If you are designing a switching power supply, filtering audio, or just trying to understand why your relay coil sparks when you open the switch, you need to know how current behaves inside a coil of wire. The fundamental equation for current through an inductor in the time domain is not a simple algebraic ratio like Ohm's Law. Instead, it is an integral: the current at any given moment is the accumulated area under the voltage curve over time, scaled by the component's inductance, plus whatever current was already flowing.
In this guide, we will break down the time-domain integral equation, track units through real solved problems, and look at a bench failure that happens when you misapply this math in a switching regulator.
The Core Equation for Current Through an Inductor
When a voltage is applied across an ideal inductor, the current cannot change instantaneously. The relationship between the voltage across the terminals and the current flowing through the component is defined by the time-domain integral equation:
i(t) = (1/L) ∫t₀t v(τ) dτ + i(t₀)
| Symbol | Description | Standard SI Unit |
|---|---|---|
| i(t) | Instantaneous current at time t | Amperes (A) |
| L | Inductance of the component | Henries (H) |
| v(τ) | Voltage across the inductor as a function of time | Volts (V) |
| t₀ | Initial starting time of the observation window | Seconds (s) |
| i(t₀) | Initial current flowing at time t₀ | Amperes (A) |
| τ | Dummy variable of integration (time) | Seconds (s) |
Assumptions and Realistic Magnitudes
This equation assumes a linear inductor. It assumes the core material does not saturate (meaning L remains constant regardless of current) and ignores parasitic elements like DC winding resistance (RDCR) and parallel parasitic capacitance. For a deep dive into non-ideal parasitics, refer to the All About Circuits inductor guide.
What does a realistic answer look like? In hobbyist and prototype DC-DC converters, motor drives, and audio filters, you should expect calculated transient currents in the 10 mA to 5 A range. If your integral yields 400 A, you likely have a unit conversion error or a dead short. If it yields 0.00002 A, your inductor is likely too large for the applied voltage pulse, or the circuit is effectively open.
Rearranged Forms and Unit Pitfalls
Depending on what you are trying to design or measure, you will need to rearrange the core equation. Here are the most common algebraic forms derived from the integral:
- Solving for Inductance (L): L = (1 / Δi) ∫ v(t) dt
Use this when you know your allowable current ripple and need to select a physical component. - Solving for Voltage (v): v(t) = L (di / dt)
The derivative form. Use this to calculate the voltage spike when a switch opens and current drops to zero in microseconds. - Solving for Time (t) with Constant Voltage: Δt = (L × Δi) / V
Use this to find out how long it takes for a constant DC voltage to ramp an inductor's current from zero to a specific trip point.
The Micro-Henry Trap: Unit Mistakes That Break the Math
The single most common reason this equation fails on the bench is unit mismatch. The formula strictly requires base SI units: Henries, Volts, Seconds, and Amperes. Datasheets and schematic symbols almost never use base units. They use milliHenries (mH), microHenries (µH), and microSeconds (µs).
If you plug 47 µH into the equation as "47" instead of "0.000047", your calculated current will be off by a factor of one million. Always convert your prefix multipliers to base scientific notation (e.g., 47 µH = 47 × 10-6 H) before running the integral.
Solved Problems: Tracking Units from Bench to Paper
Let's run through two practical scenarios, explicitly tracking units at every step to ensure the dimensional analysis holds up. According to the Electronics Tutorials inductor reference, tracking the Volt-second product is the key to mastering magnetic components.
Problem 1: DC Step Response (Relay Coil Ramp-Up)
Setup: You apply a constant 5V DC step across a 10 mH inductor. The initial current is 0 A. What is the current after 2 milliseconds?
- Convert to base SI units: L = 10 × 10-3 H; V = 5 V; t = 2 × 10-3 s; i(t₀) = 0 A.
