The direct answer for the energy stored in an inductor equation is E = ½ L I². In this formula, E is the energy in Joules (J), L is the inductance in Henries (H), and I is the instantaneous current in Amperes (A). Whether you are calculating the flyback spike of a relay coil or sizing the output filter for a 2026-era GaN buck converter, this single equation dictates your component selection, thermal limits, and saturation margins.
The Core Energy Stored in an Inductor Equation
When current flows through an inductor, it establishes a magnetic field. The power supply does work to build this field against the inductor's self-induced back-EMF. That work is stored as potential energy in the magnetic flux. When the circuit opens or the current decreases, the field collapses, returning that energy to the circuit.
Symbol Definition Table
| Symbol | Parameter | Standard SI Unit | Common Sub-Units in Practice |
|---|---|---|---|
| E | Stored Energy | Joules (J) | Millijoules (mJ), Microjoules (µJ) |
| L | Inductance | Henries (H) | Millihenries (mH), Microhenries (µH), Nanohenries (nH) |
| I | Instantaneous Current | Amperes (A) | Milliamperes (mA) |
Rearranged Forms for Circuit Design
In practical design, you rarely solve for energy just to know the energy. You use the energy relationship to find the maximum allowable current before core saturation, or to size the inductance required to buffer a specific energy pulse. Here are the algebraic rearrangements you will use on the bench:
- Solving for Inductance (L):
L = (2 × E) / I²
Use case: Sizing an inductor to absorb a known flyback energy pulse without exceeding a target peak current. - Solving for Peak Current (I):
I = √(2 × E / L)
Use case: Determining the absolute maximum current a given inductor can handle if its core saturation energy limit is known. - Solving for the ½ Constant Origin: The factor of ½ arises from the integral of power over time (
∫ v·i dt) as current ramps linearly from 0 to I. It is mathematically identical to the kinetic energy equation (½ mv²) or capacitive energy (½ CV²).
Worked Examples with Strict Unit Tracking
The most common point of failure in inductor math is unit misalignment. Datasheets specify inductance in µH and current in mA, but the base equation demands Henries and Amperes. Below are two solved problems demonstrating strict unit tracking.
Problem 1: SMPS Inductor Energy Buffer
Scenario: You are designing a synchronous buck converter. The selected power inductor has a nominal value of 22 µH. Under maximum load, the peak instantaneous current (including ripple) reaches 4.5 A. How much energy is stored in the magnetic field at the peak of the ripple cycle?
- Convert to base SI units:
L = 22 µH = 22 × 10-6 H
I = 4.5 A (already in base units) - Square the current:
I² = 4.5 × 4.5 = 20.25 A² - Apply the equation:
E = ½ × L × I²
E = 0.5 × (22 × 10-6) × 20.25 - Calculate and format:
E = 222.75 × 10-6 Joules
E = 222.75 µJ (or 0.222 mJ)
Problem 2: Relay Flyback Diode Sizing
Scenario: A 12VDC automotive relay coil has a measured inductance of 120 mH and draws a steady-state current of 85 mA. When the driving MOSFET turns off, the collapsing field dumps its stored energy into the flyback diode. What is the total energy the diode must absorb?
- Convert to base SI units:
L = 120 mH = 120 × 10-3 H = 0.12 H
I = 85 mA = 85 × 10-3 A = 0.085 A - Square the current:
I² = 0.085 × 0.085 = 0.007225 A² - Apply the equation:
E = 0.5 × 0.12 × 0.007225 - Calculate and format:
E = 0.0004335 Joules
E = 433.5 µJ (or 0.433 mJ)
Note: While 0.433 mJ seems small, if the MOSFET lacks a flyback diode, this energy forces a voltage spike high enough to avalanche the silicon junction in nanoseconds, destroying the transistor.
When the Formula Applies (and Where Unit Mistakes Break It)
The equation E = ½ L I² is elegant, but it relies on strict physical assumptions. Blindly applying it to real-world magnetic components without checking these boundaries will result in failed prototypes.
