How the ADJ board sets its voltage

1.25 V, held across R1

The ADJ chip does one thing: it keeps 1.25 V between its OUT pin and its ADJ pin. On this board those two pins are joined by R1, a 220 Ω resistor on the back. 1.25 V across 220 Ω is 5.7 mA, so 5.7 mA flows through R1 whatever else is going on.

That current has one way to ground: through the trimmer, R2. Current through a resistance makes a voltage, so the trimmer adds its own voltage on top of the 1.25 V. Turn it up and the output rises:

As a sum, with R2 in ohms:

OUT = 1.25 V × (1 + R2 ÷ 220)

The chip's ADJ pin leaks about 60 µA into R2 as well, which adds about a tenth of a volt at the top of the range. The figure includes it; the sum above does not need to.

Why R1 is only 220 Ω

A smaller R1 means more current wasted, so why not 2.2 kΩ? Because the adjustable chip needs some current flowing out of OUT at all times to stay in regulation. UMW's datasheet does not give a figure; the original AMS1117 datasheet puts it at 5 mA typically, 10 mA at worst, and says R1's current is normally what provides it. 220 Ω gives 5.7 mA. A little load on OUT covers the rest.

The trimmer

It is a 3 mm single-turn trimmer: about 270° from one stop to the other, three quarters of a turn. Up to a tenth of the travel at each end does nothing, and its 2 kΩ can be 25% out either way. So the top of the range is somewhere from about 9.9 V to 15.6 V, depending on the board.

The input decides first, though. OUT cannot rise within about a volt of VIN, and VIN is recommended to stay at or below 12 V — which puts the practical top near 11 V on every board, whatever its trimmer could do.