Pololu 5V, 2.5A Step-Down Voltage Regulator D24V25F5
by Pololu

This small synchronous switching step-down (or buck) regulator takes an input voltage of up to 38 V and efficiently reduces it to 5 V. The board measures only 0.7″ × 0.7″, but it allows a typical continuous output current of up to 2.5 A. Typical efficiencies of 85% to 95% make this regulator well suited for powering moderate loads like sensors or small motors. High efficiencies are maintained at light loads by dynamically changing the switching frequency, and an optional shutdown pin enables a low-power state with a current draw of a few hundred micro amps.

Using the regulator
Connections:

This buck regulator has five connection points for four different connections: enable (EN), input voltage (VIN), 2x ground (GND), and output voltage (VOUT).

The input voltage, VIN, powers the regulator. Voltages between 4.5 V and 38 V can be applied to VIN, but for versions of the regulator that have an output voltage higher than 4.5 V, the effective lower limit of VIN is VOUT plus the regulator’s dropout voltage, which varies approximately linearly with the load (see below for graphs of dropout voltages as a function of the load).

The output voltage, VOUT, is fixed and depends on the regulator version: the D24V25F3 version outputs 3.3 V, the D24V25F5 version outputs 5 V, the D24V25F6 version outputs 6 V, the D24V25F7 version outputs 7.5 V, and the D24V5F9 version outputs 9 V.

The regulator is enabled by default: a 100 kΩ pull-up resistor on the board connects the ENABLE pin to reverse-protected VIN. The ENABLE pin can be driven low (under 0.6 V) to put the board into a low-power state. The quiescent current draw in this sleep mode is dominated by the current in the pull-up resistor from ENABLE to VIN and by the reverse-voltage protection circuit, which will draw between 10 µA and 20 µA per volt on VIN when ENABLE is held low. If you do not need this feature, you should leave the ENABLE pin disconnected.