# Understand the Joule Thief circuit

I quote from Wikipedia (https://en.wikipedia.org/wiki/Joule_thief):

... This self-stroking/positive-feedback process almost instantly turns the transistor on as hard as possible (putting it in the saturation region), making the collector-emitter path look like essentially a closed switch (since VCE will be only about 0.1 volts, assuming that the base current is high enough).

Referred to this schematic:

My question is: How the transistor can go in the saturation region? If the transistor is in this region, it means that Vce will be around 0.4 V. That's impossible, because on the primary coil there will be Vcc-0.4 (assuming Vcc=1.5V, it will be 1.5-0.4=1.1 V) and because of the low coil resistance (lower than 1 ohm for a typical homemade toroid transformer for this circuit) the Ic current should be excessively high (for 0.7 ohm resistance of the coil it should be I=V/R=1.1/0.07=15.71 A).

It shouldn't be more probable that the core of the transformer will saturate and thus the primary coil won't induce any voltage into the secondary (so the transistor will be forced to turn off)?

Here is the circuit being discussed:

To understand how this works, consider each phase of operation over a whole cycle.

Let's start with everything off and power newly applied. the transistor is off, so no collector current. However, there is a current path thru the resistor, the transformer secondary (the left coil), and the base of the transistor. This turns on the transistor, at least somewhat.

The transistor turning on brings the bottom side of the primary (the right coil of the transformer) to a lower voltage than the power. At this point, the primary acts mostly like a inductor, so current builds up linearly over time.

This change in current flowing out of the dot end of the primary causes the secondary to try to make current flowing into its dot end. This additional current thru the secondary also means more base current, since the two are in series. More base current means the transistor turns on harder, which causes more current thru the primary, which causes more base current, which turns the transistor on harder, etc.

Due to this positive feedback, the transistor is turned on so hard that its collector voltage can't go any lower, usually about 200 mV. That's called saturation. At this point, the transistor C-E looks like a closed switch, except for the small saturation voltage.

With the fixed voltage of the supply minus the saturation voltage applied to the primary, it's current ideally builds up linearly. However, it can't do that indefinitely. One of two things happen: The current levels off due to the inherent equivalent series resistance of the primary winding, or the the transformer core magnetically saturates. Either way, the secondary is no longer driven to create higher current. Since a transformer works on the change of the magnetic field, the secondary stops providing additional base drive at all.

Now the positive feedback works to quickly turn off the transistor. At first when the magnetic field levels off, the secondary no longer provides additional base drive, and the base current is just what the resistor allows to pass. That is not enough to sustain the same collector current, so the transistor starts to turn off a bit. This now reduces the magnetic field. The secondary now actively opposes the base drive provided by the resistor. The transistor allows less current, which causes more active negative base drive, which causes even less collector current, etc.

Due to this positive feedback, the transistor turns off hard quickly. The primary now acts like a inductor. Its current can't be turned off instantly. The inductor makes whatever voltage it has to to maintain the same current instantaneously. The voltage rises quickly, and when it gets to the LED forward voltage, the current flows thru the LED and the voltage stays roughly fixed.

The inductor now has reverse voltage across it, so the current thru it is decreased. Eventually it gets to 0, the magnetic field stops changing, the secondary stops opposing the current thru the resistor, and we're back to about the state where power was first applied.

• There is also the internal resistance of the battery, here in my circuit it is the cause of the transistor turning off. Jul 31, 2018 at 14:38
• This is the only explanation of the joule thief circuit that actually made any sense to me. The key point here is that without the ADDITIVE effect of the secondary, the transistor receives insufficient base current to saturate, and therefore is not passing as much collector-emitter current as it would otherwise. And the additive effect of the secondary tapers off once the core is saturated. Feb 8, 2021 at 1:33

Yes, the transistor saturates almost instantly, aided by the positive feedback. The coil current rises linearly until it too saturates, removing the base drive and cutting off the transistor. At this point, the energy stored in the coil is dumped into the LED. But once the energy is depleted, the field collapses, the base drive returns, and the cycle starts over.

• So why wikipedia say that the transistor saturate when the saturation of the transformer core occurs before the saturation of the transistor? Feb 24, 2018 at 16:12
• The inductance of the coil slows down the rate of rise of collector current. The transistor turns on much faster than that. Feb 24, 2018 at 16:16
• I've finally understood it! I didn't consider this thing. Thanks! Feb 24, 2018 at 16:44
• @Wolf It is either the saturation of the BJT and the inadequacy of $\beta$ to meet the rising need or else the inductor core saturation. It would be a total accident of circumstances for both to be involved simultaneously. So one or other starts the turn-around. If you were to use an air-core transformer, there would be no possible core saturation and it would be $\beta$ that does the trick.
– jonk
Feb 24, 2018 at 18:22