2013-04-27 Low Current, High Performance Joule Thief

Paul got me thinking about the Joule Thief with low battery current.  In his nineteenth JT he used the very high performance Fairchild KSD5041, same as the 2SD5041 (Japanese pinout – the center pin is the collector).  He thought it was his brightest, and that could very well be.  His theory is that the coil should be wound with the maximum number of turns of fine wire.

My assumptions are based on the theory that less is more, less turns allows more current and since the stored energy is equal to the current squared, the lower winding resistance means higher peak current and therefore lower losses,    I decided to make up this proof of theory JT using the KSD5041 and the following parts.

Toroid core = YJ41003TC, core O.D. = 3/8″ or 9mm.  This is a high permeability core, which, with very few turns, gives the optimum inductance of 100 uH or more. Both windings are  7 turns of solid wire, the primary winding 24 AWG with plastic insulation, the feedback winding 24 AWG enameled.  Each winding measured 120 uH.

LED = 5mm Blue, with a 1 ohm resistor in series to measure the current.

The resistor was a 4.7k with a 50k trimpot in series.  I found that a 68 pF capacitor across both resistors kept the LED lit even at the higher pot settings, and looked brighter than when I used a 1000 pF.  The LED would not light at the higher setting without the capacitor.  These were connected between the coil and the base lead of the transistor.

I put a 33 uF bypass capacitor across the plus and minus terminals.  I built the circuit on a 1-1/2 by 2 inch piece of scrap wood.  I pounded some 5/8 inch brass brads into the wood for terminals and wrapped and soldered the wire leads to them.

I set the supply at 1.5V and adjusted the trimpot to give 25 mA supply current.  I measured the voltage across the 1 ohm resistor and found that there was 7.2 mA flowing through the LED.

I removed power and measured the resistors and found the total to be 14.7 k.

I calculated the efficiency at (.007A * 3.3V) / (1.5V * .025A) = 61.6 percent.  This is one of the best, if not the best efficiency that I’ve measured for a conventional Joule Thief.  A typical JT is around 50%, sometimes 55%.  But I have seldom seen one go above 60 percent.

I can adjust the trimpot and get the LED to light from very bright at minimum setting to almost dark at maximum setting.  This gives complete control over the battery current, which can make the battery last a lot longer, with some sacrifice in LED brightness.

This shows that it’s possible to get excellent efficiency and performance at low current from a Joule Thief using a high performance transistor.

Update Apr 30 – I experimented with the value of the 68 pF resistor bypass capacitor.  I connected an adjustable capacitor in parallel with the 68 pF, and adjusted it until the LED current peaked at a very broad peak.  I removed the capacitor and measured it and it was about 130 pF.  So I found a 120 pF ceramic disk capacitor and soldered it across the 68 pF, for a total of about 190 pF.  I checked the LED current and it was almost 8 mA.  But the supply current had also gone up slightly.  So I readjusted the trimpot to set the supply current at 25 mA.  The LED current was then about 7.8 mA, so the JT circuit was slightly more efficient than it was in the original test above. The total resistance then measured 14.85k.

I’m doing a few more measurements, but the circuit has been optimized.

17 Responses

  1. kalle says:

    can u give us a circuit diagram?

  2. John Watson says:

    Formula to get nh I found was AL*N^2 and that coil is AL=2196 so 7^2=49 2196*49=107604nH or 107.604uH which is dang close to your measured inductance.

    Is the 120uH the sweet spot or is that variable depending on the circuit?

    • admin says:

      In reply to your earlier comment, the problem is the 2N3904. You said 2N3905 but that’s a PNP, I’m assuming you meant 2N3904. This transistor was never made to handle more than 100 mA, with 200 its rated max. You need to get some 2N4401, PN2222A or BC337-25 to get decent performance from a conventional Joule Thief with a 5mm standard size LED. These transistors can handle a half amp or more, peak current.

      For higher performance, you need a transistor that can handle really serious current, 1 amp or more, with very low Vce(sat), and good gain holdup at high current. These are the 2SD5041 or KSD5041, 2SC2500D, ZTX1048A, NTE11 (too expensive). Check out Quantsuff’s web pages for some higher power circuits.

      The inductance of the coil determines the switching speed or frequency. I think it’s a good idea to keep this down to under 100 kHz, just to prevent interference to the Am broadcast band. But I’ve had JTs running at well over 1 MHz, but the efficiency is somewhat less. 100 uH gives a comfortable frequency, somewhere in the 50 to 100 kHz range, but some use lower, some use higher. The critical factor is the DC resistance of the coil windings. You want minimal loss due to the resistance of the primary winding (connected to the collector). The feedback winding is not important. I try to keep the primary’s resistance well below 1 ohm, preferably less than 1/4 ohm. A high permeability core makes it easy to get the inductance you want with a short length of wire, sometimes only 6 to 10 turns. That’s just a few inches long.

