This question already has an answer here:

I wanted to experiment a Smartphone Fast Charger. My source would 5V 25A, and output would be in 4 channels, all in parallel. Would that be safe to charge a smartphone?

If no, what can I use to bring down the ampere in one channel? Say, I'll be lowering the ampere down to 3.1A - like most fast charger does.


marked as duplicate by winny, Dave Tweed Aug 8 '18 at 10:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ Are you aware that charge control is done by the battery management system inside the phone? \$\endgroup\$ – Transistor Aug 8 '18 at 10:33

The device that you connect to the charger port will take the current that it needs at the voltage provided. (I am indicating that the source voltage can have an impact on how much current a device load will take but the device still determines how much current it will draw from the source).

If some other charger is specifying 3.1A that is a specification of the maximum current that the charger can provide while still keeping its output voltage in range and operating in a safe condition such as not burning up or shutting down. If a device attached to that charger needed to draw more than 3.1A then that charger would not be suitable for that device.

Let us look at it from the aspect of your proposed charger. Just because your charger could deliver equal loads at 6.25A per port there is no value to that if the devices that you intend to attach will never draw that much current. You may also consider adding a current detector on each output to take action if a load device is ever attached that wants to try to draw more current than your ports are rated for. The actions that you take could be to either shut the output off till the load is removed or lower the output voltage to a level that the attached device stops drawing so much current.


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