Peter asked in an email what I thought about the schematic he gave a link to in his email. With his permission I’m quoting his email:
I found this on Youtube: http://www.youtube.com/watch?v=xO0V6foiiy4
I wanna build this (or something like this) as a surprise gift for my wife – the schematic is here: http://i123.photobucket.com/albums/o298/RODALCO/Joulethiefwithflipflop.gif
May I beg you to give me your opinion on this? I think the transistors should be changed but I´m not exactly an expert on things like that!! Or maybe you´ve got somehing better than this – I searched your blog but didn´t find a “Joulethiefwithflipflop”.
I get Google Alerts and I recently received one for this Youtube video. As he says in the video, it’s an astable multivibrator. A flip-flop doesn’t oscillate by itself. If you watch the video, you will see that the two LEDs speed up, then they both go on. He doesn’t say what he’s doing, but I have seen this effect with astable mvbrs when the supply voltage drops too low. I think this is the real challenge, and I’ll try to explain why.
The Joule Thief is optimized to put out the maximum current to the LED when the battery voltage is 1.5V. When the cell’s voltage drops down to 1.2 volts, for example, the LED current drops a lot more, and the LED may have much less current than at 1.5V. The current may drop by more than 1/3.
When the Joule Thief’s output is rectified and filtered, as it is in the schematic, the DC power output will also drop by a large amount as the cell voltage drops. But the astable mvbr needs enough current to keep it running, since it’s a load that does not change. The Joule Thief’s output voltage drops to reduce the current it has to supply, and soon the astable’s flash rate increases as it becomes current starved. Then the LEDs both come on when the astable can no longer sustain oscillation.
One way to solve this is to design the Joule Thief to put out enough power at a lower voltage, 1 volt for example. But then the Joule Thief will draw excessive current at 1.5V. What has to be added is a circuit that starts to take effect when the cell voltage rises above 1 volt. At higher cell voltage, the circuit turns on and shunts the base bias current from the Joule Thief’s 1k resistor. As the cell voltage drops, this shunting is reduced and the Joule Thief keeps putting out nearly the same current.
In order to get the Joule Thief to put out enough at 1V cell voltage, the 1k base bias resistor must be reduced. The lower value depends on the transistor, but I would guess 470 ohms or even less. And the transistor must be a much better choice than the BC547, which can’t do the job at 1.5V. I would try a BC337-40 as a starting point.
Next is the voltage, current and power demands of the astable itself. You will notice in the schematic there are two resistors, each 560 ohms and each in series with the LED. These resistors cause a voltage drop, and waste power. It would be better if we could reduce their value a lot, or even eliminate them (zero ohms). But the astable needs some resistance to develop some voltage so the capacitors will feed back some voltage to keep the circuit oscillating. I do this by putting a resistor across the LED. I have not found any information on what this resistance should be, but I use a value that is 1/10 of the base bias resistor. If the base bias resistor is 47k, for example, then I use 4.7k.
More to come.