If you are designing a switch-mode power supply, a filter, or a snubber network, you cannot rely on rule-of-thumb component selection. You need the foundational physics. The core equation for inductor behavior dictates how energy is stored in a magnetic field and how voltage is generated when current changes. Getting this math right is the difference between a clean 5V rail and a MOSFET that violently avalanches on your bench.

The Core Equation for Inductor Voltage and Inductance

The fundamental relationship governing an inductor in a time-domain circuit is Faraday’s law of induction, expressed in circuit terms as:

V = L × (di / dt)

This equation states that the voltage across an inductor is directly proportional to its inductance and the rate of change of current flowing through it. Here is the strict definition of every symbol:

Symbol Definition and SI Base Units
Symbol Parameter SI Base Unit Engineering Prefix Commonly Used
V Voltage across the inductor Volts (V) Volts (V) or millivolts (mV)
L Inductance Henries (H) Microhenries (μH) or millihenries (mH)
di Change in current Amperes (A) Amperes (A) or milliamperes (mA)
dt Change in time Seconds (s) Microseconds (μs) or nanoseconds (ns)

Assumptions and Realistic Magnitudes

This equation applies under the assumption of a linear core. It assumes the inductor has not reached magnetic saturation. Once the core saturates, the permeability (μ) drops drastically, L effectively collapses toward zero, and di/dt spikes uncontrollably, limited only by the parasitic DC resistance (DCR) of the wire.

It also ignores parasitic elements: winding capacitance (which causes high-frequency ringing) and core losses. In practical 2026 power electronics, a realistic inductance magnitude for a buck converter is 1 μH to 47 μH. For line-frequency filtering or audio crossovers, you will see 1 mH to 100 mH. A 1 Henry inductor is physically massive and largely restricted to low-frequency lab demonstrations or heavy industrial chokes.

Rearranged Forms for Circuit Design

You will rarely use the base formula exactly as written. Depending on your design goal, you need to isolate specific variables. Here are the rearranged forms you will use on the bench:

  • Solving for Inductance (Sizing L for a target ripple):
    L = (V × dt) / di
    Use case: You know your switching frequency and input/output voltages, and you want to limit the ripple current to a specific percentage of your load.
  • Solving for Ripple Current (Calculating di):
    di = (V × dt) / L
    Use case: You have a fixed inductor on hand and need to verify if the peak-to-peak ripple current will exceed the inductor's saturation rating or cause excessive output voltage ripple.
  • Solving for Time (Calculating on-time or dt):
    dt = (L × di) / V
    Use case: Designing a constant-on-time controller or calculating the exact pulse width needed to ramp a solenoid current to a specific threshold.
  • Solving for Voltage (Calculating flyback spikes):
    V = L × (di / dt)
    Use case: Determining the snubber requirements when opening a switch in series with an inductive load (like a relay coil or motor winding).

Worked Examples with Strict Unit Tracking

The most common point of failure in these calculations is mixing engineering prefixes with SI base units. The golden rule: convert everything to base SI units (Volts, Amperes, Seconds, Henries) before calculating, then convert the result back to engineering prefixes.

Problem 1: Sizing a Buck Converter Inductor

Scenario: You are designing a 12V to 5V buck converter. The switching frequency is 500 kHz. Your maximum load current is 3A, and you want to limit the inductor ripple current (di) to 30% of the load current. What inductance do you need?

  1. Identify Knowns & Convert to SI:
    • VL (Voltage across inductor during on-time) = Vin - Vout = 12V - 5V = 7 V
    • Duty Cycle (D) = Vout / Vin = 5 / 12 = 0.4167
    • Switching period (T) = 1 / 500,000 Hz = 2 × 10-6 s
    • On-time (dt) = D × T = 0.4167 × 2 × 10-6 = 8.33 × 10-7 s
    • Target ripple (di) = 30% of 3A = 0.9 A
  2. Apply the Rearranged Formula:
    • L = (V × dt) / di
    • L = (7 × 8.33 × 10-7) / 0.9
    • L = 5.831 × 10-6 / 0.9
    • L = 6.48 × 10-6 H
  3. Convert to Engineering Prefix:
    • 6.48 × 10-6 H = 6.48 μH

Result: You would select the next standard value up, typically 6.8 μH or 10 μH, to guarantee the ripple stays below your 30% target.

Problem 2: Calculating a Relay Flyback Spike

Scenario: You are switching a 12V automotive relay coil using a MOSFET. The coil has an inductance of 47 mH and draws 150 mA at steady state. The MOSFET turns off in 20 nanoseconds. What is the theoretical voltage spike at the MOSFET drain if no flyback diode is present?

