Step 2: Technical Information

The principle of operation is very simple.  The circuit is a basic boost switching converter that converts a lower voltage to a higher voltage through the use of switch and inductor.


The Joule Thief circuit uses a programmable boost converter IC. The boost converter is advertised as a 350mA LED Driver, although it’s really just a simple boost controller IC without an external voltage feedback input. The circuit is powered via a single 1.5V AA battery.   L1 is the 3.3uH boost inductor while C2 is a 4.7uF ceramic capacitor used as the boost output filter which provides a DC voltage to the 1W Cree LED.  Potentiometer R2 is used to vary the output brightness of the LED, while C1 is added to provide soft turn-on and turn-off of the LED.

Parts List:
  • (2) Resistor, 0.0 ohm, 0805 (R1,R3)
  • (1) Potentiometer, 1Meg (R2)
  • (1) Capacitor, 1uF, 0805 (C1)
  • (1) Capacitor, 4.7uF, 1206 (C2)
  • (1) Cree 1W LED (D1)
  • (1) Inductor, 3.3uH, 2.5A (L1)
  • (1) Programmable Boost Controller IC
  • (1) Switch
  • (2) AA Battery Clips
I'm seriously considering using this design for an IR LED project I'm working on...one problem...where do I get the LTC3490? I can't find it anywhere, not even on the 'internet bidding site'.
vc pode usar um joule com 2 transistores que pode ligar led de 1watt
Interesting. Main driving IC is the LTC3490 from Linear (http://www.linear.com/product/LTC3490); looks like the schematic is the example(s) given in their datasheet. Useful layout.
very nice :)
I noticed an error in your schematic. The 4.7uF cap should be connected to the "CAP" pin on the IC, and not to the "LED" pin. Just a heads-up! :)

About This Instructable


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Bio: My name is Allison and I am an engineer co-op student working for Eastern Voltage Research creating fun and exciting educational kits for students and ... More »
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