How to cut down the Voltage???

Hello guys it's electricnewbie again,

I have another newbie question that I think you can help me out with.

OK so go easy with me since I'm a beginner. . .

So, I have a laser/light (from walmart) that takes three 1.5 volt batteries that gives it a total of 4.5 volts. So, I want to hook it up to USB cable. Now, a USB cable supplies about 5.1 volts. How do I drop the current from 5.1 volts down the 4.5? How could I use the same principle in my other projects? Say I have a 9v battery and a 3.4v buzzer. I know how to use resistors with an LED. [R=V1-V2/mAH] Since I know the mAH of the LED, then I know what resistor to use. But how use that equation if at all on other things??

Thanks guys,

The .5V difference between the laser's battery and the 5V from USB shouldn't make much of a difference so you can directly connect the laser to the USB port.

But when you want to drop your 9V to 3.4V you will need a voltage regulator. If your looking to use the 9V to power one or more LEDs then you use a LED calc like this one. LEDs are more concerned with the amount of current that is flowing and less about the input voltage. In this case if your LED has a forward voltage of 3.4V and a forward current of 20mA then all you need is a 330 Ohm resistor in line with the LED and you can run it directly off the 9V battery.
HavocRC (author)  mpilchfamily4 years ago
I understand how to do it with an LED but how can I use it practical? Say a 12v power source and a 3 volt speaker or something else like that thats not just and LED
Any other circuit that isn't a simple LED will need a voltage regulator to bring the voltage down to the appropriate level. BTW you won't be hooking up a 12V source directly to a speaker. You may need to adapt a 12V source to work on a 6V amp for a speaker.
HavocRC (author)  mpilchfamily4 years ago
Haha ok thank you I'm just a little confused and trying to figure things out!