All of us are probably used to using our phones while they are charging. This is why manufacturers make longer cables – so that the consumer can have the facility of using and charging the question at the same time.

My question is this: Assuming a cellphone consists of battery with the 'phone' (camera, screen, etc) connected in series, how can we charge it (which would require that current flow in the opposite direction of the battery terminals – we were taught this in physics class – you need to apply a voltage more than that of the cell and in the opposite direction to charge it) and use it (which would require that the current flow in the same direction as battery terminals) at the same time.

My attempt: (two possible cases)

  1. The above thinking is wrong and somehow the manufacturers managed to use and charge the battery at the same time.
  2. While charging, the battery does not provide current at all and the current from the wall socket is divided into two currents – one for charging and one for operating the phone.

Which one of the above explanations is correct? Or, is there a third, more complicated explanation?

  1. If explanation #1 is correct – is this achieved in a simple way or do I need to get a degree in electrical engineering to understand this?
  2. If explanation #2 is correct – when the phone is being charged and used at the same time, the rate of charging is noticeably slower. Shouldn't the current drawn from wall socket be increased to make it charge at the same rate?
  • 2
    \$\begingroup\$ Your case 2 is what's done in modern cell phones. See fig.1 in this TI application note. This arrangement is often called PowerPath. \$\endgroup\$ Apr 30, 2018 at 17:00
  • \$\begingroup\$ The battery and charger are in parallel, not series. \$\endgroup\$
    – zeta-band
    Apr 30, 2018 at 17:03

4 Answers 4


Explaination 2 is correct. And your intuition about drawing more current from the wall socket is correct, but there's a step in between: the USB charging system. There's hardware in the wall-wart which has to convert 120V AC power to the 5V DC power that USB uses (and thus is used to charge your phone). That hardware could heat up and break (or even cause a fire) if too much power is drawn. The designers of USB limit the current that can be drawn across a USB cable. The initial spec limited you to 500mA of 5V current, though later upgrades to the USB specification permit the 1A and 2A "fast" chargers that we see today. Regardless, the phone is not permitted to draw more power from the wall, even if it wanted to.

To meet these rules, the phones have regulatory hardware which ensure it doesn't draw more power than it is permitted to. If it can't get enough power to do everything it wants, it will simply charge the battery more slowly. In some extreme cases, with power hungry phones with GPS turned on and applications using the CPU, this can actually consume so much power that the battery doesn't get to charge at all!


Basically, the charging source, battery, and load are all connected in parallel.

If the charger supplies more current than the load requires, the excess current will be used to charge the battery.

If the load requires more current than the charger can supply, the battery will supply the excess.

While it is true that you cannot charge and discharge the battery at the same time, it "looks like" you are doing so, as the battery will change between charge and discharge automatically as the load demand or charging supply vary.

  • \$\begingroup\$ Nowadays the battery is not quite in parallel with the load. The charger IC has a system or load connection, a battery connection, and a VBUS input connection. Three separate connections. TI calls this "power path". This provides many benefits which I won't get into in a comment. \$\endgroup\$
    – mkeith
    May 2, 2018 at 19:01

Your explanation #2 of the process of charging while working is correct. A battery is either in charge mode, or in supply mode. This is achieved via complicated electronic circuitry containing analog switches and up-down DC-DC converters.

Regarding your concern about the rate of charging while working, the #2 is also valid. However, a phone knows that the supply (wall charger or else) has limited abilities, and simply can't draw more current than the charger can supply by a standard. The intelligent IC has a special means to restrain its current needs. The input current is split between system supply (used for phone functions), and battery charge.

One typical line of intelligent chargers used in portable devices is produced by Texas Instruments (formerly Benchmarq), the BQ2xxxxx line. Here is an example of typical block-level architecture of the BQ24296 IC

enter image description here

Now you need to decide yourself what kind of engineering profess is needed to make this kind of circuit.

  • 4
    \$\begingroup\$ Transmitting with the mobile radio can require bursts of more power than a basic-rate (USB spec) charger can deliver, so it's not just a question of only the leftover being usable to charge; the battery actually has to be used as a power reservoir, even while nominally running off the adapter. \$\endgroup\$ Apr 30, 2018 at 17:10
  • \$\begingroup\$ @ChrisStratton, there are things called "capacitors". A battery on long inductive leads can't supply short spikes for 2-Amp burst of RF transmitter. \$\endgroup\$ Apr 30, 2018 at 17:59
  • \$\begingroup\$ While it is possible to make something that solves the problem with capacitors (radios in USB dongle form exist afterall) it's non-trivial and more expensive. Phones traditional use the battery; it's right there, the leads aren't long, and they are engineered to work that way. The current pulse is not all that fast, which is part of what makes it a hard problem to solve with capacitors alone. When people try to use feature-phone-derived GSM modules in projects, they'll typically fail if they try to use a power supply - the radio is really designed to be connected to a phone's battery. \$\endgroup\$ Apr 30, 2018 at 18:03
  • \$\begingroup\$ @ChrisStratton, could you please explain how a 100-uF capacitor can be "non-trivial and more expensive"? Example: nimbelink.com/Documentation/Skywire/4G_LTE_Cat_M1_Telit/… Pay attention to Section 3.1.1 \$\endgroup\$ Apr 30, 2018 at 18:23
  • \$\begingroup\$ As your own link points out, that capacitor is in addition to a 2 amp rated power supply. Please explain how a legacy USB-spec charger or port can provide 2 amps... hint, it cannot. That's why phones are designed to rely on the battery. \$\endgroup\$ Apr 30, 2018 at 18:27

you need to apply a voltage more than that of the cell and in the opposite direction to charge it

No, you have to apply a voltage more than the voltage of the cell but in the same direction as the voltage the cell produces.

Getting back to the question of how devices can be used and charged at the same time.

The simplest approach is simply to connect the load, battery and charger output in paralell. If the charger produces more current than the device consumes then current flows into the battery. If the charger produces less current than the device consumes then current flows out of the battery.

Afaict this is what most phones do. It does have a couple of downsides though.

  1. Current taken by the device will reduce the current available to charge the battery. At peak currents the battery may even start to discharge with a charger connected.
  2. The device generally will not work without a battery connected. Lithium ion chargers have to have undervoltage protection for safety reasons. So if no battery is present the charger will detect an initial voltage on the battery of zero and will refuse to supply the system.

The other option is to have a bypass path that bypasses the battery completely. Afaict this is what most laptops do.

Note that the term "charger" in this post refers to the device that controls the charge voltage/current so that the battery is charged in a safe manner. In devices with lithium ion batteries this is nearly always built into a device. The external brick that users often reffer to as a "charger" is just a DC power supply.

  • \$\begingroup\$ I can attest to the last sentence regarding my MacBook Pro. You can remove the battery entirely and it will still work when connected to a power source (I had to run like this for a week when my battery failed and I was waiting for a replacement). \$\endgroup\$
    – Barmar
    Apr 30, 2018 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.