Instructables

Make a SUPER Joule Thief Light!

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A joule thief is a simple circuit that can take 1.5 volts and put out as much as 5 volts. It can light an LED super bright! But have you ever heard about getting 120 volts out of a AA battery? The Super Joule Thief can do just that! It is perfect for lighting lights during power outages or just for a desk lamp. It can even charge cameras and cell phones! Watch the video for an overview.

I entered this instructable into the Make it Glow contest, so if you like it, please vote for me!! 
 
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Step 1: The Parts

Picture of The Parts
This circuit requires very few parts:
-TIP31 transistor [HAS to be a TIP31 because it works the BEST :) ]
-10k POT
-0.1uf-1uf capacitor (I used a .22uf)
-Ferrite toroid or rod
-20 to 30 feet of 30 gauge magnet wire
-3 feet of 22 gauge magnet wire
-heatsink? (Do you want to power it with more than 1.5 volts?)
-AA battery and holder
-AC output adapter (A 2 prong to 3 prong adapter works great)
-Perfboard
-Epoxy

Scissors, solder, and other tools will also help.

Step 2: The Circuit

UPDATE!! I changed the .22uf capacitor to a .68uf capacitor and I am getting a higher and more stable output.                                                                                                                                                                                                         Here is the circuit. As you can see, the transistor acts as an oscillator and pulses the current into the primary coil. This induces a large voltage into the secondary coil. The two prongs of the AC adapter can be connected to the ends of the secondary coil, or you can solder the ends to a light bulb directly. This setup has a 10 kilo-ohm potentiometer. This controls the current that flows to the base of the transistor. More current means a higher voltage in the secondary, and ultimately a brighter light.  The .22uf capacitor seemed to improve the output of my toroid, so I left it in. You may have to change the value depending on the resonant frequency of your toroid, or you can just omit it and the circuit will still work great. It is a very simple circuit as you can see.

I'm starting to build this. How do you think 22 gauge wire would go with building this instead of 22 and 30? Would it increase or decrease voltage? Also, how would a 2N3904 transistor work instead? Would I need a beefier one or a smaller one if I can't use that one? Lastly, how would a 1uf capacitor work instead of a 0.68?

amueslim1 month ago

how about the transistor may i change to the other type....? please help me

clevelandstorms (author)  amueslim1 month ago
Yes you can. What type do you have?
AS673 months ago
is there a way to make a 13.4v version to jump start a car with a dead battery with (or similar) method?
PhilKE3FL AS671 month ago

When clevelandstorms says several amps he means around 650 cold cranking amps, meaning in cold temperatures. Car batteries are rated in Cold Cranking Starting Amps or CC: 650 CC or 700 CC for cold cranking amps.

Let's look at using an old standard 4 Watt 120 Volt night-light bulb. This bulb requires about 33 mA at 120 Volts. The circuit we make would still have to supply 4W but, from your 1.5V battery, which means about 2.67 AMPS of current. Which, to say the least, is requiring far more than we normally require from a single AA battery! The thing to remember is POWER, how many Watts does it take? P=I*E or in words, Power (Watts) = Current (Amps) times (*) Electromotive force (Volts) so 4W = I * 1.5V => 4W / 1.5 V = 2.67 Amps yet at 120 Volts 4W / 120V = 0.033 Amps or 33mA. This means to get the 33mA at 120Volts the 1.5 Volt battery (if supplying 1.5 volts) needs to supply about 2.67 Amps of current, and as the voltage drops the battery has to supply MORE current to produce the same power, Watts, to keep the bulb going.

Simply stated, to use this circuit to run an LED night light is, sort of, counter productive. Don't get me wrong, this is a wonderful demonstration of running a very small 120Volt device from a 1.5 Volt battery. But, if all we want is a night light, then using the standard Joule Thief circuit without that large coil (with its shock hazard) to simply run an ultra bright 3V LED directly would be far more efficient. You'll get about a week or more of operation from a dead (~ 1.2 Volt) AA battery instead of only 2.5 hours on a new AA battery for the 120Volt LED night light. It would be the same for charging a cell phone, if we could do it directly with a simple Joule Thief we'd get far more power transferred to the phone battery and far less power wasted just running the conversion circuits.

clevelandstorms (author)  AS673 months ago
Unfortunately not with this circuit. You need something that could supply several amps and this circuit will supply a few milliamps at most.
arghya92 months ago
i tried it many times,but it did not work. plz plz plz help me.

Note, the two shorter coils need to use the opposite ends of the wires: The starting end of one wire and the ending end of the second wire get tied together and they go to the + side of the battery. The diagram above makes it look like only one wire is going to the + side of the battery but it is actually two wires from the two short coils. I usually wind the two wires together then take the starting end of one & the ending end of the other, solder them together and that goes to the + side of the battery. Hope that helps, it is usually the easy mistake to make using the two ends which are together but this produces a non-inductance circuit and so won't work. Look at: http://www.bigclive.com/joule.htm to see how he shows which wire to use.

asdfrtyvbn3 months ago
hi!thank you for your nice project.can you tell more about output frequency and how can i change it?
dudes3 months ago
Awesome design
clevelandstorms (author)  dudes3 months ago

Thank you

infanati4 months ago
im a little confused with one part of your schematics
the arrow at the end of the base of the transistor connecting to the resistor. Is that supposed to be ground?
clevelandstorms (author)  infanati4 months ago
That's the variable resistor's wiper terminal
bneo994 months ago
About this :

REMEMBER! THIS DEVICE OUTPUTS A HIGH VOLTAGE SO IF YOU TOUCH THE SECONDARY LEADS WHILE THE DEVICE IS ON, YOU WILL GET SHOCKED!

