Mastering Inductor Equations: Worked Examples and Kickback Math

Inductors are the bouncers of the electronics world: they resist any sudden change in current. Whether you are designing a high-frequency buck converter or just switching a 24V solenoid valve with a microcontroller, predicting exactly how an inductor will behave requires moving beyond abstract theory and getting comfortable with the math. In this guide, we will break down the fundamental inductor equations, track the units through real-world solved problems, and look at a bench failure that proves why this math matters.

The Core Inductor Equations: Symbols, Units, and Rearrangements

There are two primary equations that govern inductor behavior in DC and transient circuits. The first defines the voltage induced by a changing current, and the second defines the energy stored in the magnetic field.

Dynamic Voltage Equation: V = L × (Δi / Δt)
Stored Energy Equation: E = ½ × L × I²

Rearranged Forms

On the bench, you rarely solve for V. You are usually sizing a component or calculating a time constant. Here are the practical rearrangements:

• Solving for Inductance: L = V × (Δt / Δi)
• Solving for Time: Δt = L × (Δi / V)
• Solving for Current Ripple: Δi = (V × Δt) / L
• Solving for Steady Current (from Energy): I = √(2E / L)

When the Formula Applies (and Its Assumptions)

These equations assume an ideal inductor. In reality, every physical inductor has parasitic elements: DC resistance (DCR) in the copper windings, parallel parasitic capacitance between the coils, and a core material with a non-linear permeability (µ). The equation V = L × (Δi / Δt) holds true only as long as the core is not saturated. Once the magnetic core saturates, its permeability drops toward that of air, L collapses to a tiny fraction of its nominal value, and Δi spikes uncontrollably, often destroying the driving switch.

Realistic Magnitudes and the Unit Traps That Break Your Math

Before plugging numbers into your calculator, you must understand the realistic magnitude of inductance in modern circuits and the prefix traps that cause 90% of calculation errors.

• Nanohenries (nH, 10⁻⁹ H): Used in RF matching networks, VHF antennas, and high-speed digital decoupling. (e.g., 4.7 nH).
• Microhenries (µH, 10⁻⁶ H): The standard for switch-mode power supplies (SMPS) like buck, boost, and flyback converters operating between 100 kHz and 3 MHz. (e.g., 4.7 µH to 47 µH).
• Millihenries (mH, 10⁻³ H): Found in 50/60Hz line filters, audio crossover networks, and heavy relay coils. (e.g., 10 mH to 500 mH).
• Henries (H): Extremely rare in physical form outside of superconducting magnets or massive laboratory variacs. If your math results in a 2H inductor for a PCB, you have made a unit error.

Worked Problem 1: Sizing a Buck Converter Inductor

Scenario: You are designing a 12V to 3.3V buck converter running at a switching frequency of 1 MHz. You want to limit the inductor ripple current (Δi) to 0.6 A. What inductance value do you need?

1. Identify the ON-state voltage across the inductor (V):
   When the high-side MOSFET is ON, the voltage across the inductor is V_in - V_out.
   V = 12V - 3.3V = 8.7 V
2. Calculate the ON-time (Δt):
   First, find the duty cycle (D) = V_out / V_in = 3.3 / 12 = 0.275.
   The switching period (T) = 1 / f_sw = 1 / 1,000,000 = 1 µs (1 × 10⁻⁶ s).
   Δt = D × T = 0.275 × (1 × 10⁻⁶ s) = 2.75 × 10⁻⁷ s (or 275 ns).
3. Rearrange the formula to solve for L:
   L = V × (Δt / Δi)
4. Plug in the base SI units and solve:
   L = 8.7 × (2.75 × 10⁻⁷ / 0.6)
   L = 8.7 × (4.583 × 10⁻⁷)
   L = 3.987 × 10⁻⁶ H
5. Convert to practical units and select a standard part:
   3.987 × 10⁻⁶ H = 3.98 µH.
   Bench Decision: You would select a standard 4.7 µH shielded ferrite inductor (like the Wurth Elektronik 7447742047) rated for at least 2A saturation current to handle the DC load plus the ripple without saturating.

Worked Problem 2: Calculating Inductive Kickback Voltage

Scenario: You are switching a 24V automotive relay coil with a BJT transistor. The coil has an inductance of 250 mH and draws a steady-state current of 150 mA. If the transistor turns off in 200 ns and there is no flyback diode, what is the theoretical voltage spike at the transistor's collector?

