— David Wagner 2007/07/07 10:33

                                   //   
                                +-->|--+
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+------------------------+-->|--+--||--+
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|  =======             |/ C     +--Z<--+
+--UUU+UUU--+-/\/\/-+--|               |
      |             | B|\|E            |
      |             |    +-------------+
      |             |                  |
      |             +--/\/\/-----------+
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      +-----||-------------------------+
         +1V|                           

It is usually easier to wind this inductor than to find or salvage one. The circuit is very tolerant of the transformer used and can even provide fairly efficient power conversion when the inductance is moderately high. About 300 µH in each coil seems to be ideal for most of the circuits presented here, and this is fairly easy to produce using ferrite toroids or beads larger than ¼”. Larger iron powder toroids may work even better, but require more windings.

So far, it seems a ½” high-permeability ferrite (not iron powder) toroid wound with a pair of about 30 gauge wires, both wound together around 20 turns, works with most transistors. This results in a 1::1 transformer with about 300 µH inductance in each coil. Lower permeability ferrite or iron powder cores need more turns to achieve the inductance required for efficient operation. For example, a medium-permeability ½” T50-43 ferrite core needs 24 turns, and a lower-permeability T50-61 ferrite will require at least 66 turns for each coil.

The smaller T37-43 needs about 27 turns, which should be barely achievable with both wires in a single layer using 32 gauge wire, and thicker wire will require two-layer winding. You would need 34 gauge wire to fit the 75 turns needed for each coil wound around a T37-61 in two layers.

There are two basic ways to wind the inductor. The first requires half the winding, but you'll need to figure out which wires are which when you're done.

1. Fold the wire in half, wind the two strands together, then cut the loop and sort out the wires.
2. Wind half the wire, leave a loop loose, then wind the rest in the same direction.

With the second method you can mark the wires in the loop before winding the rest to avoid confusion later, and for some variations of this circuit it isn't even necessary to do this.

Most of the coils I use are approximately a meter of magnet wire wound about twenty times around a small ferrite torus.

• A tiny FT23-43 core saturates over 1/8 A, but it may be able to handle the 100mA of 1.2V needed to provide 20mA to a white LED. To do this, wind two full layers (about 25 turns of 30-gauge for about 300 µH each), and use the outer layer for the feedback coil. You can probably double the power it can handle by winding two cores together.

It would be nice if a coil printed on the PCB would work, but it's tough to get the inductance needed for regular transistors. For example, a 4” diameter coil printed at 16 turns/inch with a 0.5” center is only 0.035 mH. Even 20 turns/inch only yields 0.055 mH. ¹⁾ However, I did get 57% efficiency using a low-threshold FET with a small bar ferrite providing only about 0.017 mH in each coil, so this is still possible.

However, FETs and fast transistors work best with lower inductances, as low as a couple of microhenries from an iron powder torus wound with 25 turns.

• Toroids, Parts, Kits, Toroid King, QRP Erector Set also has a nice assortment

Haven't tried these suppliers yet.

• Toroids from Surplus Sales of Nebraska
• CWS Bytemark also has beads(website looks squirrelly)

To start itself, the transistor must be able to begin conducting current with only the supply voltage (V_in > V_BE)), and to run efficiently should switch fully on (V_BE(sat)) at about double the minimum supply voltage. For load currents (I) around 20 mA, typical silicon transistors will start from input voltages as low as 0.7V, while germanium transistors should get going around 0.3V (verified 0.33V) and perhaps as low as 0.15V (verified 0.175V). Transistors with faster switching times should lose less power while switching, and in general, a transistor with a higher hFE rating will provide more power than one with a lower value. NPN transistors are slightly more efficient than PNP. A transistor with a higher hFE should operate slightly more efficiently than one with a low hFE. Low input capacitance seems to help also. Elaborate on parasitic capacitance.

Using a low-threshold FET (such as a ZVNL110A which will start at 0.85V) with a low (20-200Ω) base gate resistor improves efficiency significantly. (ZVNL120A should start at 0.5V!)

