Verified examples
Six calculations that can be reproduced by hand.
These cases isolate a fundamental relationship by setting unrelated dynamic terms to zero. The browser runs the same checks at startup and reports PASS only when its result matches the independently evaluated equation within 1 ppm, with a minimum absolute tolerance of 10⁻⁸.
6 / 6 REFERENCE CHECKS PASSED
1. Fully-on MOSFET conduction
Given: I = 10 A, RDS(on) = 5 mΩ, Ta = 25 °C, Rθ = 2 °C/W.
Formula Pcond = I² × RDS(on)
Substitute Pcond = (10 A)² × 0.005 Ω
Answer Pcond = 0.50000 W
Formula Tj = Ta + Pcond × Rθ
Substitute Tj = 25 °C + (0.50000 W)(2 °C/W)
Answer Tj = 26.00000 °C
2. Manual PWM conduction and switching
Given: V = 48 V, I = 10 A, D = 0.5, RDS(on) = 5 mΩ, tr = tf = 20 ns, fs = 100 kHz, Qg = 60 nC, Vg = 10 V.
Formula Pcond = I² × D × RDS(on)
Substitute Pcond = (10 A)² × 0.5 × 0.005 Ω
Answer Pcond = 0.25000 W
Formula Psw = ½ × V × I × (tr + tf) × fs
Substitute Psw = 0.5 × 48 V × 10 A × (20 ns + 20 ns) × 100 kHz
Answer Psw = 0.96000 W
Formula Pgate = Qg × Vg × fs
Substitute Pgate = 60 nC × 10 V × 100 kHz
Answer Pgate = 0.06000 W
Answer Ptotal = 0.25000 + 0.96000 + 0.06000 = 1.27000 W
3. Synchronous buck in CCM
Given: VIN = 48 V, VOUT = 12 V, IOUT = 10 A, L = 100 µH, fs = 100 kHz, RDS(on) = 5 mΩ for both FETs.
Formula D = VOUT / VIN
Substitute D = 12 V / 48 V
Answer D = 0.25000
Formula ΔIL = (VIN − VOUT)D / (Lfs)
Substitute ΔIL = (48 V − 12 V)(0.25) / (100 µH × 100 kHz)
Answer ΔIL = 0.90000 A
Formula Pcond,total = (IOUT² + ΔIL²/12)RDS(on)
Substitute Pcond,total = ((10 A)² + (0.9 A)²/12)(0.005 Ω)
Answer Pcond,total = 0.5003375 W
4. Boost converter in CCM
Given: VIN = 24 V, VOUT = 48 V, IOUT = 5 A, η = 1, L = 100 µH, fs = 100 kHz, RDS(on) = 5 mΩ.
Formula D = 1 − VIN/VOUT
Substitute D = 1 − 24 V/48 V
Answer D = 0.50000
Formula IIN = (VOUT × IOUT)/(η × VIN)
Substitute IIN = (48 V)(5 A)/(1 × 24 V)
Answer IIN = 10.00000 A
Formula IFET,rms = √[D(IIN² + ΔIL²/12)]
Substitute IFET,rms = √[0.5((10 A)² + (1.2 A)²/12)]
Answer IFET,rms = 7.075309 A
Answer Pcond = IFET,rms²RDS(on) = 0.250300 W
5. Full-bridge current sharing
Given: VIN = 400 V, VOUT = 48 V, IOUT = 10 A, η = 1, RDS(on) = 5 mΩ for four FETs.
Formula Ipri = (VOUT × IOUT)/(η × VIN)
Substitute Ipri = (48 V)(10 A)/(1 × 400 V)
Answer Ipri = 1.20000 A
Formula IFET,rms = Ipri/√2
Substitute IFET,rms = 1.2 A/√2
Answer IFET,rms = 0.848528 A
Formula Pcond,total = 4 × IFET,rms² × RDS(on)
Substitute Pcond,total = 4 × (0.848528 A)² × 0.005 Ω
Answer Pcond,total = 0.014400 W
6. Totem-pole PFC line current
Given: VAC = 230 Vrms, VOUT = 400 V, IOUT = 2.5 A, η = 1, PF = 1, RDS(on) = 5 mΩ for four FET positions in the simplified sharing check.
Formula Iline,rms = (VOUT × IOUT)/(η × VAC × PF)
Substitute Iline,rms = (400 V)(2.5 A)/(1 × 230 V × 1)
Answer Iline,rms = 4.347826 A
Formula IFET,rms = Iline,rms/√2
Substitute IFET,rms = 4.347826 A/√2
Answer IFET,rms = 3.074377 A
Formula Pcond,total = 4 × IFET,rms² × RDS(on)
Substitute Pcond,total = 4 × (3.074377 A)² × 0.005 Ω
Answer Pcond,total = 0.189036 W
What this verification does—and does not—prove
These checks confirm that selected calculator paths reproduce their stated equations. They do not validate a MOSFET model against laboratory hardware, certify a topology, or replace datasheet curve extraction, simulation, thermal testing, and measured switching waveforms.
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