MOSFETs

Switch something a GPIO cannot

Every board you have designed so far has driven loads a microcontroller pin can handle directly: one LED, a few milliamperes. This capstone crosses that line. You will size a low-side MOSFET switch for a load well beyond a GPIO's rating, specify the two resistors beginners forget, and then build the board that carries it.

1. The load is an eight-LED WS2812 strip drawing up to $480$ mA at $5$ V when all pixels are white. Confirm from Lesson 3's numbers that this is far beyond the RP2040's per-pin limit of roughly $12$ mA, so a switch is required rather than optional.
2. Choose a logic-level n-channel MOSFET with $R_{DS(on)} = 0.04\ \Omega$ specified at $V_{GS} = 2.5$ V. Compute the conduction loss and the voltage dropped across the switch at full load, and confirm the strip still sees adequate voltage.
3. Size the gate resistor and the gate pull-down: pick a gate resistor of $100\ \Omega$ and compute the peak current the GPIO must deliver into a $1200$ pF gate through it; pick a $100$ kΩ pull-down and compute its steady current from $3.3$ V (confirming it is negligible).
4. Compute the switching-edge time constant from CA-5: $\tau = R_G C_{iss}$ with your values, and state roughly how long the gate takes to settle ($5\tau$). Compare that with a WS2812 data bit period of about $1.25$ µs and explain why the gate resistor cannot be made arbitrarily large.
5. Open PCB Studio and build the board: Pico 2 W, the ws2812_strip8 part, and the motor_driver_header (which brings out a switched load connection). Generate the schematic and identify which nets correspond to your switch's drain, source, and gate.
6. Run "Generate + PCB layout" and export the Gerbers. As you look at the routed board, note which traces carry the $480$ mA load current versus the milliamperes of gate drive — on a real board those two need very different trace widths, which is a PS-track topic waiting for you.