If you want to predict how an inductor will behave on your workbench, you have to look past the abstract physics and focus on the time-domain math. The fundamental equation for current through an inductor dictates everything from the ripple in your buck converter to the flyback spike that just blew your MOSFET.
Below, we will derive the core integral and exponential forms, track the units through two solved bench problems, and walk through a real-world power design failure where ignoring the math's assumptions resulted in a melted prototype.
The Core Equation for Current Through an Inductor (and What It Actually Means)
The foundational physics definition relates voltage to the rate of change of current: v(t) = L(di/dt). But on the bench, you usually know the voltage you are applying and need to find the resulting current. By integrating both sides, we get the primary equation for current through an inductor:
i(t) = (1/L) ∫t0t v(τ) dτ + i(t0)
For practical DC circuits where an inductor is in series with a resistor (an RL circuit) and a step voltage is applied, the integral resolves into the exponential step-response formula:
i(t) = (V/R) × (1 - e-Rt/L)
Symbol Definition and Assumptions
| Symbol | Parameter | Standard Unit | Practical Bench Context |
|---|---|---|---|
| i(t) | Instantaneous current at time t | Amperes (A) | The value you are trying to find or measure with a current probe. |
| L | Inductance | Henries (H) | Assumed constant. Assumption: The core has not reached magnetic saturation. |
| v(τ) | Voltage across the inductor over time | Volts (V) | The driving force. In a buck converter, this switches between (Vin-Vout) and (-Vout). |
| i(t0) | Initial current at time t0 | Amperes (A) | Crucial for continuous conduction mode (CCM) power supplies; rarely 0A in steady state. |
| R | Series resistance | Ohms (Ω) | Includes load resistance plus the inductor's internal DCR (DC Resistance). |
When this applies and its assumptions: These equations assume an ideal inductor. They assume the core material operates in its linear region (permeability is constant), and they ignore parasitic parallel capacitance. In reality, every inductor has a DC resistance (DCR) and a saturation current limit (Isat). If your calculated i(t) exceeds Isat, the inductance L collapses toward zero, the math breaks down, and current spikes uncontrollably.
Rearranged Forms and the Unit Mistakes That Break Your Math
On the bench, you rarely solve for current in isolation. You are usually sizing a component or calculating a timing window. Here are the rearranged forms derived from the base differential equation V = L(di/dt):
- Solving for Inductance (Sizing):
L = (V × dt) / di
Use case: Sizing a buck converter inductor for a target ripple current. - Solving for Time (Timing):
dt = (L × di) / V
Use case: Calculating how long a MOSFET must stay on to reach a peak current limit. - Solving for Ripple Current:
di = (V × dt) / L
Use case: Verifying if your chosen inductor will cause excessive output voltage ripple.
The Unit Traps
The most common reason makers blow up prototypes is a unit mismatch in the exponent or the denominator.
- The Micro/Milli Trap: Inductors are rated in μH or mH. Time is often in μs or ms. If you plug 47 μH directly into the equation as "47" instead of "47 × 10-6", your calculated current will be off by a factor of a million. Bench trick: If your voltage is in Volts, time in μs, and inductance in μH, the 10-6 terms cancel out perfectly, leaving you with Amps.
- The Exponent Trap: In the RL step response
e^(-Rt/L), the exponent must be unitless. If R is in Ohms, t is in seconds, and L is in Henries, it works. If L is in millihenries, you must convert it to base Henries first, or your exponential decay curve will be completely wrong.
What a Realistic Answer Magnitude Looks Like
Before you trust your calculator, sanity-check the magnitude. For signal and RF inductors (0402 to 1206 SMD packages), realistic currents are 10 mA to 500 mA. If your math yields 12A for a tiny SMD inductor, you forgot to convert μH to H. For power inductors (shielded drum or toroidal), realistic currents range from 1A to 50A. If you calculate 0.002A for a massive 22mm through-hole inductor in a power supply, you likely inverted a fraction.
Solved Bench Problems: Tracking Units Step-by-Step
Let us run two common scenarios, explicitly tracking the base SI units to prevent calculator errors.
Problem 1: Constant Voltage Pulse (Integral Form)
Scenario: You are driving a 47 μH inductor with a 12V DC pulse that lasts for 5 μs. The initial current is 0A. What is the peak current at the end of the pulse?
- Convert to base SI units:
L = 47 × 10-6 H
V = 12 V
dt = 5 × 10-6 s - Apply the simplified integral (since V is constant):
Δi = (V × dt) / L - Substitute and track units:
Δi = (12 V × 5 × 10-6 s) / (47 × 10-6 H) - Cancel the 10-6 terms:
Δi = (12 × 5) / 47 = 60 / 47 - Result: 1.276 A. (The units resolve to Amperes because Volts × Seconds / Henries = Amps).
