# Electronic – Getting 1.2V from USB

currentusbvoltage-regulator

I have FIIO E3 headphones amplifier which is powered by 1.5V AAA battery. Am I using it just with my PC, so I got an idea: why don't I power it using USB port and stop buying new batteries ?

Anyway, I have few questions regarding that:

1) I know that USB port can give 500mA, but somewhere I've read that you can get only 100mA without enumeration. Is that true ?

2) Will 100mA be enough for replacing 1.5V AAA battery ? What is the maximum current that 1.5V AAA battery can give ?

3) Let's say that 100mA is enough. Now I need to build something 🙂 I've thought about using LM317T. In it's datasheet, there's a foruma like this: Vout = 1.25V*(1+R2/R1)+Iajd(R2).
If I'm reading the datasheet correctly, Iadj will be really small, so I don't have to worry about that. R1 is fixed at 240 ohms. So I only need to worry about R2. After calculating, it should be around 50 ohms. Or did I do something wrong ?

4) If I get 100mA @ 5V, is it true that I can get about 300mA @ 1.5V ? I don't know why this could be right or not, so I'm sorry if I've said something really stupid 🙂

1) Typically yes, but USB hosts are sometimes very loosely implemented. I've got a cheap Sweex USB hub which doesn't do anything when I short circuit the +5V USB power..

2) Depends on the headphones and the amplifier. You can probably figure out how much it could take to look at the maximum power, and divide it by the 1.5Volts. It's probably a maximum with maximum decibels.

I don't know the exact limits of a AA battery, but if you got a 2200mAh battery, I would say they should be able to deliver 2.2Amps (1C discharge rate).

3&4) A LM317T is a simple solution. It dissipates the 'left over' volts into heat. So if the input to output difference is 3,5V at 100mA, it will dissipate 100mA*3.5V=350mW. Note that if you take 100mA at 1.5V, it will also take about 100mA at 5V. This also means that theoretically you can only power up to 100mA @ 1.5V. If you assume that the enumeration thing is not a problem, it would be 500mA @ 1.5V.

So, question 4: no, if you want that you need a switch mode power supply (SMPS). A linear supply (like a LM317) will have Current in = Current out, (leaving quiescent current out for now). The switching power supply will try to be Power in = Power out (without it's efficiency taken into account). So 5V 1A could be 2.5V 2A, 1.2 4.166A, as they all equal 5W. If you take into account an efficiency of 80%, you would probably see something more like 1.2 3A or something.

A SMPS is more complex to build up as it needs a inductor and a flyback diode. Also note that it may create a hum in the sound if the switching frequencies would 'leak' into your audio signal. So I think it's best to see if a LM317 is capable of powering your circuit.

@xsari3x: A current mirror is not used to deliver power. It's used to bias transistor amplifiers within opamps or other signal amplifiers. Furthermore, those outputs are constant current , where we need a constant voltage output here.