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This one has a supply current of 70 mA at 1.5V and a LED current of 25 mA at 3.3V (actually 25 millivolts measured across the 1 ohm [2 resistors] in series with the LED’s green wire).  This calculates to an efficiency of 78.6 percent.  The frequency is 250 kHz.

The circuit is almost as simple as the conventional Joule Thief; it requires a diode and 680 pF capacitor in addition to the 1k resistor.  The end of the feedback winding that was normally connected to positive is instead connected to the 1k and 680 pF as shown in the picture.  I used a SS8050 transistor, which is a Fairchild equivalent to the C8050.  It can handle up to 1.5 amp collector current.

The circuit will give more LED current for about the same supply current, or the resistor can be increased to 1.5k to give about the same LED current for less supply current.  The two current sensing resistors that are in parallel on the lower right are optional and can be removed, and the LED’s green wire connected directly to the heavy negative wire.

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