The FastCharge 2.0 standard has these charging specifications in the exact same charger (a VoltCraft car charger):

  • 5 V up to 2.0 A (10 W max.)
  • 7 V up to 2.5 A (17.5 W max.)
  • 9 V up to 2.0 A (18 W max.)
  • 12 V up to 1.5 A (18 W max.)

So the 7 V, 9 V and 12 V modes deliver pretty much the same power. The phone automatically adjusts to the supplied voltage. Some smartphones such as UlePhone Armor2 support 9V2A and 12V×1.5A.

For cables with higher resistance (i.e. long cables), the phone can 'request an increase of power' from the charger to compensate for the increased cable losses. So it's, "I want 1.66A, please increase the voltage until 1.66A is reached, I will deal with resistance control".

My question is:

When is utilizing 7 V × 2.5 A, when 9 V × 2 A and when 12 V × 1.5 A more suitable? The latter two deliver the same Wattage.

  • 2
    \$\begingroup\$ different battery voltages \$\endgroup\$
    – jsotola
    Jan 17, 2018 at 7:52
  • 4
    \$\begingroup\$ @jsotola unlikely, this is mostly used in the domain of phones, which as far as I know use a single cell. \$\endgroup\$
    – Arsenal
    Jan 17, 2018 at 7:58
  • \$\begingroup\$ @jsotola The UlePhone Armor 2 supports both 9V2A and 12V×1.5A. When does it use 9V, when 12V? Both equal 18W. \$\endgroup\$
    – neverMind9
    Jan 17, 2018 at 8:00
  • 1
    \$\begingroup\$ I guess it could switch to 12V and 1.5A if the voltage drops too much at 9V and 2A (because of resistance), but I'm only guessing. \$\endgroup\$
    – Arsenal
    Jan 17, 2018 at 8:03
  • 1
    \$\begingroup\$ Does every fast charger have to support every combination of voltage and current? Maybe some power sources are not capable of delivering 2A at 9V, but they are capable of delivering 1.5A at 12V. Or the other way around. So the optimal thing for the power consumer is to support as many configurations as possible. \$\endgroup\$
    – user57037
    Jan 17, 2018 at 8:21

1 Answer 1


So it's, "I want 1.66A, please increase the voltage until 1.66A is reached, I will deal with resistance control".

No, it doesn't work like that. To be able to efficiently (fast) charge a single cell (3.6 - 4.2 V) battery from more than 5 V, a switching charging controller is needed.

A switching charging controller is basically a switching converter (a DCDC converter) where voltage and current are controlled using feedback and a microcontroller to charge the battery.

Only if that switching charging controller finds that it is unable to provide a certain charging current to the battery, due to the input voltage being too low, then it will request a higher input voltage.

Due to the switching nature of a switching controller, the current at the input side (coming from the adapter) and the current charging the battery, are not the same. A 100% efficient switching charger can for example use 9 V at 1 A and using that charge a battery at 4 V at 2.25 A. Both equate to 9 Watt of power. Note how "only" 1A flows through the cable while charging happens with 2.25 A !

  • \$\begingroup\$ Thanks. But why do some phones support both then? When do they use 9V and when 12V? (question text updated) \$\endgroup\$
    – neverMind9
    Jan 17, 2018 at 11:18
  • 1
    \$\begingroup\$ I do not know why that is so. My guess is that it allows for some flexibility regarding the electronics in the phone, if a manufacturer does not want to support 12 V (yet) they can limit the voltage to 9 V and still support QuickCharge 2.0. Supporting 12 V will require a circuit supporting that. \$\endgroup\$ Jan 17, 2018 at 11:26
  • \$\begingroup\$ It could also be because of efficiency. E.g. suppose the DC-DC converter the phone utilizes has 95% efficiency at Vin=9V, and 90% at 12V. So at 12V you'd have to deal with dissipating 1.8W of waste heat in the phone, as opposed to just 0.9W at 9V. \$\endgroup\$
    – anrieff
    Aug 31, 2019 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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