# How can I convert 12v down to 5v?

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unixbigot says: May 8, 2009. 2:31 AM
Forget using a 7805 regulator unless you have one on hand, and have room for a fat heatsink. Go to a "dollar store", thrift store, or flea market, and buy an obsolete car charger for a cellular phone. These typically contain an a small circuit consisting of about a dozen components, primarily a MC34063 switchmode-regulator chip (an 8-pin integrated circuit) that is much more efficient than a 7805 (and therefore waste less energy as heat). If you take it out of the case (insulate it with a piece of heatshring or bicycle tube), a typical 12->5v phone charger circuit is no larger than a 9v battery and can handle about 1 ampere of current. You can't build one for the price you'll pay at a junk store. You can change the output voltage to anywhere from 12v to 1v by changing one of the resistors (google mc34063 datasheet for details).
BobS in reply to unixbigotJun 6, 2009. 8:24 AM
Great answer! This is why I love instructables!
Guldarious says: May 18, 2013. 2:44 PM
I'm looking to run a usb port on an ebike to run pc speakers. The battery is 12 volts and the speaker takes 5 volts through usb port. How can I run a line off of one battery (bike uses 4 of them) and make this happen so I don't have to keep using a battery pack that only lasts 1/2 an hour
james34602 says: Oct 29, 2012. 9:13 AM
I suggest you us a lm317 voltage regulator .
Have a simple circuit (LM317 , resistor , Pot) .
You can also use Pot to replace all resistor
Input Voltage 1.5 - 40v
John2013 says: Aug 30, 2012. 9:49 PM
One related scenario is bringing down 12v power source to a 5v device.

In one situation, a user wanted to use a 5v DC power digital voltmeter or ammeter. His power source could be either a 9v dry cell or use in 12v vehicle battery. The quickest way is to use a serial resistor to bring down the voltage for 9v or 12v to around 5v. The actual voltage does not need to be exactly 5v. There is usually some tolerance. I guess around 4.75v to 5.25v is a good target range.

One could use a 1k Ohm potentiometer to dial in required resistor value. (VERY IMPORTANT) Set the potentiometer to the HIGHEST value. (VERY IMPORTANT) Put the pot in series with the meter and the voltage source. Turn the potentiometer down very very slowly till the voltage across the voltmeter is at 5v. Turn off the circuit. Isolate the potentiometer and use an Ohm meter to determine the resistance (Ohms). Now use the resistor and confirm final voltage.

If not already offered, ask seller if he could provide free resistor and shrink wrap to their 5v meters. Let seller know what is your source voltage.

- John
weecoo says: Mar 23, 2012. 2:25 AM
Some of the "dollar store" car chargers / USB chargers promise to provide a maximum current of 2A which seems to be far from the truth. If you need more than 500 mA it is a lot safer to buy one of the many UBEC converters which have a low power loss and stronger current (some up to 5A). Search for UBEC in one of the many online gadget stores - there are plenty of them available.

Dr.Bill says: May 7, 2009. 10:25 PM
Or you could use a voltage divider made with resistors.
evildoctorbluetooth in reply to Dr.BillJun 1, 2009. 12:52 PM
True, however we do need to realise what load we are going to apply this 5 volts to.
for example, so we use a voltage divider, say, a 5 ohm in series with a 7 ohm. now if we measure accross the 5 ohm resistor we get 5 volts. ace, now lets just connect it to my load. this load is also 5 ohms.
now are we applying 5 volts to the load.
NO.

now we have connected the 5 ohm load in paralel to the 5 ohm part of the potential divider, its total ohmage is now 2.5 ohms (product over sum) now if we use the potential divider sum, Vpd = 12 volts *2.5/(2.5+7) = 3.16 volts.

I hope that helped. another thing, assuming that we were only chargng one thing, and took that into account when calculating what resistors to use. (assuming we are still having a 5 ohm load.) that would give us a potential divider ratio of 2.5ohm:3.5ohm (with the 2.5 reprisenting a 5 ohm resistor in paralel with our load) we would get 5 volts. The problem with this is the wasted energy diserpated. The current drawn by this circuit is 12/6 = 2 amps. that is assuming the connecting wires have no resistance. now if we work out how much energy is disarpated at the 3.5 ohm resistor the formula is
Watts = I2*R = 10.5 watts, wasted. a percentage of usefull energy used (efficiency)
total wattage = IV = 2*12 = 24 watts
the useful current is 1 amp, as the current is split between the 5 ohm resistor and load)
useful = I2*R = 1*5 = 5 watts.
efficiency = 21%

i hope that this made sense, that i was polite, not teaching to suck eggs, not too advanced.

any questions message me, if i cant answer you il know someone who can.
Great answer..not only the solution, but thank you for taking the time to also explain the process...that helped a lot..thank you again
HADJISTYLLIS says: Aug 28, 2009. 5:01 AM
(removed by author or community request)
resistors drop milliamps (IC), not voltage. Diodes drop anywhere from 0.6 to 1.6 volts
raykholo says: Jul 17, 2009. 9:58 PM
voltage regulator really is the way to go here. Simplest answer is simply this

pin 1 is input +, pin 2 is ground, and pin 3 is output +. this guy is also useful for a variety of voltage inputs and uses
mattccc says: Jun 2, 2009. 4:57 PM
use a voltige regulator
evildoctorbluetooth says: Jun 1, 2009. 12:56 PM
There are several things to do depending how effiecient you want to be, or how smooth you want the output to be. i have posted below against using a voltage divider as its load regulation is very poor. I have had a thought that i would like to thing about.
Berserk87 says: May 29, 2009. 3:30 AM
LM7805 is a 5V regulator and maxes at 1Amps output. its the simplest and cheapest for you, unless you can find the parts unixbigot is talking about. the LM7805's dissipate the power through heat. so its a cheap and easy solution, but there very inefficient.
sboy365 says: May 25, 2009. 2:35 AM
Get (a) resistor(s) out of radio shack or maplin, they should be able to help you with which kind you need.
HADJISTYLLIS says: May 24, 2009. 10:37 AM
Use a 5V voltage regulator
whittey says: May 22, 2009. 6:39 PM
Wouldn't ten silicon diodes in circuit drop 7 volts?
M4industries says: May 20, 2009. 3:18 PM
Try a mosfet transistor. It works wonders.
billbrown says: May 19, 2009. 4:30 AM
subtract seven volts just kidding
AndyGadget says: May 6, 2009. 9:37 AM
You need a 7805 regulator to do the job - HERE'S a circuit which will give you up to an amp, although if you're going near that you'll need a heat sink on it.
There are similar regulators around for higher currents.
kelseymh says: May 6, 2009. 9:30 AM
With a voltage regulator (Google Is Your Friend) or with a resistor.