The reference voltages of standard low-dropout (LDO) linear regulators are usually in the 1.2- to 1.25-V range. But that's not low enough to meet the latest CPU and logic voltage requirements, which can be as low as 0.9 V. That is, the reference voltage isn't low enough to use a simple feedback divider at the feedback node.
There is a way, however, to use an LDO linear regulator with a reference that's higher than the output voltage. If the input voltage is well regulated (±5% or less), then it can be used as a simple current source to pull up the feedback voltage to the reference voltage at the feedback node. This makes the LDO usable for lower output voltages.
The figure shows the technique applied to an LT1763 LDO micropower linear regulator. The circuit regulates a 1-V output voltage from a 3.3-V, ±5% input voltage, over an output current range from no load to 500 mA. The tolerance of the 1-V output is ±2%, with the 3.3-V input supplying the feedback bias current through R1. The reference (feedback) voltage of an LT1763 regulator is 1.22 V, so approximately 10 µA flows thru R1 and R2. That 10 µA of bias current, together with R2, generates the 0.22-V drop from the feedback to the output, producing the 1-V output. R3 is chosen to carry twice the feedback bias current, ensuring that the LT1763 power transistor is always on, even at no load.
The following equations were used to select the component values. Choose a feedback current or use 10 µA as a starting point:
R1 = (VIN − VFB)/IFB
R2 = (VFB − VOUT)/IFB
R3 = VOUT/2IFB
Of course, the smaller the resistance tolerances, the tighter the output voltage.
This method of generating the feedback reference voltage only works well if the input voltage is tightly regulated. If tighter output tolerances are required, use an external voltage reference instead of the input voltage.