Inductor Energy Storage Calculator
Find the energy stored in an inductor or the peak current needed to reach a target energy storage level. Essential for SMPS, filter, and power electronics design.
🔌 What is Inductor Energy Storage?
Inductor energy storage is the energy held in the magnetic field of an inductor when current flows through it. The relationship is E = 0.5 times L times I squared, where E is stored energy in joules, L is inductance in henries, and I is the current in amperes. Because energy scales with the square of current, doubling the current quadruples the stored energy, making peak current the dominant factor in inductor and core selection.
Understanding inductor energy storage is critical in three main areas of electronics design. First, in switch-mode power supplies (SMPS) such as buck, boost, and flyback converters, the inductor is the primary energy-storage element. Each switching cycle, the transistor charges the inductor to a peak current and the energy is then transferred to the output. Second, in filter inductors for AC power lines and motor drives, the stored energy determines how much current ripple the inductor will smooth. Third, in pulsed power systems and wireless charging transmitters, the inductor is deliberately charged to a precise energy level and then discharged into a load at a controlled rate.
A common misconception is that a larger inductance always stores more energy. In practice, many SMPS designs use smaller inductance at higher switching frequency, which actually reduces stored energy per cycle while maintaining the same output power. The designer optimises inductance to balance core size, switching losses, and output ripple simultaneously.
The peak flux linkage output (L times I, in webers) is equally important for core selection. Ferrite cores used in switching power supplies saturate at 300 to 400 mT; exceeding this causes the inductance to collapse suddenly, leading to uncontrolled current rise and potential transistor failure. This calculator shows flux linkage alongside stored energy so both design constraints can be checked in one step.
📐 Formula
E = ½ × L × I²
E = energy stored in the magnetic field (joules, J)
L = inductance (henries, H; enter in μH above)
I = peak current through the inductor (amperes, A)
Reverse (find I): I = √(2E / L)
Flux linkage: λ = L × I (webers, Wb) - used to check core saturation
Example: L = 100 μH, I = 5 A → E = 0.5 × 100×10²−&sup6; × 25 = 1250 μJ = 1.25 mJ
📖 How to Use This Calculator
Steps
1
Choose the calculation mode - select Find Energy to compute stored energy from inductance and peak current, or Find Current to calculate the peak current needed to store a target energy in a given inductor.
2
Enter inductance in microhenries - type the inductor value in the inductance field (in μH). Use the slider for quick adjustment. For millihenry values, multiply by 1000 first: 2.2 mH = 2200 μH. For nanohenry values, divide by 1000: 470 nH = 0.47 μH.
3
Enter peak current or target energy - in Find Energy mode, enter the peak (not average) inductor current in amperes. In Find Current mode, enter your target stored energy in microjoules (μJ).
4
Read and apply the results - the primary result shows stored energy in the most appropriate unit. The flux linkage result (L × I in Wb or mWb) lets you verify that the core will not saturate by comparing against the core's rated saturation flux times its winding area.
💡 Example Calculations
Example 1 - Buck Converter Filter Inductor (100 μH at 5 A)
L = 100 μH, Peak Current I = 5 A
1
Convert inductance: L = 100 μH = 100 × 10-6 H = 0.0001 H
2
Apply the formula: E = 0.5 × 0.0001 × 5² = 0.5 × 0.0001 × 25 = 0.00125 J
3
Convert to common units: E = 1.25 mJ = 1250 μJ
4
Flux linkage: λ = 0.0001 × 5 = 0.0005 Wb = 0.5 mWb. Verify against core saturation spec before finalising.
Result: E = 1250 μJ (1.25 mJ), λ = 0.5 mWb
Try this example →Example 2 - High-Current Boost Converter Inductor (47 μH at 10 A)
L = 47 μH, Peak Current I = 10 A
1
Convert: L = 47 × 10-6 H = 0.000047 H
2
Energy: E = 0.5 × 47e-6 × 100 = 0.00235 J = 2350 μJ = 2.35 mJ
3
Flux linkage: λ = 47e-6 × 10 = 0.47 mWb. Core must handle at least 0.47 mWb without saturating.
Result: E = 2350 μJ (2.35 mJ), λ = 0.47 mWb
Try this example →Example 3 - Low-Current Filter Inductor (220 μH at 2 A)
L = 220 μH, Peak Current I = 2 A
1
Convert: L = 220 × 10-6 H = 0.00022 H
2
Energy: E = 0.5 × 0.00022 × 4 = 0.00044 J = 440 μJ
3
Flux linkage: λ = 0.00022 × 2 = 0.44 mWb. Despite the higher inductance, the lower current stores less energy than the 47 μH at 10 A example.
Result: E = 440 μJ, λ = 0.44 mWb
Try this example →Example 4 - Find Current for 500 μJ in a 50 μH Inductor (Find Current Mode)
Target E = 500 μJ, L = 50 μH
1
Use the reverse formula: I = √(2E / L) = √(2 × 500e-6 / 50e-6)
2
Simplify: I = √(1000e-6 / 50e-6) = √(20) = 4.472 A
3
Flux linkage: λ = 50e-6 × 4.472 = 0.2236 mWb. Select a core with this flux capacity and a saturation current above 4.5 A.
