The relationship between current change and induced electromotive force is the defining characteristic of inductive components. Whether you are sizing a choke for a buck converter or calculating the flyback spike that will destroy your switching MOSFET, the governing equation is non-negotiable. Below is the complete breakdown of the formula for voltage across an inductor, complete with rearranged design forms, unit-tracked worked examples, and the real-world edge cases that textbooks often gloss over.
The Core Formula and Symbol Definitions
The instantaneous voltage across an ideal inductor is directly proportional to its inductance and the rate of change of current flowing through it. The formula for voltage across an inductor is expressed as:
VL = L · (di / dt)
According to Georgia State University's HyperPhysics, this is a direct consequence of Faraday's Law of Induction, where the induced EMF opposes the change in current (Lenz's Law). In practical circuit analysis, we often drop the negative sign when calculating magnitude and assign polarity based on the passive sign convention.
| Symbol | Parameter | Standard SI Unit | Practical Description |
|---|---|---|---|
| VL | Inductor Voltage | Volts (V) | The potential difference measured across the inductor terminals at a specific instant. |
| L | Inductance | Henrys (H) | The component's ability to store energy in a magnetic field. Usually specified in mH or μH on datasheets. |
| di | Change in Current | Amperes (A) | The delta between final and initial current (Ifinal - Iinitial) over the measured time window. |
| dt | Change in Time | Seconds (s) | The duration over which the current change (di) occurs. Often in μs or ns in switching circuits. |
| di/dt | Current Slew Rate | Amperes per second (A/s) | The first derivative of current with respect to time. Represents how 'fast' the current is ramping. |
Rearranged Forms for Circuit Design
On the bench, you rarely solve for VL in isolation. You are usually given a voltage constraint (like your microcontroller's GPIO limit or a MOSFET's VDS rating) and need to find the required component value or timing. Here are the rearranged forms of VL = L · (di / dt) solved for each variable:
- Solving for Inductance (L):
L = VL / (di / dt)
Use case: Sizing a buck converter inductor to limit current ripple (di) given a known switching frequency and input/output voltage differential. - Solving for Current Change (di):
di = (VL · dt) / L
Use case: Calculating the peak-to-peak ripple current in a power supply filter when the inductance and switching period (dt) are fixed. - Solving for Time (dt):
dt = (L · di) / VL
Use case: Determining the minimum on-time required for a switching regulator to reach its peak current limit threshold.
Worked Examples with Unit Tracking
The most common point of failure in these calculations is unit misalignment. Datasheets specify L in microhenrys (μH) and switching times in microseconds (μs), but the formula demands base SI units. Below are two solved problems demonstrating strict unit tracking.
Problem 1: Buck Converter Inductor Sizing (Linear Ramp)
Scenario: You are designing a synchronous buck converter. During the switch ON time, the voltage across the inductor is a steady 2.4V. You need the current to ramp from 2.0A to 5.0A over a 15μs ON-time. What inductance (L) is required?
- Identify knowns and convert to base SI units:
- VL = 2.4 V
- di = 5.0A - 2.0A = 3.0 A
- dt = 15 μs = 15 × 10-6 s
- Select the rearranged formula:
L = VL / (di / dt) - Calculate the slew rate (di/dt):
di/dt = 3.0 A / (15 × 10-6 s) = 200,000 A/s - Solve for L:
L = 2.4 V / 200,000 A/s = 0.000012 H - Convert to practical units:
0.000012 H × 106 = 12 μH
Result: You need a 12μH inductor. In practice, you would select a standard 10μH or 15μH shielded ferrite inductor with a saturation current rating well above the 5A peak.
Problem 2: Relay Coil Flyback (Inductive Kickback)
Scenario: A 12V automotive relay coil has an inductance of 50mH and draws a steady-state current of 100mA. When the driving BJT transistor switches off, the current collapses to 0A in 10μs. What is the peak voltage spike across the inductor?
