RustyBolt.Info/wordpress

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.

The conventional Joule Thief uses a single cell at 1.5V for power.  It also typically uses a coil that has both windings the same number of turns.  This means that the voltage across the primary, which is also the voltage across the LED, is the same as the voltage across the feedback winding.  Almost all LEDs have a forward voltage of 3.3 volts down to 1.8V for red, so a single cell will not light the LED.  This is the purpose of the Joule Thief – to light the LED with a lower voltage.

We can put two or more LEDs in series on the Joule Thief, which will have a forward voltage of more than 3.3 volts: from 3.6 volts for two red LEDs up to 6.6 volts for two white or blue LEDs.  But with two white or blue LEDs, the voltage across the primary winding will be at least 6.6 volts, and this is reflected back to the feedback winding, since it has the same number of turns.  The typical transistor has an absolute maximum reverse voltage across the emitter to base junction of 5 to 6 volts, and the peak voltage from the feedback winding will exceed this.  So what do we do?  Reduce the number of turns of the feedback winding, so the voltage stay be  able s below the maximum.  If the feedback winding has half the turns of the primary winding, then the voltage will be half, and 6.6 volts on the primary will be 3.3 volts on the feedback winding.

Note: QS reminded me that the primary voltage has 1.5V or whatever the supply voltage is already added to it, so this voltage must be subtracted from the feedback voltage.  With a 1.5V cell, the 3.3V LED should have 1.5V from the cell plus the rest of the voltage from the primary winding, which should be about 1.8V peak.  But I’ve seen the voltage across  the LED climb to 4.5 volts peak, so I would guess that the actual voltage the primary  adds is about 3 volts.

This will allow us to put two LEDs in series on the primary.  Since the LED forward voltage is higher, it will also allow us to use two cells in series for a 3V supply.  Normally, with a single cell, the starting voltage would be 1.5V, and the ending voltage would be 0.6V, which is about 0.9 V total.  With two cells in series, the starting voltage would be 3.0 V, and the ending voltage 0.6 V, for a total of 2.4 V, a much greater change overall.  Each cell should be able to run down to 0.3 V, half the typical JT value.

One other benefit is that since the supply voltage is higher, the current can be lower for the same power to the LEDs.  This also puts less demand on the transistor and the losses will be lower.  More than  two LEDs could be connected in series across the transistor, as long as the ratio of the coil windings is changed to keep the maximum voltage across the emitter to base junction below 5V, and the maximum voltage across the emitter to collector less than the transistor’s maximum rating, which might be 40V for a PN2222A or 2N4401, but could be as high as three hundred volts for a high voltage transistor such as the MPSA42.  The coil windings might have a 5 or 10 to 1 ratio.

Another Circuit

Another way to boost a low voltage without using a specially wound coil is to use a two transistor circuit like the one in my previous blog.  The coil is a choke or inductor with a single winding.  A small amount of the current is fed back through a small capacitor to the first transistor to keep the oscillations going.  But the voltage is no longer limited by the first transistor, it is limited by the second transistor’s maximum collector voltage and the load, which might be several LEDs in series.  One important point I should make is that the demands on this inductor are greater with a greater ratio between the input voltage, which was 3 volts in our earlier example, to the output voltage, which might be 20 or 30 volts.

A question a Youtuber brought up made me think about going to the opposite extreme from the one I normally pursue: trying to get a Joule Thief to run at very low power.  Normally I would try to maximize both the light output and efficiency.

I started out with a run-of-the-mill Joule Thief having a coil, a BC337-25 transistor, a 1k resistor and a blue LED.  The coil was a T231212T core only a quarter of an inch (6.4 mm) O.D., with four windings of 7 inch lengths of 30 AWG enameled wire wound quadrifilar, with three of the four windings connected in parallel for the primary winding.  The blue LED lit up very brightly, and the supply current was the typical 60 or so milliamps.  In order to reduce the LED brightness and the supply current, I chose to increase the 1k resistor.  I added a 100k pot in series to allow me to adjust the brightness.  As I adjusted this pot from 0 to 100k, the brightness dropped a lot, but the LED was still putting out quite a bit of light.

I decided I would try a higher resistance, so I put a 51k in series with the pot, and as I adjusted the pot to a resistance above 110k, the LED went dark.    Well, I figured that the problem was being caused by the loss of drive from the feedback winding, which also had to go through this 110k or more resistor.  I clipped a small capacitor across the resistors so that it was in parallel with the total 151k of the pot and 51k resistor, but it was still in series with the 1k resistor.  The capacitor was a .0047 uF or 4.7 nF.  The LED lit up again, but not brightly.  So now that I had the LED working, I could again increase the total resistance, so I put a 470k in series with the 150k and put the capacitor across the 470k and 150k, but leaving the 1k resistor still in series.  The LED still stayed lit, so I measured the supply current, and it was only 60 microamps, which was very low.

I removed the 4.7 nF capacitor and replaced it with a 470 pF, which was 1/10 the capacitance of the 4.7 nF.  I couldn’t tell if the LED was brighter, so I put the 4.7 nF across the 470 pf, and the LED got slightly dimmer.  Weird.  This meant to me that there was some sensitivity to the size of the capacitor, so I removed both caps, and connected a variable capacitor that could be adjusted from 10 to 150 pF.  When I adjusted this capacitor I found that there was a point where the LED lit up brightest.  So I left the capacitor at that spot, and disconnected it and measured its capacitance, and found that it was 37 pF.  But this peak was very broad, so I got a 47 pF capacitor from the spare parts box and soldered it in, and the LED lit up not very brightly, but it was clearly visible.  I measured  the supply current, and it was 310 microamps, or slightly less than a third of a milliamp.  That is very low power: 1.5V times 0.00031 amp is about 0.000465 Watt, or 465 microwatts, not even a half milliwatt.  Comparing that to the usual 120 milliwatts for the Joule Thief, it was about 250 times lower in power.  Wow, I now had a very low power  Joule Thief!