Result: Required peak current = 4.4721 A, λ = 0.2236 mWb
Try this example →❓ Frequently Asked Questions
What is the formula for energy stored in an inductor?
The energy stored in an inductor is E = 0.5 times L times I squared, where E is in joules, L is in henries, and I is in amperes. For a 100 uH inductor at 5 A: E = 0.5 times 100e-6 times 25 = 1250 uJ = 1.25 mJ. This is the same form as kinetic energy (0.5 m v squared) but for the magnetic field instead of a moving mass.
How does increasing inductance affect the energy stored?
Energy scales linearly with inductance at constant current. Doubling L doubles E. However, energy scales with the square of current, so doubling current at the same L quadruples E. In practice, SMPS designers reduce L (and thus stored energy per cycle) at higher switching frequency to shrink core size, relying on faster cycling rather than more energy per cycle to deliver the same output power.
What is peak flux linkage and why does it matter?
Peak flux linkage is lambda = L times I, measured in webers (Wb) or milliwebers. It equals N times phi, where N is the number of turns and phi is the magnetic flux through the core. If this value exceeds the core's saturation flux (B_sat times A_core), the inductance drops sharply and current rises uncontrolled. Always compare the flux linkage output against the core's rated lambda or compute B = lambda divided by (N times A_core) and check against the material's B_sat (typically 300 to 400 mT for MnZn ferrite).
What happens when an inductor current is suddenly interrupted?
When current is suddenly stopped, the stored energy (0.5 L I squared) must be dissipated instantly. The inductor generates a large voltage spike (V = L times dI/dt) in an attempt to maintain current flow. This spike can easily reach hundreds of volts even for small inductances at moderate currents, destroying unprotected switching transistors. Flyback diodes, snubber networks, or TVS clamps are mandatory in any circuit where inductor current can be interrupted.
How do I choose an inductor for a buck converter?
Calculate the minimum inductance using L_min = (Vin minus Vout) times Vout divided by (Vin times ripple_current times Fs). Then use this calculator to find the peak stored energy at I_avg plus half the ripple current. Select an inductor with a saturation current above this peak, a DCR that limits conduction losses to an acceptable level, and a core loss specification that suits your switching frequency. Always leave a 20 to 30 percent margin on the saturation current rating to account for temperature and tolerance variation.
What is the saturation current rating of an inductor?
Saturation current (Isat) is the peak current at which the inductance drops by a defined percentage from its nominal value, typically 20 or 30 percent depending on the datasheet convention. Operating above Isat causes the core to saturate: inductance collapses, current rises rapidly, and the switching transistor can be destroyed. Always confirm that the peak current computed by this calculator falls below the inductor's Isat rating with margin to spare.
How does switching frequency affect inductor energy requirements?
At higher switching frequency, less inductance is needed for the same output ripple current. A smaller inductance at the same peak current stores less energy per cycle (E = 0.5 L I squared with smaller L). This allows a physically smaller core. The trade-off is increased switching losses in transistors and diodes at higher frequency. Common SMPS design ranges are 100 to 300 kHz for ferrite-core inductors, balancing core size against switching losses.
Can I use millihenry inductors with this calculator?
Yes. The calculator uses microhenries (uH). To enter millihenry values, multiply by 1000: 2.2 mH = 2200 uH, 10 mH = 10000 uH. Millihenry inductors appear in AC power line filters, motor drive output filters, and audio crossover networks. They typically operate at much lower switching or signal frequencies than microhenry SMPS inductors and handle proportionally lower peak currents.
What is the difference between inductor energy storage and capacitor energy storage?
An inductor stores energy in a magnetic field (E = 0.5 L I squared, driven by current). A capacitor stores energy in an electric field (E = 0.5 C V squared, driven by voltage). Inductors resist changes in current; capacitors resist changes in voltage. In switching power supplies, the two are complementary: the inductor limits current ripple and stores energy between switching cycles while the output capacitor maintains a stable output voltage. In resonant converters, they exchange energy back and forth at the resonant frequency.
How do I find the current required to store a specific energy in an inductor?
Use the Find Current mode on this calculator, or rearrange the formula: I = square root of (2E divided by L). For a target of 500 uJ in a 50 uH inductor: I = sqrt(2 times 500e-6 / 50e-6) = sqrt(20) = 4.47 A. This is useful when you have a fixed inductor and need to know what current level is needed to store a given amount of energy, for example in a pulsed power or wireless charging design.
What is the energy stored in a 10 mH inductor at 1 A?
E = 0.5 times 10e-3 times 1 squared = 0.005 J = 5 mJ = 5000 uJ. Enter 10000 uH and 1 A in the Find Energy mode to verify. A 10 mH inductor is relatively large; it would typically appear in a power line filter or audio circuit, not an SMPS, and would carry much less than 1 A in most applications.
Does winding resistance affect the energy stored in an inductor?
Winding resistance (DCR) dissipates energy as heat but does not directly change the stored magnetic energy formula E = 0.5 L I squared. However, DCR causes a voltage drop that reduces the effective current (and therefore the stored energy) under real operating conditions. Higher DCR also increases I squared R losses, reducing converter efficiency. Choose inductors with the lowest DCR consistent with your size and cost constraints, and use the actual peak current (not the rated current) in this calculator.