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.
I noticed this circuit because I have a lot of 3.7v rechargable batteries from laptop’s. These will have the current needed for many devices. Just have to decide on the right circuit for the job, like power for a DMM.
I haven’t tried using the cells out of laptop batteries because they are very high current and could catch fire if shorted. I don’t know how much charge they lose just sitting and not being used, but if I used one in my DMM, I’m sure that most of the charge would be wasted because the DMM uses a very small amount of current – under 10 mA – and even if left on a lot, the cell would last for a long time. You say that you “Just have to decide on the right circuit for the job…” but I would also say that one has to decide on the right battery for the job, and a cell from a laptop is a bit of an overkill for a DMM, in my opinion.
I fastened three CR123 cells to the back of one of my DMMs, and connected them in series for 9V. They work just fine, and should last for a very long time. Check my blog for more info.
I tried this circuit, it works but had to replace the 390 ohm resistor with 584 ohms to get 9.19v. 620 ohms put the voltage at 8.85v. I put a green led with a 680 ohm resistor across the output and the voltage only dropped to 9.17v. I used a toroid to wind the coil. Max volt out is 28v at 3.7 in. But I also don’t think the output current will be much even though I used a 2n4401 and 22awg wire. Will have to try the one in your link.
The important point of these circuits is to regulate the output but at the same time, shut down the JT so that at low output currents, the circuit is using very little current from the battery. Most circuits that run off a 9V battery can tolerate lower voltage, down to 7 volts or so. In these cases the DC-DC converter does not have to have very good regulation, and the resistive voltage divider works just fine. But zeners are cheap and do a very good job of regulation. I also thought about putting a 1N4148 diode in series with the base, to give a sharper ‘knee.’ It should help somewhat. Perhaps you could try this and let us know what the difference is.
Problems now, with the led and 680 ohm the 2n4401 gets hot and voltage dropes to 3v, but with just a 1k resistor it runs fine. Will work with this some more latter, the idea was to have a dependable rechargable adjustable battery, these laptop batteries seem to like the slow solar charging and are free I must have 50 by now so it’s time to use them.
Seems like the ‘4401 is handling too much current. You might try using two of them. Connect the emitters and collectors together. Use a second 1k resistor to the base of the second ‘4401. This should double the current, and both should not get hot. Let us know, and if you send a photo, I can put it in the blog. I said I was going to do this earlier but now that I have more free time I can get to those things.