The Thevenin source your LED actually sees
Every theorem in this unit compresses a network down to two numbers. In this capstone you compress a real one: the circuit your CA-1 capstone board presents to its LED — then check the compression against the maximum-power ideas, and see why the design deliberately refuses to match.
- On paper: from the LED's two terminals, the rest of the CA-1 capstone board is the 3.3 V rail in series with the 330 Ω resistor (wiring resistance negligible). Write down the Thevenin equivalent the LED sees, then convert it to its Norton form.
- Compute the maximum power this source could deliver to a perfectly matched load, and the load resistance that would capture it. Then compute the power the LED (≈2.0 V at ≈3.9 mA) actually takes, and its share of the maximum.
- Explain in two or three sentences why matching would be the WRONG goal here: what the resistor's real job is (current limiting for a device that must not see more than ~20 mA), and what would physically happen to an LED connected straight to an ideal 3.3 V rail.
- Open PCB Studio, rebuild the one-LED board (Pico 2 W + one 5 mm LED), generate, and read the series resistor value out of the spec JSON. Recompute the Thevenin equivalent with the engine's chosen value and compare with task 1.
- Superposition preview of things to come: your board's 3.3 V rail also feeds the Pico itself. Sketch (mentally or on paper) why analysing the LED branch separately was legitimate — each branch hangs in parallel on an ideal rail, so branches don't disturb each other. Then note the fine print: the rail is only NEARLY ideal (its own R_Th is small but not zero), which is why boards with many LEDs need the power budgeting you'll meet in ME-6.