: Should VBE be in there (VBE≈0.4V for germanium transistors)

Explain why peak voltage is about twenty times the input voltage.

Usually, V_CE determines the primary transistor power loss, but in this circuit all the power through the emitter is lost. The average voltage going through the transistor is the input voltage less a tiny drop around 0.05V (partly due to V_BE and V_CE, but no matter). Much more important is the current shunted to ground when the transistor is switched on. The average current through the transistor is equal to the current into the circuit less the current through the diode (the output current). Since the output power is at a higher voltage, its current is proportionally smaller so most of the input current flows through the collector.

In practical terms, this means the transistor should be rated to handle the full input current. But keep in mind that, up to a point, a larger transistor will have less input impedance and a lower voltage drop for a given current. Thus, using transistors rated for higher currents will generally be able to provide more power at greater efficiency. But these larger transistors usually need larger inductors, at least a few hundred microhenries, and will operate at lower frequencies into the upper audio range.

For example, a typical configuration may convert input power of 100 mA at 1.2 V to output of 20 mA at 3.0 V, and an average of 80 mA at 1.2 V flows through the transistor and results in 50% power conversion efficiency²⁾. To provide a margin of safety, this transistor should be rated at least 100 mA and 24 V. But the input impedance of this circuit with a 100 mA transistor may be too high to actually achieve 80 mA of input current, and the voltage drop across this transistor at these current level may result in greater power loss, so a 200-1000 mA transistor should provide better performance.

For use with one solar cell (and a maximum useful current of 150mA, assuming a circuit impedance of 3.3Ω), the transistor should be able to handle at least 150mA and 10V. For two in series, double these values.

Once you have a transistor, you can estimate the base resistor.

From the tests I've done so far, standard silicon transistors seem to do best with a base resistor value between 1000 and 1800 Ohms. Output power decreases with increasing feedback resistance, and after peaking between 1000 and 1800 Ohms, efficiency drops from about 2000 to 3000 Ohms (perhaps due to some kind of resonance) before rising again, then tapers off until power output is insufficient to drive the load.

Startup voltage, power efficiency, and power output control selection of the base resistor. A smaller resistance will allow the oscillator to start with a lower voltage, while a larger value will dissipate less current. The transistor does not need to be in saturation to get the ball rolling, but it does need to get there when running to avoid unnecessary power loss. A smaller resistor will also increase output, but efficiency drops off rapidly at higher power output.

The following formula should give a reasonable resistor value to try first³⁾.

     (Input Voltage - Transistor VBE) * Transistor hFE 
R = ---------------------------------------------------
                    1.3 * Load Current

The 1.3 is the standard fudge factor to be certain there will be enough current to saturate the transistor.

• For a single NiCd cell using a typical silicon transistor to drive an LED .
• For a single solar cell using a good germanium transistor to drive an LED .

Note the solar cell has just barely enough voltage to get things going.

This is a common-emitter configuration, and standard circuit design recommends a value about ten times the base current-limiting resistor between the base and emitter of the transistor. However, the joule thief is known to be unreliable with a pull-down resistor in place. But a large enough resistor should stabilize the current without shutting down the circuit, and 10kΩ seems to do just this.

Note it may be somewhat better to put this in as a pull-up resistor.

Because of the fairly small amount of current they'll be passing, most diodes will operate near the low end of their specified forward voltage range. However, using an overrated diode optimized for low forward voltage will give the best results. For example, a 2A Schottky diode (21DQ06) has a forward voltage drop of about 0.4V at 20 mA, while a 1A Schottky diode (MBR1100TR) may drop about 0.6V at this current.

Of course, rectifiers designed for very low loss, low-voltage operation would likely be best. See the 1A rectifiers 1N5817 (20V), 1N5819 (30V), 1N5819 (40V) with a forward voltage drop of only 0.23V at 20mA.

It may be worth using a photodiode or phototransistor for solar charging circuits if doing so reduces losses.

