I think I wrote about this circuit in my late Watsonseblog. This circuit is essentially a Joule Thief with an added transistor to regulate the output – somewhat. It doesn’t have very good regulation because the 6.8k is a resistor (see Note). If this resistor is changed to a Zener diode then it regulates much better and doesn’t go as high as 10.4 V. The Zener can be replaced by the emitter to base junction of a NPN transistor.
The emitter to base junction of many NPN transistors is typically rated at 5V, but can withstand 8 to 10 volts before it breaks down. When it breaks down it acts much like a Zener diode. But the voltage varies from transistor to transistor so it may require selecting one from a few transistors.
I connect the transistor and a 1k resistor in series and to a variable power supply that can supply at least 12VDC. The transistor’s base goes to negative and the emitter goes to the resistor, and then to positive. I cut off the collector lead because after this breakdown, the transistor’s characteristics, especially the current gain, can be permanently damaged. Then I turn up the power supply until I measure about 1 volt across the resistor, which is equal to 1 milliamp flowing through the resistor. Then I measure the voltage across the emitter to base. This is the Zener voltage, and for this converter circuit, we need about 8.5V. But I’ll explain later how to use other voltages.
The 8.5V Zener or transistor should give about 9.1V open circuit and somewhat lower than that when current is drawn from this circuit. The regulation will be improved. The 390 ohm resistor can be increased to 1k to reduce the amount of wasted current.
If the Zener voltage is below 8.5V, the voltage can be increased some by putting a diode in series with the transistor or zener diode. The diode should increase the voltage by about 0.6V. More than one diode may be used. Most 9V devices can run at 7 to 8V without too much decrease in performance. Some devices use a 5V regulator chip so 7V is okay as long as it doesn’t fall below 7V.
The circuit in TE can only put out 30 mA, which may not be enough for some devices such as a transistor radio. The BC338 can be replaced by a higher current transistor for more output current. Or a second BC338 could be connected in parallel with the existing one, and it should help, but not as much as doubling the output current. I would also replace the 1N4148 with a Schottky diode such as the 1N5819 for lower loss and higher current. Also the 100 uF capacitor on the output should be increased to 470 uF or more. This helps reduce the interference that this circuit can generate. I have used this kind of circuit on AM/FM radios and the AM band is barely usable because of the interference, but the FM band is okay. It may require more filtering to get the AM band to be usable.
The coil TE uses is a small cylindrical bar only 7 mm long. It’s probably difficult to wind 55 turns of wire onto this, especially if the wire is thick enough to be good for 200 milliamps supply current. I think it would be much better to use a 3/8 inch or 9 mm toroid instead, with at least 28 AWG (0.5 mm) enameled wire for both primary and feedback windings. Using a high Mu toroid, the windings can probably be reduced to 10 and 18 turns respectively.
Two other things to consider.
One may want to change the two AA cells to NiMH rechargeable cells. They put out only about 2.5 volts, so some changes may have to be made to get the circuit to work well. A typical JT draws about 80 mA supply current from 1.5V, and puts out about 18 mA to the LED. That’s 66 milliwatts to the LED. For the TE circuit, it’s 9V times 0.03 A or 270 mW. That’s FOUR times as much power as a JT, so it takes a lot more current from the battery. If you use 1.5V, the AA cell gets used up quickly. Therefore it’s much better to use two AA cells in series for 3V.
This circuit can work okay for powering lower power devices such as a DMM. I wrote a blog about that one, too. Mainly, the circuit is put on a diet to reduce the current with no load and with an output a load of about 7 to 10 mA. For this low power a single 1.5V cell can be used for the supply.
Note: With two resistors, the output voltage is divided by the ratio of the two resistors and applied to the base of the BC547. When the voltage gets up to about 0.5 to 0.6V, the transistor conducts and reduces the output of the JT. So the Zener effect is done by the transistor. Problem is that transistor is a very poor substitute for the sharp turn on point of a Zener. Using a Zener considerably improves the regulation.
Here is a link to one I made for my DMMs.