The Linearity Assumption and Core Saturation
This formula assumes L is constant. In an ideal air-core inductor, L is indeed constant. However, 95% of the power inductors you buy in 2026 use ferrite or powdered iron cores to multiply inductance. These materials exhibit a non-linear B-H (magnetic flux density vs. magnetic field strength) curve.
As current increases, the core approaches magnetic saturation. Once saturated, the relative permeability (µr) of the core drops toward that of air (µr = 1). Consequently, the actual inductance L drops drastically. If you calculate stored energy using the nominal, unsaturated inductance value at a current level that saturates the core, the equation will dangerously overestimate the stored energy. Always verify that your peak current I is below the component's ISAT (saturation current) rating.
The Unit Trap: Micros and Millis
The most frequent engineering mistake is mixing prefixes.
If you plug L = 47 (meaning 47 µH) and I = 2000 (meaning 2000 mA) directly into the formula without converting to Henries and Amperes, your calculator will output 94,000,000.
Is that Joules? Millijoules? No, it is numerical garbage.
Rule of thumb: Always strip prefixes and write the exponent (e.g., 47e-6 and 2.0) before multiplying. For a deeper theoretical breakdown of magnetic fields and calculus, refer to the foundational texts at All About Circuits.
Decision Path: Sizing a Power Inductor for a Buck Converter
Let us move from theory to the workbench. You are tasked with selecting an inductor for a 12V-to-5V buck converter. The load requires 3.0 A continuous. The switching frequency (fsw) is 500 kHz. We will use the energy equation and ripple constraints to terminate on a specific, purchasable part number.
| Decision Step | Calculation / Logic | Resulting Requirement |
|---|---|---|
| 1. Define Ripple Current (ΔI) | Target 30% of nominal load current for optimal efficiency/size trade-off. ΔI = 3.0 A × 0.30 |
ΔI = 0.9 A peak-to-peak |
| 2. Calculate Required Inductance (L) | Duty Cycle (D) = Vout / Vin = 5 / 12 = 0.416 L = [Vout × (1 - D)] / [fsw × ΔI] L = [5 × (1 - 0.416)] / [500,000 × 0.9] |
L = 6.48 µH |
| 3. Determine Peak Current (Ipeak) | Ipeak = Iload + (ΔI / 2) Ipeak = 3.0 + (0.9 / 2) |
Ipeak = 3.45 A |
| 4. Calculate Peak Energy Storage | E = ½ × L × (Ipeak)² E = 0.5 × (6.48 × 10-6) × (3.45)² |
E = 38.5 µJ |
| 5. Apply Saturation Margin | The chosen inductor's ISAT must exceed Ipeak by at least 20% to prevent the inductance drop-off described in the saturation section. | Minimum ISAT = 4.14 A |
The Concrete Component Pick
Based on the decision matrix above, we need an inductor with a nominal value near 6.5 µH, an RMS current rating (IRMS) > 3.0 A to handle thermal heating, and a saturation current (ISAT) > 4.14 A to safely contain the 38.5 µJ energy pulse without core saturation.
Final Selection: Coilcraft MSS1048-682MLB
- Inductance: 6.8 µH (Slightly higher than 6.48 µH, which reduces ripple current and lowers peak energy storage—always a safe direction).
- IRMS (Thermal Limit): 5.2 A (Easily handles the 3.0 A continuous load without exceeding a 40°C temperature rise).
- ISAT (Magnetic Limit): 6.4 A (Well above our 4.14 A minimum requirement, ensuring the core remains linear and the E = ½ L I² equation holds true at peak load).
For further validation on high-frequency magnetics selection, review the design notes published by Analog Devices on synchronous buck converter magnetics. By strictly tracking units, respecting the physical limits of core saturation, and terminating your math on a verified datasheet parameter, you eliminate the guesswork from magnetic component design.
