The Joule Thief is a simple circuit that takes a low voltage source and, through induction, turns it into a higher voltage. Voltage inputs can be as low as a considered “dead” AA battery. The circuit works by rapidly switching the transistor causing the current in the coil of wire to stop and go at high speeds. When a current flows through a coil of wire, an electromagnetic force (EMF) is generated and creates a magnetic field. When the current flow stops, that magnetic field collapses and the EMF induced in the coil reaches a much higher voltage than its primary input. The transistor is doing this well over 10s of thousands of times per second that the human eye cannot tell the LED is turned on and off so quickly.
Note, however, power is not being created, but is being transformed—the power input from the battery is the same power outputted from the circuit minus the power being absorbed by the transistor and light-emitting diode (LED), except at a much higher voltage.
Why is it necessary to use the Joule Thief to run the LED? Double A batteries are typically 1.5 volts and LEDs need at least 2-3.5 volts to work. However, LEDs do not need a significant amount of current. Since the Joule Thief uses the majority of the current to induce a higher voltage the smaller current left over can be used to run the LED. Go ahead, try and light one of the LEDs with just the AA battery—it won’t work.
See the attached schematic for the Joule Thief circuit:
Step 1: Parts & Step 1
If you are not familiar with schematics, the following step-by-step instructions will aid you in your construction of the Joule Thief. Ω
Make sure that the following components are available:
1. Solder-less Breadboard
2. AA battery holder
3. Copper magnet wire
4. Ferrite torroid
5. Transistor (2N3904)
6. 1 kOhm resistor (1000 ohms Ω)
8. Some extra jumper wires