Introduction: Joule Thief Circuit How to Make and Circuit Explanation
A “Joule Thief” is a simple voltage booster circuit. It can increase the voltage of a power source by changing the constant low voltage signal into a series of rapid pulses at a higher voltage. You most commonly see this kind of circuit used to power LEDs with a “dead” battery, but there are many more potential applications for a circuit like this.
Step 1: Gather Your Components
BUY transistor 2N3904:
BUY 1K Resistor:
Ferrite toroid core
NPN transistor 2N2222, 2N3904, or similar
1k ohm resistor
A used AA battery (if you don't have one than you can use new AA too)
Components buying link (affiliate):-
Toroid ferite core -
Resistor set -
Step 2: Circuit & Working Explanation
A Joule Thief is a self-oscillating voltage booster. It takes a steady low voltage signal and converts it into a series of high frequency pulses at a higher voltage. Here is how a basic Joule Thief works, step by step:
1. Initially the transistor is off.
2. A small amount of electricity goes through the resistor and the first coil to the base of the transistor. This partially opens up the collector-emitter channel. Electricity is now able to travel through the second coil and through the collector-emitter channel of the transistor.
3. The increasing amount of electricity through the second coil generates a magnetic field that induces a greater amount of electricity in the first coil.
4. The induced electricity in the first coil goes into the base of the transistor and opens up the collector-emitter channel even more. This lets even more electricity travel through the second coil and through the collector-emitter channel of the transistor.
5. Steps 3 and 4 repeat in a feedback loop until the base of the transistor is saturated and the collector-emitter channel is fully open. The electricity traveling through the second coil and through the transistor are now at a maximum. There is a lot of energy built up in the magnetic field of the second coil.
6. Since the electricity in the second coil is no longer increasing, it stops inducing electricity in the first coil. This causes less electricity to go into the base of the transistor.
7. With less electricity going into the base of the transistor, the collector-emitter channel begins to close. This allows less electricity to travel through the second coil.
8. A drop in the amount of electricity in the second coil induces a negative amount of electricity in the first coil. This causes even less electricity to go into the base of the transistor.
9. Steps 7 and 8 repeat in a feedback loop until there is almost no electricity going through the transistor.
10. Part of the energy that was stored in the magnetic field of the second coil has drained out. However there is still a lot of energy stored up. This energy needs to go somewhere. This causes the voltage at the output of the coil to spike.
11. The built up electricity can't go through the transistor, so it has to go through the load (usually an LED). The voltage at the output of the coil builds up until it reaches a voltage where is can go through the load and be dissipated.
12. The built up energy goes through the load in a big spike. Once the energy is dissipated, the circuit is effectively reset and starts the whole process all over again. In a typical Joule Thief circuit this process happens 50,000 times per second.
Step 3: Wind the Toroid
The transformer in the circuit is made by winding wire around a ferrite toroid. These toroids can be purchased from electronics suppliers or they can be salvaged from old electronic equipment such as power supplies.
Take two pieces of thin insulated wire and wrap them around the toroid 8-10 times. Be careful not to overlap any of the wires. Make the wires as evenly spaced as possible. To hold the wires in place while I was prototyping, I wrapped the toroid in tape.
And after that join two opposite color wires from both end together as shown in image & refer video for better understanding.
Step 4: Connections
follow the above circuit and solder the positvive of the led to collector of transistor & negative to the emitter & 1 k ohm to base then one of the single wire of toroid to the collector & other one to the 1k resistor as shown in image & video and connect a wire to the emitter then connect +ve of the battery to the two together joined wires of toroid & -ve of battery to the wire connected to emitter.
Step 5: Final Step
After this make this permanent on a pcb along with the switch to turn it on or off and reuse your old used AA battery in your mini torch made with joule thief circuit.
If having issue with circuit etc. Then refer vudeo for better understanding.
Enjoy making your own joule thief & reuse your old AA batteries.
