MOSFETs & JFETs

Field-effect transistors are voltage-controlled: the gate voltage sets the channel current with almost no gate current. Learn JFET and MOSFET structure, enhancement vs depletion, N- vs P-channel, threshold voltage, square-law and triode regions, transconductance, R_DS(on) switching, gate charge and logic-level drive.

MOSFETs are the most manufactured device in history — billions in every processor — and the dominant power switch in converters. JFETs are used for low-noise and high-impedance analog inputs.

An analogy

A FET is like a garden hose squeezed by an electric field instead of a hand: the gate voltage pinches or opens the channel without touching the water.

Types

JFET: depletion-mode, normally ON; gate reverse-biased to pinch off. MOSFET: insulated gate (oxide); enhancement-mode (normally OFF) is most common. N-channel conducts with positive V_GS; P-channel with negative V_GS.

Regions

Cut-off (V_GS < V_th), triode/linear (small V_DS — behaves like a voltage-controlled resistor, used for switches) and saturation (constant current — used in amplifiers).

MOSFET as a switch

Choose V_DS rating, R_DS(on) at the actual gate drive (logic-level parts are specified at V_GS = 4.5 V or 2.5 V), gate charge Q_g (switching speed = Q_g/I_drive), and thermal resistance. Low-side N-MOSFETs are easiest to drive.

CMOS and power devices

Complementary N and P MOSFETs form CMOS logic with near-zero static power. Power MOSFETs use vertical trench structures; GaN and SiC FETs switch faster with lower losses in chargers and EV inverters.

Formulas

• MOSFET saturation: I_D = ½ k_n (V_GS − V_th)² (V_DS ≥ V_GS − V_th)
• MOSFET triode: I_D = k_n [(V_GS − V_th)V_DS − V_DS²/2]
• Transconductance: g_m = 2 I_D / (V_GS − V_th)
• JFET (Shockley): I_D = I_DSS (1 − V_GS/V_P)²
• Switch conduction loss: P = I² R_DS(on)

Worked example: MOSFET bias point

Given: k_n = 2 mA/V², V_th = 1 V, V_GS = 2 V

Find: I_D and g_m.

• I_D = ½ × 2 × (1)² = 1 mA
• g_m = 2 × 1 mA/1 V = 2 mS

Answer: 1 mA, 2 mS

Worked example: JFET current

Given: I_DSS = 10 mA, V_P = −4 V, V_GS = −2 V

Find: I_D.

• I_D = 10 × (1 − 0.5)² = 2.5 mA

Answer: 2.5 mA

Common mistakes

• Wrong: Using R_DS(on) at 10 V when driving with 3.3 V. — Right: Check the datasheet at your actual V_GS.
• Wrong: Floating gates. — Right: Add a gate pull-down so the FET stays OFF at reset.

Applications

• Load switches: MCU-controlled LED strips, motors, heaters.
• Switching converters: Buck/boost power stages.
• Low-noise preamps: JFET inputs for sensors and audio.

Key takeaways

• Voltage-controlled, near-zero gate current.
• I_D = ½k(V_GS − V_th)² in saturation.
• Switches: R_DS(on) at real V_GS, Q_g.
• JFET: I_DSS(1 − V_GS/V_P)².

Going further

• Miller plateau: Gate charge curve and switching losses.
• GaN & SiC: Wide-bandgap power devices.
• Short-channel effects: Velocity saturation in modern CMOS.