Current of Inductor Equation: Formulas, Units, and Worked Examples

The fundamental equation governing the current of an inductor over time is the integral form of Faraday's law: i(t) = (1/L) ∫ v(t) dt + i(0). In practical DC switching and transient circuits, calculating the exact current requires the derived RL step-response formula: i(t) = (V/R)(1 - e^-t/τ), where the time constant τ = L/R. For high-frequency switching applications like buck converters, the peak-to-peak ripple current is calculated using the rearranged derivative form: Δi = (V_L × t_on) / L.

The Core Current of Inductor Equation Defined

Inductors oppose changes in current by inducing a voltage proportional to the rate of change of that current. The foundational differential equation is V = L(di/dt). To find the current at any given time, we integrate the voltage over time. According to Georgia State University HyperPhysics, this integration yields the absolute current, provided you know the initial current state.

Rearranged Forms and Unit Traps

On the bench, you rarely solve for i(t) directly from the integral. You usually know your target current ripple or transient time and need to solve for L, t, or V. Here are the practical rearranged forms:

• Solving for Inductance (L): L = (V_L × Δt) / Δi (Used for switching converter design)
• Solving for Time (t): t = -τ × ln(1 - (i(t) × R / V_s)) (Used for relay pull-in time calculations)
• Solving for Voltage (V): V = L × (Δi / Δt) (Used for calculating flyback voltage spikes)

Worked Example 1: DC Transient RL Circuit

Scenario: You are driving a 12V DC relay coil. The coil has a measured DC resistance (R) of 10Ω and an inductance (L) of 50mH. The relay requires 0.4A to reliably pull in the contacts. How long after applying 12V will the relay click?

Step 1: Calculate the time constant (τ).
τ = L / R
τ = 50mH / 10Ω = 5ms

Step 2: Calculate the theoretical steady-state maximum current (I_max).
I_max = V_s / R
I_max = 12V / 10Ω = 1.2A

Step 3: Rearrange the RL step-response equation to solve for t.
i(t) = I_max × (1 - e^-t/τ)
0.4A = 1.2A × (1 - e^-t/5ms)
0.3333 = 1 - e^-t/5ms
e^-t/5ms = 0.6667

Step 4: Take the natural logarithm (ln) of both sides.
-t / 5ms = ln(0.6667)
-t / 5ms = -0.4055
t = 0.4055 × 5ms
t = 2.027ms

Verification: At t = 2.027ms, the current reaches exactly 0.4A. The unit tracking holds: [mH]/[Ω] = [ms], and the exponent is unitless.

Worked Example 2: Switching Converter Ripple Current

Scenario: You are debugging a 500kHz buck converter stepping 12V down to 5V. The schematic specifies a 10µH inductor. You need to verify the peak-to-peak ripple current (Δi) to ensure the inductor won't saturate. Assume ideal switches and continuous conduction mode (CCM).

Step 1: Calculate the duty cycle (D) and ON time (t_on).
D = V_out / V_in = 5V / 12V = 0.4167
Switching period T = 1 / f_sw = 1 / 500kHz = 2µs
t_on = D × T = 0.4167 × 2µs = 0.833µs

Step 2: Determine the voltage across the inductor during t_on.
When the high-side MOSFET is ON, the left side of the inductor is at 12V and the right side is at 5V.
V_L = V_in - V_out = 12V - 5V = 7V

Step 3: Apply the rearranged inductor equation to solve for Δi.
V_L = L × (Δi / Δt) → Δi = (V_L × t_on) / L
Δi = (7V × 0.833µs) / 10µH

Step 4: Track units and calculate.
Convert to base SI to be safe: (7 × 0.833 × 10^-6) / (10 × 10^-6)
The 10^-6 terms cancel out perfectly.
Δi = 5.831 / 10 = 0.583A

Verification: A ripple of ~0.58A on a 5V/2A output is roughly 29% ripple, which is the industry-standard target for buck converters. The math checks out against empirical bench measurements.

Decision Path: Sizing an Inductor for a Buck Converter

Use this decision matrix to select a physical inductor part number based on the current of inductor equation. This path terminates in a concrete component recommendation for a standard 12V-to-5V, 2A output design.

Real-World Assumptions and Flyback Safety

The equations above assume an ideal inductor. In reality, every physical inductor has parasitic elements that alter the math at the extremes. According to Electronics Tutorials, you must account for three non-ideal behaviors:

1. DC Resistance (DCR): The copper wire has resistance. In the RL transient equation, R is not just your load resistor; it is R_load + DCR. Ignoring DCR in low-voltage, high-current circuits will cause your calculated steady-state current to be higher than reality.
2. Core Saturation: The equation V = L(di/dt) assumes L is constant. If the current exceeds the inductor's I_sat rating, the magnetic core saturates, and L drops precipitously (often by 90% or more). When L drops, di/dt spikes, leading to catastrophic MOSFET failure in switching converters.
3. Parasitic Capacitance: At very high frequencies (MHz range), the inter-winding capacitance creates a parallel resonant tank, invalidating the simple integral equation.

By strictly tracking your units, accounting for DCR, and respecting the saturation limits derived from the fundamental current equations, you can reliably predict inductor behavior in both transient and steady-state switching environments.