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I'm trying to replace a Li-Ion battery powering a device with a USB cord directly connected to my PC.

Which design parameters should I consider in choosing the right components? I only took in consideration voltage and (estimated) current draw.

The device uses a single cell 3.7V battery and, since I think that the full charge voltage should be 4.2V and the end-of-discharge voltage should be 3V, I'm going to supply the device with 3.3V voltage.

The way I'm going to achieve that is by a Linear Voltage Regulator: which to choose to best emulate the Li-Ion battery? Is an AMS1117 good or should I choose something different? Will some component damage my PC?

Current draw should be around 50 mA.

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    \$\begingroup\$ Do you want to actually emulate the falling voltage of the battery, or just supply the 3V3 to your device? In the second case, the AMS1117 will be fine. Your power dissipation will be 85 mW, which is well within the power limit of the regulator, and it can handle a maximum current of 1500 mA. \$\endgroup\$ – awjlogan Nov 8 '17 at 16:13
  • \$\begingroup\$ No, I'm actually using the AMS1117 as a simple DC 3.3V power supply so I should be good to go. Thanks for the answer! \$\endgroup\$ – Marco Cavinato Nov 30 '17 at 17:48
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You are on the right track. The AMS1117 at 3.3 V is perfect for your application. All that your device needs is a voltage in between approx. 3 V (better a bit more) and 4.2 V. You can safely power your system by using the LDO.

Another story is it if your system wants to charge the battery, thus, the LDO. But even then, you should be safe with the LDO and everything will work fine. It has no problem with 4.2 V at its output; that what the system would put there; at least not as long as it has 5 V at its input.

Finally, if your system expects a smart battery (i. e. a battery that can communicate via a two-wire SMB bus), it depends on its software whether it will startup if no communication is established. But in any case, you can try.

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  • \$\begingroup\$ Thanks for your answer. I didn't actually take into account the smart battery issue but by inspecting the device I found that the circuit is a simple two-terminal battery so there shouldn't be any smart circuit. Thanks for pointing it out, anyway! \$\endgroup\$ – Marco Cavinato Nov 30 '17 at 17:50

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