MOSFET as a switch

YOU WILL BE ABLE TO

Design a low-side n-channel switch driven from a logic pin.
Explain $R_{DS(on)}$ and compute conduction loss.
Explain why high-side switching with an n-channel device is difficult.
Recognise the need for logic-level parts and gate resistors.

A microcontroller pin can source only a small current — on the Pico's RP2040, a few milliamperes comfortably and around $12$ mA at the absolute limit per pin. An LED strip, a motor, a relay, a heater: all want far more. The MOSFET is the standard answer, used not as an amplifier but as a switch driven hard between cutoff (fully off) and deep triode (fully on).

The low-side switch: the GPIO drives the gate, the MOSFET completes the load's path to ground.

The topology matters. In this LOW-SIDE arrangement the MOSFET sits between the load and ground, so its source is permanently tied to $0$ V. That makes $V_{GS}$ equal to the GPIO voltage directly — drive the pin to $3.3$ V and, provided the part's threshold is low enough, the device turns fully on. Simple, cheap, and the default choice.

Fully on, the MOSFET is not a perfect short but a small resistance, $R_{DS(on)}$ — the deep-triode resistance of Lesson 2, now read straight off a datasheet. It sets the conduction loss, and because loss goes as current squared, it is what limits how much load a given part can carry.

P_{\text{cond}} = I_D^2 \, R_{DS(on)}

conduction loss

Now the difficulty. Try the same n-channel device on the HIGH side — between the supply and the load — and the source is no longer at ground; it sits at the load's top, which rises as the device turns on. To keep $V_{GS}$ above threshold you would need a gate voltage ABOVE the supply rail, which a $3.3$ V pin plainly cannot provide. This is why the practical options are a p-channel MOSFET on the high side (its gate is pulled DOWN to turn on, referenced to the supply) or a dedicated high-side gate driver with a charge pump. Low-side first, always, unless the load must be referenced to ground.

WORKED EXAMPLE

A complete low-side switch design

Drive an LED strip: $5$ V supply, $500$ mA load, from a $3.3$ V GPIO. Choose a logic-level n-channel MOSFET with $R_{DS(on)} = 0.05\ \Omega$ at $V_{GS} = 3$ V.
Conduction loss: $P = I^2 R = (0.5)^2 (0.05) = 12.5$ mW. Negligible — no heatsink, no concern.
Voltage lost across the switch: $V_{DS} = IR = 0.5 \times 0.05 = 25$ mV. The load sees essentially the full rail.
Compare the same load through a non-logic-level part stuck in saturation with, say $1.5$ V across it: $P = 0.5 \times 1.5 = 750$ mW — sixty times the heat, from the identical circuit with the wrong part number.

12.5

mW with the right part;

750

mW with the wrong one.

WORKED EXAMPLE

Two resistors the schematic needs

A GATE resistor (typically $22$ – $220\ \Omega$ ) in series with the gate limits the peak current the GPIO must deliver into the gate capacitance during switching, and damps ringing from the gate's inductance — CA-6's second-order response, on a real board.
A PULL-DOWN resistor (typically $10$ – $100$ kΩ) from gate to source guarantees the device is OFF when the microcontroller is in reset or unprogrammed and its pins float. Without it, a floating gate can drift above threshold and switch the load unpredictably.
Neither resistor is optional in production work; both are frequently missing from beginner schematics.

Gate resistor for the switching edge, pull-down for the undefined states.

PRACTICE

A low-side MOSFET switch carries $2$ A with $R_{DS(on)} = 0.02\ \Omega$ . Find the conduction loss and the voltage across the device.

Explain why an n-channel MOSFET is hard to use as a high-side switch from a $3.3$ V logic pin.

A datasheet lists $R_{DS(on)} = 0.03\ \Omega$ at $V_{GS} = 10$ V and does not specify it at $3.3$ V. Is this part suitable for direct GPIO drive, and what should you look for instead?

Why does a gate pull-down resistor matter even though the MOSFET's gate draws no steady current?