The fundamental equation of inductor voltage is V = L(di/dt). It dictates that the voltage induced across an inductor is directly proportional to its inductance and the rate at which the current through it changes. If you are designing a switch-mode power supply, driving a relay coil, or debugging a noisy motor driver, this single relationship governs your energy storage, ripple current, and destructive flyback spikes.
The Core Equation of Inductor Voltage and Symbol Definitions
At the bench, we usually write the equation of inductor voltage in its differential form. It describes the instantaneous voltage across the component at any given microsecond.
| Symbol | Parameter | SI Unit | Practical Bench Range |
|---|---|---|---|
| VL | Induced Voltage | Volts (V) | Millivolts (ripple) to Kilovolts (flyback) |
| L | Inductance | Henries (H) | nH (RF) to mH (power/relays) |
| di | Change in Current | Amperes (A) | mA (signal) to tens of Amps (power) |
| dt | Change in Time | Seconds (s) | ns (MOSFET switching) to ms (relays) |
Rearranged Forms: Solving for L, di, and dt
You rarely use the base equation of inductor voltage exactly as written. Depending on your design phase, you will rearrange it to find the required component value, the expected ripple, or the necessary switching speed. According to the Texas Instruments Magnetics Design Handbook, mastering these rearrangements is the first step in power stage design.
- Solving for Inductance (L):
L = (VL · dt) / di
Use case: Sizing the output inductor for a DC-DC converter to keep ripple current within a target percentage of your nominal load. - Solving for Current Change (di):
di = (VL · dt) / L
Use case: Calculating the peak-to-peak ripple current in an existing circuit to ensure your output capacitors can handle the RMS ripple without overheating. - Solving for Time (dt):
dt = (L · di) / VL
Use case: Determining the minimum on-time or off-time a controller must provide to achieve a specific current ramp in a motor driver or LED driver.
Assumptions, Boundaries, and Unit Mistakes That Break the Math
When the Formula Applies (and When It Doesn't)
The equation of inductor voltage assumes an ideal, linear inductor. It holds true perfectly as long as the magnetic core material is operating below its saturation flux density (Bsat). Once the core saturates, the inductance L plummets toward the air-core value (often dropping by 90% or more). When L collapses, di/dt spikes violently, and the simple linear equation fails to predict the massive current surge that will inevitably destroy your switching MOSFET.
The "Microsecond Trap" and Other Unit Mistakes
The most common reason hobbyists and junior engineers get wildly incorrect answers is failing to convert prefixes to base SI units before calculating. The formula demands Henries, Amperes, Seconds, and Volts.
If you plug 4.7 (meaning 4.7 μH) and 2.0 (meaning 2.0 μs) directly into the formula without the
10-6 multipliers, your math will cancel the micro-prefixes incorrectly depending on the rearrangement. Always convert to scientific notation (e.g., 4.7e-6) before touching your calculator.
Solved Problems with Strict Unit Tracking
Let's walk through two common bench scenarios, tracking every unit conversion to ensure the math holds up to physical reality.
Problem 1: Finding the Flyback Voltage Spike (Switch-Off Event)
Scenario: A relay coil has an inductance of 4.7 mH. It is carrying a steady-state current of 2.5 A. A mechanical switch opens, interrupting the current in 10 μs. What is the induced voltage?
- Identify and convert variables:
L = 4.7 mH = 4.7 × 10-3 H
di = 2.5 A - 0 A = 2.5 A
dt = 10 μs = 10 × 10-6 s - Apply the equation:
VL = L · (di / dt) - Substitute and solve:
VL = (4.7 × 10-3) · [ 2.5 / (10 × 10-6) ]
VL = (4.7 × 10-3) · (250,000 A/s)
VL = 1,175 Volts
Bench Reality: This 1.1 kV spike is exactly why we place flyback diodes across relay coils. Without the diode, this voltage will arc across the mechanical switch contacts, causing pitting and premature failure, or blow out the silicon junction of a driving transistor.
Problem 2: Sizing an Inductor for Target Ripple Current
Scenario: You are designing the output filter for a buck converter. During the MOSFET's on-time, the voltage across the inductor is 7 V. The on-time is 2 μs. You want to limit the ripple current (di) to 0.5 A. What inductance do you need?
- Identify and convert variables:
VL = 7 V
dt = 2 μs = 2 × 10-6 s
di = 0.5 A - Rearrange for L:
L = (VL · dt) / di - Substitute and solve:
L = (7 · 2 × 10-6) / 0.5
L = (14 × 10-6) / 0.5
L = 28 × 10-6 H = 28 μH
Bench Reality: You would select a standard 27 μH or 33 μH shielded ferrite inductor from a supplier like Coilcraft or Würth Elektronik, ensuring its saturation current rating (Isat) exceeds your peak load current plus half the ripple.
Real-World Bench Scenario: The Flyback Diode Failure
Abstract math is clean; physical silicon is messy. Here is a classic failure mode that occurs when engineers trust the ideal equation of inductor voltage without considering parasitic limits.
The Setup
A hobbyist is driving a standard 12V automotive relay using a 2N2222 NPN bipolar junction transistor (BJT). The relay coil measures approximately 50 mH and draws 75 mA at 12V. To save board space and cost, the builder omits the reverse-biased flyback diode across the coil, assuming the 2N2222 can handle the 12V rail.
The Numbers
When the microcontroller pulls the BJT base low, the transistor turns off rapidly. Let's assume the current drops from 75 mA to 0 A in just 50 nanoseconds (a typical storage/fall time for a saturated BJT).
- L = 50 × 10-3 H
- di = 75 × 10-3 A
- dt = 50 × 10-9 s
Plugging these into V = L(di/dt):
V = (50 × 10-3) · [ (75 × 10-3) / (50 × 10-9) ]
V = (50 × 10-3) · (1,500,000)
V = 75,000 Volts
The Outcome
The math predicts a 75 kV spike. In reality, the voltage does not reach 75,000V. Instead, it rises until it finds a path to discharge. The collector-emitter breakdown voltage (VCEO) of a standard 2N2222 is typically 30V to 40V. The inductor forces the voltage up to ~40V, triggering an avalanche breakdown inside the silicon die. The magnetic energy stored in the relay coil (E = ½LI², roughly 141 μJ) dumps entirely into the tiny semiconductor junction of the transistor.
What Went Wrong
The equation of inductor voltage assumed an ideal open circuit where dt could stretch out infinitely to keep the current flowing. The physical reality was that the transistor's parasitic capacitance and avalanche limits clamped the voltage, violently altering dt and destroying the component. The energy had nowhere to go but through the silicon lattice, melting the junction and shorting the transistor. As noted in fundamental DC circuit theory, inductors resist changes in current, and if you force an instantaneous change without a snubber path, the inductor will break whatever is stopping it. Always provide a freewheeling path for inductive kickback.
