The Current Through Inductor Formula: Derivation & Worked Examples

The Core Current Through Inductor Formula

If you are designing a switch-mode power supply, tuning an audio crossover, or just trying to understand why a relay coil takes a few milliseconds to pull in, you need to know exactly how current behaves in an inductor. Unlike resistors, where current responds instantly to voltage, an inductor resists changes in current. The fundamental relationship governing this behavior is derived from Faraday's law of induction.

The foundational differential equation is v(t) = L * (di/dt). To find the actual current at any given moment, we rearrange and integrate both sides with respect to time. This yields the definitive current through inductor formula:

i(t) = (1 / L) * ∫ v(τ) dτ + i(0)

For practical bench work where a constant DC voltage is applied across the inductor for a specific time window (like the ON-time of a MOSFET in a buck converter), the integral simplifies to a linear ramp. This is the version you will use 90% of the time in power electronics:

ΔI = (V * Δt) / L

This tells us that the change in current (ΔI) is directly proportional to the applied voltage (V) and the time it is applied (Δt), and inversely proportional to the inductance (L). According to All About Circuits, this linear relationship is the cornerstone of energy storage in magnetic fields.

Variable Definitions and Rearranged Forms

Before plugging numbers into your calculator, you must lock in the SI units. A single decimal error here will result in a saturated core or a blown switching FET.

Rearranged Forms

Depending on what you are trying to design or troubleshoot, you will need to isolate different variables. Here are the rearranged forms of the constant-voltage simplification:

• Solving for Inductance (L): L = (V * Δt) / ΔI
  Use case: Sizing a buck converter inductor to achieve a specific ripple current.
• Solving for Time (Δt): Δt = (L * ΔI) / V
  Use case: Calculating how long a MOSFET must stay ON to reach a peak current limit.
• Solving for Voltage (V): V = (L * ΔI) / Δt
  Use case: Calculating the inductive kickback (flyback) voltage spike when a switch opens rapidly.

Worked Examples: Tracking Units from Bench to Board

Let's apply the formula to two real-world scenarios. We will explicitly track the units to prove that the math resolves to Amperes. Remember the base SI definition: 1 Henry = 1 Volt-second per Ampere (V·s/A). Therefore, (V * s) / H = A.

Problem 1: Buck Converter Inductor Ripple Current

Scenario: You are designing a 12V-to-3.3V synchronous buck converter switching at 500 kHz. The output inductor is 4.7 μH. The duty cycle (D) is roughly Vout/Vin = 3.3/12 = 0.275. What is the peak-to-peak ripple current (ΔI) through the inductor during the high-side MOSFET's ON-time?

Step 1: Determine the voltage across the inductor (V).
When the high-side FET is ON, the left side of the inductor is at 12V, and the right side is held at the output voltage of 3.3V.
V_L = 12V - 3.3V = 8.7V

Step 2: Determine the ON-time (Δt).
Δt = Duty Cycle / Switching Frequency
Δt = 0.275 / 500,000 Hz = 0.00000055 seconds = 550 ns (or 0.55 μs)

Step 3: Convert units to base SI and solve.
L = 4.7 μH = 4.7 × 10⁻⁶ H
Δt = 0.55 μs = 0.55 × 10⁻⁶ s
ΔI = (8.7 V * 0.55 × 10⁻⁶ s) / (4.7 × 10⁻⁶ H)
Notice how the 10⁻⁶ micro prefixes cancel each other out perfectly:
ΔI = (8.7 * 0.55) / 4.7 = 4.785 / 4.7 ≈ 1.018 A

Bench Reality Check: A ~1A ripple on a 3.3V rail is typical. If your load draws 3A DC, your inductor current will ramp between 2.5A and 3.5A. Ensure your inductor's saturation current rating (I_sat) is at least 20% above the peak (3.5A * 1.2 = 4.2A minimum).

Problem 2: Time to Reach Target Current in a Laboratory Choke

Scenario: You are testing a large 50 mH air-core filtering choke. You apply a constant 24V DC from a bench supply. Assuming the starting current is zero, how long will it take for the current to ramp up to 5A?

Step 1: Identify knowns and convert to base SI.
V = 24 V
L = 50 mH = 50 × 10⁻³ H
ΔI = 5 A - 0 A = 5 A

Step 2: Rearrange formula to solve for Δt.
Δt = (L * ΔI) / V
Δt = (50 × 10⁻³ H * 5 A) / 24 V
Δt = 0.25 (V·s) / 24 V = 0.010416 seconds

Step 3: Convert to practical units.
Δt ≈ 10.42 ms

Unit tracking verification: (Henries * Amperes) / Volts = ((V·s/A) * A) / V = (V·s) / V = seconds. The math holds.

