3.3 V LDO voltage regulator for microcontroller (STM32F103x)

Question

I am looking to replace a 1 A switch mode supply with a cheaper (low-drop) linear voltage regulator to power a microcontroller with 3.3 V from 5 V supply (which is stepped-down from 24 V).

According to its datasheet, an LD1086 has 5 mA quiescent current that "flows into the load, increasing efficiency.". What happens if the microcontroller is haltet and only draws microamps?

Also, assuming max. current consumption (70 mA for the CPU without active outputs, plus active outputs) to be 150 mA at 3.3 V, could I use the DFN8 package of this LDO when powering from 5 V?

Approx. 1.7 V drop at 150 mA means 255 mW (1/4 W) loss. DFN8 package has 33 °C/W junction-ambient thermal resistance, so it should be 8.4 K above ambient temp, correct?

Competitor LM1117 in SOT-223 has 61.6 °C/W resulting in theoretical 15.7 K above ambient. The datasheet suggests it needs a tantalum capacitor at the output for stability, which I would like to avoid. What needs to be considered in order to use a ceramic capacitor instead (e.g. murata GRM31CR71C106KA12L)?

Answer 1

The question about quiescent current looks legit, so I'll answer that. Yes, your circuit should accommodate for quiescent current which can be as high as 10mA in the worst case. If you don't, you can't rely on output voltage anymore.

Clearly LD1086 is not a good choice to power the microcontroller from. It is designed to provide high current (up to 1.5A) which you don't need, and has this minimum current limitation which matters to you. You can probably get away with a much smaller and cheaper regulator, something like a NCV663.