FETs and MOSFETs: Field Effect Transistors Explained

A field effect transistor, or FET, is controlled by an electric field rather than an input current. In a bipolar transistor you must push current into the base, but in a FET you simply apply a voltage to the gate and almost no gate current flows. This voltage control makes FETs easy to drive and gives them enormously high input impedance.

The MOSFET is the most important transistor ever made: billions of them form every microprocessor and memory chip. Understanding FETs and MOSFETs, how they differ from bipolar transistors, and where each shines is essential for modern electronics, from tiny logic gates to huge motor controllers.

Key Concepts

• Terminals: gate (control), source (carriers enter), drain (carriers leave). Current flows through a channel between source and drain.
• JFET: a reverse-biased pn junction at the gate squeezes the channel. Always depletion-mode: conducts at zero gate voltage, turns off as VGS goes negative (n-channel).
• Pinch-off voltage (VP): the gate-source voltage at which the JFET channel closes and drain current stops.
• MOSFET: the gate is insulated by silicon dioxide, so DC gate current is practically zero. Comes in n-channel (NMOS) and p-channel (PMOS).
• Enhancement-mode: normally off; a gate voltage above the threshold Vth creates the channel. The type used in digital chips.
• Depletion-mode: normally on; gate voltage depletes the channel to turn it off.
• Transconductance (gm): change in drain current per change in gate-source voltage, in siemens. The FET equivalent of β.
• CMOS: complementary MOS pairing NMOS and PMOS; draws almost no power when idle, which is why it dominates digital electronics.

Formulas

• JFET (Shockley): ID = IDSS × (1 − VGS/VP)2, valid in the saturation region.
• Enhancement MOSFET: ID = k × (VGS − Vth)2 in saturation.
• Voltage gain: a common-source stage gives about −gm × RD.
• Gate switching current: I = C × dV/dt; the gate is a capacitor, so fast switching needs brief current pulses.

Worked Example

An n-channel JFET has IDSS = 10 mA and VP = −4 V. Find ID at VGS = −1 V.

1. ID = 10 × (1 − (−1)/(−4))2 = 10 × (1 − 0.25)2 = 10 × 0.5625 = 5.625 mA.
2. Check: at VGS = 0 the current is the full 10 mA; at −4 V it is zero, so 5.625 mA at −1 V is sensible.

Common Misconceptions

• The gate draws no current at all. DC gate current is essentially zero, but the gate is a capacitor that must be charged and discharged on every switch; weak drivers cause slow, hot switching.
• FETs cannot amplify. Amplification needs a control effect, not input current: gate voltage strongly modulates drain current via transconductance.
• All MOSFETs are enhancement-mode. Depletion-mode MOSFETs, on at zero gate voltage, exist too; check the datasheet symbol.

Real-World Applications

• CPUs, GPUs, and memory chips are built from billions of CMOS MOSFETs acting as switches.
• Power MOSFETs switch large currents in motor controllers, LED drivers, and laptop DC-DC converters.
• JFETs suit audio preamplifiers and instruments because their high input impedance does not load delicate sources.
• Touch sensors exploit the insulated-gate principle: tiny voltage changes on an electrode control a signal.

To recap: FETs control drain current with gate voltage instead of base current, JFETs are normally-on devices described by the Shockley equation, and enhancement MOSFETs are normally-off switches that made CMOS logic and the digital age possible.