The fundamental current formula for an inductor defines how current changes over time when a voltage is applied. Unlike a resistor, where current is instantly determined by Ohm’s Law, an inductor resists changes in current. The direct answer for calculating inductor current over time is the integral form: i(t) = (1/L) ∫ v(t) dt + I₀. For practical switching power supplies where voltage is constant over short intervals, this simplifies to the algebraic delta form: ΔI = (V × Δt) / L.

Below, we break down the exact derivation, track units through two real-world solved problems, and provide a concrete decision path for selecting a physical inductor part number for a switch-mode power supply (SMPS).

The Core Current Formula for Inductors (and Its Rearranged Forms)

The relationship between voltage and current in an inductor is governed by Faraday’s Law of Induction. The differential form states that the voltage across an inductor is proportional to the rate of change of current through it. By rearranging and integrating, we get the formula for current.

Bench Tip: In DC steady-state (after a long time), the rate of change of current (di/dt) is zero. Therefore, the voltage across an ideal inductor is zero, making it act as a short circuit. Current is then limited only by the parasitic DC resistance (DCR) of the wire.

Symbol Definition Table

SymbolParameterStandard SI UnitCommon Practical Unit
i(t)Instantaneous current at time tAmperes (A)Milliamps (mA)
LInductanceHenries (H)Microhenries (µH), Millihenries (mH)
v(t)Instantaneous voltage across the inductorVolts (V)Volts (V)
tTimeSeconds (s)Microseconds (µs), Nanoseconds (ns)
I₀Initial current at t = 0Amperes (A)Amperes (A)
ΔIChange in current (ripple)Amperes (A)Amperes (A)

Rearranged Forms for Constant Voltage Intervals

When dealing with switching circuits (like PWM or buck converters), the voltage across the inductor is effectively constant during the ON and OFF states. We drop the calculus and use algebra:

  • Solve for Current Change (ΔI): ΔI = (V × Δt) / L
  • Solve for Inductance (L): L = (V × Δt) / ΔI
  • Solve for Time (Δt): Δt = (L × ΔI) / V
  • Solve for Voltage (V): V = (L × ΔI) / Δt

Operating Assumptions, Realistic Magnitudes, and Unit Traps

When This Formula Applies (and When It Breaks)

The linear formulas above assume the inductor is operating in its linear magnetic region. This means the current must remain below the component's Saturation Current (Isat). If current exceeds Isat, the magnetic core saturates, permeability drops to near that of air, inductance (L) collapses toward zero, and the current spikes violently (ΔI approaches infinity), often destroying the driving MOSFET.

Furthermore, these formulas assume an ideal inductor. In reality, you must account for parasitic elements like DC Resistance (DCR) and Equivalent Series Capacitance (ESC), which alter the waveform at high frequencies or high DC currents.

Realistic Answer Magnitudes

  • Signal/RF Inductors: 1 nH to 10 µH. Currents are typically 10 mA to 500 mA. Used for filtering and tuning.
  • Power Inductors (SMPS): 1 µH to 100 µH. Currents range from 1 A to 50 A. Used for energy storage in buck/boost converters.
  • Line Frequency Chokes: 1 mH to 100 mH. Currents from 1 A to 20 A. Used in 50/60 Hz AC filtering.

Unit Mistakes That Break the Math

The most common way to fry a board on the bench is a unit conversion error in the ΔI = (V × Δt) / L formula.

The Micro-Cancellation Trap: If your time (Δt) is in microseconds (10-6) and your inductance (L) is in microhenries (10-6), the 10-6 terms cancel out perfectly. You can just plug in the raw numbers (e.g., 5V × 2µs / 10µH = 1A). However, if your time is in nanoseconds (10-9) and inductance is in microhenries (10-6), you will be off by a factor of 1,000. Always convert time to seconds and inductance to Henries first if you are unsure.

Worked Example 1: DC Transient Step Response

Scenario: You apply a constant 5V DC step from a microcontroller GPIO (buffered by a driver) across a 10 µH inductor. The initial current is 0 A. How much current flows after 2 microseconds?

