How does this battery discharger work?

Question

I'm trying to figure out how this rather simple discharging circuit works and what the SOT23 component is. What kind of component am I looking for? Some sort of transistor, right? But what kind? It doesn't match the common pinouts of anything I've looked at and I can't wrap my head around the functional principle of this circuit to derive it from that.

Specifically the 5.1 MOhm resistor confuses me because while it was obviously chosen to ensure a negligible standby current, it also drops the divider's output voltage into the mV range.

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Functional description: It is designed exclusively for a specific battery type/voltage (4S LiIon/LiPo), which it discharges to approx. 15.5 V before it disconnects the load.

Component description: A voltage divider consisting of a 500 Ohm (R1) and 5.1 MOhm (R2) resistor, connected to pin 2 of an unidentified SOT23 component, a 430 Ohm load resistor connected to pin 1 of said SOT23 component, a fuse¹ and—in parallel with the load resistor—an activity indicator LED (+resistor).

¹based on the silkscreen label on a virtually identical product

Schematic (sans LED and mystery SOT23):

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

This might be many things – from exploiting the exponential current vs voltage curve of a BJT (so that there's a sharp decline in current through the collector-emitter junction when the base-emitter voltage drops below a certain value) to simply a battery protection IC whose job is to cut off a battery when its voltage drops below a specific value.

That's the thing with unmarked ICs, nobody can know what's going on inside.

If you want to design a circuit with a purpose yourself, you might want to ask about that design; of course you can refer to existing circuits, but generally, there's usually good reason to not copy commercial designs 1:1 in your own designs; the parts you choose when you produce many of a thing can be pretty different.

it also drops the dividers output voltage into the mV range.

no, that's not true. Your R1 is labeled "105", that's 10·10⁵ Ω , so 1 MΩ. So this is a 1:5 voltage divider, so 1/6 of the 16V fall of over R1, and 5/6 over R2, in other words, at 16V in, the voltage at the node between these two resistors is 2.67 V.

The load resistor says "4300", so that's 430·10⁰ = 430 Ω, not 100Ω. That makes a lot of a difference in practice, because assuming there's 16V across that resistor, there's 0.6W of heat being produced when it's 430 Ω, and 2.56 W if it was 100 Ω; that's a lot of heat.