Assumptions, Unit Traps, and Realistic Magnitudes

The linear formula ΔI = (V * Δt) / L is incredibly useful, but it is an idealization. If you take this formula to the workbench without understanding its boundaries, your circuit will fail. Here is what you need to know to bridge the gap between textbook theory and physical reality.

When the Formula Applies (and When It Doesn't)

This linear ramp formula assumes an ideal inductor with zero parasitic resistance (DCR). In the real world, every inductor has wire resistance. The true current curve is exponential, governed by the RL time constant (τ = L / R).

The linear formula is only accurate when the time interval (Δt) is much shorter than the time constant (τ). Specifically, it holds true for roughly the first 10% of the exponential curve. In switch-mode power supplies switching at hundreds of kilohertz, Δt is in the nanosecond/microsecond range, while τ is often in the millisecond range, making the linear approximation perfectly valid. However, in Problem 2 above, if that 50mH choke had 10 ohms of DCR, the time constant would be 5ms. Trying to ramp to 5A in 10.4ms would fail because the current would asymptotically approach V/R (24V / 10Ω = 2.4A) and never reach 5A.

Unit Mistakes That Break the Math

According to standard reference texts like Electronics Tutorials, the most common engineering errors occur during unit conversion. Watch out for these traps:

• The Micro/Milli Mix-up: Confusing μH (10⁻⁶) with mH (10⁻³). A 4.7mH inductor is 1,000 times larger than a 4.7μH inductor. Plugging the wrong prefix into your calculator will result in a calculated current that is off by three orders of magnitude.
• Frequency vs. Time: Forgetting to invert the switching frequency to find the period before applying the duty cycle. Δt must be in seconds, not Hertz.
• Ignoring Net Voltage: Using the input voltage (e.g., 12V) instead of the net voltage across the inductor (e.g., 12V - 3.3V = 8.7V). The inductor only 'sees' the potential difference across its own terminals.

What a Realistic Answer Magnitude Looks Like

When you finish your calculation, sanity-check the result against these industry benchmarks:

• Signal/RF Inductors (nH to low μH): Currents are typically in the 10 mA to 100 mA range. If you calculate 5A through a 0402-size RF choke, you've made a math error or the part will instantly vaporize.
• Power Supply Inductors (μH range): Ripple currents (ΔI) are usually designed to be 20% to 40% of the maximum DC load current. For a 10A CPU VRM, expect a calculated ΔI of 2A to 4A.
• Filter Chokes / Motor Drives (mH to H range): Currents range from 1A to 50A+, but the ramp times (Δt) are correspondingly much longer (milliseconds to seconds).

Frequently Asked Questions

How does parasitic resistance alter the current through inductor formula?

Parasitic resistance (DCR) turns the linear current ramp into an exponential curve. The exact formula becomes i(t) = (V/R) * (1 - e^(-R*t/L)). For short switching intervals where t << L/R, the Taylor series expansion of the exponential term collapses back into the linear ΔI = (V * Δt) / L formula. However, for long-duration DC applications (like a relay coil holding steady), the resistance dominates, and the steady-state current is simply determined by Ohm's Law (I = V / R), rendering the inductance irrelevant to the final magnitude.

Can the current through an inductor change instantaneously?

No. Looking at the rearranged formula V = (L * ΔI) / Δt, if the change in time (Δt) approaches zero (an instantaneous change), the required voltage (V) approaches infinity. In the physical world, infinite voltage is impossible. If you try to force an instantaneous change by opening a mechanical switch or turning off a MOSFET without a flyback diode, the inductor will generate whatever voltage is necessary to keep the current flowing—often resulting in an arc across the switch contacts or the avalanche breakdown of the semiconductor. The current must always ramp, never step.

How do I apply this formula to an AC sine wave?

For AC circuits, we abandon the time-domain integral and move to the frequency domain using phasors. Instead of calculating instantaneous ramp times, we use the concept of inductive reactance (X_L = 2 * π * f * L). The AC equivalent of Ohm's law becomes I_rms = V_rms / X_L. The time-domain integral formula is still mathematically true—the integral of a sine wave is a negative cosine wave (meaning current lags voltage by 90 degrees)—but using reactance is vastly more practical for AC steady-state analysis.

Why does my bench measurement show a curved ramp instead of a straight line?

If you are probing a switching node with a current probe and the 'linear' ramp looks bowed or curved, you are likely seeing one of two physical realities. First, the parasitic DCR and the output capacitor's ESR are introducing an RC time constant that bends the line into an exponential curve. Second, and more dangerously, your inductor core may be entering saturation. As the core saturates, the inductance value (L) drops dynamically during the pulse. Since L is shrinking in the denominator of ΔI = (V * Δt) / L, the slope of the current ramp (di/dt) gets steeper as the pulse progresses, creating a distinct 'hockey stick' upward curve on your oscilloscope. If you see this, you need a larger inductor or one with a higher saturation current rating.