This capacitor does influence the both the frequency and efficiency of the oscillator. You want this capacitor to be big enough to to store most of the current coming through the diode, but small enough to discharge enough through the LED before the next spike comes to leave room for this current. Also, if the zener diode is connected to a smaller capacitor, it will shunt less more? current when the capacitor has less charge left in it when the spike comes.

Though fairly large capacitors with low leakage are probably not the best choice for driving an LED, a 2µF tantalum works well for driving a white LED from a single rechargeable cell. A larger value tantalum or electrolytic (10μF works but is probably too large) works better for recharging a battery cell. Check this using smaller coils.

Figure out how to estimate the capacitor value.

It's best to choose a zener diode with its reverse-current breakdown voltage rated above the load voltage, but below the voltage that would damage the LED. The common 5.6V zener seems to be adequate to protect LEDs with a 10uF capacitor and appears to shunt negligible current to ground during normal operation even with smaller capacitors in place.

From LED

White LEDs exhibit about 50Ω resistance.

Due to their finicky charging requirements, I do not recommend trying to charge lithium batteries using this circuit, and you need to be careful even with more forgiving cell chemistries. Of the small rechargeable battery types commonly available, the venerable nickel-cadmium (NiCd or NiCad) battery cell is a better choice than nickel-metal-hydride (NiMH) cells for most applications, since nicads usually last through more charge cycles and have better power storage efficiency.

The standard charge current for a nickel-based cell is “C/10”, one-tenth the rated capacity. This is the amperage needed to charge the cell in ten hours. Since this circuit has no overcharge protection, do not exceed this charging current, or try to charge nickel cells faster than this. For example, if you are using a 2000 mA-hr cell, do not provide it with more than 200 mA charging current. In fact, you should probably limit the charge current to a trickle charge to avoid damaging the cell.

These figures assume about 80% efficiency in each stage. Overall, you get out about half the sun's energy captured by the solar cell. The area is approximated considering 2-5 mW/cm². ⁴⁾

• AAAA NiMH batteries are typically 200-400 mAhr, 1.626” tall, 0.276” diameter, 0.19 oz. (5.2 g), $1.50-$2. (Find these inside rechargeable 9V batteries sometimes.)
• Surplus button cells: A single Varta rechargeable Ni-MH battery cell is about 0.6”D x 0.25” thin, 70 mAhr. ($1/6)
• Tiniest surplus buttons: 0.4”D x 0.16” thick, 15mAhr. ($1/2)

It would be interesting to try using a single 2V sealed lead-acid cell.

The maximum total power dissipation with a 0.45V source appears to be about 50mW, corresponding to a total equivalent circuit impedance of about 4 Ohm. ⁵⁾ Removing the 1-ohm resistor used to measure the current provided to the circuit increased the power to the load by about a third⁶⁾, confirming the total circuit impedance is about 3 Ohm.

Also, the breadboarded test circuit performs worse than the fabricated circuit should at very low voltages, and significant performance improvement should occur simply by moving the circuit from the breadboard to soldered connections. However, another limitation of how much power this circuit can deliver is the impedance in the circuit from the center of the coil and through the primary coil and transistor to ground when the transistor is conducting. I have already noticed power output gains (in some cases) by simply adding another transistor in parallel to the existing one, and perhaps some additional minor improvement can be made by using a smaller toroid or thicker wire. (A typical coil uses about a meter of 30-gauge wire with approximately 0.3 Ohm resistance.)

The bottom line is that even if another Ohm of resistance can be shaved out of this circuit, and assuming 50% efficiency, the largest single solar cell size that can be used effectively (with a germanium transistor) is about 40 mW (90 mA), perhaps 2½ square inches (16 cm²) of a lower-performing (25 W/m²) photovoltaic (PV) cell. This should provide about enough power for a C/10 charge to a 150 mAHr cell, which can then run a white LED at 20 mA for almost two hours. This is sufficient for a solar-charged LED throwie, especially if the run time is increased by using a lower-voltage LED, by decreasing the LED drive current, or by flashing the LED.

Two PV cells in series (each about five square inches or rated about 180mA each) should be able to provide at least four times the charging current and allow the use of a 600 mAHr cell to run an LED for eight hours.