Buck regulators find wide application as step-down regulators for converting large positive input voltages into a smaller positive output voltages.  shows a simplified buck regulator that operates in continuous-conduction mode—that is, the inductor current always remains positive. The output voltage, V_OUT, is equal to D×V_IN, where D is the duty-cycle ratio of the buck switch, Q₁, and V_IN is the input voltage. The duty cycle, D, is equal to T_ON/T_S, where T_ON is the on-time of Q₁ and T_S is the switching-frequency period.

You can reconfigure a buck regulator into a buck-boost circuit to convert a positive voltage into a negative voltage (). The basic component configurations of both circuits are similar, and the inductor and the rectifier diode are transposed. Because the main switch, Q₁, remains in the same location for both configurations, you can use an IC buck regulator for either topology. Switching on Q₁ applies input voltage V_IN across power inductor L₁, and current in the inductor ramps up while Q₁remains on. When Q₁ switches off, inductor current continues to flow through C₁, the load resistance and D₁, producing a negative output voltage. During Q₁'s next on-time interval, the output capacitor supplies current to the load.

 shows a low-cost buck-boost converter based on the LM5010 buck-regulator IC that converts a 10 to 50V positive supply voltage into –12V. Although many applications use a fixed switching frequency and modulate the output pulse width, this design features a constant-on-time approach in which the IC's internal output transistor turns on for an interval that's inversely proportional to the difference between the circuit's input and output voltage.

Inside IC₁, a regulation comparator monitors the output voltage from voltage divider R₁ and R₂ and a 2.5V internal reference, and, if the output voltage falls below the desired value, the comparator switches on IC₁'s output transistor for an interval that an on-timer determines:

Providing that current through L₁ remains continuous, V_OUT remains regulated. Because R₃ and K are constants, switching frequency F_S remains constant. This relationship holds true provided that the current through the inductor remains continuous. At lighter loading, the current in the inductor becomes discontinuous—that is, the inductor current drops to zero for a portion of the switching cycle. At the onset of discontinuous operation, the switching frequency begins to drop and thus brings V_OUT back into regulation.

Operating a buck-boost regulator in fixed-frequency mode without an oscillator eliminates loop compensation and stabilization components and, as a bonus, offers fast transient response unlimited by feedback-network lag time. With the component values in , the regulator operates at approximately 400 kHz, delivering 12V at approximately 0.5A for 10V input and approximately 1A of output current for 50V input. Resistor R₄ ensures that the minimum amount of output-ripple voltage necessary for regulation—approximately 25 mV—is available.

Fixed-frequency operation without an oscillator offers a low-cost, easily implemented regulator with no loop-compensation or stability issues to worry about. The transient response is fast, because there are no bandwidth-limiting feedback components. The regulator operates at approximately 400 kHz. The output-current capability varies with the input voltage. When you apply 10V input voltage, the output-current capability is approximately 0.5A, and, at 50V input, the output current is approximately 1A.