1. Convert all values to base SI units:
   L = 250 mH = 0.250 H
   Δi = 150 mA to 0 A = 0.150 A
   Δt = 200 ns = 200 × 10⁻⁹ s
2. Calculate the stored energy (optional but good for context):
   E = ½ × L × I² = 0.5 × 0.250 × (0.150)² = 0.00281 J (2.81 mJ).
   This is enough energy to pit the contacts of a mechanical switch or punch through a silicon junction.
3. Apply the dynamic voltage equation:
   V = L × (Δi / Δt)
   V = 0.250 × (0.150 / 200 × 10⁻⁹)
4. Solve for the induced voltage:
   V = 0.250 × 750,000
   V = 187,500 V

Reality Check: In the real world, you will not measure 187 kV. Parasitic capacitance across the coil windings and the transistor's internal capacitance will form an LC tank, and the voltage will clamp at the transistor's avalanche breakdown voltage (likely around 60V to 100V for a standard BJT), instantly destroying the silicon lattice in the process. For a deeper dive into how magnetic fields store and release this energy, refer to the Georgia State University HyperPhysics database on inductor energy.

Bench War Story: When Δt Goes to Zero and the MOSFET Pays the Price

Equations on a whiteboard are clean; components on a workbench are not. Here is a real-world scenario that demonstrates what happens when you ignore the Δt variable in the inductor equation.

The Setup

A junior engineer was tasked with driving a 24V pneumatic solenoid valve using an ESP32 development board. The solenoid had a measured inductance of 50 mH and drew 1.0 A at steady state. To interface the 3.3V GPIO of the ESP32 with the 24V solenoid, they used an IRLZ44N N-channel logic-level MOSFET. The gate was driven directly from the ESP32 pin, and the solenoid was placed between the 24V supply and the MOSFET drain. Crucially, no flyback diode was installed across the solenoid coil.

The Numbers

When the ESP32 pulled the GPIO low, the MOSFET gate discharged rapidly. The IRLZ44N has a very fast turn-off time, roughly 50 ns under these specific gate-drive conditions. Let us look at the math for the inductive kickback:

• L = 0.050 H
• Δi = 1.0 A
• Δt = 50 × 10⁻⁹ s
• V_kick = 0.050 × (1.0 / 50 × 10⁻⁹) = 1,000,000 V

The Outcome

The voltage at the drain of the MOSFET skyrocketed the microsecond the channel pinched off. The IRLZ44N is rated for a maximum Drain-to-Source Voltage (V_DSS) of 55V. The drain voltage hit 55V almost instantly, forcing the MOSFET into avalanche breakdown. While some modern MOSFETs are rated for repetitive avalanche, the IRLZ44N is not designed to dissipate the 25 mJ of energy (E = ½ × 0.050 × 1²) repeatedly without a massive heatsink. The silicon overheated in milliseconds, the die cracked internally, and the drain shorted directly to the gate. The 24V supply then fed directly back through the failed gate oxide into the ESP32's 3.3V GPIO pin, instantly vaporizing the microcontroller's internal trace and bricking the board.

What Went Wrong (and The Fix)

The failure was a direct violation of the physics described by V = L × (Δi / Δt). By allowing Δt to be extremely small (50 ns) without providing an alternate path for the current, the induced voltage exceeded the component's dielectric limits.

The Fix: A 1N5819 Schottky diode was added in reverse bias across the solenoid coil (cathode to 24V, anode to the MOSFET drain). When the MOSFET turned off, the inductor's collapsing magnetic field forward-biased the diode. This clamped the drain voltage to roughly 24.4V (24V supply + 0.4V Schottky forward drop). The Δt for the current decay was no longer 50 ns; it was dictated by the L/R time constant of the coil and the diode loop, stretching the decay out to several milliseconds and safely dissipating the energy as heat in the coil's DCR and the diode. For more on protecting solid-state switches from inductive loads, All About Circuits provides excellent practical topologies for snubber networks and flyback routing.

Mastering inductor equations is not just about passing an exam; it is about ensuring your components survive the transition from ON to OFF. Always track your units, respect the core saturation limits, and never let an inductor's current change without a defined path.