Joule Thief The circuit is essentially a blocking oscillator somehow working without a capacitor, perhaps relying on parasitic transistor (Miller) capacitance to operate. A brief test of this bare-bones circuit thrown together from scavenged components, powered by a freshly-charged but old NiCd battery, and driving a white LED indicated 30% efficiency, though it isn't hard to improve it. And if you try, you might get up past 75% efficiency using only a few common, cheap parts.

Joule Thief ImprovementsAdding a diode and capacitor as shown below can more than double efficiency. Although this change introduces another loss due to the forward voltage drop of the diode, the capacitor captures the current during the oscillation cycle between Vf,diode and Vf,LED that would otherwise be lost in the transistor or shunted to ground. This configuration also provides a blocking diode for charging battery cells from one or two solar cells. Once started, the oscillator averages the supply voltage as it swings between the forward voltage of the diode, and a value slightly above the sum of the forward voltages of the diode and the load. For example, a 1.2V NiCad cell driving a 3.0V LED through a standard 0.7V silicon diode will average 1.2V as it oscillates between 0.7V and some value greater than 3.7V. Although the first circuit works fine and is easy to make from salvaged components, a few more improvements and protection circuitry complete the design. The lower voltage drop of a Schottky rectifier or germanium diode is more important than its reverse current leakage. The pulldown resistor does seem to stabilize the circuit. It surprised me to learn it is entirely possible to blow a white LED with a single NiCd cell. Apparently if the LED disconnects from the running circuit, as often happens when messing with components on a breadboard, the capacitor can charge up to a fairly high voltage and blow the LED when it reconnects. You can solve this problem by placing a 5V (or so) zener diode across the LED to shunt excess voltage to ground. (As an alternative, you may place the zener between ground and the transistor's collector to protect the transistor and Schottky diode from voltage spikes as well.) In either case you now have a zener-regulated power supply. Though an LED load draws so much current it drops the voltage across

it to the LED's forward voltage, a higher-impedance microamp load will get a fairly stable supply voltage. (See below for improved power-supply characteristics.) It is also a very good idea to protect the base of the transistor from reverse voltage spikes with a diode between it and ground. A normal silicon diode is fine for this. Note the diode's reverse current leakage makes the bias resistor redundant.

The third circuit may provide decent voltage and current regulation as well as component protection, though this circuit may be more appropriate for loads other than the LED shown. Note the zener diode Z1 is redundant in this circuit. Also, a more nonlinear cutoff response can be provided by putting a different zener diode (perhaps 3.0V to regulate at about 3.6 V) in series with Q2 base resistor R4. It may also be a good idea to pull the base of Q2 down to ground with a fairly large resistor. Another possibility is to put a capacitor between the collector and emitter of Q2 to try to minimize the amount of time Q1 is in the linear response region.

If the supply voltage is 1V, the maximum (saturation) voltage at the base of germanium transistor Q1 is 0.3V, and R1+R2 should be 200 during normal operation (Vout=3V), consider R1=100 and R2=100. The maximum voltage at the point between R1 and R2 is 0.65V. At this point, Q2 needs to be just below cutoff. Choose R1 and R2 so the voltage between them is at the C-E forward voltage drop of the transistor when the supply voltage is high enough to provide maximum output current ignoring Q2? Suppose Q1 is a silicon transistor with hFE=100 and provides 20mA at 3.1V while conducting 100mA at 1.2V with a base resistance of R1+R2=1000. (This is similar to test results using a 2N5192). When on, the base of Q1 is at a potential of about 0.7V and conducts 1mA, so the end of the coil must be at 0.7+0.0011000 = 1.7V. Q2 (also Si, hFE=100) is just barely at cutoff,