  1. Identify knowns: V = 5V, L = 10 µH (10 × 10-6 H), Δt = 2 µs (2 × 10-6 s), I₀ = 0 A.
  2. Select formula: ΔI = (V × Δt) / L
  3. Substitute values with units:
    ΔI = (5 V × 2 × 10-6 s) / (10 × 10-6 H)
  4. Cancel the micro (10-6) prefixes:
    ΔI = (5 × 2) / 10
  5. Calculate:
    ΔI = 10 / 10 = 1 A
  6. Add initial current:
    i(t) = ΔI + I₀ = 1 A + 0 A = 1 A

Result: After 2 µs, the current ramps linearly to exactly 1 Ampere. If you leave the 5V applied longer, the current will continue to rise until limited by the circuit's parasitic resistance or the power supply's current limit.

Worked Example 2: Buck Converter Inductor Ripple Current

Scenario: You are designing a 12V-to-5V buck converter switching at 500 kHz. You selected a 4.7 µH inductor. What is the peak-to-peak inductor ripple current (ΔIL)?

Reference: For buck converter topology math, the Texas Instruments SLVA374 application note provides the foundational derivations used by power engineers globally.

  1. Calculate Duty Cycle (D):
    D = V_out / V_in = 5V / 12V = 0.4167
  2. Calculate Switching Period (T) and ON-time (Δt):
    T = 1 / f_sw = 1 / 500,000 Hz = 2 µs (2 × 10-6 s)
    Δt (ON-time) = T × D = 2 µs × 0.4167 = 0.833 µs
  3. Determine Voltage across inductor during ON-time:
    V_L = V_in - V_out = 12V - 5V = 7V
  4. Apply the rearranged current formula:
    ΔI_L = (V_L × Δt) / L
  5. Substitute and track units:
    ΔI_L = (7 V × 0.833 × 10-6 s) / (4.7 × 10-6 H)
  6. Calculate:
    ΔI_L = 5.831 / 4.7 = 1.24 A (peak-to-peak ripple)

Result: The inductor current will sawtooth up and down by 1.24 A around your DC load current. If your load draws 3A DC, the current swings between 2.38 A and 3.62 A.

Decision Path: Sizing an Inductor for a 5V/3A Buck Converter

Knowing the formula is only half the battle; picking the physical component is where designs fail. Use this decision tree to lock in your inductor specifications. We assume a 12V input, 5V output at 3A max load, and a 500 kHz switching frequency.

Design ParameterIf / ConditionThen / ActionCalculated Value
1. Target Ripple Current Standard SMPS design practice Set ΔI_L to 30% of max DC load current (I_out) 3A × 0.30 = 0.9 A
2. Required Inductance (L) Using V_L = 7V, Δt = 0.833 µs, ΔI = 0.9A Calculate L = (V_L × Δt) / ΔI_L (7 × 0.833) / 0.9 = 6.48 µH
3. Standard Value Selection 6.48 µH is not a standard E-series value Round UP to the next standard value to keep ripple below 30% Select 6.8 µH
4. Saturation Current (I_sat) Peak current = I_out + (ΔI_L / 2) I_sat MUST be > Peak Current to prevent core saturation and MOSFET failure 3A + (0.9A / 2) = 3.45A. Require I_sat > 3.5 A
5. RMS Current (I_rms) Thermal heating limit of the wire I_rms rating must be ≥ max DC load current Require I_rms ≥ 3.0 A
6. Shielding Operating in noise-sensitive environments or high density Choose a magnetically shielded inductor to prevent EMI coupling Require Shielded topology

The Concrete Pick

Based on the decision path above, we need a 6.8 µH shielded SMD power inductor with an Isat > 3.5A and Irms > 3.0A.

Default Recommendation: Würth Elektronik 7447742068 (WE-PD series).

  • Inductance: 6.8 µH
  • Isat: 4.1 A (Safely above our 3.5A peak requirement)
  • Irms: 3.2 A (Handles the 3A continuous load without overheating)
  • DCR: 48 mΩ (Low enough to minimize I²R conduction losses)
  • Shielding: Yes (Molded ferrite core)

Do not substitute this with an unshielded drum-core inductor of the same inductance unless your layout has strict EMI clearance, as the radiating magnetic field will couple noise into nearby high-impedance feedback traces.