This is an example of efficiently driving a 1 watt white LED from a 12 volt battery using a buck converter. The LED could simply be connected with a series resistor to get the desired current, but the efficiency would be only 25% since the resistor would drop 9 volts while the LED only requires 3. The buck converter provides about 90% efficiency. The idea is to establish a circulating current through the inductor, diode and load, while the switch replenishes the lost load energy on each cycle. The duty cycle of the switch will be the output voltage divided by the input voltage, or about 3/12 (25%) in this case. It's actually a little greater since there is a small (2.2 ohm) resistor in series with the LED that drops about 0.5 volt, so the total load is about 3.7 volts and the duty cycle is around 31%. The circuit could also be used to charge AA batteries from a 12 volt source with adjustment to the duty cycle. The driver section uses a CMOS hex inverter (CD4069) where two of the in...