Light becomes a voltage

A current, then a voltage

A phototransistor is a transistor with a window. Infrared falling on it does what a current into the base does in an ordinary transistor: it lets current through from collector to emitter, and more light lets more through. On this board the collector sits on VCC and the emitter on SIGNAL, so that current comes out of SIGNAL and down the 10 kΩ to GND.

Move the slider. Below about 2 V the sum is Ohm's law and nothing else: the current times 10 kΩ. 0.1 mA makes 1 V, 0.05 mA makes 0.5 V. Switch VCC between 3V3 and 5V and nothing changes: this block is not a divider, and VCC is not in that sum.

The slider is the current, because that is what the maker specifies. Its test light is 1 mW per square centimetre of 940 nm infrared, and it makes at least 0.7 mA in this part, typically 2 mA. The irradiance under the drawing is that typical figure run backwards along a straight line: a guide, not a calibration.

VCC sets the ceiling

Push the slider to the right. SIGNAL stops at about 2.9 V with VCC on 3V3, about 4.6 V on 5V. The transistor needs a little voltage across itself, at most 0.4 V by its datasheet, and once SIGNAL is that close to VCC it has run out of room. More infrared changes nothing: the reading is pinned.

That is the reason VCC goes to your board's logic voltage. 3V3 beside an ESP32, an ESP32-S3 or a Pico keeps the ceiling under 3.3 V. 5V beside an Uno gives the whole of its 0 to 5 V range.

In the dark

With no infrared, the datasheet allows at most 100 nA through the transistor, which is 1 mV across 10 kΩ. So a covered sensor reads 0, or within a count or two of it, and the block takes almost nothing from VCC.