Both transistors are therefore strongly ON - and current therefore flows through the inductor via Q2 to ground. My theory so far: the 20K pulls down the base of the PNP to ground, so initially the PNP is on (V_BE negative, more negative than 0.6V) - which means that NPN is also on (V_BE also much higher than 0.6V). How does it do that, though? More Googling under way, now that I know the purpose. and see now that it boosts the voltage! Which makes perfect sense, now that I realize the LED needs more than 1.5V (the AAA battery I added in). I could add a link to the product, but I don't want this post to be misclassified as spam. Amateur at electronics, go easy on me :-) I am clearly missing something here - why the two transistors? voltage rising on a charging cap reaches 0.6V, transistor opens, flashes the LED, cap charges again. Unsure why one would want that - but even if that is the case, couldn't that be achieved by just a single transistor - whose base would be driven by the cap, placed between base and ground? i.e. Therefore this circuit clearly is not meant to arrive at such a steady state - it is designed (presumably) to oscillate. EDIT Was wrong here - this was battery-driven, so 1.5V, not 5V!.

If the circuit does end up in a steady (read: frozen) state, then the inductor acts as just a wire - so the LED ends up just being directly connected to the 5V (which wouldn't bode well for its longevity). I'd like to understand how it works, though - specifically, why it uses a combination of a PNP and an NPN transistor (schematic from the product page is below). I bought and soldered a dirt-cheap flashlight kit from China.

Pnp transistor schematic