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How do I convert a 3V battery powered device to be USB powered instead? Answered

I saw another article about this, but his device had 3 batteries.  My device has 2 AA batteries.

I don't know anything about how a circuit works, but I can follow a schematic.  What do I need to do to change the output from the USB of 5V to only provide 3V for my device?

I find a lot of articles about converting 12V to 5V, but I'm not sure how to translate those schematics into something that will go 5V -> 3V.


Using an LM317 is is worth a try, but you are only working with a 2V difference from one side of the reg to the other.

It may not work, you REALLY need something called a "Low Drop Out" (LDO) regulator, but they are harder to find from the like of Radio Shack, and you should look at folks like Sparkfun.



I know this question is a bit old. Steveastrouks answer works really well. I used an LM317T (purchased from radio shack) 2x100ohm 1/2watt resistor and one 40ohm 1/2 resitor. It produced 2.9v which is enough power for most double cell peices of equipment. You could also use a potentometer in place of the 140ohm resistor to fine tune the voltage output from the LM317T. I would recommend getting the 1kohm 1w 15 turn.

The schematics can be found at quite a few places. Searching for LM317 and looking on wikipedia's website will show a few. This is what I used.

Vin ---_______-------Vout----~~~~~---o--------------3v output
| LM317| 100ohm |
------------ |
Adj | | r2

USB cables are supposed to have a standardized conductor colors. Red+, Black-, White(data), Green(data), Shielding (ground). If you cut one end off of the cable you'll want to do the following:
1. Connect the red wire to the Vin pin on the LM317
2. Connect the r1 100ohm resistor to the Vout on the LM317
3. Connect the 100ohm resistor and the 40 ohm resistor to create r2.
4. Connect r2 to r1 and the adj pin on the LM317
5. Connect r2 to the shielding/ground of the USB cable
6. Add a lead from r1 for 3v output voltage.

I actually hollowed out a battery and stuffed the LM317T and the resistors in to it. I then used a wooden dowl and connected the negtive lead to a screw in it. I then connected everything to a USB micro female plug, so I could use a standard USB micro cable to power my device when I was done.

Hope this helped and good luck!


Don't try to translate them. Use 3 silicon diodes in series, in line with the positive lead, to drop approximately 2 volts from the 5 volt power supply. Each diode (use 1n4001) will drop from 0.6 to 0.7 volts, cumulatively. That means the total will be between 1.8 and 2.1 volts. Quick, easy, done.


Resistor droppers depend on a constant current draw to deliver a constant voltage drop. Real loads vary depending on what the device is doing - motors are notoriously variable loads. A dropper resistor that drops just the right voltage when a motor is running for example will drop so much when the motor is starting, it might not start.

That's the point of using a proper regulator. Its pretty nearly a perfect voltage source, and it has effectively zero output resistance.

A voltage regulator ? Any kind of electronics component store. There are fixed voltage regulators, with part numbers like 78XX where XX is the output voltage: so the 7805 is a 5V regulator, the 7812, a 12V regulator etc etc, and there are adjustable output regulators like the LM317 and 337 which need two external resistors to set the output voltage for you.

The only caveat is that these regulators need to see a difference from input to output of ~2 V. If you need to work with less difference than that (typically on a battery supply), you use a "LDO" regulator (Low Drop Out) regulator, and these work on voltage differences below 2V.


I suggested the diodes in series to a friend, and he said that the voltage drop isn't always a guaranteed amount with those.

He suggested I try to use an adjustable-voltage regulator.  Radio Shack has one, LM317T.  He said this should work, however if the draw was a lot, I may need a heatsink.

Does this sound legitimate?