  1. Identify Knowns & Convert to SI:
    • L = 47 mH = 0.047 H
    • di = 150 mA to 0 A = 0.15 A
    • dt = 20 ns = 20 × 10-9 s
  2. Apply the Base Formula:
    • V = L × (di / dt)
    • V = 0.047 × (0.15 / 20 × 10-9)
    • V = 0.047 × 7,500,000
    • V = 352,500 V

Result: 352.5 kV. In reality, the voltage will not reach this level. It will rise until it exceeds the MOSFET's breakdown voltage (e.g., 60V), causing the MOSFET to avalanche and dissipate the energy as heat, likely destroying it. Alternatively, the energy will arc across the physical switch contacts. This massive theoretical number is exactly why a freewheeling diode is non-negotiable.

⚠ Callout: The 'Micro' Trap

The most frequent unit mistake that breaks inductor math is failing to convert microhenries (μH) to Henries. If you type 10 into your calculator instead of 10e-6 for a 10 μH inductor, your calculated ripple current will be off by a factor of one million. Always write out the 10^-6 explicitly in your scratchpad math.

Unit Mistakes That Break the Math

Beyond the microhenry trap, here are the specific unit errors that lead to blown components or failed prototypes:

The Mistake Why It Breaks The Fix
Using frequency (Hz) instead of time (s) for dt Hz is 1/s. Plugging 500,000 directly into the dt slot inverts the math entirely. Always calculate dt = 1 / fsw (and multiply by duty cycle if finding on-time).
Confusing Peak-to-Peak Ripple with Peak Current di in the formula is the peak-to-peak ripple. The inductor's saturation rating must handle the peak current (Iload + di/2). Calculate di first, then add half of it to your DC load current to check against Isat.
Ignoring the Voltage Drop Across the Switch/Diode Assuming VL during the off-time is exactly Vout ignores the diode forward voltage (Vf). For the off-time calculation, use VL = Vout + Vf (e.g., 5V + 0.4V for a Schottky).

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

Let's terminate the theory into a concrete purchasing decision. You need an inductor for a 12V-to-5V, 3A continuous buck converter switching at 500 kHz. Follow this decision tree to select the exact part.

Inductor Selection Decision Tree

  1. Calculate Target Inductance: As proven in Problem 1, the math demands 6.48 μH for a 30% ripple target.
  2. Select Standard Value: Round up to the nearest E12 series value to ensure ripple stays below 30%.
    Decision: Pick 10 μH. (This lowers ripple to ~19%, improving efficiency and reducing output capacitor stress).
  3. Calculate Peak Current for Saturation Check:
    • di = (7V × 833ns) / 10μH = 0.58A.
    • Ipeak = Iload + (di / 2) = 3A + (0.58 / 2) = 3.29A.
  4. Define Isat Requirement: The inductor's saturation current (Isat) must be greater than 3.29A. To account for transient load steps and temperature derating (ferrite cores lose permeability as they approach 100°C), apply a 20% safety margin.
    Decision: Required Isat ≥ 4.0A.
  5. Define DCR Requirement: To keep I²R losses under 100mW at full load (3A), DCR must be ≤ 11 mΩ.
    Decision: Target DCR ≤ 15 mΩ (allowing slight margin as 100mW is easily managed thermally).
  6. Execute Part Selection: Query a manufacturer database (like Coilcraft's Inductor Finder or Bourns) filtering for: 10μH, Isat > 4A, DCR < 15mΩ, shielded construction (to prevent EMI coupling into nearby feedback traces).

The Concrete Pick

Based on the decision path above, the optimal part is the Bourns SRP1265A-100M.

  • Inductance: 10 μH ± 20%
  • Isat: 11 A (Massive margin over our 4A requirement, ensuring zero saturation during hard short-circuit transients).
  • Irms: 9 A (Handles the 3A DC load with minimal temperature rise).
  • DCR: 18.5 mΩ (Typical). At 3A, this dissipates P = 3² × 0.0185 = 166 mW, which is thermally negligible in a 12x12mm footprint.
  • Construction: Shielded ferrite, AEC-Q200 compliant (automotive grade, meaning rigorous vibration and thermal shock testing).

By anchoring your design in the raw physics of V = L(di/dt), tracking your SI units rigorously, and applying a safety margin to the saturation current, you eliminate the guesswork from power supply design. The math dictates the component, not the other way around.