      For kicks I have wound a 1 inch O.D. core with wire, winding every four or five turns and then running the wire out to make a tap. That way I can pick the number of turns as I experiment with the coil. Just make sure you don’t have too much voltage on the feedback winding, because it can damage the emitter to base junction.

      • John Watson says:

        Nope, 2N3905 which is just the 3904 in reverse so basically the same.

        Now the problem I am having is not that it can’t handle the power the fact is that it isn’t giving the LED the power period and why I said I thought the 2N3905 would burn up but the sucker never even gets warm. I mean with a 23ma draw for the entire circuit it really shouldn’t.

        Since my coil is bifilar 1:1 shouldn’t both the primary and the secondary read the same ohms?

        • John Watson says:

          Oh, my coil reads 1.2 ohms with my DMM so it is close to your 1 ohm.

          • admin says:

            Short the probe ends together. On cheap DMMs, it may read a half ohm or more just because they use really thin wire for the leads. Subtract whatever it is from the value you get for the coil.

          • John Watson says:

            1.2-.9=.3 ohm for my previous coil and the new coil (on this teensy weensy little toroid) is 1.0-.9=.1 ohm.

            Thanks for the heads up about these cheap DMMs (especially ones like this without even a zero function which I miss from my now blown up DMM).

        • admin says:

          Yes, they should read the same if they are both wound with the same size wire.

          The 2N3905 is PNP, so you should reverse the polarity of the battery and the LED. But even if you do, it is a lower gain version of the 2N3906 and you will never get the same performance as if you used the higher gain 2N3906. Like I said, if you want to get adequate performance with a 5mm LED, don’t use the 2N390x, x being 3, 4, 5, or 6. Use the 2N4401 or the PN2222A (same pinout), or the BC337-25 (opposite pinout). These are all NPN. If you want PNP, use the 2N4403 or the PN2907A, or the BC327-25.

  3. John Watson says:

    Well, I wound a 1.5mm O.D toroid with 2 or 3 turns of enamel wire and it made a HUGE difference. My JT now goes past 100ma for the circuit but what I found was that 29ma circuit drain is 3.4ma to the 10mm LED. Man, that blows at a little over 11% efficient.

    • admin says:

      Are you sure it’s 1.5mm O.D.?? That’s a really tiny core. The typical conventional JT is around 50% efficient.

      • John Watson says:

        Yep, I am sure as 24 ga with plastic coating I could barely get 2 wires in the core and no way to wrap anything. That is how small the sucker is but it came from a 60watt CFL.

        The typical is 50% but this isn’t typical it is draining 29ma from the battery (1.2v) and delivering 3.4ma to the 10mm 20ma LED.

        • admin says:

          The typical conventional JT uses 60 to 80 mA at 1.5V. But if you connect the milliammeter in series with the battery lead, it will have a drop of up to a quarter volt, which trows your measurement off because the JT is getting only part of the cell voltage. I made a 0.1 ohm resistor by winding fine wire onto a large resistor. This measures 1 millivolt drop for each ten milliamps. When this problem is fixed you may find that your measurements are more meaningful.

          The transistor is the main limit on the performance. Try this. Connect another transistor across the existing one, emitter to emitter and collector to collector. Then connect the base through another 1k resistor to the feedback winding. If the LED gets brighter, then you know the transistor is limiting the LED current.

          • John Watson says:

            Well, the next day after I wrote the above response I tried another transistor and viola even the 3W gets bright but I do not understand two things:

            1) The 3W LED didn’t even make the transistor hot.
            2) best I can do is 700mv to turn on a 20ma (the 700ma was a no go) LED but it is barely on.

            The transistor I salvaged (after reading it documentation spec sheet) is S8050 (NPN) and I still get this sense it can be even better with lower voltage turn on and even brighter (the 3W was still not even near its max brightness was was too bright to look into for too long).

          • admin says:

            First off, about the transistor getting hot. In a JT, the transistor acts like a switch, which is turned off or fully on. The transistor dissipates much of its heat during the time between off and fully on states. In a JT, it spends very little time between the off and fully on states, so the dissipation is very low. This is why you can use a very small transistor such as the S8050 to drive a large LED.

            The 8050 is a good choice for a Joule Thief – it can handle an amp or so. It is NPN, which is the opposite polarity of the 2N3905. However the pinout is the same: E B C when you’re holding the leads and looking at the flat side. Make sure the emitter is connected to negative. If you can get more of these 8050 transistors, you can put them in parallel to drive the LED with more current.

  4. John Watson says:

    Well, the 700mv was cleared up as I was scanning around the BJT needs .7V to turn on so I would have to go to a germanium if I needed to go below 700mv or so I read.

    • admin says:

      The silicon transistor has about 0.63V base to emitter when it is used as an amplifier. But it will conduct a small amount of current with the base as low as 0.5V. With the JT resistor at a value lower than 1k, it can start up at less than that and run down to 0.45V. Germanium transistors work down to 0.25V, but they are hard to obtain and expensive. I like to experiment with the 2N7000 MOSFET and two button cells to give it a few volts bias on the gate.

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