Instructables

How does the joule thief circuit work?

the best i can figure, the current coming through toroid turns on the transistor, then the collector drains the current to the negative, while pulling it from the emiter back through the diode to the collector? if i am correct, how would this boost the voltage?

Picture of How does the joule thief circuit work?
frollard4 years ago
http://en.wikipedia.org/wiki/Joule_thief

A better picture of the circuit is on wikipedia:
essentially the power gets switched on by the transistor, causing a magnetic field in the other coil that shuts the transistor off, and the changing electric current means the magnetic field collapses sending out a pulse of power to the led (since the transistor stopped conducting).  When that pulse ends, the transistor gets current again and conducts...etc.

http://www.instructables.com/answers/how_does_a_toroid_in_Joule_Thief_circuit_work/

Videogamer5557 months ago

Going by your diagram, here's how I think the circuit works:
First current flows through the primary of the transformer, the resistor, into base (B connection) of the transistor and out of the emitter of the transistor (E connection). When this happens, a magnetic field from the primary coil. However the current through the primary isn't instantaneously at maximum level as the inductance of the transformer's primary coil impedes the flow of current. So at the instant that power is applied, there is no current through the primary. However there is an output voltage on the secondary, as there is an increase in current (derivative of current is positive) through the primary, and inductors output voltage according the derivative of the current. Therefore the secondary coil of the trans former is outputting a voltage the moment power is applied to the device. Since the battery is also wired in series with the transformer's secondary, and with proper polarity to add to the battery's voltage. Now as the device continues to have power applied, current through the secondary of the transformer and LED increases (same phenomenon as on the primary side, inductive impedance). Now also current is slowly increasing in the primary side of the transformer and transistor. At some point the current through the primary side of the transformer and transistor becomes sufficient to switch on the transistor. When it does so, the output of the transformer is shorted to ground, by flowing from the secondary of the transformer into the collector (C connection) of the transistor and out of the emitter (E connection) of the transistor, and to the negative terminal of the battery. When this his happens, there is a sudden increase in current on the secondary of the transformer and corresponding inductive voltage surge opposing this current on the secondary side of the transformer. At the same time (because the primary and secondary are magnetically linked), there is an equivalent voltage surge on the primary side sufficient to oppose the current flowing through the primary of the transformer and the base of the transistor. This causes the transistor to shut off, and the whole cycle repeats. Of course shutting off a transformer suddenly with a transistor will produce a significant voltage spike across the transformer and transistor which can damage the transistor. You might want to have another diode in parallel with the LED in the opposite polarity as the LED so that instead of getting a voltage spike (which can damage the transistor and LED) you get a current spike through the diode. Make sure you use a diode that can handle a significant current (a big wopping 4A or 6A diode might be overkill, but I'd use one just to be sure).

acmefixer3 years ago
I tried to give a simplified explanation in my blog. There are other links to other explanations, both technical and non-technical. The term Joule Thief is not a technical term, the actual name of the circuit is a blocking oscillator, and doing a search for this will give some good explanations of its operation. Blocking oscillators have been around for a very long time and are well documented.
Look up Reverse EMI or "Inductive Kick"...same reason you need steering diodes on inductive loads which are connected to logic circuits.   When the field from the coil collapses it creates a voltage spike in the negative direction which is many times larger than the initial current. 
knektek4 years ago
it's an inverter.
faham4 years ago
If you noticed that the LED turns ON only when the transistor is OFF. And it turns OFF when the transistor is ON. The transformer is configured as to switch on and off the transistor periodically at a rate dependant on the transformer inductance at each arm, the 1 K resistor and the base to emitter resistance of the transistor. The boost convert principle uses the transformer to generate higher voltage from a lower voltage similar to this principle.
When the left xfrmr arm conducts current on power ON the transistor shuts down and creates a much higher voltage across LED than the battery because the battery voltage adds to the voltage induced at the right  xfrmr arm and the LED shines! This is no miracle. Immediately after that the left arm voltage collapses due to the RL time constant of the 1 K resistor and the left arm, and energy dissipated by the LED - again no miracle. As a result the right arm starts conducting current from the battery because the 1 K resistor now pulls up the base of the transistor high and turns on strong and the LED shuts down. Due to vision persistence the LED may look like constant but is flickering probably. The cycle repeats as long as the battery is connected and has some juice. Eventually the battery will give up depending on how much juice left inside. Remember that the dot on the transformer refers to the polarity of the voltage induced on that arm, for example the dot refers to the positive and no dot means negative relatively. This is similar to the mechanism of pulling a rubber band to some distance and let it lose to produce that intense strike. If the left arm builds up positive voltage on the dot at the 1 K resistor then the base voltage is pushed down relative to the positive side of the battery and this shuts down the transistor as long as the voltage of the left arm dot is high enough, meaning the voltage on the base theoretically is low. Pushing down means going negative and pushing up means positive – just for sake of remembering it the easy way.
buteman4 years ago
Starting from first supplying current from the battery ( no switch in the circuit shown here - but pretend there is one )
>>>>> Assuming Conventional Current Flow i.e Positive -> negative <<<<<

Current flows from the positive pole of the battery ( the long line of the cell ) through the left hand coil in the diagram -> 1k resistor ->into base of transistor -> out of the transistor emitter -> back into the negative side of the battery.

This 'turns the transistor on' that is current also starts to flow from the positive pole of the battery  through the right hand coil in the diagram ->into the collector of the transistor -> out of the emitter of the transistor -> into the negative side of the battery.

As this happens you need to realize several things:
1/ The coils try to resist any change in the current flow so current takes time to build up through the collector/emitter circuit.

2/ This is a very short time from our point of view. It could be just a few microseconds.

3/ As current flows through the right hand coil it opposes the current through the left coil.

4/ The transistor is an amplifier so the small current in the base -> emitter becomes
a much higher current collector -> emitter.

4/ This switches the transistor off quickly.

The effect on the right coil is that it now tries to keep the current flowing and then it starts to flow through the LED but because of the speed at which the transistor turned off the voltage rises,  ( this is often referred to as back emf )
so much so that is more than the voltage provided by the battery.
This repeats until you switch it off or the battery is drained of power.

Dr.Bill4 years ago
lemonie4 years ago
It's a transformer-type thing, the transistor serves to produce the fluctuating AC-like current that transformers need. (I wouldn't disagree with previous comments, but I'm offering a different way of explaining it)

L
revelae (author)  lemonie4 years ago
thanks, but maybe i was asking the wrong question: what is the process whereby the transistor flucuates the current?
Simple RL resonance oscillator.
NachoMahma4 years ago
.  Similar to an automotive ignition. Google for details.
Energy is stored in the magnetic field in the coil, and then the collapsing field induces a higher voltage than created it in the load.

Steve