It can't kill you, but it is quite uncomfortable.


Did you accidentally shocked yourself?
clevelandstorms (author)  bneo994 months ago
Yes and not accidentally! I had to see exactly what voltage was outputted and if it was enough to give a jolt. I'm basing a pen shocker on this idea
You're trying to make a pen shocker from this idea? Is it also a joule thief?
Has there been any improvement in the high pitch noise or any idea on how to inhibit it?
clevelandstorms (author)  ramirez_armando4 months ago
Coat the toroid in epoxy. Also, if you use a higher value capacitor, the noise will be less. This is not really a problem for most toroids.
This is a nice project and a well done instructable!

Just a cautionary note: I think that it's important to consider the type of load that you put on the output - I confess that I haven't built this yet myself, but I will say that (without looking at the output on a scope), you would want to be careful of the frequency, and the shape of the waveform when powering anything other than a resistive load. You could damage some devices if you're trying to power them with something other than 60 Hz in the U.S..
Mark
Yes you are right, the output is no where near 60 Hz. It is closer to 15000 Hz. You can hear it, as well as check with a scope. I'm really surprised his cell phone charger worked.
As others have also pointed out, there are a lot of unknown variables that could have a big effect on the output, both for the frequency, and the shape of the waveform (and of course, its amplitude). Some devices might not take the spikes that could be developed with too many windings on the output. Also, it's important to be aware, as also noted by others, that when trying to meter this with some DVMs, you may not get an accurate reading on your meter, so that's not always safe either. As you've done, the scope is the best way measure what's really there, for your own safety, and that of the device one is trying to power.

Regrettably, I should mention that I fried one of my Fluke meters while testing the output of just such a circuit.

Good luck to all - be safe and have fun!

Mark
Thanks for pointing this out. It can damage some devices
Tony Cicero5 months ago
Awesome project! Do you know how many amps are outputted?
clevelandstorms (author)  Tony Cicero5 months ago
The max output is about 250-300 miliamps. Thanks for viewing!
300 milliamps on 120V that is 40W even if the circuit is highly resonant on the 1.5V side this is nearly 30A. Have you ever seen a 1.5V battery delivering 30A....
clevelandstorms (author)  f.dufourg5 months ago
I've tried this circuit with a large 20 watt bulb and my multimeter said that it drew 260 miliamps. It lasted for 30 minutes. Thank you for the correction though. I don't know the output, I just know what it can draw, and the max was 260ma.
This kind of circuit does not deliver sinus voltage. It generates spikes which can be really high but for a very short period of time(milliseconds or less). So a multimeter is of no use in that case. Multimeters are designed to work with DC or AC sinus (with low frequency up to a few hundreds of Hertz) apart from that they give erroneous informations.
An AA battery is 1.5V with around 2500mAh capacity. That is 3.75Wh let round it to 4Wh for ease of calculation. During 1/2 an hour it can only deliver 2W.
In 1/2 an hour it can deliver 8W in your example,
4Wh devided by 0,5h is 8W
Oupsss you're right
A small correction. Many multimeters give wrong answers when measuring non-sinusoidal waveforms. That's because many multimeters are designed for low manufacturing cost and a cheap sales price.  As a result, they are only somewhat trustworthy.

But well-designed quality multimeters will correctly measure non-sinusoidal waveforms up to 100kHz or so. For example, Fluke has been designing and manufacturing "True RMS" voltmeters for decades. I have two Fluke meters. Both are from the 80's. Despite decades of use and abuse, they still work admirably.
clevelandstorms (author)  f.dufourg5 months ago
So you are saying that the output is equivalent to a fraction of the voltage spikes?
First, the output is high voltage during very short time.
Second, the circuit works at high frequency.It is difficult to make any computation as the toroid is unknown but you can expect the circuit working in the 10kHz range (or higher).
With high energy spikes at relatively high frequency,the devices you plug in the converter don't behave as usual.
With this kind of circuit, if you have enough turn to secondary, you can light a fluorescent tube without a starter circuit for example.
I would point out that there is no control on the output so the voltage can be anything but what a phone charger will expect to safely do its job. For a light bulb it is not really a problem.
clevelandstorms (author)  f.dufourg5 months ago
So if you would wanted a higher current and a more stable voltage, then you would use a two transistor oscillator with maybe 5 volts on the input. And you would need to calculate the output wattage to find out how much current the battery must supply.
(removed by author or community request)
36W is 36W if a 120V power supply gives 36W of cource this is 0.3A of current.
On this circuit, the power source is the 1.5V battery. If you take 36W from a 1.5V source this is 24A as the circuit in not 100% efficient, there is power lost in the inductor, the transistor, ... so I estimate the current the battery should deliver is around 30A.
wingerr f.dufourg5 months ago
Aw, bringing in physics and spoiling the party!
I am amazed that only one person picked up on this. At that current the toroid primary would burn not to mention the fact that the TIP31 max Ice would be exceeded 10-fold.
Exactly. Most people reading aren't doing the math. The transistor (assuming it's well heat sinked) is rated at about 3A current. 3A*1.5V=4.5 watts INPUT and you won't see anything like that on the output. It'll be closer to 30ma on the output. Since the frequency of the AC will be dependent on the toroid and the pot, whatever is on the other end better not be picky about input voltage or frequency. His example isn't since the LED light would likely have a transformer/rectifier/cap and a buck voltage regulator internally. I've heard of cheaper LED "bulbs" that just use a half wave bridge/cap/resistor,
agis685 months ago
hi, nice instructable....any idea how the circuit change for european standards....230V and about 50Hz?
angelesdm5 months ago
may i know what voltage to use for the .22uF capacitor? thank you. I will try to make this. I will update if i would be successful and vote for you.
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