Inductor Energy Calculator

Inductors store energy in a magnetic field, releasing it when current falls — the complementary behavior to capacitors storing energy in electric fields. Buck and boost converters, flyback transformers, relay drivers, and defibrillator circuits all depend on calculating how much energy an inductor holds at peak current. Undersizing leads to saturation and runaway current; oversizing adds bulk and cost.

Stored energy in an inductor:

E = ½ × L × I²

Where E is energy in joules, L is inductance in henries, and I is the current flowing through the inductor in amperes. Energy scales with the square of current — doubling peak current quadruples stored energy and heat stress on switching MOSFETs.

Inductor voltage during change follows V = L × di/dt. When current is interrupted abruptly (opening a switch), the resulting voltage spike V = L × ΔI/Δt can destroy semiconductors unless clamped by freewheel diodes or active snubbers. This is why relay drivers and DC-DC layouts place catch diodes across inductive loads.

Worked example: A buck converter uses a 47 µH inductor with 2 A peak ripple current (average 1 A, ΔI = 2 A). Stored energy at peak I = 2 A: E = 0.5 × 0.000047 × 2² = 0.5 × 0.000047 × 4 = 94 µJ per switching cycle. At 500 kHz, average power transferred relates to energy per cycle times frequency — but saturation depends on peak I, not average. A 10 mH relay coil at 0.1 A holds E = 0.5 × 0.01 × 0.01 = 50 µJ — tiny, yet opening the contacts without a diode generates kilovolt spikes from the same L di/dt.

Compare with the capacitor charge calculator for dual energy-storage perspectives in filter and timing design.