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I have a large LiFePo4 battery cell made by Headway (model 40152S 15AH single CELL, 3.2v). I'd like to use it to power my Samsung Galaxy S3 phone. On the original phone battery it says that it is a 3.8v. The phone has a micro USB input to charge the phone.

I bought a cable that has a Micro-B connector at one end. I haven't cut it yet, but after cutting the cable I'm hoping that there will be about 4 or 5 wires, where one of them would be for the positive and one for the negative power.

Can I use this 3.2v battery to power the phone even though the battery that comes with the phone works with 3.8v? Will using just the two power leads be enough to make the phone work?

If my headway battery is too low of a voltage to work, would I need to use a 3.7 volt Li-ion battery instead? Is 3.7v close enough? I haven't been able to find any batteries that are exactly at 3.8 for a single cell.

edit after searching for a DC-DC Regulator Module:

I found a module that is designed to do this:

Step up DC-DC regulator Designed for convert 1 cell Li-Ion / LifEPO4 battery pack to 5.0VDC output Running / charging most popular 5.0VDC devices such as cell phone/MP3/MP4/PDA/iPod/PSP/GPS from Li-Ion / LifEPO4 battery pack

Input Voltage
2.0 ~ 4.2VDC via "VIN+" / "VIN-" Soldering spot

Output Voltage
5.0VDC 2.0A Max via USB Female Jack 5.0V 1.0A Max each

Will this module be enough to place between my 3.2V battery and the cable that I can use to go from the USB in the regulator to the micro USB on the phone?

Thanks very much for your help.

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closed as off-topic by Leon Heller, JYelton, Keelan, Daniel Grillo, Matt Young Sep 15 '14 at 15:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Questions on the use of electronic devices are off-topic as this site is intended specifically for questions on electronics design." – Leon Heller, JYelton, Keelan, Daniel Grillo, Matt Young
If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ The answers in this remotely related tread "Is it possible to charge one phone off another?" may help you get better understanding about what a charging circuit in a cell phone expects. \$\endgroup\$ – Nick Alexeev Sep 14 '14 at 22:12
  • \$\begingroup\$ Maybe it will work, but probably it will not. Even when internal battery would be disconnected and the phone would not care about input voltage, the highest internal power rail is probably 3.3V which is higher than LiFePo voltage. \$\endgroup\$ – venny Sep 14 '14 at 22:15
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The USB port on the phone expects 5 volts in. I doubt the phone will do anything if you give it less than 4.5 volts or so, though it might be possible.

But if you were to use a boost regulator to generate 5 volts from the external battery, at least that would work for powering the phone. Connecting to the +5V and ground lines of the USB connection should be enough, but phone chargers also connect some resistance to the data lines (or connect them together) to indicate that they're a charger instead of a real USB host.

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  • \$\begingroup\$ I found a module that is designed to do this: Step up DC-DC regulator Designed for convert 1 cell Li-Ion / LifEPO4 battery pack to 5.0VDC output Running / charging most popular 5.0VDC devices such as cell phone/MP3/MP4/PDA/iPod/PSP/GPS from Li-Ion / LifEPO4 battery pack Input Voltage 2.0 ~ 4.2VDC via "VIN+" / "VIN-" Soldering spot Output Voltage 5.0VDC 2.0A Max via USB Female Jack 5.0V 1.0A Max each Will this module be enough to place between my 3.2V battery and the cable that I can use to go from the USB in the regulator to the micro USB on the phone? Thanks very much for your help. \$\endgroup\$ – Leo Sep 14 '14 at 23:23
  • \$\begingroup\$ Yes, that sounds perfectly suited for this use. You don't even need to modify a USB cable :) \$\endgroup\$ – Rennex Sep 15 '14 at 2:51
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It is an USB port. USB specifications say 5.00±0.25 V (pre-3.0) for high-power ports and 5.00+0.25-0.55 V for low-power ports; 5.00+0.25-0.55 V (USB 3.0) for the supply voltage (taken from Wikipedia and usb.org respectively). So anybody designing the charging circuit in a cell phone would stick to those values. Your approach is most likely going to fail. Besides the fact that Li-ion batteries require special charging you cannot expect a 3.2V source to charge a 3.8V battery (think of water not flowing up the hill...).

Before throwing out your new LiFePo battery right now however you might consider getting a DC-DC step-up boost converter (variable input to 5V output). There are pre-assembled modules available (but I cannot recommend any).

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  • \$\begingroup\$ The Wikipedia article must be mistaken. The original USB power specification is 5.0 V +/- 10%, which is +/- 0.5V. \$\endgroup\$ – Dave Tweed Sep 14 '14 at 22:37
  • \$\begingroup\$ USB 2.0 Spec (from usb.org/developers/docs/usb20_docs/#usb20spec) state 4.75 to 5.25 V for high-power ports and 4.4 to 5.25 V for low-power ports. USB 3.0 is afaik a little more generous. But no matter what the values are, they will be higher than the voltage provided by the above mentioned battery! \$\endgroup\$ – Ghanima Sep 14 '14 at 22:52

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