It works! (JT/multivibrator mash-up) - comments welcome

[carbonzit]

OK, folks, I'm giving this one last go, to find some way to eliminate those intermediate transistors (Q5 & Q6 in this circuit). If someone here doesn't come along soon with some suggestions, then I guess I'll just build the circuit as-is, since if all the best minds on Electro Tech can't come up with a solution, then I guess there must be none ...

 

[colin55]

**broken link removed**
The only simple answer to your requirement is the 2-transistor circuit above.
This circuit alternately flashes two white LEDs, on a 3v supply and produces a very bright flash. The circuit produces a voltage higher than 5v if the LED is not in circuit but the LED limits the voltage to its characteristic voltage of 3.2v to 3.6v. The circuit takes about 2mA and is actually a voltage-doubler (voltage incrementer) arrangement.
The 1k charges the 100u and the diode drops 0.6v to prevent the LED from starting to illuminate on 3v. When a transistor conducts, the collector pulls the 100u down towards the 0v rail and the negative of the electro is actually about 2v below the 0v rail. The LED sees 3v + 2v and illuminates very brightly when the voltage reaches about 3.4v. All the energy in the electro is pumped into the LED to produce a very bright flash.

 

[carbonzit]

Thanks, Colin. I'll keep that around for future reference.

Unfortunately, it's not really what I'm looking for, since it produces a short intense flash. I prefer the action of my multivibrator/JT mash-up. Perhaps I shouldn't use the word "flash", as what I want is just alternately-blinking LEDs. And I'm actually pretty happy with what I have now. Even with the two extra transistors, it can be built on a very small board and doesn't consume much power.

 

[colin55]

You can simplify your 6-transistor circuit enormously by using the following design:
**broken link removed**

This circuit alternately flashes two white LEDs, on a 1.5v supply and produces a very bright flash. The circuit produces a voltage of about 25v when the LEDs are not connected, but the LEDs reduce this as they have a characteristic voltage-drop across them when they are illuminated. Do not use a supply voltage higher than 1.5v. The circuit takes about 10mA.
The transformer consists of 30 turns of very fine wire on a 1.6mm slug 6mm long, but any ferrite bead or slug can be used. The number of turns is not critical.
The 1n is important and using any other value or connecting it to the positive line will increase the supply current.
Using LEDs other than white will alter the flash-rate considerably and both LEDs must be the same colour.

 

[TechnoGyppo]

Great stuff Colin55

I take my hat off to colin55 for his work with the Joule thief and LED flasher circuits.

I've searched long and hard for design formula for the self oscillating boost converter and have had no luck at all. Seems like it has vanished and been replaced with the 'get a chip to do it' mentality. Does anyone have any design equations for it? If not I can share what I've managed to come up with but it needs to be verified.

I'm beginning to wonder why some people need to simulate circuits like this instead of the good old fashioned design and test method. While simulation has it's place, for something like a Joule Thief, the words sledgehammer and nut spring to mind.

 

[colin55]

The circuit is a fly-back arrangement and the voltage developed is due to the QUALITY of the transformer, called the "Q-factor." There is no formula for this phenomenon.

 

[TechnoGyppo]

Joule Thief Operation and Formula

I've been researching the self oscillating boost converter for a few days now. The technical name for it's operation is boundary conduction mode / Critical conduction mode.

The absolute value of the inductor is not critical as long as it is big enough (> 50uh). The inductor will only affect the frequency that the circuit runs at - bigger = lower frequency.

The important thing is to get a suitable switching transistor with a low Vce(sat) at the peak current the circuit operates at.

If you know what load you want to drive then you can calculate the peak current easily enough. You only need to know the wanted output voltage, output current, saturation voltage of the transistor, forward voltage of the rectifier (for a smoothed output) and the input voltage.

Output power = output voltage x output current

To get the peak current the formula is:

Ipk = Po ( ( 2/(Vs - Vce )) + ( 2/(Vo - Vs + Vd) ) )

Where:
Ipk = Peak current.
Po = Output power.
Vs = Supply Voltage.
Vce = Transistor saturation voltage.
Vo = Output voltage.
Vd = Output smoothing diode forward voltage.

I have built a joule thief type circuit, measured the voltages and currents and the formula gives roughly the right answer. I say roughly due to the use of a not to accurate old oscilloscope used for the peak current / saturation voltage measurements. At low voltage, a small error in the transistor saturation voltage will have a large impact on the calculation of the peak current. A small re-arrangement of the above formula can give you the likely maximum output power if you can measure the peak current.

Trying to set the peak current in a practical circuit is the difficult bit as every transistor has it's own value of gain, meaning that the base drive resistor needs to be adjusted for a particular transistor. The power supplied to the circuit from the battery will be higher due to losses. Using a good switching transistor and a fast diode will help minimise the losses.

Hope this is of use to someone....