The instantaneous power p absorbed or delivered by an ideal inductor is calculated using the formula p = L · i · (di/dt). While beginners often confuse this with the formula for stored magnetic energy (W = ½Li²), power represents the rate at which energy is transferred into or out of the magnetic field at any exact microsecond. Understanding this distinction—and tracking your units rigorously—is the difference between a functional switch-mode power supply and a blown MOSFET.
The Instantaneous Inductor Power Formula Defined
To derive the formula, we start with the fundamental definition of electrical power: p(t) = v(t) · i(t). According to Faraday’s law of induction, the voltage across an ideal inductor is proportional to the rate of change of current through it: v(t) = L · (di/dt). Substituting this voltage expression into the power equation yields the core inductor power formula.
p(t) = L · i(t) · (di/dt)
| Symbol | Parameter | Standard Unit | Unit Abbreviation |
|---|---|---|---|
| p(t) | Instantaneous Power | Watts | W |
| L | Inductance | Henrys | H |
| i(t) | Instantaneous Current | Amperes | A |
| di/dt | Rate of Current Change | Amperes per second | A/s |
Assumptions and Limitations
This formula applies strictly to an ideal inductor. It assumes zero DC winding resistance (DCR), no parasitic parallel capacitance, and no magnetic core losses (hysteresis or eddy currents). In a real-world component, the total instantaneous power dissipated as heat is the sum of the reactive power calculated above and the resistive copper loss (I²R). For high-frequency switching circuits, core losses also become significant. You can model these real-world losses using tools like the Coilcraft Inductor Analyzer to pull exact DCR and core loss data for specific part numbers.
Rearranged Forms and Variable Isolation
On the bench or in simulation, you rarely have all variables handed to you. Here are the algebraic rearrangements of the instantaneous power formula to solve for any single missing variable, assuming the other three are known at a specific snapshot in time.
- Solving for Inductance (L):
L = p / (i · (di/dt))
Use case: Sizing a choke when you know the maximum allowable power spike and your current slew rate. - Solving for Slew Rate (di/dt):
di/dt = p / (L · i)
Use case: Determining how fast your gate driver must switch a MOSFET to keep inductive kickback power within safe limits. - Solving for Current (i):
i = p / (L · (di/dt))
Use case: Finding the instantaneous current threshold that triggers a specific power dissipation event. (Note: This is an algebraic snapshot; finding current over time requires integrating the power function).
Worked Examples with Unit Tracking
The most common point of failure in these calculations is dropping a decimal place during unit conversion. Below are two real-world scenarios with every intermediate step and unit explicitly tracked.
Problem 1: The Turn-On Transient
Scenario: A 47 mH smoothing inductor in a DC-DC converter ramps its current linearly from 0 A to 3 A over a 5 ms window. What is the instantaneous power absorbed by the inductor at the exact moment the current reaches 1.5 A?
- Convert to base SI units:
L = 47 mH = 47 × 10⁻³ H = 0.047 H
Δt = 5 ms = 5 × 10⁻³ s = 0.005 s - Calculate the slew rate (di/dt):
di = 3 A - 0 A = 3 A
di/dt = 3 A / 0.005 s = 600 A/s - Identify the instantaneous current (i):
i = 1.5 A - Apply the formula:
p = L · i · (di/dt)
p = (0.047 H) · (1.5 A) · (600 A/s)
p = 0.0705 · 600 W
p = 42.3 W
Answer: The inductor absorbs 42.3 Watts of instantaneous power at that specific moment.
Problem 2: The Flyback Turn-Off Spike
Scenario: A 100 µH relay coil carrying a steady 5 A is suddenly interrupted by a mechanical switch. The current collapses to 0 A in 2 µs. What is the peak instantaneous power at the very beginning of the interruption?
- Convert to base SI units:
L = 100 µH = 100 × 10⁻⁶ H
Δt = 2 µs = 2 × 10⁻⁶ s - Calculate the slew rate (di/dt):
di = 0 A - 5 A = -5 A(Current is decreasing)
di/dt = -5 A / (2 × 10⁻⁶ s) = -2,500,000 A/s - Identify the instantaneous current (i) at t=0:
i = 5 A - Apply the formula:
p = (100 × 10⁻⁶ H) · (5 A) · (-2,500,000 A/s)
p = (500 × 10⁻⁶) · (-2,500,000) W
p = -1,250 W
Answer: The inductor delivers -1,250 Watts (1.25 kW) back into the circuit. This massive negative power spike is exactly why flyback diodes or snubber networks are mandatory to prevent arcing and semiconductor destruction.
Real-World Magnitudes and Common Unit Traps
What a Realistic Answer Looks Like
In steady-state DC, an ideal inductor's power is exactly 0 W because di/dt = 0. During slow transients (like Problem 1), you will see realistic magnitudes in the tens of Watts. During fast switching events or fault conditions (like Problem 2), instantaneous power can easily spike into the kiloWatts for microseconds. If your steady-state calculation yields 500 W for a small signal inductor, you have made a math error; if your transient calculation yields 0.001 W for a collapsing relay coil, you have missed a microsecond conversion.
Unit Mistakes That Break the Math
The most frequent bench mistake is failing to convert prefixes to base SI units before multiplying.
- The milliHenry Trap: Plugging '47' into L instead of '0.047' inflates your power answer by a factor of 1,000.
- The Microsecond Trap: Dividing by '2' instead of '0.000002' for dt shrinks your slew rate by a factor of 1,000,000, making a lethal flyback spike look harmless on paper.
- The Mixed Prefix Trap: Multiplying milliHenrys directly by Amperes and dividing by milliseconds. While the 'milli' prefixes might intuitively seem to cancel out, (10⁻³) / (10⁻³) only works if they are in the numerator and denominator correctly. Always convert everything to Henrys, Amps, and Seconds first.
Frequently Asked Questions
Does an ideal inductor consume real power in a steady DC circuit?
No. In a steady DC circuit, the current is constant, meaning the rate of change of current (di/dt) is zero. When you multiply by zero, the instantaneous power p becomes 0 W. The inductor maintains a static magnetic field but does not continuously draw or dissipate power. (Note: A real inductor will still dissipate a small amount of heat due to the DC resistance of its copper windings, calculated via P = I²R).
What is the difference between the inductor power formula and the inductor energy formula?
Power (p = L · i · di/dt) is the rate at which energy is transferred, measured in Watts (Joules per second). Energy (W = ½Li²) is the total capacity stored in the magnetic field at a given current, measured in Joules. Think of energy as the amount of water in a tank, and power as the rate at which water is currently flowing through the pipe to fill or drain that tank.
Why does my calculated inductor power show as a negative number?
A negative power value indicates that the inductor is acting as a power source rather than a load. When the current through an inductor decreases (di/dt is negative), the collapsing magnetic field induces a voltage that pushes current back into the circuit. This is the fundamental operating principle behind boost converters and flyback transformers, where energy stored during the 'on' time is delivered to the load during the 'off' time.
How do I calculate inductor power dissipation for a real-world component?
For a physical component, the ideal reactive power calculated by our formula does not turn into heat. To find the actual thermal dissipation (which dictates if the part will overheat), you must calculate the copper loss and core loss. Copper loss is I_RMS² × R_DCR (where R_DCR is found on the manufacturer datasheet). Core loss depends on the switching frequency, flux density, and core material (e.g., ferrite vs. powdered iron), and is typically estimated using the manufacturer's core loss graphs or online calculators like those provided by Electronics Tutorials or specific vendor tools from TDK and Würth Elektronik.
