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For Students · Power Supply Designer

Power Supply Designer for a Class-D amp

Built for students learning it for a class or exam. See the concept move instead of memorizing formulas — and check your homework intuition. Simulate a Class-D amp live below — adjust the inputs and watch it respond, right in your browser.

RLC Circuit

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RLC Circuit StudioLive

series RLC step response

Controls

Apply a step voltage to a series RLC circuit and watch charge and current respond — under-, over-, or critically damped.

Presets

Resistance R

Inductance L

Capacitance C

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Data Inspector

Damping ζ0.750

Regimeunderdamped

ω₀1.000

Governing equation

1 \overset{q}{¨} + 1.5 \overset{q}{˙} + \frac{q}{1} = V_{0}, ω_{0} = \frac{1}{L C} = 1.000, ζ = \frac{R}{2} \frac{C}{L} = 0.750

Reading this result: ζ = 0.750 < 1: underdamped. With ω₀ = 1.000 rad/s the charge overshoots and rings, oscillating while it decays toward its steady value. Lower ζ means more ringing.

Runs locally in your browser — free forever. Scale to the cloud when reality gets heavy.

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Filter Designer

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Analog Filter DesignerLive

RC & RLC frequency response

Controls

Presets

Resistance R (Ω)

Capacitance C (nF)

Passive filters shape a signal's frequency content with resistors, capacitors, and inductors. An RC low-pass or high-pass has a cutoff at 1/(2πRC) where the response drops 3 dB; an RLC bandpass resonates at 1/(2π√(LC)) with sharpness set by its quality factor Q.

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Data Inspector

Cutoff fc1.6 kHz

Resonance f₀5.0 kHz

Quality factor Q0.32

Governing equation

∣ H (f) ∣ = \frac{1}{1 + ( f /1592 ) ^{2}}

Reading this result: Low-pass: signals below the 1.6 kHz corner pass, above it they roll off at 20 dB/decade, and the response is exactly 3 dB down right at the corner.

Runs locally in your browser — free forever. Scale to the cloud when reality gets heavy.

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Thermal Resistance

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Composite Wall Heat TransferLive

R-value & U-value

Controls

Presets

Drywall (mm)

Insulation (mm)

Brick (mm)

Inside temp (°C)

Outside temp (°C)

Heat flows through a wall like current through resistors in series. Each layer's thermal resistance is its thickness over its conductivity; adding them gives the total R-value, and the heat flux is the temperature difference divided by it. Insulation, with very low conductivity, dominates the R-value — the steep temperature drop shows where it does its work.

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Data Inspector

Total R-value2.71 m²K/W

U-value0.37 W/m²K

Heat flux9.6 W/m²

Governing equation

R = \sum \frac{L _{i}}{k _{i} A} = 2.71 \frac{m ^{2} K}{W}, Q = \frac{Δ T}{R} = 9.6 \frac{W}{m ^{2}}

Reading this result: The insulation layer carries the most thermal resistance, so the temperature falls most steeply across it. A total R-value of 2.71 m²K/W lets 9.6 W/m² flow for this 26°C difference.

Runs locally in your browser — free forever. Scale to the cloud when reality gets heavy.

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Frequently asked questions

Is this good for students?: Yes — this version of "Power Supply Designer for a Class-D amp" is framed for students learning it for a class or exam. See the concept move instead of memorizing formulas — and check your homework intuition.
Do I need to install anything?: No. It runs in any modern browser, free, with no account required.