**Ωne** of the readers of my late, great watsonseblog sent me an email and asked some questions about the Joule Thief coil.

But before I get up to speed I think it would be a good idea to read the warning at the end of this blog.

my Q was: what is the relationship of the coil and/or wire to the performance of the circuit (assuming other components are constant)?

from what i remember in your answers, the fatter the wire, the more current flows to the led. also, more turns makes higher inductance but lower frequency. what then is the factor for voltage?

my understanding is that we only need enough turns to make the circuit kick into oscillation, and as long as the circuit puts out at least 3.5v, the led will light up.

additional Q’s:

– which is the primary winding? secondary? is there a naming convention for this?

– which coil should be fatter/more turns?

**I think I should start** with the last two Q’s to clear up any confusion about terminology. It will also help the neophyte experimenter understand the Joule Thief, which I will abbreviate to JT.

**Coil** The typical JT coil consists of two windings; each winding consists of from less than 10 to more than 25 turns, wound on a ferrite toroid which looks like a tiny donut (but remember that other types of coils can be used, too). The primary winding carries almost all of the current and is the heavier of the two windings; the feedback winding can be very fine wire because it carries very little current (but it doesn’t harm anything if it’s the same size wire as the primary). Both windings are wound onto the core at the same time and have the same number of turns; this is called bifilar wound. Winding both at the same time is easier, but the windings can be wound separately.

**Windings ** The primary winding’s leads are connected to the transistor’s collector and to the positive end of the battery (when the transistor is NPN). One end of the feedback winding is also connected to the battery positive; the other end is connected to one end of a 1000 ohm (1k ohm) resistor; the other end of the resistor is connected to the base of the transistor. The two ends connected to the positive are not at the same end of the coil; the feedback winding’s leads should ‘cross over.’ The second winding is called the feedback winding because that is all it does. There is no secondary winding because the voltage conversion is all done in the primary winding. The feedback just keeps the transistor turning on and off. A winding would be secondary if it was a transformer, but it is not a transformer.

That answers the 2 additional questions at the end. Now for the rest, starting with the coil’s relationship question.

The coil is important because all of the current flows through it, and most of the current flows through the primary winding. The battery current can reach a half amp or more, so the DC resistance of the primary winding must be very low, to let the current freely flow. There is going to be some voltage drop across the transistor and the battery voltage is going to be less than 1.5V, so at most the primary winding will have a volt or less to push the current through it. If the DC resistance of the primary wire is 1 ohm and 1 volt is across it, and 1 amp of current is flowing, then the wire will dissipate 1 watt; that is wasted energy that does not go to the LED, but gets wasted as heat in the wire. So the DC resistance of the primary wire must be much less than 1 ohm to prevent waste and inefficiency. And this last sentence above answers the second question, “what is the factor for voltage?” Again, the DC resistance of the primary must be low so that the voltage drop across the primary winding is kept low.

**Wire Size** Let’s say I’m using 30 AWG copper wire (countries that use metric wire have a wire table in ohms per thousand meters). The copper wire table says that 30 AWG has 104 ohms per thousand feet, which is about the same as 1/10 of an ohm per foot (actually 0.104 ohms per foot but we’ll round it off). Suppose the primary winding has a length of 10 feet. Then the total DC resistance will be 1 ohm, and that is too high. Instead we use 24 AWG, which has a DC resistance of 25.7 ohms per 1000 feet, which is about the same as .0257 ohms per foot, and is one quarter the resistance of 30 AWG. A ten foot length will have a DC resistance of about 1/4 ohm, which is getting down to (or we could say approaching) the point where the power loss will be low and the efficiency will be reasonably good.

**“…we only need enough turns…”** is the last question and it involves some math. The coil is required because it is where the energy is stored in the magnetic field. Think of the battery as a lake or stream, and the LED is like a 5 gallon or 20 liter container in the bed of your truck, parked next to the lake or stream. And you have to fill this container by scooping up water in your ‘filling container’ and emptying it into this 5 gallon container in the truck. You start out dipping a drinking glass that holds a pint or half liter in the lake and emptying it onto the container. After a dozen or so of that, you say to yourself, this is going to take a long time! I need a bigger filling container. So you go get a wide mouth bottle that holds 2 liters or half gallon, and start dipping it into the lake and filling the container. Things are going much better and the job gets done reasonably fast.

### The math

The amount of energy stored in the coil is equal to the one half of the inductance times the current squared, or L*I^2/2. So we can double the inductance and increase the energy stored to double. Or we can double the current and increase the energy stored to four times, because it’s the current squared, or the current times itself – current times current. The voltage remains the same across the JT coil. We can increase the energy stored by using fewer turns, and we lower the inductance by half, but the DC resistance is half, the current doubles, and the energy is doubled and doubled, or four times (the current squared or current times itself). We sacrifice inductance to get more current and even more energy stored.

