Mastering the Voltage Formula for an Inductor: Worked Examples and Real-World Spikes

The Core Voltage Formula for an Inductor

The fundamental voltage formula for an inductor defines the relationship between the voltage across its terminals and the rate of change of current flowing through it. Unlike a resistor, which opposes current magnitude, an inductor opposes changes in current. The direct answer to the governing equation is:

v(t) = L × (di / dt)

To use this equation correctly on the bench or in simulation, every variable must be mapped to its strict SI base unit. Below is the definitive symbol table for the formula.

Inductor Voltage Formula Symbol Definitions

Symbol	Parameter	SI Unit	Unit Abbreviation
v(t)	Instantaneous voltage across the inductor	Volts	V
L	Inductance of the component	Henries	H
di	Change in current (Delta I)	Amperes	A
dt	Change in time (Delta t)	Seconds	s

Assumptions and When the Formula Applies

This formula assumes an ideal inductor. In real-world applications, it applies accurately only when parasitic effects are negligible. Specifically, it assumes:

• Zero DC Resistance (DCR): The copper windings have no inherent resistive voltage drop (IR drop).
• Linear Core Material: The magnetic core has not reached saturation. Once a core saturates, permeability drops sharply, L is no longer constant, and the inductor behaves more like a short circuit with a small series resistance.
• No Parasitic Capacitance: At very high frequencies (e.g., VHF/UHF), inter-winding capacitance creates a self-resonant frequency (SRF), invalidating the pure inductive model.

Realistic Answer Magnitudes

What does a "normal" answer look like? It depends entirely on the application domain. In a low-power RF filter using a 10 nH inductor with a 5 mA current change over 2 ns, the calculated voltage is a benign 25 mV. However, in power electronics, magnitudes scale aggressively. A 47 μH inductor in a buck converter switching 2 A in 100 ns yields roughly 940 V. This is why switch-mode power supplies (SMPS) require careful snubber design and fast-recovery diodes to manage the di/dt induced spikes.

Rearranged Forms of the Inductor Equation

When debugging a circuit or selecting a component, you rarely solve for voltage directly. More often, you need to find the required inductance for a target ripple current, or determine the switching time limit. Here are the algebraically rearranged forms solving for each variable:

• Solving for Inductance: L = (v × dt) / di
  Use case: Sizing the output inductor for a buck converter to maintain a specific peak-to-peak ripple current.
• Solving for Current Change: di = (v × dt) / L
  Use case: Calculating the peak inductor current during the MOSFET on-time to ensure the core won't saturate.
• Solving for Time: dt = (L × di) / v
  Use case: Determining the minimum on-time required for a gate driver to ramp a relay coil to its pull-in current.

Worked Examples with Strict Unit Tracking

The most common point of failure in these calculations is unit mismatch. The golden rule: convert all prefixes (micro, milli, nano) to base SI units (Henries, Amperes, Seconds) before multiplying.

Problem 1: Buck Converter Inductor Sizing

Scenario: You are designing a 12V-to-5V buck converter using a TPS5430 IC. The switching node applies 7V across the inductor during the on-time. You need to limit the ripple current (di) to 0.8 A, and the controller's on-time (dt) is measured at 3.5 μs. What inductance (L) is required?

1. Identify the target variable: We need to solve for L.
2. Select the rearranged formula: L = (v × dt) / di
3. Convert to base SI units:
   • v = 7 V (already base)
   • dt = 3.5 μs = 3.5 × 10^-6 s
   • di = 0.8 A (already base)
4. Substitute and calculate:
   • L = (7 × 3.5 × 10^-6) / 0.8
   • L = (24.5 × 10^-6) / 0.8
   • L = 30.625 × 10^-6 H
5. Convert to standard engineering prefix: 30.625 μH. (Select the next standard E24 value, which is 33 μH, to ensure ripple stays slightly below the 0.8 A target).

Problem 2: Relay Coil Flyback Spike

Scenario: A 12V automotive relay coil has an inductance of 150 mH and draws a steady-state current of 120 mA. When the driving NPN transistor switches off, the current collapses from 120 mA to 0 A in 2 μs due to stray circuit capacitance. What is the magnitude of the inductive kickback voltage spike?

1. Identify the target variable: We need to solve for v.
2. Select the base formula: v = L × (di / dt)
3. Convert to base SI units:
   • L = 150 mH = 150 × 10^-3 H
   • di = 0 A - 0.120 A = -0.120 A
   • dt = 2 μs = 2 × 10^-6 s
4. Substitute and calculate:
   • v = (150 × 10^-3) × (-0.120 / 2 × 10^-6)
   • v = 0.150 × (-60,000)
   • v = -9,000 V
5. Interpret the result: The negative sign indicates the voltage polarity has reversed to maintain current flow. The magnitude is 9,000 V (9 kV). Without a flyback diode (like a 1N4007) clamping this spike to roughly 12.7V, this 9 kV transient will instantly avalanche and destroy the driving transistor. For deeper theory on inductive transients, refer to the All About Circuits chapter on inductors and calculus.

Common Unit Mistakes That Break the Calculation

When the math yields an answer that is off by a factor of 1,000 or 1,000,000, the culprit is almost always a unit conversion error. Watch for these specific traps:

• The "Micro" Trap: Confusing milliseconds (ms, 10^-3) with microseconds (μs, 10^-6). Oscilloscope timebases are often in μs, while datasheet switching times might be in ns (10^-9). Always write out the exponent explicitly.
• Treating di/dt as a Single Variable: Beginners sometimes look for a "di/dt" parameter on a datasheet. It is not a single component spec; it is a ratio derived from the circuit's operating conditions. You must calculate the change in current and divide it by the time interval.
• Ignoring the Sign of di: If current is decreasing, di is negative. If you drop the negative sign, you will calculate a positive voltage spike, which leads to incorrect polarity assumptions when placing clamping diodes or TVS (Transient Voltage Suppressor) arrays.
• Forgetting Core Saturation Limits: The formula will happily calculate that a 10 μH inductor will support a 500 A current change. In reality, the core will saturate at perhaps 5 A, the inductance will drop to near zero, and the actual voltage will collapse while current skyrockets, destroying the circuit. Always cross-reference your calculated di with the component's I_sat (saturation current) rating. See Electronics Tutorials for more on magnetic saturation limits.