Physics/Option Modules/Electronics/MOSFET as a Switch
Physics · Electronics · A2 Level

MOSFET as a Switch

A microcontroller pin or a delicate sensor might only be able to safely supply a few milliamps — nowhere near enough to drive a motor, a relay, or a bank of LEDs directly. The MOSFET bridges that gap: a tiny control voltage, and almost no current at all, is enough to switch a much larger load on and off.

8 min read

A simple MOSFET switching circuit

Take the enhancement-mode MOSFET from the previous lesson and wire it into the simplest possible switching arrangement. The drain connects through a load — a resistor, or a real load such as a relay coil or an LED with its own current-limiting resistor — up to a supply voltage. The source is connected to the circuit's ground (0 V). A control voltage, the input signal, is applied between the gate and the source.

  • Gate voltage below $V_{th}$: the MOSFET is off. No channel exists between drain and source, so the drain-source resistance is very high and essentially no current flows through the load. With negligible current flowing, there is negligible voltage dropped across the load, so almost all of the supply voltage appears across the MOSFET itself (between drain and source).
  • Gate voltage above $V_{th}$: the MOSFET turns on. The drain-source resistance drops to a very low value, and current flows freely through the load. Now most of the voltage drop occurs across the load, and the voltage across the MOSFET itself (drain to source) falls to a small residual value.

In other words, the MOSFET behaves like a voltage-controlled relay with no moving parts: a small control voltage at the gate decides whether the load circuit is effectively open or closed.

Worked exampleOutput state of a MOSFET switching circuit
1
Set up: a MOSFET switch has its drain connected through a $470\ \Omega$ load resistor to a $12\ \text{V}$ supply, and its source grounded. The gate is driven either at $0\ \text{V}$ (device off) or well above $V_{th}$ so that the on-state drain-source resistance is negligibly small (device fully on). Find the voltage across the load in each state.
2
Off state ($V_{GS}=0$): no channel exists, so the drain-source path carries negligible current.
With $I_D \approx 0$, the voltage dropped across the load resistor is $V_{load}=I_D R \approx 0\ \text{V}$.
3
The full $12\ \text{V}$ supply therefore appears across the MOSFET's drain-source terminals: $V_{DS}\approx 12\ \text{V}$.
This is exactly what "off" means for a switch — nearly all the supply voltage is dropped across the open switch, not the load.
4
On state ($V_{GS} \gg V_{th}$): the channel is strongly conducting, so the drain-source resistance is very small compared with the $470\ \Omega$ load.
Almost all of the $12\ \text{V}$ now drops across the load resistor instead: $V_{load}\approx 12\ \text{V}$, giving a load current of roughly $I_D = V_{load}/R \approx 12/470 \approx 0.026\ \text{A} = 26\ \text{mA}$.
5
The voltage remaining across the MOSFET itself in the on state, $V_{DS}$, is correspondingly small — only a small residual fraction of a volt for a well-designed switching MOSFET carrying this current.
This is why a MOSFET used as a switch loses very little power itself when on: nearly all the load's power is dissipated in the load, not the switch.
result

This is the essential value of the MOSFET switch: it lets a small, low-power control signal — from a microcontroller output pin, a logic circuit, or a sensor — switch a much larger current through a load that the control signal itself could never drive directly. The controlling circuit only has to supply enough voltage to raise $V_{GS}$ above $V_{th}$; it doesn't have to supply the load current itself.

Why the gate's high input resistance matters for sensors

The previous lesson noted that the gate is insulated from the channel by a thin oxide layer, giving the gate an extremely high input resistance — typically far higher than almost any other circuit element it might be connected to. That fact does more than just make the MOSFET convenient to drive; it makes the device particularly well suited to being controlled directly by a sensor.

Many sensors produce only a very small output current, and can only supply that current without their own output signal being disturbed, or "loaded down." If a sensor's output is connected to something that draws a significant current from it, the extra current drawn can pull the sensor's output voltage away from the true value it's trying to represent — the measurement becomes distorted by the act of connecting to it. A MOSFET gate, because it is separated from the channel by an insulating layer, draws negligible steady current from whatever is driving it. Connecting a sensor's output directly to a MOSFET gate therefore lets the sensor control the switching of a much larger load current, without the sensor's own delicate output signal being affected by the connection at all.

Doesn't load the source signal

Because the gate draws essentially no current, connecting a sensor to a MOSFET gate does not "load" (disturb) the sensor's own output. This is precisely why MOSFETs, rather than devices that require a continuous base or input current to switch on, are so widely favoured for interfacing delicate sensors with circuits that need to switch larger currents or power real loads.

Key takeaways

  • In a simple MOSFET switching circuit, the load sits between the supply and the drain, the source is grounded, and the gate receives the control voltage.
  • Below $V_{th}$: the MOSFET is off, drain-source resistance is high, negligible current flows, and nearly all the supply voltage appears across the MOSFET.
  • Above $V_{th}$: the MOSFET is on, drain-source resistance is very low, current flows through the load, and nearly all the supply voltage now appears across the load.
  • A small control voltage at the gate can therefore switch a much larger load current — the controlling circuit doesn't need to supply the load current itself.
  • The gate's very high input resistance (from the insulating oxide layer) means it draws negligible current from whatever drives it.
  • This matters for sensor interfacing: a delicate sensor with limited output current can control a MOSFET switch without its own signal being disturbed or loaded down.
  • This "doesn't load the source" property is a key practical reason MOSFETs are preferred over devices needing a continuous input/base current for many sensor-driven switching applications.
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