The circuit in Figure 1 is similar in principle to that of a previous Design Idea (Reference 1) but offers improved, more reproducible performance. The output current is almost constant over an input-voltage range of 1.2 to 1.5V and is insensitive to variations of transistor gain. Transistors Q₁ and Q₂ form an astable flip-flop. R₁ and C define the on-time of Q₂. During that time, Q₁ is off, and the voltage at the base of Q₁ and the current in inductor L ramp up. When the voltage at the base of Q₁ reaches approximately 0.6V, Q₁ turns on, and Q₂ turns off. This switching causes "flyback" action in inductor L. The voltage across the inductor reverses, and the energy stored in the inductor transfers to the LED in the form of a down-ramping pulse of current. During flyback time, voltage across the LED is approximately constant.

The voltage for yellow and white LEDs is approximately 1.9 and 3.5V, respectively. When the current through the LED falls to zero, the voltage at the collector of Q₂ falls sharply, and this circuit condition triggers the next cycle. Assuming the justifiable approximation that the saturation voltage of Q₂ is close to 0V and that the LED's forward voltage, V_D, is constant, you can easily derive the expression for the average dc current through the LED:

At first glance, I_AVE depends strongly on V_IN. But close examination of the logarithmic term reveals that, with a proper selection of V_B, the logarithmic term can become a sharply declining function of V_IN. The logarithmic term thus fully compensates for the term V_IN² in the expression. That compensation is precisely the purpose of the diode, D₁, in series with the base of Q₁. The circuit drives a high-brightness yellow or white LED. Table 1 shows the proper component selection for both colors. Table 1 also shows some measured results at V_IN=1.35V. Because the voltage across the white LED falls from 3.9 to 3.1V during flyback, capacitor C subtracts current from the amount available to the base of Q₁. This subtraction might retrigger the circuit before the current in L falls to zero. The addition of R₃ and D₂ solves this problem. During flyback, the current that flows through R₃ compensates for the current withdrawn through C.