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I have 6V lead-acid battery as a host to charge smartphones lithium ion battery.

Standard USB charger charges the smartphone at 5V/1A-2A but here if I directly connect battery to the smartphone does it hurt or blow up smartphone's battery? Or does the smartphone have protection circuit or buck convert or something else to make charge at CV 4.2V/CC 1A?

I know there are lots of thing to safely charge smartphones but i am asking for curiosity.

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closed as off-topic by duskwuff, Leon Heller, Enric Blanco, PeterJ, laptop2d Jun 12 '17 at 17:05

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." – duskwuff, Leon Heller, Enric Blanco, PeterJ, laptop2d
If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ The battery charger and supervisory circuitry is in the phone not in the external (dumb) power supply. If you are charging by USB port then you will probably destroy the USB circuit and the phone as the USB port is specified for 5 V power. Look up the specifications for USB. \$\endgroup\$ – Transistor Jun 12 '17 at 6:20
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The battery charging circuit is in the phone not in the "charger" which is just a dumb 5V power supply.

The issue is that the circuit in the phone will have a maximum input voltage before things break. This circuit will be designed to assume a maximum normal input voltage of around 5.5 V (USB is up to 5.25 and you need to allow a little margin). Your 6 V lead-acid battery will output around 7 V when fully charged.

Is 7 V enough to break things or will the phone cope with it just fine? Impossible to say, it depends on the internal components used and what the designers considered a sufficient safety margin.

Assuming the phone doesn't break it is going to charge slowly, unless the data lines are held in the correct state the phone will assume it's in a standard USB port and not exceed the power ratings of the USB spec. In order to pull 1 A charge current you need to set the data lines to the correct fast charging configuration, this is normally a couple of resistors. I can't remember the exact values needed but a search this site will find the values, it's been asked before. Take care, going significantly over 3.3 V on the USB data lines will break things.

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