MOSFET as Switch

A MOSFET switch controls kilowatts of load power with a microcontroller signal of just 3.3V. Every motor driver, LED driver, and power converter uses this fundamental circuit.

Core Concept

When used as a switch, a MOSFET operates in two extreme regions: cutoff and triode. In cutoff, VGS is below the threshold voltage Vth, the channel does not exist, ID is essentially zero, and VDS equals VDD. The MOSFET looks like an open switch.

In triode (also called the linear or ohmic region), VGS is driven well above Vth. The MOSFET channel is fully open and behaves like a small on-resistance RDS(on). For the IRF540N, RDS(on) is as low as 44 mΩ at VGS = 10V. The voltage drop across the device is ID*RDS(on), which is negligibly small. The MOSFET looks like a closed switch.

The key design rule: always drive VGS at least 4V to 5V above Vth to guarantee full enhancement and minimum RDS(on). Logic-level MOSFETs like the 2N7000 have Vth below 2V, so a 3.3V logic output can drive them. Power MOSFETs like the IRF540N need VGS of 10V for rated RDS(on), requiring a gate driver IC like the IR2110.

Key Equations

Condition for OFF: V_GS < V_th, ID = 0, VDS = VDD

Condition for ON (triode/ohmic): V_GS >> V_th and V_DS < V_GS - V_th

On-state voltage drop: V_DS(on) = I_D * R_DS(on)

Power dissipation in ON state: P = I_D^2 * R_DS(on)

Power dissipation in OFF state: approximately 0 (leakage current is nanoamps)

Switching losses (simplified): P_sw = 0.5 * VDD * I_D * (t_on + t_off) * f where f is switching frequency.

Given:
  VDD     = 12 V
  RL      = 100 Ω  (load resistance)
  VGS     = 10 V   (gate drive, switch ON)
  RDS(on) = 0.044 Ω  (IRF540N at VGS=10V)
  Vth     = 3.5 V

Why this formula:
  Switch is ON (VGS >> Vth). Find ID and VDS(on).

Step 1 - Check ON condition:
  VGS = 10V > Vth = 3.5V: confirmed ON.

Step 2 - Find load current (RDS(on) << RL, so VDS(on) ≈ 0):
  ID ≈ VDD / RL = 12 / 100 = 120 mA

Step 3 - Find VDS(on):
  VDS(on) = ID * RDS(on)
           = 0.120 * 0.044
           = 0.00528 V ≈ 5.3 mV

Step 4 - Check triode condition:
  VDS = 5.3 mV < VGS - Vth = 10 - 3.5 = 6.5V : confirmed triode.

Step 5 - Power in MOSFET:
  P = ID^2 * RDS(on) = (0.120)^2 * 0.044 = 0.0144 * 0.044 = 0.634 mW

Final Answer:
  ID = 120 mA, VDS(on) = 5.3 mV, Power dissipated in switch = 0.634 mW

Exam Tip: GATE problems on MOSFET switches ask you to verify which region the device is in. Always check both conditions: VGS > Vth (not cutoff) AND VDS < VGS - Vth (triode, not saturation). A MOSFET in saturation is NOT a good switch; it drops significant voltage and wastes power. Drive VGS high enough to push VDS into the triode region.

Key Properties

• OFF state: VGS < Vth, ID = 0, VDS = VDD. Device acts as an open circuit.
• ON state: VGS >> Vth, VDS small, device in triode region with resistance RDS(on).
• IRF540N has RDS(on) = 44 mΩ at VGS = 10V, allowing currents up to 33A with minimal loss.
• Logic-level MOSFETs like 2N7000 turn on fully at VGS = 5V, suitable for direct microcontroller drive.
• Power dissipation in ON state is ID^2 * RDS(on), which is very small for low RDS(on) devices.
• Switching losses increase with frequency; at high frequencies (above 100 kHz) they dominate over conduction losses.
• Gate charge Qg determines how fast the switch turns on; gate driver circuits must supply Qg quickly for fast switching.

Quick Revision

• OFF: VGS < Vth, ID = 0, VDS = VDD.
• ON: VGS >> Vth, VDS = ID*RDS(on) (millivolts).
• Triode condition for ON state: VDS < VGS - Vth.
• Power in switch: P = ID^2 * RDS(on).
• IRF540N: RDS(on) = 44 mΩ, Vth = 3.5V, ID(max) = 33A.
• Logic-level FETs: full enhancement at VGS = 5V or 3.3V.
• Switching loss = 0.5*VDD*ID*(ton+toff)*f; grows with frequency.
• Exam trap: Students verify VGS > Vth but forget to check VDS < VGS - Vth, missing that the MOSFET is in saturation rather than triode and not a good switch.