for 1.9A max.

enter image description here

The main advantage of the PPS is that the smartphone can remote-program the supply voltage so that a simple hysteresis buck converter in CC mode with some ripple can be current regulated then CV regulated with an extra comparator. Adjusting the voltage difference then limits the frequency range of the converter with a lower voltage difference yields a lower switching frequency. This one ranges from 50 kHz to 500 kHz at max 9Vin roughly but regulating the voltage difference allows one to optimize the switching frequency.

Each trace can have unique time scales and initial conditions were 3V after reset using 3V RRIO Op Amps. All high current paths are affected by component resistance values.

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    \$\begingroup\$ Thank you for your explanation of PPS! With the knowledge gathered so far, the point still remains, that the cell is in fact never charged with voltage above 5V, usually with charging voltage never above 4.2v, despite the charger outputting 9V over the cable? Please do confirm or deny, to get a definite answer, even when PPS is considered. \$\endgroup\$
    – olokos
    2 days ago
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    \$\begingroup\$ Yes the PPS voltage is slightly above the Yellow battery voltage which NEVER exceeds ratings and the current is constant but raised in steps until CV. The voltage difference is controlled by a switch with dI(t)/dt=ΔV/L Thus ΔV could rise to 9V if L is large enough or dt is small enough. So it depends on the design for ΔV=LdI/dt \$\endgroup\$ 2 days ago
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Most cell phone charging works the same as EV level 1-2 charging.

That external thing is not a charger!

https://youtu.be/RMxB7zA-e4Y?t=122

The connection (USB cable; J1772 cord) is just a fat pipe to bring raw power into the device. Conversion must be done. No phone can charge direct on 5V, and no EV battery can charge direct on 120/230/240V. That would be bad.

Onboard the device is the actual battery charger - which is responsible for managing and guarding the unique battery chemistry to that phone/car. That does the conversion from "whatever the input power is" to "exactly what the battery needs". It is built into the phone/car, so there is no chance of getting mismatched.

What PPS does is let the phone ask for a particular voltage which makes its job easier and requires less conversion and less inefficiency in conversion. It's like if you have a 450V pack in a Tesla and the charger could tell the level 2 EVSE "hey can you give me 363V? That rectifies into an ideal voltage for me and I won't have to buck-boost."

Now to follow the EV metaphor further, EV DC fast charging is hot-wiring the external charger to the battery, the way you think PPS is. But PPS is not that - quite. The phone's battery management system is still controlling the power. For good reason: a level 3 DC fast charger is a $100,000 piece of equipment. A PPS USB block is what, $30? The phone has to be put on guard from a cheap/knockoff PPS that might misbehave. A Tesla doesn't.

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  • \$\begingroup\$ Also, a phone charger is pulling power through a small cable at low voltage. Using higher voltage and lower current could be less total loss even if bypassing a conversion step inside the phone was something they wanted to try. (e.g. 1A @ 4.2V instead of 0.466A @ 9V has more than 4x the I^2 R losses. \$\endgroup\$ yesterday
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I think the trick with the higher voltage on the USB cable is there to prevent the wires of the cable to become too hot. If a power supply delivers 27 Watt, that would be more than 5 Amperes at 5 Volt; more than the cable can handle. At 9V that would still be 3A, requiring a good cable. My guess is that the voltage is even more than 9V as recent phones charge with 33 or even 66W.

So the charger in the phone converts the (higher voltage, lower current) back to (lower voltage, higher current) to match the battery cell.

See also (for example) USB Power Delivery explained: What you need to know about ubiquitous charging

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