(a) Buck converter. (b) Boost converter. (c) Buck-boost converter.
To keep the converter simple, it contains no preset controls, which means that the level of the output voltage is dependent to some degree on the load. In theory, the output level is twice the level of the input voltage, but, owing to losses in the converter, that level cannot be attained. The main losses occur at the semiconductor junctions of the transistors and the rectifier diodes. Since the drop at these junctions is constant at about 0.6 V, the losses are proportionally larger with an input voltage of 6 V than with one of 18 V. Oscillator IC1 generates a signal at a frequency of about 10 Depending on the output level of the IC, either T1 or T2 is switched on. This results in C2 being charged during a one-half period; during the other half period, the charge of C2 is transferred to C3. This results in an output voltage of twice the input voltage less the )1 losses mentioned. The circuit contains no critical components: ICS may be any version of the 555. bipolar or CMOS, while the transistors may be inexpensive LF types. Although the present circuit uses 1N4004 diodes, Types 1N4001 will do just as well. Although the switching frequency is of the order of 10 kHz, and the diodes are designed for lower frequencies, no problems were experienced in the prototype, primarily because the voltages and currents are relatively small. The converter draws a current of 5 mA (555) plus twice the output current.