Problem 2: RL Circuit Step Response (Exponential Form)
Scenario: A 5V DC source is switched into a series circuit with a 10 Ω resistor and a 50 mH inductor. What is the current at exactly t = 2 ms?
- Convert to base SI units:
V = 5 V, R = 10 Ω, L = 50 × 10-3 H, t = 2 × 10-3 s - Calculate the time constant (τ = L/R):
τ = (50 × 10-3) / 10 = 5 × 10-3 s(or 5 ms) - Calculate the exponent (-t/τ):
- (2 × 10-3) / (5 × 10-3) = -0.4(Notice how the 10-3 terms cancel, leaving a unitless exponent). - Apply the step response formula:
i(t) = (5 / 10) × (1 - e-0.4)
i(t) = 0.5 × (1 - 0.6703)
i(t) = 0.5 × 0.3297 - Result: 0.1648 A (or 164.8 mA).
Real-World Scenario: Sizing a Buck Converter Inductor (And What Went Wrong)
Math on paper is clean; the workbench is not. Here is a teardown of a real design failure where the equation for current through an inductor was applied correctly, but the assumptions of the formula were ignored.
The Setup: A maker was designing a 12V-to-3.3V buck converter to power an ESP32 and a strip of WS2812B LEDs, requiring a steady 2A output. The switching frequency (fsw) was set to 500 kHz. They used the standard ripple current equation (derived from our core integral):
ΔIL = [Vout × (1 - D)] / (L × fsw)
The Numbers: Targeting a 30% ripple (0.6A), they calculated the required inductance:
- Duty Cycle (D) = 3.3V / 12V = 0.275
L = [3.3 × (1 - 0.275)] / (0.6 × 500,000) = 7.97 μH
Since 7.97 μH is not a standard value, they rounded down to a standard 6.8 μH shielded SMD power inductor. Looking at the datasheet, the inductor had an Irms (thermal current rating) of 2.8A and an Isat (saturation current) of 2.2A. The maker reasoned: "My load is 2A, and the thermal rating is 2.8A. I have plenty of headroom."
The Outcome: Upon powering up, the inductor emitted a high-pitched squeal, the ESP32 browned out, and the oscilloscope showed massive, jagged spikes on the switch node. The converter's IC kept tripping its internal Over-Current Protection (OCP).
What Went Wrong: The maker confused average DC current with peak inductor current. By dropping to 6.8 μH, the ripple current increased:
ΔIL = [3.3 × 0.725] / (6.8 × 10-6 × 500,000) = 0.70A- The peak current is the DC load plus half the ripple:
Ipeak = 2.0A + (0.70A / 2) = 2.35A.
The peak current (2.35A) exceeded the saturation current (2.2A). When the core saturated, the inductance L effectively dropped to near-zero. According to di = (V × dt) / L, as L approaches zero, the change in current (di) approaches infinity. The current spiked violently, tripping the OCP. The fix: They replaced it with a 10 μH inductor rated for 3.5A Isat, which lowered the ripple and kept the peak current well below the saturation threshold. For deeper reading on power stage magnetics, Analog Devices provides excellent guidelines on inductor selection that emphasize checking Isat over Irms for peak transient survival.
Bench Verification: Measuring What the Math Predicts
You cannot measure inductor current directly with a standard multimeter in a switching circuit; the bandwidth is too low and the multimeter's internal shunt will disrupt the loop. To verify your math on the bench:
- Use an AC/DC Current Probe: A Hall-effect current probe (like the Tektronix TCP0030A or a budget Hantek CC-65) clamps around the wire and outputs a voltage proportional to the current (e.g., 10mV/A). This allows you to see the exact triangular ripple waveform predicted by the integral equation.
- The Shunt Resistor Trick: If you lack a current probe, insert a low-value, high-wattage shunt resistor (e.g., 0.1 Ω) in series with the inductor's ground return path. Measure the voltage across the shunt with your oscilloscope and use Ohm's law (
I = V/R) to reconstruct the current waveform. Keep the shunt on the low-side to avoid blowing out your scope's common-mode voltage limits. - Check the DCR Drop: For DC steady-state verification, measure the voltage drop directly across the inductor's terminals. Divide this millivolt reading by the datasheet's DCR spec to find your true DC current. If your math predicted 2A, but the DCR drop indicates 2.4A, your load is drawing more than expected, and you need to re-evaluate your thermal margins.
Mastering the equation for current through an inductor is not just about passing an exam; it is about knowing exactly why a component fails when you push it past its physical limits. Track your units, respect the saturation threshold, and always verify with a scope.
