A learning project by Huy · AI-assisted content · Updated
Devices11 min read
MOSFET as a Switch
Drive loads efficiently by understanding gate voltage, RDS(on), switching loss and protection.
Mental model
The gate is like an insulated control plate. Applying voltage rearranges charge inside the device and opens a low-resistance path between drain and source.
The load connects VDD to drain D; source S and the gate driver share ground. Rg limits gate current, and Rpd holds the gate low when undriven. D1 has its cathode at VDD so inductive current can circulate after turn-off.Scroll sideways to view the complete circuit.UNDERSTAND THE BEHAVIOR
01
An N-channel MOSFET used as a low-side switch sits between the load and ground. The load receives current when gate-to-source voltage is high enough to create a low-resistance channel.
02
The gate draws almost no steady DC current, but it behaves like a capacitor during switching. The driver must repeatedly move gate charge, and slow transitions increase the time spent in the high-loss linear region.
03
A drain-voltage spike appears when inductive current is interrupted. A flyback diode, TVS or snubber provides a controlled path for that stored magnetic energy.
Why it matters
MOSFET switches drive motors, LEDs, heaters and power converters with much lower loss than a resistor or linear transistor stage.
Core ideas
VGS(th) only indicates the beginning of conduction, not full turn-on.
Use RDS(on) specified at your actual gate-drive voltage.
Inductive loads need a flyback path to control the voltage spike when switched off.
Key equationsConduction loss: P ≈ I²RDS(on)Gate energy per transition: E ≈ QgVdrive
EXAMPLE
Worked example — switching a 2 A load
1A logic-level MOSFET has RDS(on) = 40 mΩ at VGS = 4.5 V.
2Conduction loss is P = 2² × 0.040 = 0.16 W.
3At 3.3 V gate drive, the datasheet may specify a higher resistance—or none at all.
4Check temperature rise, add a gate resistor and add a flyback diode if the load is inductive.
Common mistake
VGS(th) is not the recommended ON voltage. Use the RDS(on) table and curves at the actual gate voltage.
RememberVGS(th) only indicates the beginning of conduction, not full turn-on.
Practice and check your understanding
Try each question before opening the worked answer.
1. A fully enhanced MOSFET has RDS(on) = 0.05 Ω and carries 2 A continuously. Estimate conduction loss.
Show answer and reasoning
P = I²R = 2² × 0.05 = 0.2 W. This excludes switching loss and the increase of resistance with temperature. In the linked voltage-based resistor solver, use VDS = IR = 0.1 V and R = 0.05 Ω.
2. Is a 2 V gate threshold enough evidence that a MOSFET works well from a 3.3 V GPIO?
Show answer and reasoning
No. Threshold specifies the start of conduction at a small test current. Check a guaranteed RDS(on) at an appropriate gate voltage, gate charge, driver capability, and thermal limits.
A related calculation tool, not a complete simulation of the circuit. Check its model and units.
Continue learning
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