Wednesday, January 27, 2010
Joule Thief: a simple DC voltage booster
A joule thief circuit lit from a 0.5V AAA battery. Click to enlarge.
A Joule thief is a simple circuit that acts as a DC to DC booster, raising a supply voltage by several volts. In this iteration it uses the exhausted voltage of a alkaline battery and boosts it enough to light a blue LED that requires 2.8V to light. With an otherwise dead 0.5V AAA battery, it will light a blue LED and run for days, using (at the moment) just under 2mA of current. It's much much dimmer than using a fresh battery or running the LED with a proper current through it. Giving the Joule thief circuit 3V from two fresh batteries pulls 75mA through the LED, making it very, very bright and probably short-lived.
A schematic is below. The circuit works like this: When first turned on, current flows into the inductor and produces a magnetic field in the toroid. While this is happening, no voltage appears at the base of the transistor, so the transistor remains off. The LED sees at first no voltage and while the inductor fills up, it only sees a maximum voltage of the battery, which is not enough to pass the diode. Once the inductor is charged, the battery voltage appears at the base of the transistor, turning it on. This allows the right side of the inductor to want to dump the energy it has stored in its magnetic field as quickly as possible, and this gives us a high-voltage that appears across the inductor. When that voltage exceeds 2.8V the LED turns on and lights up until the voltage drops below, triggering the sequence to begin again. I measured the frequency of the on/off oscillation and it seems to run at about 34kHz; the multimeter said between 68 and 72kHz but a radio showed there was 34kHz signal as well, which I assume was the fundamental (and the 68kHz one a harmonic). It did change in frequency a bit while on.
Bre Pettis/Windell Oskay