A low-dropout regulator (LDO regulator) is a type of a DC linear voltage regulator circuit that can operate even when the supply voltage is very close to the output voltage. The advantages of an LDO regulator over other DC-to-DC voltage regulators include: the absence of switching noise (in contrast to switching regulators); smaller device size (as neither large inductors nor transformers are needed); and greater design simplicity (usually consists of a reference, an amplifier, and a pass element). The disadvantage is that linear DC regulators must dissipate heat in order to operate.

The adjustable low-dropout regulator debuted on April 12, 1977 in an Electronic Design article entitled "Break Loose from Fixed IC Regulators". The article was written by Robert Dobkin, an IC designer then working for National Semiconductor. Because of this, National Semiconductor claims the title of "LDO inventor". Dobkin later left National Semiconductor in 1981 and founded Linear Technology where he was the chief technology officer.

The main components are a power FET and a differential amplifier (error amplifier). One input of the differential amplifier monitors the fraction of the output determined by the resistor ratio of R1 and R2. The second input to the differential amplifier is from a stable voltage reference (bandgap reference). If the output voltage rises too high relative to the reference voltage, the drive to the power FET changes to maintain a constant output voltage.

Low-dropout (LDO) regulators operate similarly to all linear voltage regulators. The main difference between LDO and non-LDO regulators is their schematic topology. Instead of an emitter follower topology, low-dropout regulators consist of an open collector or open drain topology, where the transistor may be easily driven into saturation with the voltages available to the regulator. This allows the voltage drop from the unregulated voltage to the regulated voltage to be as low as the saturation voltage across the transistor.

In Fig. 1, the opamp's non-inverting input will have a voltage of:

${\displaystyle V_{\text{-}}={\frac {R_{2}}{R_{1}+R_{2}}}V_{\text{out}}\ .}$

When this voltage is less than Vref, the opamp will turn on the pass element. The feedback loop keeps ${\displaystyle V_{\text{-}}}$ about equal to ${\displaystyle V_{\text{ref}}.}$ Solving for the output voltage yields:

${\displaystyle V_{\text{out}}=\left(1+{\frac {R_{1}}{R_{2}}}\right)V_{\text{ref}}\ .}$