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LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
Knowledge / Understand components
LNGR · Learning workshop

The MOSFET

The gate controls the current path. “Logic-level” and “low threshold voltage” are not the same promise.

Try the experiment ↓

Example package · not a pinout diagram

Start without prior knowledgeUnderstand the model limitsDE / EN
01

The mental picture

Think of an electrically controlled valve. Moving its gate requires moving charge; ideally there is no continuous gate DC current in steady state. Drain-source current can still be large. Gate charge becomes important again during fast switching.

02

The principle

RDS(on) describes the sufficiently enhanced MOSFET at specified conditions. VGS(th) only marks the start of conduction at a small test current. Check the guaranteed RDS(on) at the gate-source voltage your circuit actually provides.

03

Try it yourself

Interactive experiment

The MOSFET

Double load current at fixed RDS(on). Conduction losses quadruple. Halve duty cycle and watch average conduction loss.

Model limits: Already fully enhanced MOSFET represented by constant resistance. Conduction loss only; no gate or switching simulation.

04

A worked example

5 A through 20 mΩ produces 0.5 W in the continuous on-state. At 10 A it becomes 2 W. Voltage drop rises from 0.1 to 0.2 V. That heat must leave through the package and PCB.

A closer look · equations and assumptions
Formula notes
P_{\mathrm{cond}}=I_{\mathrm{on}}^2 R_{\mathrm{DS(on)}}D
I_{\mathrm{RMS}}=I_{\mathrm{on}}\sqrt{D}
U_{DS}=I_{\mathrm{on}}R_{\mathrm{DS(on)}}

The equation assumes rectangular current with a constant on-value. It excludes switching loss, gate-driver loss, body-diode conduction and recovery. RDS(on) typically rises with temperature. Datasheet current ratings often assume better cooling than a small hobby PCB provides.

A common pitfall. Always consider gate voltage relative to source. With a moving source potential, gate voltage measured to ground is not automatically the actual drive voltage.

05

Workshop takeaway

Low RDS(on) helps only with suitable drive and cooling. Do not choose a switch solely by its largest ampere rating.

Sources & transparency

Original illustrations and educational calculations. No measurements of a specific product. Edited 11 October 2026.

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

14 · An on-state MOSFET has 0.05 Ω on-resistance at the applied gate voltage. Load current 2 A. Find conduction loss.

Show worked answer
  1. P = I² · R.
  2. Square current: 4 A².
  3. P = 0.2 W.
Calculation
P=I^2R=2^2\cdot0.05=0.2\,W

15 · At higher temperature this resistance doubles. What happens to conduction loss at constant current?

Show worked answer
  1. New resistance: 0.1 Ω.
  2. P = 4 · 0.1.
  3. Conduction loss doubles to 0.4 W.
Calculation
P=4\cdot0.1=0.4\,W

16 · MOSFET with 0.2 W conduction loss while ON and 40% duty cycle. Find mean conduction loss; ignore switching loss.

Show worked answer
  1. Multiply ON dissipation by duty cycle.
  2. 0.2 · 0.4.
  3. Mean = 0.08 W.
Calculation
P_{avg}=DP_{on}=0.08\,W

17 · A datasheet says VGS(th) = 1 V. Does that prove proper drive at 3.3 V?

Show worked answer
  1. No, threshold is specified at a very small test current.
  2. On-resistance must be specified at a suitable gate voltage.
  3. Also check current, temperature and driver.

18 · Gate charge 10 nC, switching frequency 20 kHz. Find mean gate-driver charging current.

Show worked answer
  1. Each cycle supplies QG.
  2. Mean current = QG · f.
  3. Mean = 0.2 mA; brief peaks can be much larger.
Calculation
I_{avg}=Q_G f=10\,nC\cdot20\,kHz=0.2\,mA

19 · 10 nC gate charge must transfer in 100 ns. Estimate mean current during this edge.

Show worked answer
  1. Current = charge/time.
  2. 10 · 10⁻⁹ / 100 · 10⁻⁹.
  3. I = 0.1 A = 100 mA.
Calculation
I\approx\frac{Q_G}{t}=100\,mA

20 · A MOSFET turns a load on unintentionally while its microcontroller starts. What simple measure helps with a floating gate?

Show worked answer
  1. Use a suitable gate-source pulldown.
  2. It provides a defined OFF state when the driver is high-impedance.
  3. Check pin startup state, leakage and driver capability.

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20 exercises · PDF ↓Teacher solutions · PDF ↓