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Inductor Energy & Time Constant Calculator

Calculate energy stored in an inductor, L/R time constant, and current rise time

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Formula

E=1/2⋅LI2,τ=L/R,i(t)=Ifinal×(1−e(−t/τ))E = 1/2·LI², τ = L/R, i(t) = I_final × (1 - e^(-t/τ))
E— Stored energy (J)
L— Inductance (H)
I— Current (A)
τ— Time constant L/R (s)
R— Series resistance (Ω)

How It Works

Inductor energy calculator computes stored magnetic energy using E = ½LI² — essential for switching power supply design, energy harvesting, and motor drive applications. Power electronics engineers, SMPS designers, and magnetics specialists use this to size inductors, calculate peak currents, and prevent core saturation. Per Horowitz & Hill 'Art of Electronics' (3rd ed., p.40), energy storage is proportional to inductance and the square of current — doubling current quadruples stored energy. The L/R time constant τ = L/R governs energy transfer rate; current reaches 63.2% of final value after one time constant. Core saturation occurs when B_peak exceeds material limits (0.3-0.5T for ferrite, 1.2-1.5T for powdered iron per Magnetics Inc. specifications), causing inductance to collapse and current to spike destructively.

Worked Example

Design an inductor for a 12V-to-5V buck converter at 500kHz, 2A output with 30% ripple current. Average inductor current: I_avg = 2A. Ripple current: ΔI = 0.3 × 2A = 0.6A. Peak current: I_peak = I_avg + ΔI/2 = 2.3A. Required inductance: L = V_out × (1 - D) / (f × ΔI) = 5V × 0.583 / (500kHz × 0.6A) = 9.7μH. Select 10μH inductor rated for I_sat > 3A (30% margin). Stored energy at peak: E = ½ × 10μH × (2.3A)² = 26.5μJ per cycle. At 500kHz, power handling: P = E × f = 26.5μJ × 500kHz = 13.2W — verify thermal rating. Coilcraft XAL5030-103ME (10μH, 3.1A I_sat, 20mΩ DCR) meets these requirements.

Practical Tips

  • ✓Select inductors with I_sat > 1.5× peak current — saturation causes catastrophic current runaway in boost and buck converters
  • ✓Core losses dominate above 100kHz; use ferrite cores with loss < 100mW/cm³ at operating frequency per TDK material specifications
  • ✓For energy harvesting, maximize L/DCR ratio — Coilcraft LPS6235 series achieves 10,000 H/Ω ratio for low-power applications

Common Mistakes

  • ✗Ignoring saturation current (I_sat) — exceeding I_sat causes inductance to drop 30-50%, increasing ripple current exponentially and potentially destroying the switch
  • ✗Confusing L/R time constant with switching period — τ should be 5-10× longer than switching period for continuous conduction mode (CCM) operation
  • ✗Neglecting DCR power losses — a 10μH inductor with 50mΩ DCR at 2A dissipates P = I²R = 0.2W, reducing efficiency by 1.7% in a 12W converter

Frequently Asked Questions

τ = L/R is the time for current to reach 63.2% of final value (charging) or decay to 36.8% (discharging). A 100μH inductor with 10Ω total resistance has τ = 10μs. Full settling (>99%) requires 5τ = 50μs. This governs response time in filters and transient performance in regulators.
E = ½LI² — energy scales linearly with inductance but quadratically with current. A 100μH inductor at 1A stores 50μJ; at 2A it stores 200μJ (4×). Physical size scales with energy: 50μJ inductors fit 0805 packages; 5mJ inductors require 20mm+ diameter cores.
No — energy dissipates through DCR (copper loss) and core loss. A 100μH/50mΩ inductor with 1A stored current loses 50% energy in τ = L/R = 2ms. Superconducting inductors achieve τ > 10⁵ seconds by eliminating resistive losses.
E_max = ½L × I_sat² where I_sat is the saturation current rating. A Coilcraft MSS1210-103 (10μH, 4.9A I_sat) stores E_max = 0.5 × 10μH × 24 = 120μJ. Exceeding this causes core saturation and energy storage collapse within microseconds.
Saturation current (I_sat): limits peak energy storage. DCR: causes I²R losses (0.1-5% of transferred power typically). Core loss: dominates above 100kHz (1-10% losses). Self-resonant frequency: limits useful bandwidth (1-100MHz range for SMD inductors).

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