8 months ago
A few points to improve the description of how the circuit works:
In the beginning the voltage of the battery is high enough to turn the transistor ON. This allows current to flow through the collector-emitter junction and the magnetic field in the transformer increases and creates a voltage in the base winding that turns the transistor ON more. This continues until the transistor is fully saturated OR the magnetic flux in the core creates SATURATION.
At this point the magnetic flux is no longer EXPANDING and the extra voltage and current being passed to the base of the transistor suddenly STOPS and only the originating current into the base is present. This is not enough to keep the transistor turned on FULLY and the magnetic field starts to COLLAPSE. The collapsing magnetic field cuts all the turns on the transformer and produces a voltage in the base that has an opposite polarity to the originating voltage. This turns the transistor FULLY OFF.
The transistor effectively COMES OUT OF CIRCUIT and the lead connected to the collector produces a voltage that is opposite in polarity to previously and this means the voltage is HIGHER than the voltage of the battery by a value that can 10v to 20v.
This voltage is high enough to illuminate the LED and even 2 or 3 LEDs can be connected to the circuit.
But a LED has a characteristic voltage depending on its colour and for a white LED this is 3.2v. The voltage across the LED can never rise higher than this and the output voltage of the transformer is limited to this value and all the energy from the collapsing magnetic field is converted to light by the LED.
At the same time the magnetism is cutting the turns of the base winding and producing an output voltage that is negative and keeping the transistor fully turned OFF.
When the magnetic field has fully collapsed, the negative base voltage reduces and the battery voltage starts to take over, via the resistor and when it is above 0.6v, the cycle starts again. This process occurs at the rate of about 100,000 cycles per second.
Persistence of Vision of your eye means you cannot see the flashing of the LED and it appears to be ON all the time.
Here is an improvement to the circuit:
The resistor connected to the feedback winding should be connected between the positive rail and the feedback winding and the lower end of the winding connected directly to the base.
A 10n capacitor should be connected directly to the base.
This will make the circuit 300% more efficient.
Let me explain.
I have already explained how the whole circuit works but the reason for the improvement in efficiency with the 10n capacitor is due to the fact that the capacitor prevents the spikes from the feedback winding emerging from the top of the winding and all the spike is passed to the base.
This is not important for 90% of the cycle but when the transistor is getting turned on as hard as possible at the peak of its cycle, the magnetic flux in the transformer may be a maximum but its rate of increase is a minimum (INCREASING FLUX) and thus the energy produced by the feedback winding is at a minimum and the transistor is not being tuned on as hard as we want.
If a capacitor is placed at the top of the winding, the output from the feedback will only be able to come out the bottom of the winding and into the base.
This will turn ON the transistor a little bit harder and increase the EXPANDING FLUX and this will be passed back into the transformer as COLLAPSING FLUX and eventually illuminate the LED to a brighter level.
With the original "inefficient" circuit, energy is lost in the base resistor. And this "lost energy" is what we are using to make the circuit more efficient.
As the transistor is turned on more and more, it takes more current to do this. By adding a capacitor to the top of the feedback winding you are asking the feedback winding to deliver very little more voltage but convert the energy it is producing to create a HIGHER CURRENT. This is one of the amazing things a transformer can do.
Reply 2 months ago
Colin, it is not clear to me where the little capacitor goes.
I understand that the resistor is moved to the other side of the coil. So, it is now connected between anode and coil.
I understand that the 10nF capacitor is connected to the base of the transistor. Where does the other end go? Is it parallel to only the coil? Or is is parallel to coil+resistor?
I did find other circuits where a little cap is placed across the resistor, but the RC pair was kept connected to the base of the transistor.
Question 2 years ago on Introduction
How long will this led work?
Answer 5 months ago
That depends on the life of the battery
7 months ago
Circuit is not correctly explained. For example he says electricity is able to travel through the second coil only when transistor opens, why, there is an open path through the LED independent of the transistor.
Tip 10 months ago
It would be nice to do joule thief which is taking energy from earth anode and kathode instead of battery, and bosts it up like tariel kapanadze generator. Maybe add capacitor to store earth energy.