Reducing amperage but not voltage...possible?

Question

I am trying to build a USB charger from a 8v dc rechargeable battery. I need to achieve 5v with <= (less than or equal to) 500mA for the USB end.

So far what I have is: I have connected the 8v battery to a voltage regulator (input: 0-37v; output: 5v) which effectively brought the voltage down to 5v. However the problem I am having now is that the battery outputs 2 amps and the voltage regulator reduces it down to 1 amp (max) but I still need to bring it down to 500 mA (0.5 amps).

So my question is, what kind of component can I use to acheive this (resistor, etc).

Thanks in advance,

Jeffrey

Answer 1

any device will  draw what amperage it needs,,if the usb needs 5milliamps and you have a 10 amp power supply there is no problem,,, the device will only take what it needs,, however a 5 milliamp fuse would be needed to protect the power supply and device if for some reason the device shorted and tried to draw more than its rating...thats why you can put only so much on a 15 house plug in,,even though there is 15 amps there you can plug in a hairdryer that draws 10 amps and its ok,,,  oh yes,, 0.5 equals 1/2 amp NOT 500 MILLIAMPS!!!!!!   YOUR DECIMAL IS OUT OF WACK!!!!!

Answer 2

The amperage drawn from the battery is dependant upon the resistance of the device connected to it.  Consider that your local power plant produces mega watts of power (1000&#039;s of amps) . That doesn&#039;t mean that you have 1000&#039;s of amps passing throught your homes wiring....   Instead your homes electrical load draws the few amps it needs and ia limited to what it can safely draw by fuses or circuit breakers

THe 2 amp rating on the battery mean that you candraw upto two amps from the battery  while maintaining the rated voltage.    

If you want to limit the current drawn from a power source you have several options:

1.) Use a fuse to prevent drawing current above the fuse rating. By far the easiest and cheapest  means to limit current reliably....

2.) use a crowbar circuit

3.)Place a current limiting resistor in series with the supply output.

4.) Add a current limiter circuit