- Identify knowns and convert to base SI units:
- L = 50 mH = 50 × 10-3 H = 0.05 H
- di = 0A - 0.1A = -0.1 A (magnitude is 0.1 A)
- dt = 10 μs = 10 × 10-6 s
- Select the core formula:
VL = L · (di / dt) - Calculate the slew rate:
di/dt = 0.1 A / (10 × 10-6 s) = 10,000 A/s - Solve for VL:
VL = 0.05 H · 10,000 A/s = 500 V
Result: The inductor generates a 500V spike. This is why a flyback diode (like a 1N4007 or a faster Schottky like the 1N5819) is mandatory across the coil; without it, this 500V spike will instantly avalanche and destroy a standard 2N2222 transistor or a 5V microcontroller GPIO pin.
Assumptions, Unit Traps, and Realistic Magnitudes
While VL = L · (di / dt) is mathematically elegant, applying it blindly to physical hardware leads to blown components. You must understand the boundaries of the model.
When the Formula Applies (and Its Assumptions)
This formula assumes an ideal, linear inductor. It assumes the core material's permeability remains constant regardless of the magnetic flux density. In reality, ferrite and powdered iron cores exhibit non-linear B-H curves. It also ignores parasitic elements: the DC resistance (DCR) of the copper windings and the parallel parasitic capacitance between the winding layers. For low-frequency or slow-ramp calculations, these assumptions hold. For high-frequency switching (above 1MHz) or nanosecond transients, the parasitic capacitance creates an LC tank circuit, and the simple derivative formula no longer predicts the ringing waveform accurately.
Which Unit Mistakes Break the Calculation
The 'milli-micro' trap is the most frequent error on the workbench.
If you plug L in millihenrys (mH) and dt in microseconds (μs) directly into the formula without converting to base units (Henrys and Seconds), your result will be off by a factor of 1,000.
Rule of thumb: Always write out the scientific notation (e.g., 10 × 10-6) during the intermediate steps. If you use a calculator, ensure you are using the EXP or EE button, not just typing 'e-6', which many basic calculators interpret as a variable.
What a Realistic Answer Magnitude Looks Like
- Steady-State AC (50/60Hz): VL is typically in the range of millivolts to a few volts, governed by the inductive reactance (XL = 2πfL) rather than transient switching.
- Switching Power Supplies (100kHz - 2MHz): VL during the ON/OFF phases is usually equal to the input or output DC bus voltage (e.g., 5V to 48V).
- Unsnubbed Transients (Flyback): VL can easily reach 100V to 10,000V. As All About Circuits notes in their calculus chapter, as dt approaches zero, VL approaches infinity. In physical reality, the voltage will rise until it finds a path to discharge—usually by arcing across a mechanical switch contact or breaking down the silicon junction of a semiconductor.
Frequently Asked Questions
Why does the formula for voltage across an inductor show infinite voltage when a switch opens?
Mathematically, if a switch opens instantaneously, dt = 0. Dividing di by zero yields an infinite slew rate, meaning VL approaches infinity. Physically, infinite voltage is impossible. What actually happens is that VL rises rapidly until it exceeds the dielectric breakdown voltage of the air gap in the switch (causing an arc) or the parasitic capacitance of the inductor windings begins to absorb the energy, creating a high-frequency ringing waveform. This is why we use snubber networks or flyback diodes to artificially increase dt and clamp VL to a safe level.
How do I use the formula for voltage across an inductor with AC sine waves?
For a sinusoidal current where i(t) = Ipeak · sin(ωt), you must take the calculus derivative of the current function. The derivative of sine is cosine, so the formula becomes VL(t) = L · Ipeak · ω · cos(ωt), where ω is the angular frequency (2πf). This proves that inductor voltage leads inductor current by exactly 90 degrees. For practical RMS measurements with a multimeter, you bypass the calculus and use Ohm's law for reactance: Vrms = Irms · XL, where XL = 2πfL. For deeper AC circuit analysis, referencing Electronics Tutorials provides excellent phasor diagram breakdowns.
What happens to the inductor voltage formula if the core saturates?
When an inductor core saturates, its effective inductance (L) drops precipitously—often by 90% or more. Looking at the rearranged formula di = (VL · dt) / L, if L approaches zero while VL and dt remain constant, the change in current (di) spikes massively. In a switching power supply, core saturation doesn't just change the voltage; it causes an uncontrolled, near-vertical spike in current that routinely destroys the driving MOSFET due to overcurrent or thermal runaway. Always check the manufacturer's DC bias curves to ensure your peak current remains below the saturation threshold (typically defined as the point where L drops by 20% to 30%).
