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).
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You are a hero man! I have a bunch of these around and they're exactly what I needed!
You, my friend, are a genius! I just ordered a case of 36 from Dollar Store for a dollar each. Beets paying almost $3 for each input!
Great answer! This is why I love instructables!
Buy a car usb charger and go from there.
im making led turn indicators for my bike...problem is that the leds i want to use burn out if used above 5v....bikes battery is 12v and 4 amps....how should it be done?attaching pic of leds i want to use
that is easy, connect 4 led in series (3v x 4 = 12v) to make a 12v packet. Now make as many packet as you need and connect them in parallel to increase your brightness. As bike generator generate variable current so connect a 16v or greater capacitor before led. that's it, no need to bring voltage down to 5v or 3v
I'm doing that too! Not sure what to do just yet. Someone was saying use a MOSFET? But they are all like 60V? I dunno.
A mosfet has a rating of about 60v depending on the one you get. that isnt how much it outputs. usually N-Channel mosfets are used to power something high voltage while controlling it with a low voltage source. eg 12v lights powered by a 12v battery controlled by an arduino.
For the flashing part, use this circuit. I see the example has a resistor in already for the LEDs so they don't blow:
The part doing the flashing/timing is the 555 timer chip.
Lm7805,10uf 50v capacitors (2pcs.)0.01uf for final (ceramic cap.)
i hope this will help,,,just use a heat sink ,,,fat is better than nothing ✌?️?
Well done! what a great simple answer...You probably wrote the acronym KISS...I was struggling with resistors and heat sinks....just went to my scrap box and presto! one redundant ciggy socket phone plug and my 12v solar panel now charges phones!
@unixbigot, You are the man. Had an old one of those chargers that I still used. Got a new one from Wallmart for less than $5.00. Opened up the old one and hooked up a 12v 3 cell lipo pack @ 610mah. Then I thought Hmmm... what if I used a 9v battery? And it works at 9vdc. Then I took it a step further. I had an extra 12vdc battery that's used in the keyless car clickers. and it also worked. My steampunk costume is going to rock
if a voltage regulator is generating heat when connected to 12v i suggest to use 2 voltage regulator 7808 that transform 12v to 8v then connect a 7805 voltage regulator to transform it to 5V ;)
I have 12 v dc I want to 5 v which ohm resister have to use
And one more question
One ohm resistor oppose to how many volt
Hi guys. I read all your comments about dopping the voltage from 12 to 5. The problem with that, which I have already experienced, is t that using resistors as potential dividers generates a lot of heat for the circuit! The best on would be to use a regulator that dissipates the least heat and so obviate the need for a heatsink.
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
try a 12v to 5v Usb adapter like this:
could also try buying a bottle holder to clip to the frame and put a battery inside you could have an 11.1V li-po just for your speaker coupled with the 12V to 5V USB adapter stated above...this way no loss of run-time from your e-bikes battery life but a VERY LONG RUN-time on your speakers which would probably be able to outlast the e-bike battery!
try this battery:
put the battery inside the bottle and boom......new power supply!
How do supply 3.7v fom a 7.5v power bank? The voltage divider that I setup for my BlackBerry works only if the BlackBerry battery is less than half-full, else it just reboots. I used equal rsistors of 176 ohms each and that gave me 3.75v. I suspect that these resistors give current that is too little to charge the BlackBerry. My cslculations show that I can get 1Amp from 50 ohm resistors.
open the powerbank and their should be two lithium cells inside there... just swap only ONE of them around so that the battery has got ++ & -- of the batteries joined your voltage will be halved but you amperage will double.....meaning 3.7V but the powerbank will last two times longer when charging your phone as a benefit.
I suggest you us a lm317 voltage regulator .Have a simple circuit (LM317 , resistor , Pot) .You can also use Pot to replace all resistorInput Voltage 1.5 - 40v
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
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.
Or you could use a voltage divider made with resistors.
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 wattsthe 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
resistors drop milliamps (IC), not voltage. Diodes drop anywhere from 0.6 to 1.6 volts
voltage regulator really is the way to go here. Simplest answer is simply thispin 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
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.
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.
Get (a) resistor(s) out of radio shack or maplin, they should be able to help you with which kind you need.
Use a 5V voltage regulator
Wouldn't ten silicon diodes in circuit drop 7 volts?
Try a mosfet transistor. It works wonders.
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.
With a voltage regulator (Google Is Your Friend) or with a resistor.