Author Options:

12v joule theif from 1 AA Answered

i need to make a joule theif that can output 12v to power some rgb leds. any help would be great. but preferably in the form of a schematic. i have all of the parts needed.?



Increase voltage means decreasing the current available. If you could do what you want why not bump up the voltage to 120 and run the entire house off a single AA cell.

Is it possible to build a joule thief that use a 9v radio battery to supply 12VDC 3A? I need a portable energy source to power my netbook (not into a car...). Thanks

It all about power - Watts - Watts = Volts x Amps so no matter what the voltage unless the current is increased then your not getting the power you will need.

A 9 volt battery assuming you mean the square type battery hasn't the current available to drive much at all - if you could do this then lap tops would already have this circuitry in place so they could get more from very small battery packs.

Sorry the answer is no.

You'll probably not get enough power out of 1AA, these devices are meant to bump up enough to run ~1 LED not several.


actually if you run whatever you are using in parallel you could draw as much current as you want, but it would lower the voltage as you increase the amperage, as AA batteries have a fair bit of resistance, i have seen AA put out as much as 2 amps, not at 1.5v though. and about running out 120v AC, its not at the right frequency although im sure it could run your average lightbulb, for a short short time.

in addition i tried the way where you make a secondary winding on top of the normal joule thief toroid, but it didn't work. im not sure why.

No you cannot run whatever you want.
If you know how do do this why did you ask the question?


im asking this question to see if there is a simple solution, i already have found a more complex one, i was just wondering if there was one.

Im not trying to argue with you but if you want one here it is and yes the current that the battery can output is limited, but only by the internal resistance of the battery and the voltage of the battery it self. so if you look at this here link http://data.energizer.com/PDFs/BatteryIR.pdf and look at the chart labeled "total effective resistance fresh AA battery" it shows the resistance to range from 0.9 ohms at -40c to 0.1 ohms at 40c. so by using the formula v=ir we can surmise that the most that a fresh AA battery can output is 15 amps if it is fully charged and its at 40c, but to be more reasonable lets say a fresh AA at room tempurature it would be closer to .175 ohms, therefore around 8 to 9 amps is the maximum throughput. and that is just a normal AA not a lithium cell which has even less internal resistance.

It's not just the internal resistance that affects output, notice that in the pdf a "heavy drain" is only half an amp.
What happens to AA batteries if you connect the terminals together with a short piece of wire?


ok but still you could probably pull out at least 2-3 amps. so like 2 watts of watt ever you want to power.

IF you AA can output 2 watts - IF - Then it isn't going to light a household bulb.

i am sorry but this just doesn't work in the way you seem to think. If it did we could all run our houses off a set of batteries. (using an inverter to get AC) and we can't.

IF you increase the voltage then you have to decrease the current available (Ohms law) The joule thief increases the voltage of a single 1.5. volt cell to a high enough state to drive a very low current device such as an LED.

It works BECAUSE the LED is such a low current drain. If you boost that then it isn't going to work. and lifting the voltage to 12 volts reduces the current available to little or nothing.


Best Answer 6 years ago

I assume you have a common cathode RGB LED. This does not require 12V, each of the LEDs require a different voltage. Red is lowest at about 2 volts, green close to 3V and blue at a bit over 3V.  If you put them in parallel across the same JT circuit, the red will hog most of the current.  A resistor or diode on series with the red will help equalize the current.  I have also used a separate winding on the same core for the different colors.  Without more info such as LED type, I would not be able to offer a correct schematic.

You will need 20 mA at 12V for each of the colors, or about 1/4 watt each. A conventional JT puts out about 60 mW, so you'll need a beefed up JT with a high current transistor such as the ones found in a disposable camera flash - 2SD965, KSD5041 or NTE11 (all have Japanese pinout - E C B). The AA cell will have to supply about 2/3 amp or maybe more, so it will run down fast. A low ESR capacitor, maybe 1000 uF across the AA cell will be necessary.

Remember that the higher 12V output makes a greater demand on the JT. The additional winding on the coil will help. I would try using 3 or 4 times as many turns as the primary winding. Some would recommend rectifying and filtering the winding, but that may not be necessary. Remember that the winding is polarity sensitive so try it connected both ways. Also, the DC resistance of the primary winding must be very low in order to allow peak current of well over 1 amp - closer to 2A - to flow through it.

My recent blog might give you some other ideas.


Answer 6 years ago

oh and i will try to do a secondary coil on a working joule theif.