I had a Joule Thief already assembled, using point-to-point wiring. The transistor was a BC338. The coil was a 3/8″ (9mm) toroid core with a primary winding of about 20 turns of 26 AWG and 666 uH, and a feedback winding of 190 uH. The resistor was 1000 ohms.
Modifications
I disconnected the feedback (base) winding from the +1.5V and connected it to the collector of a BC560C PNP transistor. I connected the emitter to the +1.5V, and the base to one end of a 22K resistor. I connected the other end of the 22k to negative. I soldered the anode end of a 1N5817 Schottky diode to the collector of the BC338 and I removed the anode of the LED. I removed the other lead of the LED from the negative (no more LED). I connected the positive lead of a 470 uF capacitor to the cathode (banded) end of the 1N5817, and the capacitor’s negative lead to the negative.
So far, I have a Joule Thief that rectifies and filters the pulses to a DC output, but there is no load so the voltage could go very high. I need to regulate the voltage somehow to prevent it from doing damage. I connected a 5.6V Zener across the 470 uF capacitor, and applied power. The voltage was 5.39 with no load and 5.27 with a 1k load or 5.27 milliamps. The supply current was about 50 mA. Now I have a shunt regulated DC output of about 5V, but the supply current is still 50 mA with no load. I need to get the supply current to drop when the load is light, and that is the reason why I added the second transistor.
I disconnected the 5.6V zener from supply negative and connected it to the base of the PNP transistor. Now, when the load is light, the 5V DC voltage rises, the zener conducts, the voltage on the base rises to 1.5V and this shuts off the transistor, and there is no more current to the feedback winding, reducing or stopping the oscillations. The supply current is greatly reduced, down to less than 5 milliamps.
With no load, the output voltage is 4.95V, and with a 1k load, it’s 4.79V, which is the same as 4.79 mA load. The supply current is conserved when the load is light and the efficiency and performance is much better.
This converter can now be used to give 5V power to any device that draws no more than 5 mA. I don’t know how much current an Arduino draws, but it it’s less than 5 mA, then this could do the job. Remember that LEDs can draw more than 5 mA, so if you use an LED, it should be limited to a milliamp or less, to allow the converter to maintain the 5V. Too much current and the converter’s output voltage will drop.