- Set up the integral: Because voltage is constant, it moves outside the integral.
i(t) = (1 / 10 × 10-3 H) × 5 V × ∫02×10⁻³ dt + 0 A - Evaluate the integral: The integral of dt from 0 to 2ms is simply 2 × 10-3 s.
i(t) = (100 A/V·s) × (5 V) × (2 × 10-3 s) - Multiply and cancel units:
i(t) = 100 × 5 × 0.002 [ (A / (V·s)) × V × s ] = 1.0 A.
Problem 2: Switching Node Ripple (Buck Converter ON Time)
Setup: A 47 µH inductor in a buck converter has 1.2 A flowing through it at the start of a switching cycle. During the MOSFET's ON time, a constant 7V is dropped across the inductor for 3 µs. What is the peak current at the end of the ON pulse?
- Convert to base SI units: L = 47 × 10-6 H; V = 7 V; Δt = 3 × 10-6 s; i(t₀) = 1.2 A.
- Set up the equation for Δi:
Δi = (1 / 47 × 10-6 H) × 7 V × (3 × 10-6 s) - Notice the micro-prefix cancellation: The 10-6 in the denominator (Henries) and the 10-6 in the numerator (Seconds) cancel each other out perfectly.
Δi = (7 × 3) / 47 = 21 / 47 = 0.446 A (This is the peak-to-peak ripple). - Add to initial current:
i(t) = 1.2 A + 0.446 A = 1.646 A.
Real-World Scenario: Buck Converter Inductor Sizing Gone Wrong
Math on paper is clean; the workbench is not. Here is a scenario where misinterpreting the output of this equation destroyed a prototype.
The Setup
A hobbyist was designing a buck converter to step 24V down to 5V to power a 2A servo rail, using an LM2596 switching regulator (150 kHz switching frequency). They used the rearranged inductance formula to calculate the required inductor value to keep the ripple current (Δi) manageable, settling on a 33 µH power inductor from a bin of spare parts.
The Numbers
The duty cycle (D) is roughly Vout / Vin = 5 / 24 = 0.208.
The ON time (ton) is D / f = 0.208 / 150,000 = 1.39 µs.
The voltage across the inductor during ON time is Vin - Vout = 24 - 5 = 19V.
Using our equation: Δi = (19V × 1.39 × 10-6s) / (33 × 10-6H) = 0.8 A of ripple.
The Outcome
The builder powered the board. Within three seconds, the LM2596 internal switch shorted out, the chip vented magic smoke, and the 24V rail slammed directly into the 5V servos, destroying them.
What Went Wrong?
The builder confused average current with peak current, and misunderstood inductor datasheet ratings. The equation correctly calculated the ripple (0.8 A). However, the inductor current is a triangle wave riding on top of the 2A DC load. The peak current the inductor experienced was Iload + (Δi / 2) = 2A + 0.4A = 2.4 A.
The spare 33 µH inductor they grabbed was a small, unshielded drum core with a thermal RMS rating of 3A, but a saturation current (Isat) of only 1.5A. When the current hit 1.5A, the magnetic core saturated. At saturation, the permeability of the core drops to that of air, and the inductance (L) effectively collapses toward zero. Looking back at the derivative form (v = L di/dt), if L approaches zero while voltage is still applied, di/dt approaches infinity. The current spiked violently, exceeding the LM2596's internal current limit response time, and blew the silicon. Always check the Isat rating against your peak calculated current, not your average load.
When to Use AC Steady-State vs. Time-Domain Transient
The integral equation we derived above is strictly for transient analysis and time-domain switching circuits (like PWM motor drives, buck/boost converters, and spark suppression). It tells you exactly what the current is doing microsecond by microsecond.
However, if you are designing an analog audio crossover, an RF filter, or an AC mains choke, you are dealing with continuous sinusoidal waveforms. In that domain, calculating the integral for every cycle is unnecessary. Instead, you use the AC steady-state magnitude equation based on inductive reactance (XL):
Irms = Vrms / (2πfL)
Use the time-domain integral when switches are flipping and waveforms are square or triangular. Use the AC reactance formula when you are analyzing continuous sine waves and phase shifts. Mastering both, and knowing exactly when to deploy the integral versus the algebraic ratio, is what separates a parts-swapper from a true circuit designer.