We have the LED and a JT coil. The coil has thinner wire and a large inductance, but a lot of wire is needed to make the inductance so the DC resistance is high. We have less than 1 volt to push the current through the high DC resistance, the amount of energy stored is small because the current is low, and every time the transistor switches on and off, it is like using the small glass to fill the big container: the amount of energy transferred or the amount of water moved is low, and the container takes a long time to fill, and the LED is dim because of the low current.

We remove the thin wire and put heavier wire on the coil, and now we have half the DC resistance. Since the voltage stays the same, half the resistance allows twice the current. The inductance is half so the energy stored is half. BUT the current has doubled and the energy stored has doubled and doubled (current squared). We have sacrificed half the inductance to get twice more current and four times the energy stored! And the LED is much brighter. It’s like we are filling the container with the bigger 2 liter bottle.

**Gotcha time.** But now the transistor has to switch twice the current. And the transistor is on for part of the time, let’s say half the time. That means when it is on, it has to handle twice as much current when it’s on. The JT demands that the transistor handle a high current for a short time, and the transistor must be very low DC resistance and capable of handling high current. The transistors that are made for amplifiers that run at a few milliamps have too high resistance and can’t handle the high current (like the 2N3904 and BC547). We have to use a transistor that can handle a half amp or more.

The typical Joule Thief must put out about 66 milliwatts to the LED, which is 20 milliamps times 3.3 volts. If the JT is 50 percent efficient, then the battery has to supply twice that or 132 milliwatts. That’s 88 milliamps at 1.5 volts. The typical JT runs in the 75 to 90 milliiamps at 1.5VDC. But that’s average battery current. The peak current is at least twice that much when the transistor is turned on.

Another question that I’ve been asked is how many LEDs can I connect in series across the transistor? Assuming that the LED is white or blue, then its forward voltage is about 3.3 volts at full brightness. If we connect two of these LEDs in series the total will be about 6.6 volts. That’s okay for the voltage across the transistor from emitter to collector, which is typically 20 volts maximum or more. But we have to remember that the feedback winding has the same number of turns as the primary winding, so it will have the same voltage, but in the opposite polarity. For two LEDs, the negative voltage acrpss the feedback winding will then be -6.6V. We subtract the 1.5V supply from -6.6V and it is 5.1V. The absolute maximum emitter to base voltage for most silicon transistors is typically only 5 volts, with some (very few) transistors rated at 6 volts. So this -5.1V can cause damage to the transistor. If you add LEDs in series, then the number of turns in the feedback winding should be reduced so that the negative voltage on the base will be lower. Or else put the LEDs in parallel so the voltage will stay at about 3.3 volts.

Another way to do this is to add another winding (which will really be a secondary winding) with many more turns for much higher voltage to accommodate many LEDs in series. But this is deviating away from a simple JT so it will have to wait for another blog in the future.

I hope I’ve answered the questions in an understandable manner. Back to experimenting…

### Warning

I, and many others often use Wikipedia to get up to speed when reading about something on the ‘Net. But the Wikipedia article on Joule Thief couldn’t be a worse place to start reading about a Joule Thief.

The Wikipedia article on Joule Thief is erroneous, incomplete and in a constant state of flux because of many well-meaning but misguided individuals who try unsuccessfully to explain what it is and how it works. One example is the recurring use of the term transformer. This in an incorrect term for the coil. The JT coil has the primary winding which is the only winding needed for operation. The feedback winding is there just to invert the signal to keep the transistor switching, and can be eliminated by replacing it with another transistor. There are a number of so-called “Joule Thiefs” found on the net that use a two transistor circuit and a coil with a single winding. The circuit originally named Joule Thief has only a single transistor. As for the Wikipedia article, my best advice is to avoid it. If you try to understand it now, then come back later when you think you have learned enough about a JT, you may be disappointed to find that the article is even worse than it was the first time you read it.

Great read, thanks a lot again !

I miss an answer on one point after the read:

– what factor allows to calculate the voltage which allows the battery source of < 1.5 V to be complemented to 3.3 V ?

I understand it is related to the amount of current in the primary winding, and thereby to the amount of energy stored in the coil and the frequency of switching, but I feel like I still miss a piece of the jigsaw…

Thanks in advance for your help !

If in the case that the resistance must be low. Then instead why should there be a resistor in the circuit.

Can we directly connect the feedback winding to the base of the transistor?

There has to be a resistor to limit the current through the base to emitter junction. The amount of this resistor depends on the gain of the transistor – higher resistance for higher gain transistors. The resistor also determines the total battery current and the brightness of the LED. A 100k resistor may reduce the battery current to, say, 10 mA and the battery will last much longer. The 100k resistor may need a small value capacitor across it to help sustain oscillation. This might be a few hundred pF.