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I am making a portable phone charging battery for fun. I have a micro USB input that ties to a 5V - 4.2V buck converter to charge a li-ion battery. The IC is made to charge batteries so I think that part is fine. I then go from the battery to a 5V boost converter that connects to a USB receptacle so you can charge your phone. I use a dedicated charger port IC between the USB data lines to make it charge most devices (correct me if I'm wrong).

Basically I want an explanation on what happens if I have a wall outlet plugged in to charge the battery while I have a phone plugged into the output also charging. Is that okay? How does the battery create the 5V rail (from the 4.2 to 5V boost) if charge is being pumped into it? Wouldn't charge need to leave it to make +5VDC?!

Is there any precautions I need to make to ensure it can charge a device while also being charged?

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A 4.2V buck converter isn't good for charging a lithium ion battery unless it can't source too much current. There's three stages that a li-ion battery needs to be charged correctly. Messing any of those up can result in exploding batteries. See charging lithium ion batteries for more info on that.

Your question about the charge entering and leaving a battery isn't really an issue. I think in that regard, your question is a dupe of another on here a month or so ago that asked what happens when you charge and discharge a battery at the same time (I can't find it now...). At any rate, the point of the answer was that any extra current you need from the 4.2V buck will be sourced from it, and any current from the battery that it can source will be sourced.

If the battery is discharged lower than 4.2V, then the buck converter will supply both the battery some current and it will source the boost some current. This assumes that you have connected the battery in parallel to the boost converter.

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  • \$\begingroup\$ Hi Horta, the charger isn't just a 4.2V buck. It is a TI integrated circuit designed to charge li-ion batteries. I should've been more clear. \$\endgroup\$ – ACD Jun 9 '14 at 15:08
  • \$\begingroup\$ Ah that's good. Then the main issue you may have is if you ever drain the battery down, when you plug it back into the power supply, you may not have enough current from the 4.2V charger to properly supply the boost converter until the battery voltage rises. This all depends on your current requirements. In any case, it's not something that will break your circuit, it's just something to know. So you may have a delayed ability to recharge batteries from the 5V boost charger. \$\endgroup\$ – horta Jun 9 '14 at 15:25
  • \$\begingroup\$ This is starting to make sense haha. Thanks for your help. Let me throw an example by you. The IC charger can charge up to 2A. Scenario 1: The battery is dead and I plug in my wall charger capable of 3A output. I also have a phone connected to the output that can charge up to 2A. When connected the battery will charge at 2A and the phone will charge at a slower rate of 1Ah even though it can do more, it won't be able to until the battery is charged up enough to make up the difference? \$\endgroup\$ – ACD Jun 9 '14 at 15:28
  • \$\begingroup\$ I'm confused by your comment because we have two charger's here: the 4.2V li-ion charger, and the 5V usb charger. Which are you referring to here? \$\endgroup\$ – horta Jun 9 '14 at 15:38
  • \$\begingroup\$ The 4.2V input charger. I guess I am confused on what decides the power priority. Say the battery wants 2A and the phone output wants 2A but the input charger can only supply 3A. How does it get distributed? \$\endgroup\$ – ACD Jun 9 '14 at 16:11

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