You might think why did I leave the 1k resistor in there.  Well, I connected a jumper across the 1k, and I couldn’t see any change in brightness of the LED.  So I figured that it didn’t make any difference.  All these resistors added up to 620k, so I replaced them with a single 1 meg resistor.  I measured the supply current, and it was 240 microamps.  The frequency was 29 kHz.

When I put the 150k in parallel with the 1 Meg, the supply current jumped up to 1.7 milliamps and the LED got a lot brighter.  I figure that with the 1 Meg resistor and a quarter milliamp battery current, a fresh alkaline AA cell running 24 hours a day should last  for several months.  With the 150k resistor, the battery should last for about two months.  But this assumes that the battery current will remain the same during that time.  We all know, from our Joule Thief experiments, that the battery current tapers off as the battery voltage drops, so the LED doesn’t go out, it just gets dimmer and dimmer.  So in these cases, the LED could still remain lit for weeks more.

I left the blue LED pointing up toward the ceiling and the battery connected, and with the battery current at a quarter of a milliamp I can clearly see the spot of blue light on the ceiling when the lights are out.  That’s not bad for a half a milliwatt of power.

Conclusion

Using the conventional Joule Thief with a resistor of a much higher value, and a small capacitor in parallel with it, the experimenter can control the battery current down to a fraction of a milliamp and still have a LED  that is bright enough to see clearly.  The battery lifetime will be greatly extended, and the LED can still put out enough light to be useful.  By using a 1 meg pot in series with a 1k resistor to limit the maximum current, and the 47 pF capacitor across them, the experimenter can make a Joule Thief that is adjustable from very low light up to full brightness, and anywhere in between.  The pot should be a logarithmic taper audio pot to give better control at the brightest end.  And the left or lower resistance end of the pot should be connected to the coil winding.

This very low power technique could be applied to the twenty LED strings I recently blogged.  The light output is much lower but the battery life could be extended to a month or more.  Try this very low power Joule Thief out and see what happens; you might be pleased with the results.

Back to experimenting…

Update Apr 17 – In an email, Paul said, “I note you used quad winding but with such low currents surely that is not needed”.  With such low currents, that might hold true for the low current in the transistor, which gets pushed to its limit at low voltage and high current.  But with the core windings, which have the resistance of copper wire, the losses don’t change, percentage wise, as the current goes lower.  If you have 100 milliamps or 100 microamps current, the DC resistance doesn’t change, and still wastes the same percent of power, even though the amount may be very small.  Another point is that at higher frequencies, the Skin Effect takes effect.  That’s why Litz wire is better than solid conductor wire.  So having three conductors instead of a single conductor gives more surface area and the skin effect has more surface to give better conduction.

I took a look at the waveform with the o’scope, and saw that the waveform is a much narrower pulse than the typical JT.  It is somewhat lower amplitude, but a good part of the lower light output is from the lower duty cycle (on time) of the pulse.  When I put a 22k in parallel with the 1 Meg, the pulse amplitude gets higher, but the pulse gets a lot wider, and the transistor stays turned on longer.  The circuit has been running on a ‘heavy duty’ (not alkaline) cell for several weeks, and the cell voltage is 1.435 volts.  Looks like it will run at least a month more on this cell.

Let’s assume, for the reason of eliminating it as a factor, that the pulse height didn’t change when the 22k was put in parallel.  What we then have is a change only of the duty cycle; the on time of the 1 Meg is much lower than the 22k’s on time.  But remember that the only time there are losses in the transistor is when it is switched on.  Now the coil has losses in the resistance when the transistor is turned on and charging the coil.  When the transistor is turned off and the coil is transferring its energy to the LED, there is current flowing, so I assume there is also loss in the coil.  But the current is much greater during the on time which leads to the conclusion that most of the loss occurs during the on time.  My point is that due to the high current during  the on time, there is a justifiable reason to minimize the DC resistance of the coil’s primary winding so the losses will be minimal.

Update May 12 – I connected the low power JT up to a 50 Farad supercapacitor.  The base resistor is a 470k resistor in series with a 1k resistor.  There’s a 47 pF capacitor across the 470k only.  I connected a fresh AA cell across the 50F cap, charging it up to 1.56 volts.  I connected it to the JT, and set it aside.  After 8 hours, the capacitor voltage was 1.137 volts, and the LED was still glowing, a bit weaker but still very visible.  At 12 hours, the cap voltage was down to 1.05V, and the LED is still going, not as bright as it was, but still fully visible.  At the 20 hour point, the voltage has dropped to 0.890 volts, and the LED is still lit, slightly dimmer than earlier.  At the 24 hour point, the voltage is now at 0.865 volts, and the LED is getting dimmer, but still plainly visible.  Next morning, thirty hours later, the voltage has dropped to 0.809 volts, and the LED is getting dimmer.  And finally at 44 hours, the voltage has dropped to 0.704 volts, and the LED has dimmed to where it looks like it’s ready to go out – there’s very little light, just a faint glow.

Note:  One thing that could be added is a switch to bypass part of the resistance and increase the brightness when it gets dim.  This could also be a variable resistor or potentiometer, but just two switch settings should suffice.  The switch could be a SPDT center off switch so the on/off switch serves a dual purpose.  The idea is to allow the user to ‘turn up’ the light when the capacitor has discharged.