The difference in voltage between the input and output that is required to maintain the set output voltage is called the dropout voltage.
This is the potential difference between the input and output that can maintain the set output voltage and is read as ▲▲mV is required to draw ●●mA.
The output voltage decreases when the voltage difference between input and output is less than the dropout voltage.
Generally, the dropout voltage is proportional to the “driver FET on resistance” and output current. For this reason, the dropout voltage must be calculated in accordance with the output current that is actually used.
Let’s consider the following example.
|Dropout Voltage||Vdif||IOUT=300mA, VOUT=5.0V||-||120||200||mV|
For a product with a set output voltage of 3.0V, the dropout voltage to maintain 3.0V at 100mA is 240mV. In other words, an input voltage of at least 3.24V is required for an output voltage of 3.0V.
The dropout voltage has a linear relationship with the output voltage, so when only an output current of 50mA, which is half of 100mA, flows, the dropout voltage is also half at 120mV. In this case, the input voltage required to maintain 3.0V is at least 3.12V.
In other words, when the input voltage is under 3.12V and an output current of 50mA flows, the 3.0V output voltage cannot be maintained and so decreases.
The following relationship exists between the dropout voltage and the on resistance.
On resistance Ron = dropout voltage Vdif ÷ Iout (Iout value for the dropout voltage conditions)
The lower the on resistance is for a product, the smaller the dropout voltage. So, for applications where you want to maintain the output voltage up until just before the battery dies, for example, a product with a small dropout voltage is optimum.